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LINEAR
CIRCUITS
TIME DOMAIN, PHASOR, AND LAPLACE
TRANSrORM APPROACHES
THIRD
EDITION
Raymond A. DeCarlo
Purdue University
Pen-Min Lin
Purdue University
Kendall Hunt
p u b l i s h i n g
c o m p a n y
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Cover image (^^J^ikiaui
^ Used under license from Shutterstock, Inc.
Kendall Hunft
p u b l i s h i n g
c o m p a n y
www.kendallhunt.cpm
Send all inquiries to:
4050 Westmark Drive
Dubuque, lA 52004-1840
Copyright © 2001, 2009 Raymond A. DeCarlo and Pen-Min Lin
Copyright © 1995 Prentice-Hall, Inc.
ISBN 978-0-7575-6499-4
All rights reserved. No part of this publication may be reproduced,
stored in a retrieval system, or transmitted, in any form or by any means,
electronic, mechanical, photocopying, recording, or otherwise,
without the prior written permission of the copyright owner.
r^
Printed in the United States of America
10 9 8 7 6 5 4 3
O
TABLE OF CO N TEN TS
Preface......................................................................................................................................................................vii
Chapter 1 • Charge, Current, Voltage and Ohm’s Law ............................................................................ 1
Chapter 2 • Kirchhoff’s Current & Voltage Laws and Series-Parallel Resistive C ircu its..............51
Chapter 3 • Nodal and Loop Analyses....................................................................................................... 107
Chapter 4 • T he Operational Amplifier..................................................................................................... 155
Chapter 5 * Linearity, Superposition, and Source Transform ation................................................... 191
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems.................................... 227
Chapter 7 • Inductors and C apacitors....................................................................................................... 269
Chapter 8 • First Order RL and RC Circuits...........................................................................................321
Chapter 9 • Second Order Linear Circuits................................................................................................379
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods .................................................431
Chapter 11 • Sinusoidal State State Power Calculations.......................................................................499
Chapter 12 • Laplace Transform Analysis L Basics................................................................................. 543
Chapter 13 • Laplace Transform Analysis II: Circuit Applications................................................... 603
Chapter 14 • Laplace Transform Analysis III; Transfer Function Applications.............................683
Chapter 15 * Time Domain Circuit Response Computations: The Convolution M ethod...... 763
Chapter 16 • Band-Pass Circuits and Resonance....................................................................................811
Chapter 17 * Magnetically Coupled Circuits and Transformers........................................................ 883
Chapter 18 • Tw o-Ports...................................................................................................................................959
Chapter 19 • Principles o f Basic Filtering ............................................................................................. 1031
Chapter 20 • Brief Introduction to Fourier Series .............................................................................. 1085
In d ex................................................................................................................................................................... 1119
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PREFACE
For the last several decades, EE/ECE departments o f US universities have typically required two
semesters o f linear circuits during the sophomore year for EE majors and one semester for other
engineering majors. Over the same time period discrete time system concepts and computer engi­
neering principles have become required fare for EE undergraduates. Thus we continue to use
Laplace transforms as a vehicle for understanding basic concepts such as impedance, admittance,
fdtering, and magnetic circuits. Further, software programs such as PSpice, MATLAB and its tool­
boxes, Mathematica, Maple, and a host o f other tools have streamlined the computational drudg­
ery o f engineering analysis and design. MATLAB remains a working tool in this 3'''^ edition o f
Linear Circuits.
In addition to a continuing extensive use o f MATLAB, we have removed much o f the more com­
plex material from the book and rewritten much o f the remaining book in an attempt to make the
text and the examples more illustrative and accessible. More importantly, many o f the more diffi­
cult homework exercises have been replaced with more routine problems often with numerical
answers or checks.
Our hope is that we have made the text more readable and understandable by today’s engineering
undergraduates.
C
H
A
P
T
E
R
Charge, Current, Voltage
and Ohm’s Law
CHAPTER O U TLIN E
1.
2.
3.
4.
5.
6.
7.
8.
Role and Importance o f Circuits in Engineering
Charge and Current
Voltage
Circuit Elements
Voltage, Current, Power, Energy, Relationships
Ideal Voltage and Current Sources
Resistance, Ohm’s Law, and Power (a Reprise)
V-I Characteristics o f Ideal Resistors, Constant Voltage, and
Constant Current Sources
Summary
Terms and Concepts
Problems
CHAPTER O B jEC TIV ES
1.
Introduce and investigate three basic electrical quantities: charge, current, and voltage,
and the conventions for their reference directions.
2.
3.
Define a two-terminal circuit element.
Define and investigate power and energy conversion in electric circuits, and demonstrate
4.
that these quantities are conserved.
Define independent and dependent voltage and current sources that act as energy or sig­
5.
6.
7.
8.
nal generators in a circuit.
Define Ohm’s law, v{t) = R i{t), for a resistor with resistance R.
Investigate power dissipation in a resistor.
Classify memoryless circuit elements by dieir terminal voltage-current relationships.
Explain the difference between a device and its circuit model.
ch ap ter 1 • Charge, Current, Voltage and O hm ’s Law
1. ROLE AND IM PORTANCE OF CIRCUITS IN ENGINEERING
Are you curious about how fuses blow? About the meaning o f different wattages on Hght bulbs?
About the heating elements in an oven? And how is the presence o f your car sensed at a stoplight?
Circuit theory, the focus o f this text, provides answers to all these questions.
W hen you learn basic circuit theory, you learn how to harness the power o f electricity, as is done,
for example, in
•
an electric motor that runs the compressor in an air conditioner or the pump in a dish­
washer;
•
•
•
•
a microwave oven;
a radio, TV, or stereo;
an iPod;
a car heater.
In this text, we define and analyze common circuit elements and describe their interaction. Our
aim is to create a modular framework for analyzing circuit behavior, while simultaneously devel­
oping a set o f tools essential for circuit design. These skills are, o f course, crucial to every electri­
cal engineer. But they also have broad applicability in other fields. For instance, disciplines such
as bioengineering and mechanical engineering have similar patterns o f analysis and often utilize
circuit analogies.
W H A T IS A C IR C U IT ?
A circuit is an energy or signal/information processor. Each circuit consists o f interconnections o f
“simple” circuit elements, or devices. Each circuit element can, in turn, be thought o f as an ener­
gy or signal/information processor. For example, a circuit element called a “source” produces a
voltage or a current signal. This signal may serve as a power source for the circuit, or it may rep­
resent information. Information in the form o f voltage or current signals can be processed by the
circuit to produce new signals or new/different information. In a radio transmitter, electricity
powers the circuits that convert pictures, voices, or music (that is, information) into electromag­
netic energy. This energy then radi­
ates into the atmosphere or into
space from a transmitting antenna.
A satellite in space can pick up this
electromagnetic energy and trans­
mit it to locations all over the
world. Similarly, a T V reception
antenna or a satellite dish can pick
up and direct this energy to a T V
set. T h e T V contains circuits
(Figure 1.1) that reconvert the
information within the received
signal back into pictures with
sound.
FIG U RE 1.1 Cathode ray tube with surrounding circuitry for
converting electrical signals into pictures.
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
2. CH A RGE AND CU RREN T
CH A R G E
Charge is an electrical property o f matter. Matter consists o f atoms. Roughly speaking, an atom
contains a nucleus that is made up o f positively charged protons and neutrons (which have no
charge). T he nucleus is surrounded by a cloud o f negatively charged electrons. Th e accumulated
charge on 6.2415 x 10’^ electrons equals -1 coulomb (C). Thus, the charge on an electron is
-1 .6 0 2 1 7 6 X 10-19 C.
Particles with opposite charges attract each other, whereas those with similar charges repel. The
force o f attraction or repulsion between two charged bodies is inversely proportional to the square
o f the distance between them, assuming the dimensions o f the bodies are very small compared
with the distance o f separation. Two equally charged particles 1 meter (m) apart in free space have
charges o f 1 C each if they repel each other with a force o f 10“^ c^ Newtons (N), where c = 3 x
10^ m/s is the speed o f light, by definition. The force is attractive if the particles have opposite
charges. Notationally, Q will denote a fixed charge, and q or q{t), a time-varying charge.
Exercise. How many electrons have a combined charge o f -5 3 .4 0 6 x 10
C?
AN SW ER; 333,3 9 1 ,5 9 7
Exercise. Sketch the time-dependent charge profile q{t) = 3 (l-^ ^ 0 C, ? > 0, present on a metal
plate. M ATLAB is a good tool for such sketches.
A conductor refers to a material in which electrons can move to neighboring atoms with relative
ease. Metals, carbon, and acids are common conductors. Copper wire is probably the most com­
mon conductor. An ideal conductor offers zero resistance to electron movement. Wires are
assumed to be ideal conductors, unless otherwise indicated.
Insulators oppose electron movement. Common insulators include dry air, dry wood, ceramic,
glass, and plastic. An ideal insulator offers infinite opposition to electron movement.
C U R R EN T
Current refers to the net flow o f charge across any cross section o f a conductor. T he net move­
ment o f 1 coulomb (1 C) o f charge through a cross section o f a conductor in 1 second (1 sec)
produces an electric current o f 1 ampere (1 A). The ampere is the basic unit o f electric current
and equals 1 C/s.
The direction o f current flow is taken by convention as opposite to the direction o f electron flow,
as illustrated in Figure 1.2. This is because early in the history o f electricity, scientists erroneously
believed that current was the movement o f only positive charges, as illustrated in Figure 1.3. In
metallic conductors, current consists solely o f the movement o f electrons. However, as our under­
standing o f device physics advanced, scientists learned that in ionized gases, in electrolytic solu­
c h ap ter 1 • Charge, Current, Voltage and O hm ’s Law
tions, and in some semiconductor materials, movement o f positive charges constitutes part or all
o f the total current flow.
One Ampere
of Current "
One
;
;
Cloud o f \
se co n d ^ ....... |---- 6.24x10’® 1
later
i
;
k electrons
J
Boundary
FIG U RE 1.2 A cloud o f negative charge moves past a cross section of an ideal conductor from right
to left. By convention, the positive current direction is taken as left to right.
One Ampere
of Current
One
Coulom b
One
of positive
'second
later
charge
Boundary
FIGURE 1.3 In the late nineteenth cenmry, current was thought to be the movement of a positive charge
past a cross section of a conduaor, giving rise to the conventional reference “direction of positive current flow.”
Both Figures 1.2 and 1.3 depict a current o f 1 A flowing from left to right. In circuit analysis, we
do not distinguish between these two cases: each is represented symbolically, as in Figure 1.4(a).
The arrowhead serves as a reference for determining the true direction o f the current. A positive
value o f current means the current flows in the same direction as the arrow. A current o f negative
value implies flow is in the opposite direction o f the arrow. For example, in both Figures 1.4a and
b, a current o f 1 A flows from left to right.
1A
-1A
>
<
(a)
(b)
FIG U RE 1.4 1 A of current flows from left to right through a general circuit element.
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
In Figure 1.4, the current is constant. The wall socket in a typical home is a source o f alternating
current, which changes its sign periodically, as we will describe shortly. In addition, a current direc­
tion may not be known a priori. These situations require the notion o f a negative current.
E X A M P L E 1.1.
Figure 1.5 shows a slab o f material in which the following is true:
1. Positive charge carriers move from left to right at the rate o f 0.2 C/s.
2. Negative charge carriers move from right to left at the rate o f 0.48 C/s.
Given these conditions,
a) Find
and /^;
b)
Describe the charge movement on the wire at the boundaries A and B.
B
A
1
,
Connecting
© o
— 0
©
0
Connecting
wire
wire
Sem iconductor iVlaterial
F IG U R E 1.5 Material through which positive and negative charges move.
S o lu tio n
a)
The current from left to right, due to the movement o f the positive charges, is 0.2 A. The
current from left to right, due to the movement o f the negative charges, is 0.48 A.
Therefore, /^, the total current from left to right, is 0.2 + 0.48 = 0.68 A. Since ly is the
current from right to left, its value is then -0 .6 8 A.
b)
T he wire is a metallic conductor in which only electrons move. Therefore, at boundaries
A and B, negative charges (carried by electrons) move from right to left at the rate o f 0.68
C/s.
Exercise. In Example 1.1, suppose positive-charge carriers move from right to left at the rate o f 0.5
C/s, and negative carriers move from left to right at the rate o f 0.4 C/s. Find
and
AN SW ER: /, = - 0 .9 A; ^ = 0.9 A
If a net charge
crosses a boundary in a short time frame o f At (in seconds), then the approxi­
mate current flow is
/=
Aq
At
( 1 . 1)
where I, in this case, is a constant. The instantaneous (time-dependent) current flow is the limit­
ing case o f Equation 1.1, i.e.,
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
dq{t)
dt
( 1. 2)
Here q{t) is the amount o f charge that has crossed the boundary in the time interval [tQ, t] . The
equivalent integral counterpart o f Equation 1.2 is
q{t) = J i{r)dr
(1.3)
E X A M P L E 1.2
The charge crossing a boundary in a wire is given in Figure 1.6(a) for ? > 0. Plot the current i{t)
through the wire.
(a)
(b)
FIG U RE 1.6 (a) Charge crossing a hypothetical boundary; (b) current flow
associated with the charge plot o f (a).
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
S o lutio n
As per Equation 1.2, the current is the time derivative o f q{t). The slopes o f the straight-Une seg­
ments o f q{f} in Figure 1.6(a) determine the piecewise constant current plotted in Figure 1.6(b).
■
■
•
•
l-cos(co?)
Exercise. The charge crossing a boundary in a wire varies as q[t) = ---------------- C, for t >Q.
Compute the current flow.
A N SW ER: sin(cof) A, for f > 0
Exercise. Repeat the preceding exercise if q{t) = 5e
A N SW ER:
C, for t > 0 .
A, for f > 0
E X A M PLE 1.3
Find q{t), the charge transported through a cross section o f a conductor over [0, f], and also the total
charge Q transported, if the current dirough the conductor is given by die waveform o f Figure 1.7(a).
-l-*-t(se c)
FIG U RE 1.7 (a) Square-wave current signal; (b) q{t) equal to the integral of i{t) given in (a).
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
S o lutio n
From Equation 1.3, for t>Q,
q{t)=p{T)clT
Thus, q{t) is the running area under the i{t) versus t curve. Since i{t) is piecevv'ise constant, the
integral is piecewise linear because the area either increases or decreases linearly with time, as
shown in Figure 1.7(b). Since q{t) is constant for ^ > 3, the total charge transported is Q = q{5) =
3 C.
Exercise. If the current flow through a cross section o f conductor is i{t) = cos(120jtf) A for ? > 0
and 0 otherwise, find q{t) for t>Qi.
AN SW ER: q{t)
‘
120jt
C for r > 0
Exercise. Suppose the current through a cross section o f conductor is given in Figure 1.8. Find
q{t) for t > 0 .
FIGURE 1.8
AN SW ER; q(t) =
C for 0 <
1; q{t) = IC for r > I
T Y P ES OF C U R R EN T
There are two very important current types: direct current (do) and alternating current (ac).
Constant current (i.e., dqldt = / is constant) is called direct current, which is illustrated graphi­
cally m Figure 1.9(a). Figure 1.9(b) shows an alternating current, generally meaning a sinusoidal
waveform, i.e., current o f the form y4sin(w? + ()>), where A is the peak magnitude, co is the angu­
lar frequency, and (|) is the phase angle o f the sine wave. W ith alternating current, the instanta­
neous value o f the waveform changes periodically through negative and positive values, i.e., the
ch a p ter 1 • Charge, Current, Voltage and O h m s Law
direction o f the current flow changes regularly as indicated by the + and - values in Figure 1.9(b).
Household current is ac.
Lastly, Figure 1.9(c) shows a current that is neither dc nor ac, but that nevertheless will appear in
later circuit analyses. There are many other types o f waveforms. Interestingly, currents inside com­
puters, C D players, TV s, and other entertainment devices are typically neither dc nor ac.
i(t) (A)
t(sec)
-H ----------------------1 -
-I-----►
3
(a)
F IG U R E 1.9 (a) Direct current, or dc; i{t) = Iq\ (b) alternating current, or ac;
i{t) = 1 2 0-^ sin (1 2 0 ?) A; (c) neither ac nor dc.
10
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
Because the value o f an ac waveform changes with time, ac is measured in different ways. Suppose
the instantaneous value o f the current at time t is A!sin(ci)i- + (j>). The term peak value refers to K
in K sin(co? + (j)). The peak-to-peak value is 2K. Another measure o f the alternating current,
indicative o f its heating effect, is the root mean square (rms), or effective value. The rms or effec­
tive value is related to the peak value by the formula
rms =
X peak-value = Q .lO llK
(i.4)
A derivation o f Equation 1.4 with an explanation o f its meaning will be given in Chapter 11.
A special instrument called an ammeter measures current. Some ammeters read the peak value,
whereas some others read the rms value. One type o f ammeter, based on the interaction between
the current and a permanent magnet, reads the average value o f a current. From calculus, Fave!
the average value o f any function y(^), over the time interval [0, 7] is given by
(1.5)
For a general ac waveform, the average value is zero. However, ac signals are often rectified, i.e.,
converted to their absolute values, in power-supply circuits. For such circuits, the average value o f
the rectified signal is important. From Equation 1.5, the average value o f the absolute value o f an
ac waveform over one complete cycle with T = 2jt/co, is
K ^
2.K
Average Value = —^\s,m{wt)\dt = ----- J
0
2K -cos{(ot)
T
(O
0.5T
sin(cot)clt
^ 0
2,K
— = 0.636K
jt
( 1.6)
i.e., 0 .636 X peak value.
Exercise. Suppose i{t) - 169.7 sin(50jtr) A. Find the peak value, the peak-to-peak value, the rms
value o f i{t), and the average value o f
AN SW ER: 169.7, 339.4, 120, and 107.93 A, respectively
3. VO LTAG E
W hat causes current to flow? An analogous question might be. W hat causes water to flow in a pipe
or a hose? W ithout pressure from either a pump or gravity, water in a pipe is still. Pressure from
a water tower, a pressured bug sprayer tank, or a pump on a fire truck will force water flow In
electrical circuits, the “pressure” that forces electrons to flow, i.e., produces a current in a wire or
a device, is called voltage. Strictly speaking, water flows from a point o f higher pressure— say,
p o in ts — to a point o f lower pressure— say, point 5 — along a pipe. Between the two points and
B, there is said to be a pressure drop. In electrical circuits, a voltage drop from point A to point B
11
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
along a conductor will force current to flow from point A to point B; there is said to be a voltage
drop from point A to point B in such cases.
Gravity forces the water to flow from a higher elevation to a lower elevation. An analogous phe­
nomenon occurs in an electric field, as illustrated in Figure 1.10(a). Figure 1.10(a) shows two con­
ducting plates separated by a vacuum. O n the top plate is a fixed amount o f positive static charge.
On the bottom plate is an equal amount o f negative static charge. Suppose a small positive charge
were placed between the plates. This small charge would experience a force directed toward the
negatively charged bottom plate. Part o f the force is due to repulsion by the positive charges on
the top plate, and part is due to the attraction by the negative charges on the bottom plate. This
repulsion and attraction marks the presence o f an electric field produced by the opposite sets o f
static charges on the plates.
The electric field indicated in Figure 1.10 sets up an “electric pressure” or voltage drop from the
top plate to the bottom plate, which forces positive charges to flow “downhill” in the way that
water flows from a water tower to your faucet. Unlike water flow, negative charges are forced
“uphill” from the negatively charged bottom plate to the positively charged top plate. As men­
tioned in the previous section, this constitutes a net current flow caused by the bilateral flow o f
positive and negative charges. The point is that current flow is induced by an electric pressure
called a voltage drop.
© © © © © © © ©
© © © © © © © ©
A 0
A
Positive
negative
charge, q
Electric Field
Force on
Electric Field
charge q
charge
Negative
B © charge,-q
B
© © © 0
Force on
© © © ©
(a)
© © © © © © © ©
(b)
FIGURE 1.10 (a) Positive charge in a (uniform) electric field; (b) negative charge in a uniform elearic field.
As mentioned, in Figure 1.10, the positive charge ^ at ^ tends to move toward B. We say, quali­
tatively, that point A in the electric field is at a higher potential than point B. Equivalently, point
5 is at a lower potential than point A. An analogy is now evident: a positive charge in an electric
field “falls” from a higher potential point to a lower potential point, just as a ball falls from a high­
er elevation to a lower elevation in a gravitational field.
Note, however, that if we turn the whole setup o f Figure 1.10(a) upside down, the positive charge
q still moves from point A to point B, an upward spatial movement. Similarly, if a negative charge
- q is placed at B, as in Figure 1.10(b), then the negative charge experiences an upward-pulling
force, moving from the lower potential, point B, to the higher potential, point A.
n
12
Chapter 1 ® Charge, Current, Voltage and Ohms Law
-----------------------------------------------------------------------------------------------------------------------
—
^
n
Again, consider Figure 1. 10(a). As the charge q moves from point ^ toward B, it picks up veloci­
ty and gains kinetic energy. Just before q hits the bottom plate, the kinetic energy gained equals
the (constant) force acting on q multiplied by the distance traveled in the direction o f the force. The
kinetic energy is proportional to q and to
the “distance traveled.” Therefore,
energy converted = kinetic energy gained oc q
n
The missing proportionality constant in this relationship is defined as the potential difference or
voltage between A and B, The term “voltage” is synonymous with “potential difference.”
Mathematically,
,
. ,
voltage = potential difference =
energy converted
magnitude of charge
( 1.8)
The standard unit for measuring potential difference or voltage is the volt (V). According to
Equation 1.8, i f 1 joule {]) o f energy is convertedfrom one form to another when moving 1 C o f charge
from point K to point B, then the potential difference, or voltage, between A and B w i VTIn equation
form, with standard units of V, J, and C, we have
1V = 1 ^
(1.9)
O
The use of terms such as “elevation diflFerence,” “energy converted,” “potential difference,” or
“voltage” implies that they all have positive values. If the word “difference” is changed to “drop”
(or to “rise”), then potential drop and elevation drop have either positive or negative values, as the
case may be. The following four statements illustrate this point in the context of Figure 1.10:
The voltage between (or across) A a n d 5 is 2 V.
The voltage between (or across) B and A is 2 V.
' The voltage drop from A to B is 2W.
{
•The voltage drop from B to A Is - 2 V.
.
.
.
.
’
This discussion describes the phenomena of “voltage.” Voltage causes current flow. But what pro­
duces voltage or electric pressure? Voltage can be generated by chemical action, as in batteries. In
a battery, chemical action causes an excess of positive charge to reside at a terminal marked with
a plus sign and an equal amount of negative charge to reside at a terminal marked with a negative
^
sign. When a device such as a headlight is connected between the terminals, the voltage causes a
current to flow through the headlight, heating up the tiny wire and making it “Ught up.” Another
source of voltage/current is an electric generator in which mechanical energy used to rotate the
shaft of the generator is converted to electrical energy using properties of electro-magnetic fields.
^
All types of circuit analysis require knowledge of the potential difference between two points, say
^
^
A and B, and specifically whether point A or point 5 is at a higher potential. To this end, we speak
of the voltage drop from point A to point B, conveniently denoted by a double-subscript, as Vj^.
If the value of
is positive, then point ^ is at a higher potential than point B. On the other
hand, if
is negative, then point 5 is at a higher potential than point A. Since
stands for
the voltage drop from point B to point A,
o
n
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
The double-subscript convention is one o f three methods commonly used to unambiguously specify
a voltage drop. Using this convention requires labeling all points o f interest with letters or integers so
that
’ KiO ^12’
^13
sense. A second, more-common convention uses + and - markings
on two points, together with a variable or numerical labeling o f the voltage drop from the point
marked + to the point marked - . Figure 1.11 illustrates this second convention, where Vq denotes the
voltage drop from A (marked +) to B (marked - ) . If Vq is positive, then ^4 is at a higher potential than
B. O n the other hand, if Vq is negative, then 5 is at a higher potential than A. The value o f Vq, togeth­
er with the markings + and
stipulates which terminal is at a higher potential; neither alone can do
this. For a general circuit element, the (+, —) markings— that is, the reference directions— can be
assigned arbitrarily. A third method for specifying a voltage drop, using a single subscript, will be dis­
cussed in Chapter 2.
B
-I-
V„
FIGURE 1.11 The + and - markings establish a reference direction for voltage drop. For accuracy,
always place the (+, - ) markings reasonably close to the circuit element to avoid uncertainty.
The following example illustrates the use o f the double subscript and the (+, - ) markings for des­
ignating voltage drops.
E X A M P L E 1.4
Figure 1.12 shows a circuit consisting o f four general circuit elements, with voltage drops as indi­
cated. Suppose we know that
= 4 V, and
= 9 V. Find the values o f
V^q and
CD-
V
-I-
3V
FIG U RE 1.12 Arbitrary circuit elements for exploring the use of (+, - ) for specifying a voltage drop.
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
14
S o lutio n
T he meaiiing o f the double subscript notation and the (+, - ) markings for a voltage imply that
'DA
^ 5 C = 3I V
^CZ> = - ^ Z )C = -(-2 ) = 2 V
Exercise. In Figure 1.12, find
and Vp.^-
A N S W E R :- 3 V ; - 2 V
Exercise. T he convention o f the (+, - ) markings is commonly used as described. Figure 1.13 shows
an old 12-V automobile battery whose (+, - ) markings cannot be seen because o f the corrosion o f
the terminals. A digital voltmeter (DVM ) is connected across the terminals, as shown. The display
reads -1 2 V. Figure out the (+, - ) marking o f the battery terminals.
A N SW ER: left terminal,
right terminal, +
DVM
12V
battery
FIG U RE 1.13 Digital voltmeter connected to a 12-V (car) battery whose plus and
minus markings have corroded away.
One final note: As with current, there are different types o f voltages— dc voltage, ac voltage, and
general voltage waveforms. Figure 1.9, with the vertical axis relabeled as v{t), illustrates different
voltage types.
4. C IR CU IT ELEM EN TS
Circuits consist o f interconnections o f circuit elements. The most basic circuit element has two
terminals, and is called a two-terminal circuit element, as illustrated in Figure 1.14. A circuit eie-
Chapter 1 • Charge, Current, Voltage and O h m s Law
15
ment called a source provides either voltage, current, or both. The battery is a very common
source, providing nearly constant voltage and the usually small current needed to operate small
electronic devices. Car batteries, for example, are typically 12 volts and can produce large currents
during starting. The wall outlet in a home can be thought o f as a 110 -volt ac source. Figure 1.14(a)
shows a (battery) voltage
across a general undefined circuit element. A current z(r) flows
through the element. Recall from our earlier intuitive discussion that voltage is analogous to water
pressure: pressure causes water to flow through pipes; voltage causes current to flow through cir­
cuit elements. Total water into a pipe equals total water out o f the pipe. Analogously, the current
entering a two-terminal device must, by definition, equal the current leaving the two-terminal device.
Current
FIGURE 1.14 (a) General circuit element (connected to a battery) as an energy
or signal processor: v(i) is the voltage developed across the circuit element, and z'(r)
is the current flowing through the circuit element; (b) practical example of a
general circuit element (car headlight) connected to a car battery.
The circuit element o f Figure 1.14(a) has a specific labeling: the current i(f) flows from the plus
terminal to the minus terminal through the circuit element. Such a labeling o f the voltage-current
reference directions is called the passive sign convention. In contrast, the current iij) flows from
the minus terminal to the plus terminal through the battery; this labeling is conventional for
sources but not for non-source circuit elements.
In addition to sources, there are other common two-terminal circuit elements:
• The resistor
•
The capacitor
•
The inductor
For a resistor, the amount o f current flow depends on a property called resistance; the smaller the
resistance, the larger the current flow for a fixed voltage across the resistor. A small-diameter pipe
offers more resistance to water flow than a large-diameter pipe. Similarly, different types o f con­
ductors offer different resistances to current flow. A conductor that is designed to have a specific
resistance is called a resistor. If the device is an ideal resistor, then v(f) = Ri{i), where i? is a con­
stant o f resistance. More on this shortly.
The circuit elements called the capacitor and the inductor will be described later in the text. Also,
future chapters will describe the operational amplifier and the transformer that are circuit elements
having more than two terminals.
16
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
5. VO LTAG E, CURRENT, POW ER, ENERGY, RELATIONSHIPS
The relationship between voltage across and current through a two-terminal element determines
whether power (and, thus, energy) is delivered or absorbed. The heating element in an electric
oven can be thought o f as a resistor. The heating element absorbs electric energy and converts it
into heat energy that cooks, among other things, turkey dinners.
In Figure 1.14(a), a battery is connected to a circuit element. Figure 1.14(b) concretely illustrates
this with a 12-V car battery connected to a headlight. W ith reference to Figures 1.14(a) and
1.14(b), suppose v{t) = 12 V, and i{t) = 5 A: 5 A o f current flows through the headlight. The head­
light converts electrical energy into heat and light. Power (in watts) is the rate at which the ener­
gy is converted. At each instant o f time, the electrical power delivered to (absorbed by) the head­
light is pit) = v[t)i{t) - 12 X 5 = 60 watts. Similarly, at each instant o f time, the battery can be
viewed as delivering 60 watts o f power to the headlight. Inside the battery, the stored potential
energy o f the chemicals and metals undergoes a chemical reaction that produces the electrical
potential difference and the current flow to the headlight: chemical energy is converted into elec­
trical energy that is converted into light and heat.
Figure 1.15 depicts a more general scenario: a circuit element is connected to its surrounding cir­
cuit at points A and B. (One, o f course, could imagine that the “remainder o f circuit” is a battery,
and circuit element 1 is a headlight.) Suppose there is a constant voltage drop from A to B, denot­
ed by
Also assume that a constant current
flows from terminal A to terminal B through
circuit element 1, as shown.
FIG U RE 1.15 A general circuit in which a two-element circuit element is extracted
and labeled according to the passive sign convention.
For discussion purposes, assume
> 0 and
> 0. During a time interval o f T s, (V^g x T) C o f
charge moves through circuit element 1 from A to B. In “falling” from a higher potential, point A, to
a lower potential, point B, the charge loses electric potential energy. The lost potential energy is con­
verted within element 1 into some other form o f energy— heat or light being two o f several possibil­
ities. According to Equation 1.8, the amount o f energy converted {absorbed by the element) is
y. T) >Q. The power absorbedhj element 1 is, by definition, the rate at which it converts or absorbs
energy. This rate equals
^a b (^ab ^ T)
■^Vab I a b > 0 .
T
Chapter 1 • Charge, Current, Voltage and O h m s Law
17
Exercise. In Figure 1.15, the current ^AB - 5 niA, and
= 400 V. W hat is the energy absorbed
by circuit element 1 in one minute? W hat is the power absorbed by circuit element 1?
AN SW ER: W = 120 J; P = 2 watts
W ith respect to Figure 1.15, for constant (direct) voltages and currents, we arrive at a very simple
relationship:
P\-V ab I ab
where
0 -1 0 )
is the power (in W ) absorbed by the circuit element. Consequently, the energy, W , (in
J), absorbed during the time interval Tis
W^=P\xT
(1-11)
Now, let us reconsider Figure 1.15. One can think o f-/ ^ g as flowing from A w B through the
remainder o f the circuit. In this case,
-1 ^ ^ ^ ^ < 0 . This means that the remain­
der o f the circuit absorbs negative power or equivalently delivers |^ 5 (—
| = ^a ^AB
circuit
element 1. As such, the remainder o f the circuit is said to generate electric energy. By definition,
the electric power generated by the remainder o f the circuit is the rate at which it generates elec­
tric energy. From Equation 1.8, this rate equals
--------- ---------- - ^ ab ^ab
Observe that the rate at which the remainder o f the circuit generates power precisely equals the
rate at which circuit element 1 absorbs power. This equality is called the principle o f conserva­
tion o f power: total power generated equals total power absorbed. Equivalently, the sum o f the
powers absorbed by all the circuit elements must add to zero,
Exercise. In Figure 1.15, -Pg =
+ Pq = y^gl^B
^AB^^^AB^ ~
watts, i.e., the remainder o f the circuit absorbs - 1 0 watts o f
power. How much power does circuit element 1 absorb?
A N SW ER: 10 watts
In general, whenever a two-terminal general circuit element is labeled according to the passive sign
convention, as in Figure 1.15, then P =
whereas
P = V^b ^ab ^
> 0 means the element absorbs (positive) power,
absorbs negative power or delivers (positive) power to whatever it
is connected. As a general convention, non-source circuit elements are labeled according to the
passive sign convention. Usually, sources are labeled with the current leaving the terminal labeled
with “+”. For such labeling o f sources, if the product o f the source voltage and the current leaving
the “+” terminal is positive, then the source is delivering power to the network.
Chapter 1 • Charge, Current, Vohage and O hm ’s Law
RULE FOR C A LC U LA TIN G A B SO R B ED PO W ER
The power absorbed by any circuit element (Figure 1.16) with terminals labeled A and B is
equal to the voltage drop from A m B multiplied by the current through the element from A
to 5,
+
V AB
FIGURE 1.16
Exercise. Compute the power absorbed by each o f the elements in Figure 1.17.
-1A
_____________
-2A
_____________
2A
>
< Z3
10V
10V
10V
(a)
(b)
(c)
FIGURE 1.17
AN SW ER: (a) 10 W; (b) - 2 0 W; (c) 20 W
As mentioned, power is the rate o f change o f work per unit o f time. T he ability to determine the
power absorbed by each circuit element is highly important because using a circuit element or
some device beyond its power-handling capability could damage the device, cause a fire, or result
in a serious disaster. This is why households use circuit breakers to make sure electrical wiring is
not overloaded.
Exercise. In Figure 1.18, a car heater is attached to a 12-volt D C voltage source. How much power
can the car heater absorb before the 20 -amp fuse blows.
20 Amp Fuse
FIGURE 1.18 Car heater connected to a 12-volt car battery through a 20-amp fuse.
A N SW ER: 240 watts
19
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
As mentioned earlier, the calculated value o f absorbed power P may be negative. If the absorbed
power P is negative, then the circuit element actually generates power or, equivalently, delivers
power to the remainder o f the circuit. In any circuit, some elements will have positive absorbed
powers, whereas some others will have negative absorbed powers. If one adds up the absorbed
powers o f ALL elements, the sum is zero! This is a universal property called conservation o f
power.
PRIN C IPLE OF CO N SERVA TIO N OF PO W ER
The sum o f the powers absorbed by all elements in a circuit is zero at any instant o f time.
Equivalently, the sum o f the absorbed powers equals the sum o f the generated powers at each
instant o f time.
The 2"*^ edition o f this text contains a rigorous proof o f this principle. For the present, we will
simply use it to solve various problems. The following example will help clarify the sign conven­
tions and illustrate the principle o f conservation o f power.
E X A M P L E 1.5
Light bulbs come in all sorts o f shapes, sizes, and wattages. W a t t l e measures the power consumed
by a bulb. Typical wattages include 15, 25, 40, 60, 75, and 100 W. Power consumptions differ
because the current required to light a higher-wattage (and brighter) bulb is larger for a fixed out­
let voltage: a higher-wattage bulb converts more electric energy into light energy. In Figure 1.19,
the source delivers 215 watts o f power. W hat is the wattage o f the unlabeled bulb?
7?
100V
’ watts
watts
watts
FIG U RE 1.19. Three bulbs connected to a 100-V battery.
S o lutio n
From conservation o f power, the total power delivered by the battery equals the total power
absorbed by all the bulbs. Therefore, the power absorbed by the unknown bulb is
215 - 4 0 - 100 = 75 watts
Exercise. Determine the current / leaving the battery in Example 1.5.
AN SW ER: 2.15 amps
20
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
EXA M PLE 1.6
An electroplating apparatus uses electrical current to coat materials with metals such as copper or
silver. In Figure 1.20, suppose a 2 2 0 -V electrical source supplies 10 A dc to the electroplating
apparatus.
10A
Electroplating
Apparatus
FIGURE 1.20 Electrical source operating an electroplating apparatus.
a)
b)
W hat is the power consumed by the apparatus?
If electric energy costs 10 cents per kilowatt-hour (kW h), what will it cost to operate the
apparatus for a single 12 -h day?
S o lutio n
Step 1. From Equation 1.10, the power consumed is
/> = 220
X
10 = 2 2 0 0 W, or 2.2 kW
Step 2. According to Equation 1.11, the energy consumed per 12-h period is
2.2
X
12 = 26.4 kWh
Step 3. Therefore, the cost to operate is
26 .4
X
.01 = $ 2.64 / day
Exercise. Suppose the electroplating apparatus o f Example 1.6 draws 12 A D C at the same volt­
age. W hat is the cost o f operation for a single 12-h day? W hat is the cost o f operating for a 20
workday month?
AN SW ER: $3,168; $63.36
E X A M PLE 1.7
Each box in the circuit o f Figure 1.21 is a two-terminal element. Compute the power absorbed by
each circuit element. W hich elements are delivering power? Verify the conservation o f power prin­
ciple for this circuit.
Chapter 1 • Charge, Current, Voltage and O hm ’s
21
FIG U RE 1.21 Circuit containing several general circuit elements.
S o lutio n
Step 1. Compute power absorbed by each element. Using either Equation 1.10 or the power con­
sumption rule, the power absorbed by each element is
a)
For element 1 P i = 4 X 1 = 4 W
b)
For element 2 P l = 8 x 2 = 1 6 W
c)
For element 3 ^ 3 = 10 X 1 = 10 W
d)
For element 4
e)
For element 5 P 5 = 2
0
For element 6 Pe = 1 0 X ( - 2 ) = - 2 0 W
14
x
x
(-1)=-14W
2 = 4W
Step 2 . Verify conservation o f power. Since P 4 and Pg are negative, element 4 delivers 14 W, and
element 6 delivers 20 W o f power. T he remaining four elements absorb power. Observe that the
sum o f the six absorbed powers, 4 + 16 + 10 - 14 + 4 - 2 0 = 0, as expected from the principle o f
conservation o f power. Equivalently, the total positive generated power, (14 + 20) = 34 W, equals
the total positive absorbed power, (4 + 16 + 10 + 4) = 34 W.
Exercise. In Figure 1.22, find the powers absorbed by elements 1, 2, and 3.
FIG U RE 1.22
AN SW ER: 8 W, 20 W, - 2 8 W; element 3 equivalently delivers 28 W
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
22
Exercise. In Figure 1.22, suppose the current 2 A were changed to - 4 A. W hat is the new power
absorbed by element 3?
A N SW ER: 56 watts
If the power absorbed by a circuit element is positive, the exact nature o f the element determines the
type o f energy conversion that takes place. For example, a circuit element called a resistor (to be dis­
cussed shortly) converts electric energy into heat. If the circuit element is a battery that is being
charged, then electric energy is converted into chemical energy within the battery. If the circuit ele­
ment is a dc motor turning a fan, then electrical energy is converted into mechanical energy.
N O N -D C PO W ER A N D EN ER G Y C A LC U LA TIO N S
Consider Figure 1.23, where i{t) is an arbitrary time-varying current entering a general two-ter­
minal circuit element, and v{t) is the time-varying voltage across the element. Because voltage and
current are functions o f time, the power p{t) = v{t)i{t) is also a function o f time. For any specific
value o f ^ = ?j, the value p{t^) indicates the power absorbed by the element at that particular
time— hence, the terminology instantaneous power for p{t).
i(t)
Circuit Elem ent
Absorbing Power
p(t)
FIGURE 1.23 Calculation of absorbed power for time-varying voltages and currents for circuit ele­
ments labeled with the passive sign convention; here, power is p{t) = v{t)i{t).
Equation 1.12 extends Equation 1.10 in the obvious way.
p{t) = v{t)i{t)
( 1. 12)
i.e., the instantaneous (absorbed) power p{t), in W, is the product o f the voltage v{t), in V, and
the current i{t), in A, with labeling according to the passive sign convention. This product also
makes sense from a dimensional point o f view;
,
volts X amps =
joules
coulombs
joules
;— - x
=
coulomb
second
second
Knowing the power p{t) absorbed by a circuit element as a function o f t allows one to compute
the energy W{tQ, t) absorbed by the element during the time interval [^q, t > Iq], W[tQ, t) (J) is the
integral o f p{t) (W) with respect to t over [?q, t], i.e..
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
23
W(to,t)^ r p ir ) d r
where
the lower limit o f the integral, could possibly be -oo. For the dc case, p{t) = P (a con­
stant). From Equation 1.13,
t
t
W(tQ,t) = f p ( r ) d T = P f d r = P(t-tQ) = P x T
where T = t - t^, as given in Equation 1.11. If, in Equation 1.13, tg = -oo, then W (-co, t) becomes
a function only o f t which, for convenience, is denoted by
t
W{t)= f p ( r ) d r
L
(1.14)
W{t) = W{—00, t), in joules, represents the total energy absorbed by the circuit element from the
beginning o f time to the present time rwhen p{t) is in watts.
Exercise, a) Suppose the power absorbed by a circuit element over [0,oo) is p{i) =
W (0, oo). b) Now suppose the absorbed power o f the circuit element is
p{t) =
j
j >0
watts. Find
for t > 0 .
•
A N SW ER: 4 J; (4+t) J
Since energy is the integral o f power, power is the rate o f change (derivative) o f energy.
Differentiating both sides o f Equation 1.14 yields the expected equation for instantaneous power.
dW(t)
v m o = P ( o = ^
(1.15a)
or, equivalently, for t > (q,
Exercise. Suppose that for t > 0 , the work done by an electronic device satisfies W{t) = 10(1 —
J- If
the voltage supplied by the device is 10 V, then for t > 0, find the power and current supplied by the
device, assuming standard labeling, i.e., the passive sign convention.
AN SW ER: p{t) = \0e-‘ watts; i(f) = e'‘ A
24
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
EXA M PLE 1.8
In the circuit o f Figure 1.23, the current i{t) and vokage v{t) have the waveforms graphed in Figure
1.24. Sicetch p{t), the instantaneous power absorbed by the circuit element, and then sketch W(0,
t), the energy absorbed over the interval [0 , i\.
FIGURE 1.24 (a) Current and (b) voltage profdes with respect to t for circuit o f Figure 1.23.
S o lutio n
A simple graphical multiplication o f Figures 1.24(a) and (b) yields the sketch o f the curves in
instantaneous power shown in Figure 1.25(a). From Equation 1.13 with
= 0, we have, for 0 <
t< %
,2
p(T)dr - J — c/t = —
0
and for t> 5,
t
5
t
t
W ( 0 , t ) - J p(r)dT = J p{x)dT + J p(r)dT - 5 + J ' d T - 5 + ( t - 5 ) = l
0
0
Figure 1.25(b) presents the resulting graph.
5
5
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
25
(b)
FIGURE 1.25 (a) Profile of the instantaneous power p{t) = v{t)i{t) for the current and voltage wave­
forms of Figure 1.24; (b) associated profde of energy versus time.
6. IDEAL VO LTAG E AND CU RREN T SO URCES
Two-terminal circuit elements may be classified according to their terminal voltage-current rela­
tionships. The goal o f this section is to define ideal voltage and current sources via their termi­
nal voltage-current relationships.
The wall socket o f a typical home represents a practical voltage source. After flipping the switch
on an appliance plugged into a wall socket, a current flows through the internal circuitry o f the
appliance, which, for a vacuum cleaner or dishwasher, converts electrical energy into mechanical
energy. For modest amounts o f current draw (below the fuse setting), the voltage nearly maintains
its nominal pattern o f 120 / 2 sin(120 lit) = 169.7 sin(120 nt) V. This practical situation is ide­
26
Chapter 1 • Charge, Current, Voltage and O h m s Law
alized in circuit analysis by the ideal voltage source symbol shown in Figure 1.26(a), a circle with
a ± reference inside. The symbol is more commonly referred to as independent voltage source.
FIGURE 1.26 Equivalent representations of ideal voltage source attached to a hypothetical circuit.
The waveform or signal v{t) in Figure 1.26 represents the voltage produced by the source at each
time t. The plus and minus (+, —), on the source define a reference polarity. T he reference polari­
ty is a labeling or reference frame for standardized voltage measurement. T he reference polarity
does not mean that v(t) is positive. Rather, the reference polarity (+, - ) means that the voltage drop
from + to - is v{t), whatever its value/sign. Finally, the voltage source is ideal because it maintains
the given voltage v{t), regardless o f the current drawn from the source by the attached circuit.
voltage (V)
V,
1(A)
(b)
FIGURE 1.27 (a) Ideal battery representation of ideal voltage source; (b) v-i characteristic of ideal battery.
Figure 1.2 7 (a) shows a source symbol for an ideal battery. The voltage drop from the long-dash
side to the short-dash side is Vg, with Vjj > 0. In commercial products, the terminal marked with
a + sign corresponds to the long-dash side o f Figure 1.27(a). An ideal battery produces a constant
voltage under all operating conditions, i.e., regardless o f current drawn from an attached circuit
or circuit element, as indicated by the v-i characteristic o f Figure 1.27(b). Real batteries are not
ideal but approximate the ideal case over a manufacturer-specified range o f current requirements.
Practical sources (i.e., non-ideal); voltage sources, such as commercial dc and ac generators; and
real batteries deviate from the ideal in many respects. One important respect is that the terminal
voltage depends on the current delivered by the source. The most common generators convert
mechanical energy into electrical energy, while batteries convert chemical energy into electrical
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
27
energy. There are two general battery categories: nonrechargeable and rechargeable. A discussion
o f the dramatically advancing battery technology is beyond the scope o f this text.
Besides batteries and ideal voltage sources, devices called ideal or independent current sources
maintain fixed current waveforms into a circuit, as illustrated in Figure 1.28. T he symbol o f an
ideal current source is a circle with an arrow inside, indicating a reference current direction. An
ideal current source produces and maintains the current i{t) under all operating conditions. O f
course, the current i{t) flowing from the source can be a constant (dc), sinusoidal (ac), or any other
time-varying function.
FIGURE 1.28 Equivalent ideal current sources whose current i{t) is maintained
under all operating conditions o f the circuit.
In nature, lightning is an example o f an approximately ideal current source. W hen lightning
strikes a lightning rod, the path to the ground is almost a short circuit, and very little voltage is
developed between the top o f the rod and the ground. However, if lightning strikes a tree, the path
o f the current to the ground is impeded by the trunk o f the tree. A large voltage then develops
from the top o f the tree to the ground.
Independent sources have conventional labeling, as shown in Figure 1.29, which is different from
that o f the passive sign convention. Here the source delivers power if p{t) = v{t)i{t) > 0 and would
absorb power i f p{t) =
< 0. A complicated circuit called a battery charger can deliver ener­
gy to a drained car battery. T he car battery, although usually a source delivering power, exempli­
fies a source absorbing power from the charger.
FIG U RE 1.29 Common voltage and current source labeling.
28
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
Another type o f ideal source is a dependent source. A dependent source or a controlled source
produces a current or voltage that depends on a current through or voltage across some other ele­
ment in the circuit. Such sources model real-world devices that are used in real circuits. In the text,
the symbol for a dependent source is a diamond. If a ± appears inside the diamond, it is a depend­
ent voltage source, as illustrated in Figure 1.30. If an arrow appears inside the diamond, it is a
dependent current source, as illustrated in Figure 1.31. In Figure 1.30, the voltage across the dia­
mond-shaped source, v{t), depends either on a current, labeled
through some other circuit
device, or on the voltage
across it. If the voltage across the source depends on the voltage v^,
i.e., v{t) = p
then the source is called a voltage-controlled voltage source (VCVS). If the volt­
age across the source depends on the current z^, i.e., v{t) =
then the source is called a cur­
rent-controlled voltage source (CCVS).
FIGURE 1.30 The right element is a voltage-controlled voltage source (VCVS) if v{t) =
(p is here
dimensionless), or a current-controlled voltage source (CCVS) if v(t) = r^i (r^ here has units of ohm).
Exercise. The voltage across a particular circuit element is
element is
= 5 V, and the current through the
0.5 A, using the standard labeling.
a)
If a V CV S (Figure 1.30) with p = 0.4 were associated with the controlled-source branch,
fmd vit).
b)
If a CCV S (Figure 1.30) with
= 3 £2 were associated with the controlled branch, fmd
v{t).
ANSW ER: a) 2 V; b) 1.5 V
There is dual terminology for dependent current sources. The configuration o f Figure 1.31 shows
a voltage-controlled current source (VCCS), i.e., i{t) = g^v^, or a current-controlled current
source (CCCS), for which i{t) =
r~\
29
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
i(t) =
or
Pi
Q V
or
X
Pi:
-o
FIG U RE 1.31 The right element is a voltage-controlled current source (VCCS) if i{t) =
(g^ has
units o f siemens) or a current-controlled current source (CCCS) if i{t) = |3/^ ((3 is dimensionless).
Source voltages or currents are called excitations, inputs, or input signals. A constant voltage will nor­
mally be denoted by an uppercase letter, such as V, Vq,
cally be denoted by /, /g, /p
V^, and so on. A constant current will typi­
and so on. The units are volts, amperes, and so on. Smaller and larger
quantities are expressed by the use o f prefixes, as defined in Standard Engineering Notation Table 1.1.
Exercise. The voltage across a particular circuit element is
ment is
= 5, and the current through the ele­
= 0.5 A using the standard labeling.
a)
If a VCCS (Figure 1.31) with ^^ = 0.1 S were associated with the controlled-source branch, find
b)
If a CCCS (Figure 1.31) with P = 0.5 were associated with the controlled-source branch, find i{i).
i{i).
AN SW ER; a) 0.5 A; b) 0.25 A
TABLE 1.1. Engineering Notation for Large and Small Quantities
i
w
Name
Prefix
Value
femto
f
10-15
pico
P
10-12
nano
n
10-9
micro
P
10-6
milli
m
10-3
kilo
k
103
mega
M
lO^’
g‘ga
G
109
tera
T
1012
30
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
7. RESISTANCE, O HM 'S LAW, AND POW ER (A REPRISE)
Different materials allow electrons to move from atom to atom with different levels o f ease.
Suppose the same dc voltage is applied to two conductors, one carbon and one copper, o f the same
size and shape. Two different currents will flow. T he current flow depends on a property o f the
conductor called resistance: the smaller the resistance, the larger the current flow for a fixed volt­
age. The idea is similar to water flow through different-diameter pipes (analogous to electrical con­
ductors): for a given pressure, a larger-diameter pipe allows a larger volume o f water to flow and,
therefore, has a smaller resistance than a pipe with, say, half the diameter.
A conductor designed to have a specific resistance is called a resistor. Hence, a resistor is a device
that impedes current flow. Just as dams impede water flow and provide flood control for rivers,
resistors provide a means to control current flow in a circuit. Further, resistors are a good approx­
imate model to a wide assortment o f electric devices such as light bulbs and heating elements in
ovens. Figure 1.32(a) shows the standard symbol for a resistor, where the voltage and current ref­
erence directions are marked in accordance with xhie.passive sign convention. Figure 1.32(b) pic­
tures a resistor connected to an ideal battery.
I
+
R
V
-
(a)
FIG U RE 1.32 (a) Symbol for a resistor with reference voltage polarity and current direction
consistent with the passive sign convention; (b) resistor connected to an ideal battery.
In 1827, Ohm observed that for a connection like that o f Figure 1.32(b), the direct current
through the conductor/resistor is proportional to the voltage across the conductor/resistor, i.e., I
= V. Inserting a proportionality constant, one can write
or, equivalently,
1 = — V —GV
^
(1 .1 6a)
V = R1
The proportionality constant R is the resistance o f the conductor in ohms. The resistance R meas­
ures the degree to which the device impedes current flow. For conductors/resistors, the ohm (Q)
is the basic unit o f resistance. A two-terminal device has a 1-Q resistance i f a 1-V excitation causes
1-A o f current to flow. In Equation 1.16(a), the proportionality constant is the reciprocal o f R, i.e.,
G = HR, which is called the conductance o f the device. T he unit for conductance according to
the International System o f Units (SI) system is the siemen, S. In the United States, the older term
for the unit o f conductance is the mho ^5, that is, ohm spelled backward, which is still widely
used. In this text, we try to adhere to the SI system. If a device or wire has zero resistance {R = 0)
or infinite conductance {G = t»), it is termed a short circuit. On the other hand, if a device or
wire has infinite resistance (zero conductance), it is called an open circuit. Technically speaking,
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
31
a resistor means a real physical device, with resistance being the essential property o f the device.
In most o f the literature on electronic circuits, resistor and resistance are used synonymously, and
we will continue this practice.
O H M 'S LAW
Ohm’s law, as observed for constant voltages and currents, is given by Equation 1.16(b), with
its equivalent form in Equation 1.16(a). However, it is true for all time-dependent waveforms
exciting a linear resistor. Thus, we can generalize Equation 1.16 as
v (0 = « W
or
i{t) = —v{t) = Gv{t)
(1.17b)
R
according to Figure 1.33, whose voltage-current labeling is consistent with the passive sign con­
vention.
i(t)
AO
+
^
v(t)
----- OB
-
FIG U R E 1.33
If either the voltage or the current direction is reversed, but not both, then Ohm’s law becomes
v(t) = -Ri{t). As an aid in writing the correct v-i relationship for a resistor. Ohm’s law is stated
here in words:
For a resistor connected between terminals A and B, the voltage drop from A to B is equal to the
resistance multiplied by the current flowing from A to B through the resistor.
Exercise. Find the resistance R for each o f the resistor configurations in Figure 1.34.
AN SW ER: (a) 12 Q ; (b) 3 Q; (c) 6 Q
-1A
+
R
12V
(a)
4A
-
+
R
12V
-2A
- -
(b)
R
12V
+
(c)
FIG U RE 1.34
Once the voltage and the current associated with a resistor are known, the power absorbed by the
resistor is easily calculated. Assuming the passive sign convention, then combining Equation 1.12 for
32
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
power and Ohms law (Equation 1.17), the instantaneous absorbed power is
9
p{t) = v{t)i{t) = i' ^ { t ) R ^ ^ ^
R
(1.18a)
which for the dc case reduces to
Exercise. Find the power absorbed by each o f the resistors in Figure 1.35.
80
+
12V
(a)
4A
-
R
100
(b)
90
-
12V
+
(c)
FIGURE 1.35
AN SW ER: (a) 18 W; (b) 160 W; (c) 16 W
Equations 1.18(a) and (b) bring out a very important property; a resistor always absorbs power,
dissipating it as heat. Intuitively speaking, electrons that flow through the resistor collide with
other particles along the way. The process resembles the action in a pinball game: the pinball suecessively collides with various pegs as it rolls from a higher to a lower elevation. W ith each colli­
sion, part o f the electron’s kinetic energy is converted into heat as the voltage pressure continues
to reaccelerate the electron.
Electrical energy that is converted to heat or used to overcome friction is usually called a loss. Such
losses are termed /-squared-i? {f-R) losses because o f the form o f Equation 1.18. On the other
hand, a stove’s heating element purposely converts to heat as much electric energy as possible, in
which case, the P-R loss is desirable. This heating effect also proves useful as the basis for the oper­
ation o f fuses. A fuse is a short piece o f inexpensive conductor with a very low resistance and a
predetermined current-carrying capacity. When inserted in a circuit, it carries the current o f the
equipment or appliances it must protect. W hen the current rises above the fuse rating, the gener­
ated heat melts the conducting metal inside the fuse, opening the circuit and preventing damage
to the more-expensive appliance. Oversized fuses or solid-wire jumpers circumvent safe fuse oper­
ation by permitting unsafe operation at overload currents, with consequent electrical damage to
the appliance that may cause overheating and fire.
Resistance o f a conductor depends on the material and its geometrical structure. For a specific
temperature, R is proportional to the length I o f a conductor and inversely proportional to its
cross-sectional area A,
R= p^
(1.19)
^
33
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
where the proportionaUty constant p is the resistivity in ohm-meters (Q • m). T he resistivity o f
copper at 2 0 °C is 1.7 x 10“^ Q •m. Table 1.2 lists the relative resistivities o f various materials with
respect to copper.
Table 1.2 Resistivities of Various Materials Relative to Copper.
Silver
0.94
Chromium
1.8
Tin
6.7
Copper
1.00
Zinc
3.4
Carbon
2 .4 X 10^
Gold
1.4
Nickel
5.1
Aluminum
1.6
*The resistivity of copper at 20 degrees C is 1.7 x 10"^ Qxm.
EXA M PLE 1.9
Sixteen-gauge (16 AWG) copper wire has a resistance o f 4 .0 9 4 Q for every 1,000 feet o f wire. Find
the resistance o f 100 feet o f 16 AWG aluminum wire and 100 feet o f 16 AWG nickel wire. Then
find the voltage across each wire and the power absorbed (given off as heat) by each wire if a 10A direct current flows through 100 feet o f each wire.
S o lutio n
The resistivities o f aluminum and nickel wire relative to copper are 1.6 and 5.1, respectively.
Hence, 100 feet o f aluminum/nickel wire has a resistance o f
(aluminum) 1.6 x 0.4094 = 0.655 Q
(nickel) 5.1 X 0.4094 = 2.088 Q
Given a 10-A current flowing through 100 feet o f copper, aluminum, and nickel wire, Ohm’s law
implies
(copper) V = /?/ = 0 .4 0 9 4 x 10 = 4 .0 9 4 V
(aluminum) V = RI = 0 .655 x 10 = 6.55 V
(nickel) V = RI = 2 .088 x 10 - 2 0.88 V
Finally, from Equation 1.18(b), the absorbed power given off as heat is
(copper) P = V I ^ R I - = 0 .4 0 9 4 x 100 = 4 0 .9 4 W
(aluminum) p = VI = RI^^ 0 .655 x 100 = 6 5.5 W
(nickel) P = VI = RI~ = 2 .088 x 100 = 208.8 W
Notice that every 100 feet o f 16 AWG aluminum wire would absorb 65.5 - 4 0.9 = 2 4 .6 W more
power than copper. And nickel wire absorbs even more power:
^ ^
208.8
■ “ 4 0.94
times more power than copper per unit length. This absorbed power, given off as heat, is why
nickel wire is used for heating elements in toasters and ovens.
34
Chapter 1 • Charge, Current, Vokage and O hm ’s Law
Exercise, (a) If a constant current o f 10 A flows through 1,000 feet o f (16 AWG) copper wire, how
many watts o f heat are generated by the wire?
(b) If the wire o f part (a) were changed to (16 AWG) aluminum, how many watts o f heat would
be generated?
AN SW ER; (a) 409.4 watts; (b) 65 5 .0 4 watts
Temperature also affects resistance. For example, light bulbs have a “cold” resistance and a “hot”
resistance o f more importance during lighting. For most metallic conductors, resistance increases
with increasing temperature— except carbon, which has a decrease in resistance as temperature
rises. Since resistors absorb power dissipated as heat, they should have adequate physical dimen­
sions to better radiate the heat or there must be some external cooling to prevent overheating.
EXA M PLE 1.10
T he hot resistance o f a light bulb is 120 Q. Find the current through and the power absorbed by
the bulb if it is connected across a constant 90-V source, as illustrated in Figure 1.36.
-O 90V —
90V
R =120Q
-O -
FIG U RE 1.36 Light bulb and equivalent resistive circuit model.
S o lutio n
Step 1. From Ohm’s law. Equation 1.16(a),
V
90
/ = - = ----- = 0.75 A
R
120
Step 2 . By Equation 1.18(b), the power absorbed by the lamp is
P = 0.752
X
120 = 67.5 W
Step 3 . C/?eck conservation o f power. T he power delivered by the source is 90 x 0.75 = 67.5 W.
Therefore, the power delivered by the source equals the power absorbed by the resistor. This ver­
ifies conservation o f power for the circuit.
Exercise. In Example 1.10, suppose the battery voltage is cut in half to 60 V. W hat is the power
absorbed by the lamp? W hat is the power delivered by the battery? Repeat with the battery volt­
age changed to 120 V.
AN SW ER; 30 watts; 120 watts
35
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
The following example illustrates power consumption for a parallel connection o f light bulbs.
EXA M PLE 1.11
Figure 1.37 shows four automobile halogen Hght bulbs connected in parallel across a 12-V bat­
tery. Find the following;
(a) The effective “hot” resistance o f each bulb
(b) T he total power delivered by the source
(c) After 700 hours o f operation, the current supplied by the source drops to 11.417 A.
Discover which light bulb has burned out.
ijt )
27 watts
35 watts
50 watts
60 watts
12V
F IG U R E 1.37 Parallel connection of light bulbs.
S o lutio n
(a) From Equation 1.18(b), P - V^IR,
12^
Rxiw -
27
144
^50W =
50
= 5.33D
= 2 .88 Q
144
^ 3 5 W = ^ - 4 .1 1 4 Q
144
- ' 60
= 2.4 Q
(b) The power delivered by the source equals the sum o f the powers consumed by each bulb,
which is 172 W.
(c) Since the current supplied by the source has dropped to 11.417 A, then the power delivered
by the source drops to P;„urcenew ~
^ 11.417 = 137 watts, which is 35 watts less than the ear­
lier-delivered power o f 172 watts. Hence, the 35-watt bulb has gone dark.
Exercise. Repeat Example 1.11 (a) with the battery voltage changed to 48 V and a new set o f light
bulbs whose operating voltage is 48 V.
AN SW ER;
85.333 Q; R^c,^= 65.83 Q; R^q^ = 4 6 .0 8 Q; R(^^^= 38.4 Q..
E X A M PLE 1.12
W hen connected to a 120-volt source, halogen light bulb number 1 uses 40 watts o f power. W hen
similarly connected, halogen light bulb 2 uses 60 watts o f power.
(a)
(b)
Find the hot resistance o f each bulb.
If the two bulbs are connected in a series, as in Figure 1.38 and placed across the 120-V
source, find the power absorbed by each bulb and the power delivered by the source,
assuming the hot resistances computed in part (a) do not change.
36
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
(c)
Find the voltage
and V2 across each bulb.
b u ib l
120V
120 V ^ = i-
bulb 2
(a)
(b)
FIGURE 1.38 Series connection o f two light bulbs and equivalent resistive circuit model.
S o lu tio n
Step 1. Find the hot resistances. The hot resistances o f each bulb are given by
120^
Vt
40
‘ bulb\
= 360 and R 2 =
120 "
= 240
Step 2. Find the current through each bulb, the power absorbed by each bulb, and the power delivered
by the source. The circuit o f Figure 1.38(a) has the equivalent representation in terms o f resistanc­
es in Figure 1.38(b). By definition, in a nvo-terminal circuit element, the current entering each
resistor equals the current leaving. Therefore, the current through each resistor in the series con­
nection is the same, and is denoted /. So the new power dissipated by each bulb/resistor is
^l,new —
Snd P2^new ~ ^2^
To calculate these values, we need to know I. By conservation o f power, the power delivered by
the source is the sum o f the absorbed powers, i.e.,
^ sou rce ~
X I — P\^new
^ 1,new ~
+ -^2^
Hence, dividing through by /,
120 = R^I + R 2I = (Ri + R 2)I = 600/
Therefore,
;=™=o.2A
600
Hence,
= 360 X 0 .2 ^ = 14.4 W ,
= 240 x 0.2^ = 9.6 W,
= 24 W
37
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
Step 3. Find voltages across each bulb. From Ohm’s law,
Vl =7?,7 = 72 V and V2 = ^ 2 ^ = 48
Although involved, the solution o f this problem uses the definition o f a two-terminal circuit ele­
ment and conservation o f power to arrive at the result in a roundabout way. In Chapter 2, we can
more directly arrive at the answers by using Kirchhoff’s voltage and current laws.
A potential problem with series connections o f light bulbs is circuit failure. If one bulb burns out,
i.e., the filament in the bulb open-circuits, then all other lights are extinguished. Parallel circuits
continue to operate in the presence o f open-circuit failures and are easier to fix: only the unlit bulb
must be replaced.
8. V-l CH A RA CTERISTICS OF IDEAL RESISTOR, CO N STA N T
VO LTAG E, AN D CO N STA N T CU RREN T SO URCES
The ideal (linear) resistor is a device
that satisfies Ohm’s law. Ohm’s law is a
relationship between the current
through the linear resistor and the volt­
age across it. A graph o f this relation­
ship is known as the v-i characteristic
o f the resistor. The ideal resistor stud­
ied in this chapter has the v-i charac­
v(V )
teristic given in Figure 1.39. The slope
o f the line in the v-i plane is the value
o f the resistance.
Recall that an ideal voltage source
maintains a given voltage, irrespective
o f the current demands o f the attached
circuit. For constant-voltage sources, as
shown in Figure 1.40(a), this property is
depicted graphically by a constant hori­
FIGURE 1.39 Linear resistor characteristic in which
voltage is the constant times the current through
the resistor.
zontal line (slope equals 0 ) in the v-i plane
(Figure 1.40(b)). This means that the “internal” resistance o f an ideal voltage source is zero.
Further, if
= 0, the voltage source looks like a short circuit because the current flow, generated
by the remaining circuit, will induce no voltage across the source. For now, we must be content
with this brief discussion. Chapter 2 will reiterate and expand on these ideas.
38
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
V
+
V ( + )
•V-
/ Circuit
V
-►I
(a)
(b)
FIG U RE 1.40 (a) Constant source
attached to circuit; (b) v-i characteristic
is a constant horizontal line in the v-i plane.
Analogously, an ideal current source maintains the given current, irrespective o f the voltage
requirements o f the attached circuit. For constant-current sources, as in Figure 1.41(a), this prop­
erty is depicted by a constant vertical line (infinite slope) in the v-i plane (Figure 1.41(b)). This
means that an ideal current source has infinite “internal” resistance. Further, if
= 0, the current
source looks like an open circuit because no current will flow, regardless o f any voltage generated
by the rest o f the circuit. Again, we must be content with this brief discussion until Chapter 2 reit­
erates and expands on the ideas.
V
+
Y
/ Circuit /
(a)
FIG U RE 1.41 (a) Constant source
(b)
attached to circuit;
(b) v-i characteristic is a constant vertical line in the v-i plane.
9. SUM M ARY
Building on a simplified physics o f charge (coulombs), electric fields, and charge movement, this
chapter set forth the notions o f current, i{t) or / for dc, and voltage, v{t) or V for constant volt­
ages. A rigorous treatment would require field theory and quantum electronics. More specifically,
the notions o f current, current direction, voltage, and voltage polarity, a two-terminal circuit ele­
ment (the current entering equals the current leaving), the passive sign convention, power con­
sumption [pit) = v{t)i{t) assuming the passive sign convention], and dissipated energy (the inte­
gral o f power) were all defined. In general, we can say that every circuit element does one o f the
following:
•
•
•
•
Absorbs energy
Stores energy
Delivers energy, or
Converts energy from one form to another
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
39
T he chapter subsequently introduced ideal independent and dependent voltage and current
sources: the voltage-controlled voltage source (VCVS), the current-controlled voltage source
(CC VS), the voltage-controlled current source (VCCS), and the current-controlled current source
(C C C S). A dependent source produces a voltage or current proportional to a voltage across or a
current through some other element o f the circuit. The various types o f dependent sources are
summarized in Table 1.4.
TABLE 1.4 Summary of the Four Possible Dependent Sources.
VCVS
(Voltage-Controlled
Voltage Source, p is dimensionless)
ccvs
(Current-Controlled
Current Source,
is in
ohms)
V CC S
(Voltage-Controlled
Voltage Source,
is in S)
CCCS
(Current-Controlled
Current Source, P is
dimensionless)
-I-
40
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
The chapter keynoted a special two-terminal element, called a resistor, whose terminal voltage and
current satisfied Ohms law, v(t) = Ri{t), where v{t) is the voltage in volts, R is the resistance in
ohms, and i{t) is the current in amperes. The resistor, as defined in this chapter, is a passive ele­
ment, meaning that it always absorbs power,/>{;■) = v{t)i{t) = i?-{t)IR = Rp-{t) > 0 since R>Q. This
absorbed power is dissipated as heat. Hence, the (passive) resistor models the heating elements in
a stove or toaster oven quite well. In addition, the resistor models the hot resistance o f a light bulb.
Throughout the text, the resistor will often represent a fixed electrical load. In a later chapter, we
will discover that it is possible to construct a device with a negative resistance, R<Q, which can
generate power. However, such a device is rather complex to build and requires such things as the
operational amplifier covered in Chapter 4.
The various quantities defined and used throughout the chapter have various units. The quanti­
ties and their units are summarized as follows:
TABLE 1.5 Summary of Units
Charge
Current
Coulomb
Ampere (A)
(C)
Voltage
Resistance
Conductance
Volt (V)
Ohm (Q)
S (Siemens)
mhof3
Power
w a tt
= volt X
am p
Energy
Joule (J)
Throughout this chapter, a number o f examples illustrated the various concepts that were intro­
duced. Some simple resisrive circuits were analyzed. To analyze more complex circuits, one needs
Kirchhoff’s voltage and current laws, which specify how circuit elements interact in a complex cir­
cuit. These basic laws o f circuit theory are set forth in the next chapter.
10. TERM S AND C O N C EPTS
Alternating current: a sinusoidally time-varying current signal having the form A'sin(co?+(j)).
Battery: a device that converts chemical energy into electrical energy, and maintains approxi­
mately a constant voltage between its terminals.
Charge: an electric property o f matter, measured in Coulombs. Like charges repel, and unlike
charges attract each other. Each electron carries the smallest known indivisible amount o f
charge equal to - 1.6 x 10“ '^ Coulomb.
Conductance: reciprocal o f resistance, with siemens (S) (or formerly, mhos) as its unit.
Conductor: a material, usually a metal, in which electrons can move to neighboring atoms with
relative ease.
Conservation o f power (energy): the sum o f powers generated by a group o f circuit elements is
equal to the sum o f powers absorbed by the remaining circuit elements.
Current: the movement o f charges constitutes an electric current. Current is measured in
Amperes. One Ampere means movement o f charges through a surface at the rate o f 1
Coulomb per second.
Current source: a device that generates electrical current.
Dependent (controlled) current source: a current source whose output current depends on the
voltage or current o f some other element in the circuit.
Chapter 1 ®Charge, Current, Voltage and Ohms Law
41
'w '
Dependent (controlled) voltage source: a voltage source whose output voltage depends on the
voltage or current of some other element in the circuit.
Direct current: a current constant with time.
Ideal conductor: offers zero resistance to electron movement.
Ideal insulator: offers infinite resistance to electron movement.
Independent (ideal) current source: an ideal device that delivers current as a prescribed function
of time, e.g., {2 cos(/) + 12}A, no matter what circuit element is connected across its ter­
minals.
Independent (ideal) voltage source: an ideal device whose terminal voltage is a prescribed func­
tion of time, e.g., {2 cos{t) + 12}V, no matter what current goes through the device.
Instantaneous power: the value of p{t) =
at a particular time instant.
Insulator: a material that opposes easy electron movement.
Mho: historical unit of conductance equal to the reciprocal of an ohm.
Ohm: unit of resistance. One ohm equals the ratio of IV to lA.
Ohm’s law: for a linear conductor, the current through the conductor at any time t is proportional
to the voltage across the conductor at the same time.
Open circuit: connection of infinite resistance or zero conductance.
Passive sign convention: voltage and current reference directions, indicated by +, - , and an arrow,
which conform to that shown in Figure 1.15.
Peak-to-peak value: equals 2 K 'm K sin(co^ + (()) of the ac waveform.
Peak value: refers to K m K sin(cor + (|)) of the ac waveform.
Power: rate of change of work per unit of time.
Resistance: for a resistor, v{t) a i{t). The proportionality constant R is called the resistance, i.e.,
v{i) = Ri{t). Resistance is measured in ohms: 1 ohm means the voltage is 1 V when the
current is 1 A.
Resistivity: the resistance of a conductor is proportional to its length and inversely proportional
to its cross-sectional area. The proportionality constant p is called the resistivity of the
material. The resistivity of copper at 2 0 ^C is 1.7 x 10~^ ohm-meters.
Resistor: physical device that obeys Ohms law. There are commercially available nonlinear resis­
tors that do not obey Ohms law. Resistors convert electric energy into heat.
Root mean square (rms) or eflfective value: measure of ac current, which is related to the peak
value by the formula rms = 0.7071if, where K sin(o)^ + (|)) is the ac waveform.
Short circuit: connection of zero resistance or infinite conductance.
Siemens: unit of conductance (formerly, mho) or inverse ohms.
v-i characteristic: graphical or functional representation of a memoryless circuit element.
Voltage (potential difference): positive charge, without obstruction, will move from a higher
potential point to a lower potential point, accompanied by a conversion of energy.
Voltage is measured in volts; 1 volt between two points A and B means that the energy
converted when moving 1 Coulomb of charge between A and B is 1 joule.
V olt^e source: device that generates an electric voltage or potential difference.
Wattage: measure of power consumption.
42
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
PROBLEMS
AN SW ER: (a) -0 .1 2 1 3 C; (b) 121,3 A; (c) 3.75
X
10^'; (d) 1 - exp(-5^) for ? > 0 and 0 for ^ <
0 , from left to right; (e) line segments joining
C H A R G E A N D C U R R EN T
PRO BLEM S
(0,0), (0,1), (2,1), (2,-1), (5,-1), (5,1), (6,1),
(6,2), (7,2), (7,-2), (8,-2).
1.Consider the diagram o f Figure P I.la .
(a)
(b)
(c)
(d)
Determine the charge on 7 .5 7 3 X
10 ^^ electrons.
If this number o f electrons moves uni­
formly from the left end o f a wire to
the right in 1 ms (milli second), what
current flows through the wire?
How many electrons must pass a given
point in 1 minute to produce a current
o f 10 Amperes?
If the charge profile across the cross-sec­
tion o f a conductor from left: to riglit is
given by q{t) = t+ 0.2e'5^- 0.2 C for t >
0 and zero for ? < 0 , plot the profile o f the
current that flows across the botmdary. In
2. For the following questions, draw diagrams
whenever necessary.
(a)
(b)
Determine the charge on 6 .023 X lO '^
electrons.
If this number o f electrons moves uni­
formly from the left end o f a wire to
the right in 1 ms (milli second), what
current flows through the wire?
(c)
(d)
How many electrons must pass a given
point in 1 minute to produce a current
o f 5 Amperes?
The charge profile residing in a vol­
ume V = 10 cm^ is given by q(i) = t +
0.5 sin( 7i:^) C for t> 0 and zero for t <
what direction would the current flow?
0. Plot the current that flows across
the boundary o f the volume for 0 < ?
< 2 sec. In what direction would the
current flow at ? = 1 second? Explain.
3. Reconsider Figure 1.5 in the text in which
changed to ijyf) and
is
is changed to z'^(?). Suppose
(i)
Positive charge carriers move from left
(ii)
Negative charge carriers move from
to right at the rate o f 2cos(10z-) C/s
right to left at the rate o f 6 cos( 10 ?)
C/s
Repeat part (d) for the charge wave­
(a)
Find ij^t) and i^(?) as functions o f time.
form (in coulombs) sketched in Figure
P .l.lb .
(b)
Describe the charge movement on the
wire at the boundaries A and B.
4. (a)
Suppose the charge transported across
the cross section o f a conductor for t >
0 is q(t) = e'' sin( 12071?) C. Find the
current, z(r), t> 0 ,flowing in the con­
ductor.
(b)
The charge crossing a boundary in a
wire is given in Figure P I.4 for t > 0.
Plot the current i(t) through the wire.
See Example 1.2.
43
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
AN SW ER: Q = integral o f current. Hence, Q =
0.4 - 0.2 = 0.2.
7. (a)
(b)
Figure P 1.4 Charge crossing a hypothetical
boundary.
(c)
The current in an ideal conductor is
5. (a)
(b)
given by i{t) = 5 - 3e'^^ - 2e‘^^ A for t
> 0. Determine the charge transferred,
q{t), as a function o f time for t > 0 .
Repeat part (a) for the current plot
sketched in Figure P I .5.
The current in an ideal conductor is
given by i{t) = 2 - cos(2?) A for t
> 0 and 0 for t < 0. Determine the
charge transferred, q{i), as a function
o f time for ? > 0 .
Now suppose the charge transferred
across some surface for ^ > 0 from left
to right is q{i) = 2 - cos(2?) C.
Find the current i(t) through the sur­
face for ^ > 0 from left to right.
Repeat part (a) for the current plot
sketched in Figure P I .7. Again, the
current is zero for t< 0.
i(t) (A)
2--
-t (secs)
-
1- -
Figure P I.7
-
2- -
CHECK: (c) For ,6 > ? > 3 , g (0 = j - 4 t + 12
Figure P I.5
6 . A plot o f the current flowing past point A is
shown on the graph o f Figure P I . 6 . Find the
net positive charge transferred in the direction
o f the current arrow during the interval 0 < ^<
8 . Find i(t) when the charge transported across
a surface cutting a conductor is shown in Figure
P I.8.
6 sec, in Coulombs.
i (amps)
0.1
-t (sec)
10
-0.1--
t-*-t (sec)
Figure P I .6
Figure P I .8
44
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
9. (a)
Find the average value o f the voltage,
12.(a)
2k
T = -
(b)
(0
Hint: See Equation 1.6.
(b)
W hich elements in Figure P I. 12 are
labeled with the passive sign conven­
tion'
v{t) = K cos(cof) over one period,
Find the average value o f the absolute
In the circuit o f Figure P I. 12, volt­
ages, currents, and powers o f some ele­
ments have been measured and indi­
cated in the diagram.
value o f the voltage, v(f) = K cos(cot)
over one period, T = 2u/cl).
5A
+ 3V -
2V -
+ 5V -
-cz>
V O LTA G E, CURREN T, POW ER,
EN ER G Y
“
10. In Figure P I. 10, suppose we know that
A
4A
E
+
4V
Vab = 8 V and Vad = 18 V. Find the values of
il'
7A
3A
0 '
Figure P I. 12
(i)
If element A generates 28
power, find Va (ii) Find the power absorbed by
ment B.
(iii) I f element C generates 6
power, find Vq
(iv) I f element D absorbs 2 7
power, find //).
6 V
+
- 4 V
-
W
ele­
W
W
(v) If element E absorbs 4 W power,
Figure P I. 10
11. (a)
(b)
2A
W hich o f the three elements in Figure
P I. 11 is labeled with the passive sign
convention?
find I e (vi) Find the power absorbed by ele­
ment F.
13. Consider the circuit o f Figure P I. 13.
Find the absorbed powers for each cir­
cuit element in Figure P 1.11.
____________
2A
(a)
___________
20 V
lO V
(i)
(ii)
__ .
■
Find the power absorbed by the circuit
elements 1 and 2 .
(b) Show that the algebraic sum o f the
absorbed powers is zero. Be careful o f
sign.
4A.
3A
2A
C _ 3
20 V
(iii)
Figure PI. 11
A N SW ERS (b): (i) -40 W; (ii) 20 W; (iii) 60 W
6V
1A
3A
lOV
circuit
element 1
+
circuit
6V element 2
16v Q ) 2 A
lOV
Figure P I. 13
45
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
14. (a)
Determine the power absorbed by
+ 80V -
-25A
each o f the circuit elements in Figures
(b)
P 1.14a below.
Show that the algebraic sum o f the
absorbed powers is zero. Be careftil o f sign.
(c)
Repeat parts (a) and (b) for die circuit
of Figure P l.l4 b , where v^{t) = 3 for t>Q, Vjit) = 1 +
^^for t>Q, z'j(?)
for t>Q, and ijit) =
=2 +
for t
> 0.
9A
Figure P I. 15
14A|
+
circuit
15V
element 1
4A
5A ( ^ ^ 2 0 V
lOA
6
+ 10V-
circuit
element 2
5V
lOV
- 5V +
5A
(a)
100(1 - e * ) mA for t > 0.
(a)
How much energy does the element A
absorb for the interval [0 , t] ?
(b)
I f element B is a 5 Q resistor, deter­
circuit
element 2
-v,(t) +
2A
3.334, Rfj = 20 in ohms.
16. In the circuit shown in Figure P I. 16, i(t) =
Figure P I. 14a
I
30
CHECK: Re
(c)
mine the power absorbed at time t,
and the energy absorbed for the inter­
val [0 , t],
W hat is the energy delivered by the
source over the interval [0 , i\?
4V
(b)
Figure P 1.14b
15. In the circuit o f Figure P I. 15, there are
three independent sources and five ordinary
resistors.
(a)
Determine which o f the circuit ele­
(b)
ments are sources and which are resis­
tors.
Determine the value o f the resistance
for each o f the resistors.
25V
©
ti
Figure P I. 16
C H E C K : (b) 125 W, 1 2 5 tJ
17. Suppose energy cost in Indiana is 10 cents
per kwh.
(a)
How much does it cost to run a 100watt T V set 8 hours per day for 30
days?
(b)
How many 100-watt light bulbs run
for 6 hours a day are needed to use
$9.00 o f energy every 30 days?
A N SW ER: (a) 8 cents per day; $2.40 per
month; (b) 5 bulbs
46
Chapter 1 • Charge, Current, Vohage and O hm ’s Law
RESISTA N CE
18. Using Equation 1.19 and Table 1.2 , find
the resistance o f a nickel ribbon having these
dimensions:
length:
width:
thickness:
19. (a)
(b)
(c)
40 m
1.5 cm
0.1 cm
Compute the resistance o f800 feet o f 14gauge copper wire (2.575 Q /1000 ft).
Repeat (a) for 200 feet o f 14-gauge
nickel wire.
If one end o f the copper wire is soldered
to one end o f the nickel wire, find the
total resistance o f the 1000 feet o f wire.
Can you justify your answer?
20. The resistance o f a conductor is function of
the temperature T (in °C ). Over a range o f
temperature that is not too distant from 2 0 °C ,
the relationship between R{ T) and T is linear
Figure P I.21
22. For the circuit o f Figure P 1.22,
= 1Q , and
the input current to the circuit o f is i(^t) =
400sin(207i^) mA for ? > 0 and zero for t <Q.
(a) Com pute
the
instantaneous
power delivered by the source.
Using a graphing program, graph
the power delivered as a function
o f time for 0 < ? < 0.5 s.
(b) Now compute and graph an expres­
sion for the energy dissipated in the
resistor as a function o f for 0 < ? <
0.5 s.
and can be expressed as /?(7) = i?(20)[l + a{T-
20 )] where a is called the temperature coeffi­
cient o f the conducting material. For copper a
= 0.0039 per °C . If the resistance o f a coil of
wire is 21 Ohms at -1 0 ° C, what is the resist­
ance when the wire is operating at 10° C?
AN SW ER: 22.85 Q
A VERA G E VALU E, PO W ER, A N D
EN ER G Y C A LC U LA TIO N S
2 1 . The current through a 500 £2 resistor is
given in Figure P I .21 where /^ = 6 mA.
(a)
(b)
How much total charge is transferred
over the time interval ^ = 0 to ^ = 2 sec­
onds?
How much total energy must a source
deliver over the time interval t = 0 X.Ot
= 2 seconds?
(c)
If /(?) in Figure P I.21 is periodic, with
period equal to 2 seconds, i.e., the indi­
cated waveform is replicated every two
seconds, find the average power absorbed
by the resistor. Use intuitive reasoning.
Figure P I.22
23. The switch S in Figure P I .23 is assumed to
be ideal, i.e., it behaves as a short circuit when
closed, and as an open circuit when open.
Suppose the switch is repeatedly closed for 1 ms
and opened for 1ms.
(a) W hat is the average value o f z(^)?
(b) W hat is the average power delivered
by the source?
Kt)>'
Figure P I.23
A N SW ERS: 0.25 mA. 1.25m W
47
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
24. Repeat problem 23 when the switch is
repeatedly closed for 3 ms and opened for 1 ms.
28. In Figure P1.28, F q = 125 V.
(a)
Suppose bulb A and bulb B each use
100 watts o f power. Find /q and the
CH ECK:
= 1.875 m W
hot resistance o f each bulb.
(b)
Suppose bulb A uses 40 watts o f power
25. In Figure P1.25, Vq = 10 V, and the switch S
and bulb B 60 watts o f power. Find /q
alternately stays at position A for 4 ms and at
and the hot resistance o f each bulb.
position B for 1 ms. Find the average value o f i(tj.
5kn
Figure P I.25
C H E C K : Average current is 1.8 mA.
A PPLIC A TIO N S OF O H M 'S LAW
ANSW ER: (b) /g = 0.8 A,
26 (a)
93.75 Q
W hat is the safe maximum current o f
a 0.25 W, 2 77 k fi metal film resistor
used in a radio receiver?
(b) W hat is the safe maximum current o f
a 1 W, 130 £2 resistor?
(c) W hat is the safe maximum current o f
a 2 kW, 2 £2 resistor used in an electric
power station?
A N SW ERS: (a) 0.95 mA, (b) 87.7 niA. (c)
3 1.6 A
= 62.5 Q ,
=
29. T he power delivered by the source in the
circuit o f Figure P I .29 is 750 watts.
(a)
If /^ = 5 A, determine the value o f R.
(b)
Suppose now that R = 11 Q. Find
Hint: W hat is the power consumed in
each resistor as a function o f /^?
27. In Figure P I .27, Kq = 120 V.
(a)
T he power absorbed by the bulb in
the circuit shown in Figure P I.2 7 a is
Figure P I.29
60 watts. Find the value o f the hot
resistance o f the bulb.
(b)
The power delivered by the source to
the parallel connection o f two identi­
cal bulbs in Figure 1.27b is 150 watts.
Find the hot resistance o f each bulb.
AN SW ER: (a) 6 Q, (b) 4.33 A
30. Consider the circuit o f Figure P I.30. The
power consumed by each resistor is known to be
^2Q ^
watts, PjQ = 48 watts,
= 64 watts,
PjQ = 3075.2 watts, and
= 1944 watts.
8
(a)
(b)
(a)
(b)
Figure P I.27
AN SW ER: (b) 192 D.
For each resistor, determine the indi­
cated voltage or current.
Determine the total power delivered
by the two sources.
48
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
(ii) Determine the total power deliv­
ered by the battery.
20
+ V, -
(iii) Determine the current,
+ V,-
3n
(iv) Assuming that each bulb behaves
as a resistor, determine the hot
40
''" 6
60 <
>50
resistance o f each bulb.
(b)
Figure P I . 3 0
C H E C K : (a)
blow the 15-amp fuse in the circuit o f
= 28 V, Vg = 108 V, (b) 5523.2
Figure P I.3 2 b .
(c)
31. The power absorbed by the resistor R in
the circuit o f Figure P I.31 is 100 watts and
(a)
Find the value o f R.
(b)
Find the value o f the current flow­
ing through R and determine its
direction as per the passive sign
convention.
Find the power absorbed by the 20
Q resistor.
Repeat (b) for C C bulbs.
Ignition
Switch
Ko = 2 0 V
(d)
Determine the number o f AA bulbs in
parallel that would be required to
watts
(c)
deliv­
ered by the battery.
12V
Find the power delivered by the
source and the value o f /q.
Figure PI.32a
12V
Figure PI.32b
200
Figure P1.31.
C H E C K : /q = 6 A.
32. (a)
Consider the circuit o f Figure P I.32a,
which shows three lamps, AA, BB,
and C C in a parallel circuit. This is a
simplified example o f a light circuit on
a car, in your house, or possibly on a
Christmas tree. Halogen bulb AA uses
35 watts when lit, the Halogen Xenon
bulb BB uses 36 watts when lit, and
the incandescent bulb C C bulb uses
25 watts when lit.
(i) Determine the current through
each bulb.
AN SW ERS: (a) (i) 2 .9 167, 3, 2.08; (ii) 6.2 A;
(iii) 96; (iv) 4 .1 1 ,4 , 5.76; (b) n > 6 ; (c) n > 8
33. An automobile battery has a terminal volt­
age o f approximately 12 V when the engine is
not running and the starter motor is not
engaged. A car with such a battery is parked at
a picnic. For music, the car stereo is playing,
using 240 watts, and some o f the lights are on
using 120 watts. W ith this load, the battery will
supply approximately 3 M J o f energy before it
will have insufficient stored energy to start the
car.
(a) W hat power does the battery supply
(b)
to the load?
W hat current does the battery supply
to the load?
49
Chapter 1 • Charge, Current, Voltage and O hm ’s Law
(c)
Approximately how long can the car
remain parked with the stereo and
lights on and still start the car?
C H E C K : (c) 2.31 hours
34. In Figure P I .34, Vq = 24 V, i?, = 4 Q, the
Figure PI.35b
unknown resistance, Rj, consumes 20 watts o f
power. Find
and Rj- (How many possible
D EP EN D EN T SO U RC E PRO BLEM S
36. Consider the circuit in Figure P I .36.
solutions are there?)
(a)
-A /V ^
R,
If 1/ = 6 V, find
and the power in
watts absorbed by the load R^. W hat is
the power delivered by each source?
I'.
(b)
If the power absorbed by the load
resistor is Pj^ = 80 watts, then find V^,
/( and I/.
Figure P I.34
C H EC K : 1 A, 20 Q or ?????
R, = 20n
35. In Figure P I .35, Vq = 48 V.
(a)
Determine the value(s) o f the current
in the D C resistive circuit o f Figure
P I.3 5 a
(b)
(c)
given
that
the
unknown
Figure P I.36
devices absorb the powers indicated.
W hat value o f Vq results in a unique
C H EC K : (a)
solution for I J
(b) 1/ = 4 V
= 60 V,
= 3.6 watts;
If the circuit is modified as shown in
Figure P I.35b , determine the rwo new
values o f the current 1^.
37. For the circuit in Figure P I .37, determine
and
in terms o f /-^,
, R2 and p.
AN SW ER: (a) 10, 2 A; (b) 35.77 V
Figure P1.37
38. Consider the circuit o f Figure P I .38.
(a)
Determine an expression for
F igure P I . 3 5 a
and
the voltage gain
V
G y = ^
Vin terms o f R^, R2 , a , and V-^^.
(b)
If R.^AQ. and a = 0.8, determine the
50
Chapter 1 « Charge, Current, Voltage and Ohms Law
(c)
value of J?2 so that the voltage gain Gy
41. For the circuit o f Figure P 1.41, suppose
= 4.
Given your answer to (b), determine
the power gain, which is the ratio of
the power delivered to
divided by
= 1 0 V.
(a)
(b)
the power delivered by the source.
Find the output voltage and output
current.
Find the voltage gain
Gv =
Gp =
^out
, and the power gain
^
Pin ■
(c)
(d)
Figure P I.38
Find the power delivered by each
source.
Suppose the power absorbed by the 2
k Q output resistor were 80 watts.
Find the power delivered by the input
source,
and the voltage
39. For the circuit of Figure P I .39, suppose
- 100 mA,
50 Q,
10 Q, and
100 Q.
(a)
Find the output voltage and out­
(b)
put current.
Find the current gain,
G/ =
0.1V.
0.21
v,^
Rb
, the v olt^e gain
Figure P I.41
Gy =
, and the power gain
C H EC K : (b)
(c)
10
G p = ^
P-
42. For the circuit of Figure P 1.42, suppose
Findthepowerabsorbedbyeachresistor.
^ 3=
.
I
10i?i. Find the resistor values
so that Gy = —^
I
= 1000
Vin
2001
1000
2kO
2kn
Figure P I.39
40. In problem 39, suppose R^= 1 k Q, 7?2=
C ty
10 Q, and R^= 20 Q and P ^ = 80 watts. Find
^2.
Figure P I.42
CHECK; 4 . 1 mA
C H EC K : i?,= 5 k Q
n
C H A P
KirchhofF’s Current & Voltage
Laws and Series-Parallel Resistive Circuits
A CAR HEATER FAN SPEED -CO N TRO L APPLICATION
One use o f resistors in electronic circuits is to control current flow, just as dams control water flow
along rivers. Ohm’s law, V = RI, gauges the ability o f resistors to control this current flow: for a
fixed voltage, high values o f resistance lead to small currents, whereas low values o f resistance lead
to higher currents. This property underlies the adjustment o f the blower (fan) speed for ventila­
tion in a typical car, as represented in the following diagram.
In this diagram, three resistors are connected in series, and their connecting points are attached to
a switch. As we will learn in this chapter, the resistance o f a series connection is the sum o f the
resistances. So with the switch in the low position, the 12-V car battery sees three resistors in series
with the motor. Th e series connection o f three resistors represents a “large” resistance and heavily
restricts the current through the motor. W ith less current, there is less power, and the fan motor
speed is slow. W hen the switch moves to the Med-1 position, a resistor is bypassed, producing less
52
Chapter 2 • K irchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits
resistance in the series circuit and allowing more current to flow. More current flow increases the
fan motor speed. Each successive switch position removes resistance from the circuit, and the fan
motor speed increases accordingly.
Analysis o f such practical circuits builds on the principles set forth in this chapter.
CHAPTER O U TLIN E
1.
2.
3.
4.
5.
6.
7.
8.
9.
Introduction and Terminology: Parallel, Series, Node, Branch, and so on
KirchhofFs Current Law
Kirchhoff’s Voltage Law
Series Resistances and Voltage Division
Parallel Resistances and Current Division
Series-Parallel Interconnections
Dependent Sources Revisited
Model for a Non-ideal Battery
Non-ideal Sources
Summary
Terms and Concepts
Problems
CHAPTER OBJECTIVES
Define and utilize Kirchhoff’s current law (KCL), which governs the distribution o f cur­
rents into or out o f a node.
Define and utilize Kirchhoff’s voltage law (KVL), which governs the distribution o f volt­
ages in a circuit.
Introduce series and parallel resistive circuits.
Develop a voltage division formula that specifies how voltages distribute across series con­
nections o f resistors.
Develop a current division formula that specifies how currents distribute through a par­
allel connection o f resistors.
'
Show that a series connection o f resistors has an equivalent resistance equal to the sum o f
the resistances in the series connection.
Show that a parallel connection o f resistors has an equivalent conductance equal to the
sum o f the conductances in the parallel connection.
Explore the calculation o f the equivalent resistance/conductance o f a series-parallel con­
nection o f resistances, i.e., a circuit having a mixed connection o f series and parallel con­
nections o f resistors.
Explore the calculation o f voltages, currents, and power in a series-parallel connection o f
resistances.
Revisit the notion o f a dependent source and use a V C C S to model an amplifier circuit.
Describe a practical battery source and look at a general practical source model.
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
53
1. IN TRO D U CTIO N AND TER M IN O LO G Y: PARALLEL, SERIES,
N OD E, BRANCH, AND SO ON
The circuits studied in Chapter 1 were interconnections
o f resistors and sources that were two-terminal circuit
elements. This chapter sets forth K irchhoff’s voltage
law (KVL) and Kirchhoff’s current law (KCL). These
laws govern the voltage relationships and the current
relationships, respectively, o f interconnections o f twoterminal circuit elements.
Some new terminology underpins the statements o f
KVL and KCL. Figure 2.2a shows a series circuit con­
sisting o f a sequential connection o f two-terminal cir­
cuit elements (resistors) end to end. The common con­
FIG U RE 2.1
nection point between any elements is called a node. In
general, a node is the connection point o f one or more circuit elements. Figure 2.2b shows a par­
allel circuit, in which the top terminals and the bottom terminals o f each resistor are wired
together. The common connection point o f the top terminals is a node, as is the common con­
nection point o f the bottom terminals.
An important property o f the series connection o f Figure 2.2a is that all the rwo-terminal elements
carry the same current, in this case
because the input current for each two-terminal element
must equal the exit current. Similarly, in a parallel connection, such as Figure 2.2b, the same volt­
age, in this case, Vj^, appears across every circuit element.
+
node
d-
0
-1-
node 1
Vr
-
b
node 2
(a)
(b)
FIG U RE 2.2 (a) Series connection of resistors with the property that each resistor carries
the same current; (b) parallel connection of resistors with the property that
the same voltage appears across each resistor.
Sources interconnected with circuit elements produce currents through the elements and voltages
across the elements. For example, a voltage source connected across Figure 2.2a would generate a
current 2^ and the voltages
through v^. K irchhoff’s voltage law (KVL) governs the distribu­
tion o f voltages around loops o f circuit elements, as shown in Figure 2.2a. Similarly, a current
source connected across the circuit o f Figure 2.2b would produce the voltage
and the currents
Z] through
K irchhoff’s current law (KCL) governs the flow o f currents into and out o f a com-
54
Chapter 2 • KirchhofPs Current & Voltage Laws and Series-Parallel Resistive Circuits
mon connection point or node, as in the top and bottom connections o f Figure 2.2b. This chap­
ter sets forth precise statements o f these laws and illustrates their application.
A proper statement o f KVL and KCL requires the additional notion o f branch. A branch o f a cir­
cuit is a generic name for a two-terminal circuit element and is denoted by a line segment, as in
Figure 2.3. T he endpoints o f a branch (the terminals o f the circuit element) are called nodes, as
in Figure 2.3a. Ordinarily, however, node means a common connection point o f two or more cir­
cuit elements (branches), as shown in Figure 2.3b.
node A
(a)
(b)
FIG U RE 2.3 (a) Single branch representing a circuit element with terminals labeled as
nodes A and B; (b) interconnection of branches (circuit elements) with common
connection points labeled as nodes A through D.
The voltage polarity and current direction for the branches in Figures 2.2 and 2.3 are labeled in accor­
dance with the passive sign convention; the arrowhead on a branch denotes the reference current direc­
tion, which is from plus to minus. Recall that the + to - does not mean that the voltage is always posi­
tive if measured from the plus-sign to the minus-sign. In general, reference directions can be assigned
arbitrarily. The conventional assignment o f voltage polarity and current direction to voltage and current
soiurces is given in Figure 2.4, which is different from the passive sign convention. Note that with these
conventional assignments, the (instantaneous) power delivered by a source is/^/^) =
power absorbed by a source is
Circuit
Circuit
' 'J O ©
(a)
th^
= -pjeff)-
(b)
FIG U RE 2.4 Conventional labeling o f (a) voltage, and (b) current sources.
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
55
2. KIRCH HOFF'S CU RREN T LAW
Imagine a number o f branches connected at a common point, as at node A o f Figure 2.3b. The
current through each branch has a reference direction indicated by an arrow. If the arrow points
toward the node, the reference direction o f the current is entering the node; if the arrow points
away from the node, the reference direction o f the current is leaving. If a current is referenced as
leaving a node, then the negative o f the current enters the node, and conversely.
KIR C H H O FF'S C U R R EN T LAW (K C L )
S tatem ent 1: The algebraic sum o f the currents entering a node is zero for every instant o f
time.
S tatem ent 2 : Equivalently, the algebraic sum o f the currents leaving a node is zero for every
instant o f time.
The two statements o f KCL are equivalent because the negative o f the sum o f the currents enter­
ing a node corresponds to the sum o f the currents leaving the node. Further, from physics we know
that charge is neither created nor destroyed. Thus, the charge transported into the node must equal
the charge leaving the node because charge cannot accumulate at a node. KCL expresses the con­
servation o f charge law in terms o f branch currents. Moreover, KCL specifies how branch currents
interact at a node, regardless o f the type o f element connected to the node.
Referring to Figure 2.3b, KCL at nodey4 requires that i^{t) +
- i^ii) = 0 for all t. KCL at node
Finally, KCL at node D requires that 25(f) =
B requires that
E X A M P L E 2.1
For the node shown in Figure 2.5, find
FIG U RE 2.5 Connection o f five circuit elements at a single node.
56
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
S o lutio n
By KCL, the sum o f the currents entering the node must be zero. Hence, the current z^(^) =
9cos(2r) - 3cos(2?) - cos(2r) - 2cos(2z) = 3cos(2^) A.
Exercise. 1. Suppose the current through the voltage source in Figure 2.5 is changed to - 2 cos(2?).
Find
AN SW ER: - 4 cos(2^) A.
2. Three branches connect at a node. All branch currents have reference directions leaving the
node. If /j = /2 = 2 A, then find ly
A N SW ER; - 4 A
Two implications o f KCL are o f immediate interest. First, as a general rule, KCL forbids the series
connection o f current sources. Figure 2.6a shows an invalid connection o f two arbitrary current
sources i,(?) and i^i), where z,(?)
i^t). It is invalid because KCL requires that i^{t) = i2 {t). On
the other hand, a parallel connection o f two current sources can be combined to form an equiva­
lent source, as in Figure 2.6b, where
= i^{t) + i2 {t).
(a)
-O
(b)
FIG U RE 2.6 (a) Invalid connection of two arbitrary current sources
when Z](z)
ijii)- Avoid this violation o f KCL; (b) equivalent representation o f a parallel connection
o f rwo current sources in which
= i]{t) +
A second immediate consequence o f KCL is that a current source supplying zero current [i{t) = 0
in Figure 2.7] is equivalent to an open circuit because the current through an open circuit is zero.
An open circuit has infinite resistance, or zero conductance. This means that a current source has
infinite internal resistance. From another angle, a constant current source is represented by a ver­
tical line in the iv plane (see Figure 1.4 lb ). The slope o f the vertical line, which is infinite, deter­
mines the internal resistance o f the source.
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
57
-O
+
v(t)
-O
FIG U RE 2.7 Ideal current source with i{t) = 0 is an open circuit.
A typical application o f KCL is given in the following example.
E X A M P L E 2.2
In the parallel resistive circuit o f Figure 2.8, the voltage across each resistor is 6 cos(z) V. Find the
current through each resistor and the current,
supplied by the voltage source.
'm
6cos(t) V
IQ
<20
<3Q
FIG U RE 2.8 Parallel resistive circuit for Example 2.2.
S o lu tio n
By Ohm ’s law,
/^ l ( 0 = 6 c o s ( 0 A
.
6 c o s (/)
'R l W -------- = 3 c o s ( 0
A
, „ 3 , „ - 5 i 2 5 W = 2 c o s ( ,) A
By KCL,
iinif) = '« l ( 0 + '« 2 (0 +
= 6 cos(r) + 3cos(/) + 2cos(/) = llco s(/ ) A
Exercise. 1. In Figure 2.8, suppose the source voltage is changed to a constant, labeled
Iin in terms o f V-^.
AN SW ER:
Find
58
Chapter 2 • KirchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits
2. Suppose the source voltage in the circuit o f Figure 2.8 were changed to - 1 2 cos(2?) V. Find
AN SW ER: - 2 2 cos(2?) A
Kirchhoff’s current law holds for closed curves or surfaces, called Gaussian curves or surfaces. A
Gaussian curve or surface is a closed curve (such as a circle in a plane) or a closed surface (such as
a sphere or ellipsoid in three dimensions). A Gaussian curve or surface has a well-defined inside and
outside. Figure 2.9 illustrates the idea o f a Gaussian curve for three (planar) situations.
.............
Two
Terminal
Circuit
Elem ent /
i,(t)
i,(t) -A '/'' ............
V
(a)
o
(b)
FIGURE 2.9 Illustrations of Gaussian curves:
(a) enclosure of a two-terminal element; (b) enclosure o f a three-terminal device, such as a transistor;
(c) enclosure of a three-node interconnection with an arbitrary circuit.
For the two-terminal circuit element o f Figure 2.9a, KCL for Gaussian curves implies that i^{t) =
i2 {t), which is precisely the definition o f a two-terminal circuit element. For the three-terminal
device o f Figure 2.9b, KCL for Gaussian curves implies that i^{t) = i^{t) + i2 {t). Finally, for Figure
2.9c, i^ —i^ + if^ = 0. From these illustrations, one might imagine that the use o f Gaussian surfaces
might simplify or provide a short cut to certain branch current computations. T he general state­
ment o f KCL for Gaussian surfaces is next followed by an example that demonstrates its use for
computing branch currents.
59
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
K C L FOR G A U SSIA N CU R V ES O R SURFACES
The algebraic sum o f the currents leaving (or entering) a Gaussian curve (or surface) is zero
for every instant o f time.
E X A M P L E 2 .3
This example shows how the use o f a Gaussian curve or surface can sometimes simplify a calcula­
tion. Figure 2.10 portrays a complicated circuit whose branch currents and voltages are not solv­
able by methods learned so far. Our objective is to find the current
without having to solve a
set o f complex circuit equations.
FIG U RE 2.10 Circuit for Example 2.3, showing a Gaussian surface to compute
directly.
S o lutio n
Using KCL for the indicated Gaussian curve, - 1 .1 5 + / ^ - 0.3 + 0.95 = 0. Equivalently, /^ = 1.15
+ 0.3 - 0.95 = 0.5 A.
In the next chapter, circuits such as the one in Figure 2 .10 are analyzed using a technique called
nodal analysis.
Exercise. 1. Draw a Gaussian surface on the circuit in Figure 2.10 that is different from the sur­
face given but still allows one to compute /^.
AN SW ER; One choice is a circle enclosing the bottom node.
60
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
2. Draw an appropriate Gaussian curve to find / in the graphical circuit representation in Figure
2 . 11.
AN SW ER: 2 A
FIG U RE 2.11 Graph representation of a circuit.
3. KIRCHHOFF'S VO LTAG E LAW
Kirchhoff’s voltage law (KVL) specifies how voltages distribute across the elements o f a circuit.
Before conveying four equivalent versions o f KVL, we first set forth several necessary background
concepts. The first is the notion o f a closed path. In a circuit, a closed path is a connection o f
two-terminal elements that ends and begins at the same node and which traverses each node in
the connection only once. Figure 2.12 illustrates several closed paths. One closed path is A-B-CD-E-A, i.e., it begins at node A, moves to node B, drops to node C, moves through element 4 to
node D, down through element 6 to reference node E, and back through the voltage source to A.
A second closed path is A-B-C-E-A, and a third is B-D-C-B.
A second concept pertinent to our KVL statements is that o f a node voltage with respect to a ref­
erence. A node voltage o f a circuit is the voltage drop from a given node to a reference node. The
reference node is usually indicated on the circuit or is taken as ground. The circuit o f Figure 2.12
has branches labeled 1 through 6 and nodes labeled A through E, with node E taken as the refer­
Vg, Vq and v^. The voltage
denotes
the voltage drop from node A to node E-,
denotes the voltage drop from node D to node E, and
similarly for the remaining node voltages. Node E, being the reference node, has zero as its node
ence node. The associated node voltages are denoted by
voltage.
61
Chapter 2 • K irchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits
+
V.
v„
BD
FIG U RE 2.12 Circuit diagram illustrating (i) three closed paths {A-B-C-D-E-A)\
(ii) the concept of node voltages with respect to a given reference node E,
(iii) the concept of branch voltages
and Vj^-,
and
The concept o f a closed path and the concept o f a node voltage allow us to state our first two ver­
sions o f Kirchhoff’s voltage law.
KIRCH HOFF'S V O LTA G E LAW (K V L)
Kirchhoff’s voltage law can be stated in different ways. Following are two equivalent state­
ments o f the law.
Statem ent 1: The algebraic sum o f the voltage drops around any closed path is zero at every
instant o f time.
Statem ent 2 : For any pair o f nodes j and k, the voltage drop Vjj^ from node j to node k is
given by
at every instant o f time, where Vj is the voltage at node j with respect to the reference and
is the voltage at node ^ with respect to reference. Herey and k stand for arbitrary node indices.
For example, in Figure 2.12, j, k can be any o f the nodes A, B, C, D, or E.
62
Chapter 2 • KirchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits
Referring back to Figure 2.12, for the closed
A-B-C-D-E-A, statement 1 o f KVL requires that
^AB ^ ^BC ^CD + ^DE ^EA ^ O'
for Figure 2.12, from statement 2 o f KVL, the branch
voltages
= v^ - Vg and Vqj^ =
^D- Hence,
= -v^. Thus, by knowing the node
voltages o f a circuit, one can easily compute the branch voltages.
Exercise. 1. Find Vy^g, VgQ and V^^-for the circuit o f Figure 2.1 3 in which we have introduced
the ground symbol at node E to identify the reference node.
AN SW ERS:
= - 3 V, Vg(^=2\ V,
= 18 V
2. Again, with reference to Figure 2.13, find the node voltages V^, Vg, Vq and
AN SW ERS: 2 V, 5 V, - 1 6 V, - 6 V
3. In Figure 2.13, suppose the branch labeled 6 V is now labeled - 1 2 V. Find
AN SW ER: - 3 V
D
FIG U RE 2.13
A third concept needed for two further equivalent statements o f KVL is that o f a closed node
sequence. A closed node sequence is a finite sequence o f nodes that begins and ends at the same
node. A closed node sequence generalizes the notion o f a closed path. Finally, we define the notion
o f a connected circuit. In a connected circuit, each node can be reached from any other node by
some path through the circuit elements. Figures 2 .12 and 2.14 show connected circuits. However,
in Figure 2.14, the sequence o f nodes A-B-C-D-E-A is a closed node sequence but not a closed
path because there is no circuit element between nodes B and C.
+ 2.5V -
+ 10V D
+
V,
F IG U R E 2 .1 4 Simple dependent source circuit for illustrating the concepts o f a connected circuit and a closed node sequence.
63
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
This brings us to our last two equivalent statements o f KVL.
KIRCH HOFF'S V O LTA G E LAW (K V L)
Following are two additional equivalent statements o f KVL.
Statem ent 3 ; For connected circuits and any node sequence, say A -D -B-... -G-P, the volt­
age drop
^AP = ^AD + ^DB + - +
GP
at every instant o f time.
S tatem ent 4 : For connected circuits, the algebraic sum o f the node-to-node voltages for any
closed node sequence is zero for every instant o f time.
Referring back to Figure 2.12, statement 3 o f KVL impHes that
Vab +
=
Referring to Figure 2.14, for the closed node sequence E-A-B-E, V-^ = 10 =
+ Vg = 2.5 + Vg and Vg = 7.5 V. Now, consider the closed node sequence E-C-D-E. For this
sequence,
~
^CD
Equivalently, 30 = 10 + Vj^, in which case
= 20 V. Finally,
consider the closed node sequence, E-B-C-E, which is not a closed path because there is no cir­
cuit element between nodes B and C. Nevertheless, by statement 4 o f KVL, - Vg+ Vb £+ ^ c ~ ^
or equivalently that
= Vg -
= 7.5 - 30 = - 22.5 V.
Exercise. 1. In Figure 2.14, suppose
= 20 V, V^g = 5 V, and
and Vj^.
AN SW ER: Kg = 15 V,
60 V, V^g = 45 V and
= 40 V
10 V and Vj^ = - 3 V. Find v^ q
2. (a) In the circuit o f Figure 2.15, suppose
(b) Suppose Vg = 1 2 0
cos( 1 2 0 tc?),
Vg^ =
18
= 20 V. Find Vg, Vq
cos( 1 2 0 ti?)
and
3 2 cos(1207t^). Find
at ? =
0 .5 s.
(c) Find
when v^ = 1 0 0 V,
= - 1 0 V and
25V.
SC RA M BLED A N SW ER: 85 V, - 1 3 V, - 7 0 V
D
FIG U RE 2.15 Circuit with nodes labeled A through E. Node E is taken as the reference node.
64
Chapter 2 • K irchhoff s Current & Voltage Laws and Series-Parallel Resistive Circuits
Two further implications o f the KVL are o f immediate interest. First, as a general rule, KVL for­
bids the parallel connection o f two voltage sources— say,
and V2 (^)— for which Vj(?)
as illustrated in Figure 2.16a. O n the other hand, two voltage sources in series can be combined
to form a single source, as illustrated in Figure 2.16b, where
{t) = v-^{t) + V2 {t).
FIGURE 2.16 (a) An improper connecdon of voltage sources when v^{t) ^ i>2 (i);
(b) an equivalent representation of two voltage sources connected in series in which
= Vj{i) +
Second, a voltage source supplying 0 V is equivalent to a short circuit, as illustrated in Figure
2.17. Also, the internal resistance o f a voltage source is zero. One can see this by referring to the
fact that in the iv plane, an ideal dc voltage source is represented by a horizontal line, as was illus­
trated in Figure 1.40, The slope o f the line is zero and represents the resistance o f the source. These
ideas are dual to those expressed for current sources earlier.
-O
-o
+
ov
ov
FIG U RE 2.17 A 0 V voltage source is equivalent to a short circuit.
Finally, note that all four KVL statements can be justified using the definition and the notation
for “voltage” drop presented in Chapter 1. The justification is more readily comprehended via the
analogy o f the gravitational field, also developed in that section. Also, observe that KVL holds for
all closed node sequences, independent o f the device represented by each branch o f the connect­
ed circuit. The distribution o f voltages around closed paths can be viewed as a special case o f this
general statement.
4. SERIES RESISTANCES AND VO LTAG E DIVISION
During holidays, one often sees strings o f lights hanging between poles or trees. Sometimes these
strings consist o f a series connection o f light bulbs. Each light bulb contains a filament, a coil of
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
65
wire, that gives o ff an intense light when hot. In a circuit’s perspective, the filament acts as a resis­
tor and has an equivalent hot resistance. The series connection o f bulbs can be modeled by a series
connection o f resistors, with each resistor paired with a specific bulb. Computing the voltage
across each light (a very important type o f calculation) would then be equivalent to finding the
voltage across each o f the resistors in the equivalent circuit model. It is quite common to model
electrical loads, such as a light, by resistors.
E X A M P L E 2 .4
Figure 2.18a shows a voltage source v-JJ) connected to three resistors in series. The objectives o f
this example are to compute
the voltages Vj{t),j = 1, 2, 3, across each resistor, and
the
equivalent resistance seen by the voltage source.
ijt )
(a)
(b)
FIGURE 2.18 (a) Three series resistors connected across a voltage source. By the
definition o f a two-terminal resistor or by the KCL, the current through each resistor is
(b) equivalent resistance
= R^ ■¥Rj + R^ seen by the source, i.e., v-^{t) = Reqii„{t)-
So l u t io n
Step 1. Express the voltage across each resistor in terms o f the input current. For the circuit o f Figure
2.18a, the current through each resistor is i^JJ) by KCL. From Ohm’s law, the voltage across each
resistor is
fo t j = 1, 2, 3.
Step 2. Express v-J^t) in terms ofi-JJ), solvefor i-^<^t), and then compute an expression for v^t) in terms
o f the Rj and v-J^t). By KVL, the source voltage equals the sum o f the resistor voltages, i.e..
(2 . 1)
where we have substituted Rjii„{t) = Vj{t). Dividing Equation 2.1 by (i?j + Rj + -^3) yields
Since v-{t) = RjiiJyt) for j = 1, 2, 3,
66
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
V j{t) =
= --- ---------5
^
—
,2 2^)
Equation 2.2a is a volt^e division formula for a three-resistor series circuit. This formula imphes
that if a resistance R. is small relative to the other resistances in the series circuit, then only a small
portion o f the source voltage develops across it. O n the other hand, if a resistance R. is large rela­
tive to the other resistances, then a larger portion o f the source voltage will develop across it. One
concludes that the voltage distributes around a loop o f resistors in proportion to the value o f each
resistance. The proportion is simply the ratio o f the branch resistance R- to the total series resist­
ance.
Step 3. Compute the equivalent resistance R^^ seen by the voltage source. The equivalent resistance
seen by the voltage source for a resistive circuit is implicidy defined by Ohm s law, i.e., Vi^i) =
^eqhri^^nonzero currents, the equivalent resistance is defined as
Figure 2.18b illustrates the idea o f the equivalence. By Equation 2.1, v^^i) = R^,^ij„{t) = (7?^ +
+
implies that the equivalent resistance is R^^ = R-^ + R^ + Ry This means that from the
perspective o f the voltage source, the series connection o f resistors is equivalent to a single resistor
o f value equal to the sum o f the resistances. A formal discussion o f equivalent resistance and its
generalization (the Thevenin resistance) is taken up in Chapter 6 .
Exercise. In Figure 2.18, suppose R^ = 5R^ and R^ = 2Ry Find R
AN SW ER: R^^ = GR^,
^
^rid
,
and
Example 2.4 suggests some generalizations. Consider Figure 2.19. The first is that the equivalent
resistance R^^ seen by the source is the sum o f the resistors. This means that resistances in series add,
i.e., resistors in series can be combined into a single resistor whose resistance is the sum o f the indi­
vidual resistances.
Req = R \ + R 2+ "' + Rn
Further, since vi^t) = Rjii„{t), a general voltage division formula can be derived as
Ri
R\ +
+ ■■' + Rn
(2 .2 b)
fory = 1, ... , n.
67
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
FIG U RE 2.19 Series circuit o f n resistors driven by a voltage source.
Exercise. In Figure 2.19, suppose each resistor has value Rq. Find the equivalent resistance seen by
the source and the voltage across each resistor in terms o f the source voltages.
A N SW ER: nR^^, v jt)ln
EXA M PLE 2.5
Find the equivalent resistance seen by the source and the voltages
2.20. W hat is the power dissipated in the 14-Q resistor if
and Vj for the circuit o f Figure
= 2 V?
FIG U R E 2.20 Series circuit containing a dependent voltage source.
So l u t io n
From the preceding discussion, R^^ is defined by Ohm’s law, i.e., v-J^t) = Rg^i-Js)Step 1. Express v-^ in terms o f the remaining branch voltages. From KVL,
^in = Vi + V2 + 2vj = 3vi + V2
(2.3)
68
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
Step 2 . Express the branch voltages in terms o f
and substitute into Equation 2.3. To express
and ^2 in
terms of i^^, observe that i^^ is the current through each resistor (KCL or definition o f two-terminal circuit
element) and use Ohms law:
= 2i-^ and Vj = 14/^^. Subsrimting into Equation 2.3 yields
^ in =
Therefore,
20//„ = R e q iin
= 20
Notice that the dependent source increases the resistance o f the two series resistors by 4 Q.
Dependent sources can increase or decrease the resistance o f the circuit. W ith dependent sources,
it is even possible to make the equivalent resistance negative.
Step 3. Find the power absorbed by the 14-Q. resistor. To find the power absorbed by the l4-£2 resis­
tor when
= 2 V, first compute i^^ via Ohm s law; i^^ = v J R
=
= 0 .0 1 x 1 4 = 0.14 W.
= 2/20 = 0.1 A. It follows that P
Exercise. Suppose the dependent source in the circuit o f Figure 2 .20 has its value changed to 2[v-^
+ V2 ). Find R^q.
AN SW ER: 48 Q
5. PARALLEL RESISTANCES AND CU RREN T DIVISION
Many o f the electrical outlets in the average home are connected in parallel. W hen too many
appliances are connected to the same outlet or set o f oudets on the same fused circuit, a fuse will
blow or a circuit breaker will open. Although each appliance uses only a portion o f the maximum
allowable current for the (fused) circuit, together, the total current exceeds the allowable limit.
Because o f this common occurrence, an engineering student ought to know how current distrib­
utes through a parallel connection o f loads (resistors).
To keep the analysis simple, consider a set o f three parallel resistors driven by a current source.
E X A M P L E 2 .6
Figure 2.21a shows a circuit o f three parallel resistors driven by a current source. Our objectives are to
find expressions for v-Jyf), ij^t) in terms o f the input current i-JJ) and the circuit conductances (the recip­
rocal o f the circuit resistances) and the equivalent resistance seen by the current source.
-o
ijt )
0
-I-
- f-
v jt )
ijt )
v Jt)
-O
(a)
(b)
FIG U RE 2.21 (a) Three parallel resistors driven by a current source;
(b) equivalent resistive circuit as seen from source.
Chapter 2 • KirchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits
69
So l u t io n
Step 1. Find expressions for ij{t) in terms o f v-J^t). The variable that Hnks the branch current ij{t) to
the input current i^J^t) is the voltage
which by KVL appears across each resistor. Since v-J^t)
appears across each o f the resistors, Ohm’s law implies that each resistor current is
ij{t) = ^
^
(2.4)
where Gj = HRj is the conductance in siemens andy = 1, 2, 3.
Step 2. Compute v^^i) in terms ofi-J^t). Applying KCL to the top node o f the circuit yields
ii„{t) = /j(?) + z'2 (?) + i^{t)
Using Equation 2.4 to substitute for each ij{t) and then solving for v-^ yields
(j\ + Cj2
■
R]
Rn
^eq
^3
R^
(2.5)
Step 3. Compute ijyi) in terms ofi-J^t). To obtain a relationship between i-J,t) and ij{t), substitute
Equation 2.5 into Equation 2.4 to obtain
1
G,
.
^
± +
+
Rl R2
R3
.
G, ,
G 1 + G 2 + G3
(2 .6)
Equation 2.6 is called a current division formula. It says that currents distribute through the
branches o f a parallel resistive circuit in proportion to the conductance o f the particular branch
G. relative to the total conductance o f the circuit G^^ = G^ + G2 + Gy The greater the conduc­
tance, i.e., the smaller the resistance, the larger the proportion o f current flow through the associ­
ated branch.
Step 4. Compute the equivalent resistance
= G^ v-^, defines G^^ or, equivalently,
seen by the source. As in Example 2.5, Ohm’s law, i-^
From Equation 2.5, G^^ = G j + G2 + G3 is the equiva­
lent conductance o f the parallel circuit, and the equivalent resistance is
R.
1
± +
R,
R2
The idea is illustrated in Figure 2.21b.
R3
1
1
G 1 + G2 + G 3
G ,,
70
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
Exercise. In Figure 2.21a, suppose
through
if
= 1Oe~‘ A.
= 1 Q,
= 0-5
= 0.5 O.. Find the current
and
A N SW ER: z, (?) = 2e“^A
T he Example 2.6 suggests a very important property. Since
hn ={G\ + G2 + G 3 )v,„ = G Vj„
in addition to implying that G
= G^ + G2 + G^ is the equivalent conductance seen by the source
o f the parallel circuit, one can further interpret this to mean that conductances in parallel add to
form equivalent conductances. This parallels the property that resistors in series add to form equiv­
alent resistances. O n the other hand, resistances in parallel do not add, and conductances in series do
not add. We can conclude that from the perspective o f the source, the parallel circuit o f Figure
2.21a has the equivalent representations given in Figure 2.21b.
These ideas generalize to n resistors in parallel, as illustrated in Figure 2.22. In particular, the
equivalent resistance R
o f the parallel set o f resistors in Figure 2 .22 is
Ren = - 1
1
R,
R2
R.,
(2.7)
^eq =
+ *^2 + ••• +
equivalent conductance. Further, the current through each
branch satisfies the general current division formula
R,
G/
1 1
1
+ ---- -!-••• +
R„
^1
^2
rj A
A-•■•j-+ rG „
G
+. r
G2 +
.
G, .
~ Gr^ q
(2 .8 )
■ o
-Iijt )
( f
)
V,
(t)
-o
FIG U RE 2.22 Parallel connection o f n resistors driven by current source.
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
Exercise. Consider Figure 2.22. Suppose ten 10 Q resistors are in parallel. Find
71
and the cur­
rent through each resistor.
AN SW ER:
\ Q. and each current is Q.\i-J^t)
E X A M P L E 2 .7
Consider the circuit o f Figure 2.23 exhibiting a current source driving two parallel resistors. Show
that
R
Ry
■o
©
-O
FIG U RE 2.23 Two resistors in parallel driven by a current source.
So l u t io n
Step 1. Find the equivalent resistance seen by the current source. From Equation 2.7, with « = 2, it
follows that
R^R2
1
Ri ^ Ri
This formula, called the product over sum rule, is quite useful in many calculations.
Step 2. Find i^{t) and i^it)- From Equation 2.8, with w = 2, it follows that
R^
Gi
■
X 7X
■
Ry
and
/2 (0 =
Gi 4- G 2
Ri
,__ Ri + R'^
R2
■
72
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
Exercise. In Figure 2.23, suppose
= 12 A and
Find Rj so that
= 10
= 4 A.
AN SW ER: /?, = 20 Q
E X A M P L E 2 .8
the current ijit) through i?2> and the
For the circuit o f Figure 2.24, find the input voltage
instantaneous power absorbed by Rj when
5e ‘
^ 0
t< 0
o
y r i,(t)
^2
ijt )
U i3 (t)
SG 3
l|i,( t )
<G ,
> = 0 .0 5 U > = 0 .1 5 U > = 0 .0 2 u V = 0 . 0 3 0
-O FIG U RE 2.2 4 Parallel connection of four resistors.
So l u t io n
Step 1. Compute the equivalent conductance and equivalent resistance o f the circuit. Since conduc­
tances in parallel add,
^ e q - G j + G 2 + G 3 + G 4 = 0.25 S
and
Step 2 . Compute v^JJ). From Ohm’s law, the voltage across the current source is
2 0 e~ 'V t^ 0
0
t< 0
Step 3. Compute the current i2 (t). Using the current division formula o f Equation 2.8 yields
3e~'A t > 0
Cjeq
0.25
0
r<0
73
Chapter 2 • K irchhoff’s Current & Voltage Laws and Series-Parallel Resistive Circuits
Step 4. Compute the power absorbed by Rj- To compute the power absorbed by
P 2 (0 = v,„(O x i2 {t) = 1'2 (0 r ^2 =
for t >0,
W
Exercises. 1. For the circuit o f Figure 2.24, find i/^(t) and the power absorbed by i?4 .
A N SW ER: 0.6^-^ A, 12
W
2. In the circuit o f Figure 2.24, suppose each conductance is doubled and i-J^t) = 100 mA. Find
R , V- (?), and the power absorbed by the new Gy
AN SW ER; 2 Q, 200 mV, and 1.6 mW
6. SERIES-PARALLEL IN TERCO N N ECTIO N S
The last two sections covered series and parallel resistive networks. Suppose we take a series circuit
and connect it in parallel with another series circuit; this is a parallel connection o f two series cir­
cuits. Alternately, we could take two parallel circuits and connect them in series. This would result
in a series connection o f parallel circuits. We could also put a series connection o f two parallel sub­
circuits in parallel with a replica o f itself or some other series or parallel circuit. Many other inter­
connections are possible. Arbitrary series and parallel connections o f such subcircuits are called
series-parallel circuits. This section explores the calculation o f the equivalent resistance o f seriesparallel circuits by repeated use o f formulas for series and parallel resistance computation. Related
voltage and current computation is also explored. Example 2.11 presents a practical application o f
series-parallel concepts.
EXA M PLE 2 .9
Find the equivalent resistance, R^^, and the voltage across the source,
the voltages
V2 , the
power absorbed by the 6 kQ resistor, and the power delivered by the source for the circuit o f
Figure 2.25, when
= 20 mA.
FIGURE 2.25 Series-parallel resistive circuit.
74
Chapter 2 • KirchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits
So l u t io n
Step 1. Compute
first compute R^^y and
To compute
8 x 4.8
Kq\ = ------= —
1+ 4 .8
38.4
= 3 kQ
12.e
and
^^■^2
1
1
1 ^ 6
12
6
4
---- 1---- 1--- ------12
The resistance in parallel with the 2k£2 resistor is, say,
Reqh ~ 1000 + R^qX + R^q2 ~ 6
Finally,
R,„ = IkQ. /!R,„. =
2 +6
= 1.5 kQ
Step 2. Compute V-^ . From Ohm’s law.
= Reqhn = 20 X 1.5 = 30 V
Step 3. Compute Vj and V^- By voltage division.
'
Req-i
6
= >5 V and Vj = - ^ v ; „ = ?
6
Step 4. Compute the power absorbed by the 6
resistor.
^ (V 2 )^ ^ 1 0 0
6000
6000
1
60
Step 5. Compute the power delivered by the source.
^source ~ ^inhn — 30 x 0 .0 2 = 0 .6 W
Exercise. 1. W hat is the current through the 2 k£2 resistor firom top to bottom?
AN SW ER: 50 mA
2 . In Example 2.9, suppose the resistance o f each resistor is doubled. Find the new R^ and the
power delivered by the source.
AN SW ER: 3 k n , 1.2 watts
This example points out a very interesting fact: finding the equivalent resistance o f a series-paral­
lel connection o f resistors requires only two types o f arithmetic operations no matter the network
complexity: adding two numbers and taking the reciprocal o f a number. A hand calculator easily
executes both operations. Such is not the case with a non-series-parallel network. To find the
equivalent resistance o f a non-series-parallel network, one usually must write simultaneous equa­
tions and evaluate determinants, a topic detailed in Chapter 3 .
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
75
It is then important to recognize when a problem belongs to the series-parallel category in order to take
advantage o f the simple arithmetic operations. In the previous series-parallel examples, one— and only
one— independent source was specified on the circuit diagram. This is part o f the definition o f a seriesparallel network. The independent source must be indicated, or, equivalendy, the pair o f input termi­
nals to which the source is connected must be specified. The specification o f the input terminals deter­
mines whether or not a network is series-parallel. The following example illustrates the effect of differ­
ent input terminal designations on the computation o f equivalent resistance.
E X A M P L E 2 .1 0
For the circuit o f Figure 2.26a, determine whether or not the network is series-parallel as seen
from each o f the following terminal pairs:
1.
2.
C a s e l;( A , B)
Case 2; (A, C)
3.
Case 3: (C, D)
If the answer is affirmative, give an expression and compute the numerical value for the equivalent
resistance, using the notation // (double slash) for combining resistances in parallel, i.e.,
and
are
parallel, and R^IIR2 llR^ means
is in parallel with
means
vvhich is in parallel with Ry
(a)
>R,
A<
R1
D'
(0
FIG U RE 2.26 (a) From terminals (C, D) the network is not series-parallel. However, from terminals
{A, B) the network is a series-parallel one. (b) Redrawing of the network of (a) as seen from terminals
(A, Q; the resulting network is series-parallel.(c) Non-series-parallel network seen from (C, D).
76
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
So l u t io n
Case 1. Find equivalent resistance seen at (A, B). W ith an independent source connected to nodes
A and B, the source sees a series-parallel network. By inspection o f Figure 2.26a, the equivalent
^
resistance is
i?eq = ^ i //[(^2 +
+ ^ 5)] = 20//[(4 + 6)//(2 + 8 )] = 4
^
I
Case 2. Find equivalent resistance seen at (A, Q. W ith (A, Q as the input terminal pair, the net­
work is again series-parallel. This is made apparent by redrawing the network, as shown in Figure
2.26b, from which
i?eq =
+ [(i ?4 +
= 4//{6 + [(2 + 8)//20]} = 3.0 4 Q
Case 3. Find equivalent resistance seen at (C, U). W ith (C, D) as the input terminal pair, the net­
work is not series-parallel, as can be garnered from Figure 2.26c. T he calculation o f
for this
case requires methods to be discussed in Chapter 3 and is omitted.
Exercise. 1. In Figure 2.26b, suppose
AN SW ER: 3.11 Q
is changed to 40 Q. Find
In electrical engineering laboratories, a student often uses a meter to measure voltages associated
with a piece o f electronic equipment. In older laboratories, or when using an inexpensive meter,
the voltage reading will sometimes differ from what the student calculated or expected to meas­
ure. Typically, this results from the loading effect o f the meter. Using the concept o f series-paral­
lel resistances, the following example explores the phenomenon o f loading.
E X A M P L E 2 .il
Suppose the circuit in Figure 2.27a is part o f a laboratory experiment to verify voltage division. In
this experiment, you calculate the expected voltage Vq and then measure the circuit voltage using
an inexpensive voltmeter.
(a)
Calculate the expected voltage Vq in Figure 2.27a.
(b)
A voltmeter with a 1-kQ/V sensitivity is used to measure
V q.You
use a 0 -10-V range. In
this range, the meter is represented by a 10-kQ resistance, i.e., 10 kD = full-scale reading
meter sensitivity = 10 V x 1 kQ/V. W hat voltage will the meter read?
X
(c)
A better-quality voltmeter with a 2 0 - k H / V sensitivity is used to measure the same volt-
^0’
^ 0 -10-V scale. This better-quality meter is represented by a 2 0 0 -k tl resist­
ance. W hat new voltage will the meter read?
_
77
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
—
_
" h
10 kQ
— r-O
+
15V^
10 kO
20 ko
f-O—
-1-
—
IS v X
>10 kO
20 kO
lOkQ
— r-O-----1-
20 kQ
-o
FIG U RE 2.27 Three circuits for exploring the effect of loading on a circuit: (a) circuit for validating
voltage division; (b) circuit o f (a) with an attached voltmeter having an internal resistance o f 10 k£2;
(c) circuit o f (a) with an attached voltmeter having an internal resistance of 200 k^2.
So l u t io n
(a) Voltage division on the circuit o f Figure 2.27a yields
20
Vo =
(b)
-15 = 10 V
O n the 0-10-V range, the voltmeter internal resistance between the probes is 10 kD, as
stated. This represents a 1G-Id2 load connected in parallel with the 20-kQ resistance, as
shown in Figure 2.27b. The voltage Kg will now change because the 15-V source no
longer sees 10 ld2 in series with 20 kO. Rather, the source sees 10 kQ in series with 6 .67
k ii = 20 kX2//10 kQ. By voltage division,
6 .67
-15 = 6 V
Vo = 10 -H6 .67
This is a 4 0 % deviation from the true answer, V q = 10 V, as calculated in part (a).
(c)
O n the 0 -10-V range with the better voltmeter, the internal resistance between the probes
is 200 kD. As before, this represents a 200-kQ load connected in parallel with the 2 0 -k 0
resistance, as shown in Figure 2.27c. 20 kQ//200 klQ = 18.18 kO. By voltage division,
this yields
18.18
Vb =
-15 = 9 .6 7 7 V
10 + 18.18
This 3.23% deviation is within a reasonable tolerance o f the precise answer o f 10 V.
Example 2.11 demonstrates the effect o f loading due to a measuring instrument, emphasizing the
importance o f choosing a good voltmeter with adequate sensitivity. Although modern-day volt­
meters typically have sensitivities better than 20 kQ/V, a meter with a sensitivity o f 1 kQ./Y is used
in the example to dramatize the effect o f loading.
Exercise. 1. Repeat Example 2.11 if the 20-kQ resistance is changed to 40 kO.
AN.SWER: 12 V, 6.667 V, 1 1.538 V
78
Chapter 2 • KirchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits
2. Th e circuit o f Figure 2.28 shows a voltage divider whose voltage Kq is to be measured by a volt­
meter having an internal resistance o f 80 kO. Find Kq without the meter attached, and then find
the value o f Vq measured by the meter.
20 V
FIG U RE 2.28 Voltage divider circuit.
AN SW ER: 15 V, 13.71 V
7. D EPEN D EN T SO U RCES REVISITED
Chapter 1 introduced the notion o f a dependent or controlled source whose voltage or current
depends on the voltage or current in another branch o f the network, i.e., each source has a con­
trolling voltage or current and an output voltage or current. Figure 2.29 depicts the four types o f
controlled sources designated by a diamond containing either a ± or an arrow:
1.
2.
Voltage-controlled voltage source
(VCVS)
Voltage-controlled current source
(VCCS)
3.
Current-controlled voltage source (CCVS)
4.
Current-controlled current source (CCCS)
An arrow inside the diamond indicates a controlled current source having the reference current
direction given by the arrow. A ± inside the diamond specifies a controlled voltage source, with
the reference voltage polarity given by the ± sign. A parameter value completes the specification
o f a linear controlled source. In Figure 2.2 9 the (constant) parameters are fx, g^, r^, and |3. These
symbols are common to many electronic circuit texts and have useful physical interpretations to
practicing engineers and technicians. For consistency, a ^^-type controlled source is a V C C S and
a jO,-type source is a V C V S, and so on.
79
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
O
+
a ................... o
(a) VCVS or |j-type
O
o
(b) VCVS or g -type
o
O
Pi,
>r
6-
■o
(c )C C V S o rr -type
<;
o - ......... .........o
(d) CCCS or P-type
F IG U R E 2.29 Designations for the various controlled sources.
In practical controllecl sources, the controUing voltage (t>j in Figure 2.29a and b) or current (z'j in
Figure 2.29c and d) is ordinarily associated with a particular circuit element, but not always. For
generality, the controlling voltage
in Figure 2.29a and b is shown across a pair o f nodes. Also,
in Figure 2.29c and d, the controlling current Zj is shown to flow through a short circuit. (Strictly
speaking, neither an open circuit nor a short circuit is a circuit element.) In a real circuit, the cur­
rent may be flowing through an actual circuit element, such as a resistor or even a source.
In Figure 2.29b, once the controlling voltage v-^ is known, the right-hand source behaves as an
independent current source o f value
Since the unit for
is volt, the unit for^^ is amperes per volt, or siemens. Since
is amperes and the unit for
has units o f conductance, and the
controlling and controlled variables belong to two different network branches,
is called a trans­
fer conductance, or transconductance. The other controlled sources have a similar interpretation.
The parameter
has the unit o f resistance, ohms, and is called a transfer resistance. The param­
eter |i is dimensionless because the controlling voltage
has units o f volts and the output vari­
able
must have units o f volts. Similarly, the parameter (3 is dimensionless. The units and asso­
ciation are set forth in Table 2.1.
80
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
Table 2.1 Units and Association.
Type
vcvs
VCCS
ccvs
cccs
Unit
Appellation
dimensionless
Voltage gain
Parameter
siemens
Sm
P
Transfer
conductance
ohm
Transfer resistance
dimensionless
Current gain
Figure 2.29 portrays each controlled source as a four-terminal device. In practical circuits, the
great majority o f controlled sources have one terminal or node in common, making them threeterminal devices. The dashed lines joining the two bottom nodes in Figure 2 .29 suggest this quite
common configuration.
The controlled sources as defined in Figure 2.29 have linear v-i relationships. Controlled sources
may also have a nonlinear v-i relationship. In such cases, the element will be called a nonlinear
controlled source. This text deals only with linear controlled sources.
The next few examples describe some o f the unique features o f controlled sources.
Exercise. Find v^,
and the power delivered by each source in Figure 2.30.
FIG U RE 2.30
AN SW ER: 4 V, 0.05 A, 1.6 W, 0.05 W
E X A M P L E 2 .1 2
This example analyzes the circuit o f Figure 2.31. The independent voltage source in series with
the 3-Q. resistor represents a practical source discussed at greater length later in this chapter. The
circuit within the box o f Figure 2.30 approximates a simplified amplifier circuit by a V C C S. The
8 -Q resistor is considered a load and could, for example, model a loudspeaker. Two important
quantities o f an amplifier circuit are voltage gain and power gain, which are computed here along
with various other quantities.
(a)
(b)
Find the equivalent resistance
Compute / .
seen by the independent voltage source.
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
(c)
Compute /out'
(d)
Compute
(e)
Compute the voltage gain
(f)
Compute the power
(g)
Compute the power delivered by the dependent current source.
(h)
Compute
81
delivered to the amplifier.
the power absorbed by the 8-D. resistor.
Compute the power gain, PgJP;„-
(i)
FIG U RE 2.31 Practical source (ideal independent voltage source in series with a resistor)
driving a simplified VCCS approximation of an amplifier circuit loaded by an 8 -Q resistor.
So l u t io n
(a)
Since resistances in series add,
(b)
(c)
By Ohm’s law, /•„ =
= 0.8 mA.
To compute
one must first compute K j. Here one can use Ohm’s law directly, since
= 3 + 47 = 50 Q.
we know /-^, or one can use voltage division. Doing the calculation by voltage division,
V = — 4 0 X 10'^ = 3 7 .6 X 10'^ V
‘
50
Using this value o f
and current division on the right half o f the circuit yields
0.125
79.8 X 3 7.6 X 1 0 '^ = 2 A
“ 0.125 + 0 .0 6 2 5
(d)
V
follows by Ohm’s law
K,«.= 2
(e)
x
8 = 16V
The voltage gain with respect to the input signal is
^
= —
= 400
0.0 4
82
Chapter 2 • K irchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits
(f)
By Equation 1.18, the power delivered to the amplifier circuit is
p.^ =
(g)
4
= 47
= 47 X 0.82 >< io-<5 = 30.08
The power delivered by the dependent current source is
^VCCS = "^out ^ 79 .8 K j = 16(79.8 x 0.0376) = 48.01 W
(h)
(i)
Pout is simply the product o f voltage and current delivered to the load
The resulting amplifier power gain is the ratio o f the power absorbed by the 8 -Q load to
the power delivered to the amplifier, P-^,
^
Pin
= 1.064 X 10^
3 0 ,0 8
Exercise. Suppose the 8-f2 load resistor in Figure 2.30 is changed to 16 Q. Compute
and the power gain.
AN SW ER: 1.5 A, 24 V, 1.197 X 10<^
The analysis in Example 2.12 required only KCL, KVL, and simple voltage divider and/or cur­
rent divider formulas. More complicated linear circuits necessitate a more systematic approach. To
see this need, add a resistor between the top o f the 47-Q resistor and the top o f the dependent
current source in Figure 2.31. The methods o f solution used in the example immediately break
down because the circuit is no longer series-parallel; hence, one cannot use voltage division to
compute V j. Chapter 3 will explain more systematic methods called nodal and loop analysis.
Unlike a passive element such as a resistor, which always dissipates power as heat, a controlled
source may generate power as computed in part (g) o f Example 2 . 12, or may dissipate power in
other cases. Since a controlled source has the potential o f generating power, it is called an active
element.
In Example 2.12, the practical voltage source delivers 30.08 pW o f power to the circuit, which is
easy to accept because the source could have been a small battery On the other hand, the con­
trolled source generates 48 W. This seems a litde puzzling. Where does the power come from?
W hy not purchase a controlled source at a local electronics store and use it to power, say, a lamp?
Here it is important to recognize that a controlled source is not a stand-alone component picked
o ff the shelf like a resistor. A controlled source is usually constructed from one or more semicon­
ductor devices and requires a dc power supply for its operation. The power delivered by the con­
trolled source actually comes from the power supply. Here, we use the controlled source to math­
ematically model an amplifier and facilitate analysis o f the circuit.
Chapter 2 • K irchhoff s Current & Voltage Laws and Series-Parallel Resistive Circuits
83
W ith simple series-parallel connections o f resistors, the equivalent resistance is always positive.
When controlled sources are present, a strange result may happen, as illustrated in the next exam­
ple.
E X A M P L E 2 .1 3
Find the equivalent resistance
= 2.
for the circuit o f Figure 2.32 when (a) p = 0.5 and (b) p
•O^
FIG U RE 2.32 Calculation of
for a circuit with controlled source for two values o f p.
So l u t io n
W ith p unspecified, we can apply KVL to the single loop, noting that
Consequently, (1 - p)V^ =
= V^. Here,
and
i-x
For p = 0.5, R^q = 2R, which means that the dependent source acts like a resistor o f R Q.. In this
case, it absorbs power. O n the other hand, for p = 2, R,^ = -R , a negative equivalent resistance. In
this case, the dependent source acts like a -2R-Q. resistor and, in fact, delivers power to the inde­
pendent source. An important conclusion can be drawn from this example: in the study o f linear
circuit analysis, controlled sources allow the possibility o f negative resistances. Since a negative
resistance generates power, it is also an active element.
Exercise. In Figure 2.32, find the values o f p so that R^q - 0 .5 R and R^q - 2R.
A N S W E R :-!, 0.5
84
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
Exercise. For Figure 2.33, find
for the following three values of^^: 0.5 mS, 1 mS, and 2 mS.
FIG U RE 2.33
AN SW ER: 2 k£l, open circuit, -1 kQ
8. M O D EL FOR A NON -IDEAL BATTERY
The ideal battery o f Figure 1.30, repeated in Figure 2.34a, delivers a constant voltage regardless
o f the current drawn by a load. T he i-v plane characteristic is a horizontal line through V^, as
shown in Figure 1.40b and repeated in Figure 2.34b. Ideal batteries do not exist in the real world.
The terminal voltage always depends on the supplied current. A more accurate representation o f
a practical battery, but by no means a fully realistic one, is an ideal battery in series with a resist­
ance, say, R^, as shown in Figure 2.34c. R^ is termed an internal resistance, which crudely models
the effects o f chemical action and electrodes inside the battery.
>
Vs
(a)
(b)
FIG U RE 2.34 (a) Ideal battery; (b) i-v battery char­
acteristic; (c) battery model with internal resistance to
crudely approximate effects of chemical action and
presence o f electrodes; (d) nickel-cadmium battery.
(c)
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
85
E X A M P L E 2 .1 4
This example shows the effect o f the internal resistance o f a battery on the terminal voltage.
Suppose a nickel-cadmium battery has an open circuit terminal voltage o f 6 volts. W hen con­
nected across a 2-Q. resistor, the voltage drops to 5.97 V. Find the internal resistance o f the bat­
tery.
So l u t io n
Figure 2.35 illustrates the situation. Here, the dashed box represents the battery model with inter­
nal resistance R^. In Figure 2.35a, no load is connected to the battery. Hence, no current flows
through the internal resistance, in which case, the terminal battery voltage is 6 V.
(a)
(b)
FIG U RE 2.35 Battery model with internal resistance; (a) open circuited (Is = 0); and
(b) connected to a 2 -0 load.
Figure 2.35b shows the battery connected to the 2-Q resistive load. The measured voltage is 5.97
V. By KVL, the voltage across the internal resistance,
law, the current through
is
is Vj^ = 6 - 5-97 = 0.03 V. From Ohm’s
= (5.97/2) = 2 .985 A. Again, by Ohm’s law.
0.03
2.985
= 0 .0 1 0 0 5 Q
Exercise. In Example 2.14, suppose the internal resistance is known to be R^ = 0.005 Q and
although the load resistance is unknown, the load current is 4 A. W hat is the voltage across the
load resistance, and what is the load resistance?
AN SW ER; 5.98 V and 1.495 ^
9. NON -IDEAL SO URCES
Ideal voltage sources have zero internal resistance. Real voltage sources, such as batteries, have an
internal resistance. The value o f this resistance may change with the current load. There may also
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
86
be other effects. However, for our purposes, a more realistic model o f a voltage source contains a
series internal resistance, as illustrated in Figure 2.36a.
L .(t)
(b)
FIG U RE 2.36 (a) A non-ideal voltage source as an ideal voltage source with an internal series resist­
ance; (b) a non-ideal current source as an ideal current source with a parallel internal resistance.
Ideal current sources have infinite internal resistance. Real current sources have a finite, typically
large, internal resistance. Figure 2.36b depicts a more realistic current source model where the
internal resistance is in parallel with the ideal current source.
In the case o f constant voltage and current sources, ideal and non-ideal source models have a
graphical interpretation. The i-v (current-voltage) characteristic o f an ideal constant voltage
source {v^{t) = 1^) is a horizontal straight line. This means that the voltage supplied by the source
is fixed for all possible current loads. An ideal constant current source (z^(z) =
has a vertical
straight line characteristic, which means that the current is constant for all possible voltages across
the source. Figure 2.3 7 illustrates these relationships graphically.
Vout
V
(a)
FIG U RE 2.37 v-i characteristics o f (a) an ideal constant voltage source, and
(b) an ideal constant current source.
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
87
The non-ideal case is quite different. Because o f the internal resistance a non-ideal constant volt­
age source i-v characteristic satisfies the linear relationship
^out
^s^out
Ks'
and for a non-ideal constant current source in which
(2 . 10)
= l/R^,
hut — ^s'^oul
( 2 . 11 )
Equations 2.10 and 2.11 are illustrated by the graphs in Figure 2.38 when v^{t) =
ideal voltage source and i^{t) =
for the non­
for the non-ideal current source. For a voltage source, if the value
o f R; is very small in comparison with potential load resistances, as ordinarily expected, then the
hne in Figure 2.38a approximates a horizontal line, the ideal case. O n the other hand, for a cur­
rent source, the line in Figure 2.38b approximates a vertical line whenever
is much much larg­
er than a potential load resistance. This would then approximate the ideal current source case.
FIGURE 2.38 v-i characteristics of (a) non-ideal constant voltage source, and (b) non-ideal current source.
In a similar way, non-ideal dependent voltage sources are a connection o f an ideal dependent
source with a series resistance. A non-ideal dependent current source is a connection o f an ideal
dependent current source with a parallel resistance.
E X A M PLE 2 .1 6 Figure 2.39 shows the measured voltages o f a dc power supply found in an old
laboratory. Assuming a non-ideal model o f Figure 2.38a, find
the power supply.
and the internal resistance R^ of
88
Chapter 2 • K irch h offs Current & Voltage Laws and Series-Parallel Resistive Circuits
Vout (V)
' '
FIG U RE 2.39 Graph of measured voltages and currents for a dc power supply.
So l u t io n
= -R^
From Equation 2.10, we know that
+ V^. From the graph, when
= 0,
10 V = I/. Further, R^ = - (9.8 - 10)/(0.5 - 0.0) = 0.4 Q.
SUM M ARY
This chapter has presented the essential building blocks o f linear lumped circuit theory, beginning
with the two fundamental laws for interconnected circuit elements: KVL and KCL. KVL states
that for lumped circuits, the algebraic sum o f the voltages around any closed node sequence o f a
circuit is zero. Similarly, KCL says that for lumped circuits, the algebraic sum o f the currents enter­
ing (or leaving) a node is zero.
These laws in conjunction with Ohm’s law allowed us to develop a voltage division and a current
division formula. The voltage division formula applies to series-resistive circuits driven by a volt­
age source. The voltage developed across each resistor was found to be proportional to the resist­
ance o f the particular element relative to the equivalent resistance seen by the source. For exam­
ple, in a two-resistor series circuit, Rj in series with
we found that
Vi =
7^1 + /?2
T he current division formula applies to parallel-resistive circuits driven by a current source. Here,
the current through each resistor with conductance Gi was found to be proportional to G/ divid­
ed by the equivalent conductance seen by the source. Since conductance is the reciprocal o f resist­
ance, the idea can also be expressed in terms o f the resistances o f the circuit. For example, in a tworesistor parallel circuit, Rj is parallel with /?2>
/i =
G,
G 1 + G2
R, +Rn
In deriving the voltage division formula, we learned that the resistances o f a series connection o f
resistors may be added together to obtain an equivalent resistance, prompting the phrase “resistors
in series add.” Analogously, the derivation o f the current division formula for parallel circuits led
Chapter 2 • Kirchhofif’s Current & Voltage Laws and Series-Parallel Resistive Circuits
US
89
to conclude that a parallel connection o f resistors has an equivalent conductance equal to the
sum o f conductances. This is sometimes expressed in terms o f resistances as the inverse o f the sum
o f reciprocal, i.e., for n resistors in parallel,
p .
7^1
R„
which leads to the very special formula for two resistors in parallel,
R
=
often referred to as the product over sum rule.
Dependent sources, first introduced in Chapter 1, were re-examined in greater detail. Some prac­
tical points were described.
All o f the above ideas were applied to the analysis o f series-parallel networks that are interconnec­
tions o f series and parallel groupings o f resistors. Our analysis showed us how to compute the
equivalent resistance o f series-parallel circuits. An example was given that described the applica­
tion o f these ideas to voltage measurement. This was followed by a discussion o f battery models
and battery usage. Finally, battery modeling ideas were used to describe non-ideal source models.
12. TERM S AND C O N C EPTS
Branch: a two-terminal circuit element denoted by a line segment.
Closed node sequence: a finite sequence o f nodes that begins and ends with the same node.
Closed path: a connection o f devices or branches through a sequence o f nodes so that the con­
nection ends on the node where it began and traverses each node in the connection only
once.
Connected circuit: one for which any node can be reached from any other node by some path
through the circuit elements.
Current division: the current in a branch o f a parallel-resistive circuit is equal to the input cur­
rent times the conductance o f the particular resistor, Gj, divided by the total parallel con­
ductance o f the circuit,
= G^ + ... + G^.
Dependent (controlled) current source: a current source whose output current depends on the
voltage or current o f some other element in the circuit.
Dependent (controlled) voltage source: a voltage source whose output voltage depends on the
voltage or current o f some other element in the circuit.
Gaussian surface: a closed curve in the plane or a closed surface in three dimensions. A Gaussian
surface has a well-defined inside and outside.
Kirchhoff'’* current law (KCL): the algebraic sum o f the currents entering a node o f a circuit
consisting o f lumped elements is zero for every instant o f time. In general, for lumped
circuits, the algebraic sum o f the currents entering (leaving) a Gaussian surface is zero at
every instant o f time.
90
Chapter 2 ° KirchhoflF’s Current & Volt:«e Laws and Series-Parallel Resistive Circuits
Kirchho£F’$ voltage law (KVL); for lumped circuits, the algebraic sum of the voltage drops
around any closed path in a network is zero at every instant of time. In general, for
lumped connected circuits, the algebraic sum of the node-to-node voltages for any closed
node sequence is zero for every instant of time.
Node: the common connection point between each element; in general, a node is a connection
point of one or more circuit elements.
Node voltage: the voltage drop from a given node to the reference node.
Parallel circuit: a side-by-side connection of two-terminal circuit elements whose top terminals
are wired together and whose bottom terminals are wired together.
Series circuit: a sequential connection o f two terminal circuit elements, end-to-end.
Voltage division: each resistor voltage in a series connection is a fraction o f the input voltage equal
to the ratio of the branch resistance to the total series resistance.
// (double-slash): notation for combining resistances in parallel, i.e., R^UR2 means
and i ?2 are
in parallel, and R^IIR2 llRj^ means R^ is in parallel with
which is in parallel with Ry
r\
r\ .
r\
o
n
o
n
n
n
91
Chapter 2 • K irchh off s Current & Voltage Laws and Series-Parallel Resistive Circuits
PROBLEMS
KIR C H H O FF'S C U R R EN T LAW
1. (a)
Find the value o f /j for each o f the
node connections in Figure P2.1a and
P2.1b given that 1^ = 2 A,
Figure P2.3
1^ = 3 A, and
(b)
= 4 A.
Repeat part (a) when l 2 = I^ = 1^ = 2 A.
A N SW ER: ( b ) 4 A
4. (a)
Find the value o f /j in the circuit o f
Figure P2.4.
(b)
Find the value o f
in the circuit o f
Figure P 2.4 by a single application o f
KCL.
Figure P2.1
A N SW ERS: (b) 0, 2 A
2.
In the circuit o f Figure P2.2, each shaded
box is a general circuit element.
= 20 mA, 1-^2 = ^0 mA,
(a)
Suppose
(b)
I ini = 100 mA> and /-^4 = 0.05 A.
Apply KCL to find /j, Ij, ly and 1^.
Repeat part (a) when
= hni
=
Figure P2.4
A N SW ER: (a) 6 A
KIR C H H O FF'S V O LTA G E LAW
100 mA.
5. (a)
Consider the circuit o f Figure P2.5a
where each branch represents a circuit
element. Find Vj and 1^2-
(b)
Find Kj and
unspecified
for Figure P2.5b. Each
branch
represents
an
unknown circuit element.
.1 0 0
40V
lOOV
Figure P2.2
+ V, -
6
3. (a)
For the circuit o f Figure P2.3a,
V,
-
+
VI
Reference Node
5A
Reference Node
(a)
find the value o f the current /j using
N
lO O v C ”^
+
40V
AN SW ERS: (b) (scrambled) 200, -300,
-200, -300 mA
,
(b)
Figure P2.5
only a single application o f KCL.
(Hint: Construct a Gaussian surface.)
(b)
(c)
Repeat part (a) for 1^
True-False:
can be uniquely deter­
mined as in part (a) and part (b).
6 . (a)
For the circuit o f Figure P2.6, deter­
mine the voltages f j, ... , ^4 and the
power absorbed by each resistor.
92
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
(b)
Now determine the node voltages V^,
Vg, Vq and
with respect to the
reference
indicated
node
by
the
ground symbol.
(c)
Compute
50 V
SC RA M BLED AN SW ERS: (a) 2 V, - 2 V
K C L A N D KVL
9. (a)
Consider the circuit o f Figure P2.9a. Use
KCL and KVL to find the voltage across
each current source from the arrow head
Figure P2.6
to the arrow tail and the current through
A N SW ERS: (a) 180 V, 50 V, -110 V, 10 V
each voltage source from minus to plus.
Finally, find the power delivered by each
7. (a)
Find the values o f the voltages Vp Kj,
source and verify conservation o f power,
and Vj in the circuit graph o f Figure
(b)
P2.7, where each branch represents a
For the circuit o f Figure P2.9b, find
the voltages
and V^.
circuit element.
(b)
Now determine the node voltages
K l’ ^B’
respect
to the reference node indicated by
in
iUV
6^
4A
30V
3A
10
the ground symbol.
(c)
Compute
and
(a)
(b)
Figure P2.9
AN SW ERS: ( b ) - 1 7 V 14 V
10.
For the circuit in Figure P 2.10, calculate
the power delivered by each o f the eight inde­
pendent sources. Verify the principle o f con­
servation o f power.
Figure P2.7
SC RA M BLED AN SW ERS: (a) -65 V, 15 V, 15
V; (b) 35 V, - 4 5 V, 10 V, - 5 V
8 . (a)
Use KVL to determine the voltages
(b)
1/ and V; in the circuit o f Figure P2.8.
Now compute V^g.
Figure P2.10
A N SW ERS: - 4 , - 9 , - 3 6 . 35, 10, 0 , 10, - 6 W
93
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
11.
(b)
Four circuit elements and a dependent
Supposing that /j = 4 A, /, = 2 A, and
and the power
a = 0.25, determine
delivered to Rj^.
voltage source are shown in the circuit o f Figure
P 2 .11. The current through and the voltage
across each element are identified on the dia­
gram. However, one— and only one— voltage
(or current) value is labeled incorrectly. Mark
the incorrect voltage (or current) on the circuit
diagram and give the correct value for this volt­
age (or current).
Figure P2.14
A N SW ERS: (b) 80 Q, 20 W
15.
Consider the circuit in Figure P2.15 in
= 1 A and R^ = 84 Q. Find the value
which
+
o f R^ for each o f the following cases:
25V
(a)
The power delivered by the source is
(b)
T he power absorbed by
(c)
watts.
The power absorbed by R^ is 13.44
13.44 watts.
Figure P 2 .ll
Find the currents and voltages /^, V^, 1^,
12.
and
is 13.44
watts.
in the circuit o f Figure P2.12.
Figure P2.15
SC RA M BLED A N SW ERS: 21, 16, 336, 56
Figure P2.12
KCL, KVL, A N D O H M 'S LAW
13.(a)
16 . Consider the circuit o f Figure P2.16.
(a)
(b)
If /? = 5 Q, find Vjf
Find the value o f R when
For the circuit o f figure P2.13, sup­
= 875 mA and Rj_ = 80Q .
pose
Find V-^, /j, /j, and ly
(b)
= 40 V.
Now suppose that
= 7 A and -^2 ~ ^
A. Find V-^, ly Rj^, and the power
120V
o
4on
delivered to the load Rj^.
50V
o
O
20V
Figure P2.16
+
V.
AN SW ER: (b) 25 Q
200
,4on
Figure P2.13
C H E C K : (b) P^= 160 watts
I4.(a)
For the circuit o f Figure P2.13, deter­
mine
in terms o f /j, a and Rj^.
17. For the circuit o f Figure 2.17, find
(a)
the voltage Vj and the power absorbed
by the 10 Q resistor;
(b)
(c)
the voltage V2 ,
the power delivered by each source.
94
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
R
>J .
10 k n ,
6kO
Figure P 2.17
AN SW ER: (c) 12.5 watts and - 7 .5 watts
'+
■V,
(a)
Figure P2.20
18. Find the power absorbed by the unknown
circuit element x and the voltage
in the cir­
cuit o f Figure P2.18.
21.
500
50 V
6x 10"
ANSWERS: (b) Vi =
3V
'in
4 8 V 4 kQ
i 6xl 0' -i-/?-i-6axl0
The circuit o f in Figure P2.21 is a blower
motor control for a typical car heater. In this circuit,
6
+
0.8 A
<----
resistors are used to control the current through a
motor, thereby controlling the fan speed.
Figure P2.18
19. (a)
(b)
Ignition
Switch
JUA
Find the current 7/j and the voltage
Kuf
circuit o f Figure P 2 .19.
If a resistor o f 7? Q is placed across
the output terminals, determine the
current
and the voltage
and
12V
Chassis .
Ground
the power delivered by the 10 V
14V
44 V
6
'r
©
3R
10V
(a)
W ith the switch in the Lo position,
the current supplied by the battery is
source.
2.5 A. The voltage drops across the
Figure P2.19
C H EC K : (b) K^„,= 4 V
20. (a)
^BC= 1-5 V, Vc/j = 0.625 V, and Vq
= 3 .125 V. Consider the motor as rep­
In Figure P2.20a, Vj = 32 V and the
resented by a load resistance.
power delivered by the source is 80
(i)
mW. Compute
(b)
resistors and motor are Vjg = 6.75 V,
, V-^, and R.
In Figure P2.20b a dependent voltage
source has been added to the circuit o f
Figure P2.20a. Suppose
Determine
= 40 V.
in terms o f a and R. If
= 0.8 mA and a = 5, find K, and R.
(c)
Determine the value o f each resist­
ance and the value o f the equivalent
resistance representing the motor.
(ii) Determine the power dissipated
in each resistor and the power
used by the motor.
(iii) Determine the relative efficiency o f
For each circuit o f Figure P 2.20,
the circuit, which is the ratio o f the
determine the resistance seen by the
voltage source,
power used by the motor to the
power delivered by the battery.
95
Chapter 2 • KirchhofFs Current & Voltage Laws and Series-Parallel Resistive Circuits
(b)
W ith the switch in Med-1 position,
delivered is 1250 W. How many possible medi­
determine:
um wattages are there and what are they?
(i)
C H EC K : 10 ohms, 40 ohms
T h e voltage drop across each
resistor.
(ii) The current delivered by the battery.
(iii) The relative efficiency of the circuit.
(c)
(a)
Suppose /? = 20 Q, find the power
(b)
Suppose the power delivered by the
delivered by the current source.
Repeat part (b) with the switch in
position Med-2.
(d)
24. Consider the circuit o f Figure P2.24.
The switch is in the high position. A
current source is 120 watts. Find the
winding in the motor shorts out. The
value o f R.
fuse blows. W hat is the largest equiva­
lent resistance o f the motor that will
cause the fuse to blow?
A N SW ERS:
(a)
(i)
Rj^g = 2.7 Q.,
=0.6 0., RcD
“ (t)
^^^lOOV
= 0 .2 5 Q ,
1.25 a
Figure P2.24
= 16.875 W Pbc = 3.75 W
1.5625 W P , , „ , , = 7.8125
(ii)
C H E C K : (b) 8 </?< 15
W
(b)
(iii) 26%
(i)K^5=0> ^s c = 3 -4 3 V ,
1-43
25. Given that 4 W is absorbed by the 100-Q
=
= 7.14 V
resistor, find V} and the power delivered by the
voltage source in the circuit o f Figure P2.25.
(ii) 5.71 A
(iii) 59.5%
(c)
(i)
=0.
150 Q
2o on
Vcn = 2V,
Vmotor = 10 V
300n'
lo o n
(ii) B A
(d)
(iii) 83.3%
0.4 Q
22. Suppose one has two resistors /?j = 20 Q
and i ('2 = 20 Q that can be conected to a source,
= 100 V. By connecting the resistors to the
Figure P2.25
26. In the circuit o f Figure P2.26, suppose V2 =
60 V. Find
/^, and the power delivered by
the source.
source in different ways, what are the different
i8 o n
wattages that can be delivered by the source?
6on'
The different types o f connections represent
what might occur in an electric space heater
having a low, medium, and high setting.
''• 6
4o n
9o n
1 800
C H EC K : There are three possible connections
with medium using 500 watts.
23. In Problem 22, find the values o f R^ and
so that the lowest wattage delivered by the
100 V source is 200 W and the highest wattage
Figure P2.26
SCRA M BLED AN SW ERS: 3 360, 840, 4
96
Chapter 2 • Kirchhofif’s Current & Voltage Laws and Series-Parallel Resistive Circuits
27. Find the power delivered by each independ­
ent source and the power absorbed by each resis­
tor in the circuit o f figure P2.27. (Check; Total of
delivered power = total o f absorbed power.)
0.7 A
lOon
500
0.8 A
20V
6
C H EC K : (a) 45 <
(c) 8 0 < i? ^ < 125
200
< 65; (b) -85 < 1^2 < '6 5 ;
Figure P2.27
SC RA M BLED A N SW ERS: 59.5, 9, 8, 49.
30. Consider the circuit o f Figure P2.30.
0.5, 45, 18
(a)
28.
(b)
Write an equation for
in terms o f a
and 4
For the circuit o f Figure P2.28 with the
indicated currents and voltages, find
(a)
(b)
(c)
If 1/ = 40 V and a = 0.5, find the value
o f the current
(c)
Currents /j through
How much power is delivered by the
Voltages Vj through
independent
Power delivered by each independent
power is delivered by the dependent
source?
How
much
current source
source? Verify the principle o f conser­
vation o f power for this circuit.
5 mA
500 0
2000
Figure P2.30
C H EC K : 4 = 0.05 A
Figure P2.28
C H E C K : ^2 = 12 V, /^ = 1 mA,
= 60m W
29. For the circuit o f Figure P2.29, find
V O LTA G E A N D C U R R EN T
D IVISIO N
31. Consider the circuit o f Figure P2.31 in
(a)
Voltage drop V j, and
which
(b)
Voltage drop V2
V2 for each o f the following cases:
(c)
The value o f the unspecified resist­
(a)
ance, R
= 30 V and
= 20 V. Find
and
Switch 5j is closed and switch S2 is
open.
(b)
Switch
is open and switch S2 is
closed.
(c)
Switch
closed.
is closed and switch S2 is
97
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
34. For the circuit o f Figure P2.33,
(a)
= 120 V
= 120 Q. Find the value o f
Suppose
that is necessary to achieve V-^ = 90
V. Compute
(b)
Find the values o f R^ and Rj that are
necessary to achieve
Figure P2.31
= 100 V and
1/2 = 80 V.
32. Construct a series voltage divider circuit
wiiose total resistance is 2400 Q as illustrated
in Figure 2.32.
(a)
Suppose Vj = 0 . 7 5 and Vj = 0 .2 5 V^.
Find the values o f R-^,
(b)
^s'
Suppose V"j = 0 . 8 and V2 = 0.5V^.
Find the values o f
(c)
6on
R2 , and R^.
Suppose K, = 0.81/ and Kj = 0.5
Find the values o f 7?,, R2 , and R^.
Figure P2.34
C H EC K ; (b) 60 Q, 240 Q
35 . Figure P2.35 shows a Wheatstone bridge
circuit that is commonly used in a variety o f
measurement equipment. The bridge circuit is
said to be balanced if R J i j = Ri,R^- In this case,
the voltage
= 0 for any voltage V-^^.
(a) Use voltage division to compute the
voltages
Figure P2.32
and V^. Check:
AN SW ER: (c) R^=R^ = 960 Q, R^ = 4 80 Q
Rc + Rci
33. For the circuit o f Figure P 2.33, suppose
= 48 V.
(a)
Find
(b)
about
with the switch in position A,
i.e., the switch is open.
(b)
Find
with the switch in position B,
= 0, then what must be true
and VJ. Show that
=0
if and only if R^R^ = Rh^cSuppose that RJi^ = Rf^R^ and a 0.5D. resistor is connected across the
If
(c)
i.e., the switch is closed.
terminals. Then
6Q
(d)
find the current
through the 0.5-^^ resistor.
Suppose
= 18 V,
= 3 Q,
= 2 Q, an d R ^ = 2^ - Find
Figure P2.33
C H E C K : (b) 25 <
< 30 V
AN SW ER: (d) 3 V
=6
98
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
36. Find V^, V , and
for the circuit o f Figure
P2.36 when 1/^ = 50 V and V^ 2 = 25 V.
> ;i,
lokn
60kn
Figure P2.39
lokn
C H EC K :
= 6.4 watts
40. Find /p /2>l y V-^, and the power delivered
by the source in the circuit o f Figure P2.40
when I-^ = 120 mA.
Figure P2.36
SC RA M BLED A N S W E R S :- 5 , 10, 15
37. Consider the circuit o f Figure P 2.37 in
-o
which /.^ = 0.1 A, Gj = 2.5 mS,
(a)
i',
W ith the switch in position A, find
"^^d the power
delivered by the source.
(b)
’4kn
'• ©
Kq' h
6kn
9kn
Repeat part (a) if the switch is in posi­
tion B.
J 8 kO
-O Figure P2.40
C H E C K : V.^ = 360 V
41. Find /[ and I 2 for the circuit o f figure P2.41
when
= 10 mA, 1^^ = ^
and
= 14 mA.
Figure P2.37
SC RA M BLED AN SW ERS: (a) 50, 0 . 5 ,
5, 50, 20
'■ 7
38. In the circuit o f Figure P2.38, it is required
that /j = 0.81/^-^. Find R (in Q ),
in terms
o f /• , and V- in terms o f /■ .
r ’’ '“0f
>©
Y
30mS>
Figure P2.41
42. For the circuit o f Figure P2.42, find the cur­
rents /p /2, ly and
-O V
'
when
= 300 mA.
I
j"
3oon> 6 o o n /l
I >i 2on
-o Figure P2.38
i 2on >4on
CH ECK:
= 40/,„ and
39. In the circuit o f Figure
P 2.39,'
o
1
in
= 10 mA.
Find /p /j, V and the power delivered by the
dependent source.
Figure P2.42
SC RA M BLED AN SW ERS: - 5 0 , 40, 80 (in
mA)
N.
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
99
Rpo AND RELATED CALCULATIONS
OF SERIES-PARALLEL CIRCUITS
43.
3000
For each o f the circuits o f Figure P 2.43,
find the value o f
3000
5000,
and the power delivered „
^
" V
if a 10-V source were connected.
5000 .
1.5kO<
J 7500
i6kn
lkO<
Ik0<
Figure P2.45
(b)
SC RA M BLED AN SW ERS: (b) 29.63, 675,
0.1185
46.
Find
for each o f the circuits in Figure
P2.46. Notice that the circuit o f (b) is a modifica­
tion o f (a) and that o f (c) a modification o f (b).
2kn
15kn
Figure P2.43
A N SW ERS: 0.5R , 5 kQ, 2.6 kQ
44. Find the value o f R-^ for each o f the
circuits o f Figure P2.44.
O-
Figure P2.46
SC RA M BLED A N SW ERS: 60 kn ,
Figure 2.44 (a)
22.5 kO, 135 k n
1.5R
47. Find R^^ in the circuit o f Figure P2.47
(a)
W hen the switch is open
(b)
W hen the switch is closed
Figure 2.44 (b)
45. For each o f the circuits in Figure P2.45,
compute the equivalent resistance R seen by
the source, the input current
delivered by each source, and
the power
when
=
80 V.
C H E C K : Answers are the same.
100
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
48. This is a conceptual problem and requires
8000
no calculations for the answer. Consider cir­
>5000
cuits 1 and 2 o f Figure P2.48. All resistors are
2kO
greater than or equal to 1 Q.. We wish to deterO-
mine the relationship between R^^^ and R^ ^ 2
the presence o f the finite positive R-Q. resistor
(e)
Figure P2.49
between points a and b. W hich o f the following
SCRAM BLED ANSWERS: 4 0 0 ,7 0 0 , 500, 1500
statements is true?
> K ql
(a)
(b)
50. Consider the circuit o f Figure P2.50.
(a)
Kill = ^eq2
(d)
There
is no
general
relationship
between R , and R^^jrelation­
ship depends on the value o f R.
(b)
Explain your reasoning.
Suppose
= 320 V,
= 256 V, R^ =
R^ = 800 n . Find R,
and the
resulting R^^.
Suppose
= 320 V, V^= 192 V, R^ =
400 Q, R^ = 800 Q. Find R,
and
the resulting R
C ircuit 1
Figure P2.50
SC RA M BLED AN SW ERS: 500, 1000, 1600,
400, 170.67, 128
C ircuit 2
5 L For the circuit of Figure P2.51:
(a)
Calculate R^^q the equivalent resist­
ance seen at terminals A and C, which
would be the reading on an ohmmeter
if the two probes were connected to A
Figure P2.48
and C, respectively.
♦ 49. For each o f the circuits o f Figure P2.49,
find the value o f R that makes
= 1000 Q,.
(b)
Calculate R^q the equivalent resist­
ance seen at terminals B and C, which
would be the reading on an ohmmeter
O-
if the two probes were connected to B
and C, respectively.
5000
>3kO
O-
■7500
O-
be
mulas? State your reasons without per­
(b)
1 .2 kn ■
forming any calculations.
52Sn
o-
(c)
Can the equivalent resistance
calculated using the series-parallel for­
(a)
O-
(c)
(d)
Chapter 2 • K irchhoff s Current & Voltage Laws and Series-Parallel Resistive Circuits
(a)
101
For Figure P2.53a, how many bulbs can
be put in parallel before the 15 A fiise
blows? Given the maximum number
that can be put in parallel without blow­
(b)
ing the fuse, find R and V^.
In Figure 2.53b, bulbs BB and C C are
24 watt and 36 watt, respectively, at
approximately 12 volts. Find the inter­
52. Some physical problems have models that
nal resistances o f each bulb. How
are infinite ladders o f resistors, as illustrated in
many C C bulbs can be present before
Figure P2.52.
(a)
Find the equivalent resistance
the 15 amp fuse blows? Given this
at the
number o f C C bulbs, find R^^ and
terminals a-b in figure P2.52a. (Hint:
Since the resistive network is infinite,
the equivalent resistance seen at termi­
nals a-b is the same as the equivalent
resistance to the right o f terminals c-d\
this means that the network to
the right o f c-d can be replaced
by what???) Evaluate if
=1Q
and
= 100 Q. This type of
problem is useful for represent­
ing series and shunt conductance
(b)
in transmission lines.
(b)
Find
Figure P2.53
at terminals a-b for the ladder
network o f Figure P2.52b.
C H E C K ; (a) n = 16; (b) n = 4
- 0 --------------->
>^1
54. In the circuit o f Figure P 2.54,
V and
= 40 Q. The switch
= 330
closes at ? = 5
s, S2 closes at f = 10 s, 5^ closes at ? = 15 sec,
bO -
and 54 closes at ^ = 20 sec. Plot
and cal­
culate RAi) for 0 < /■< 25 s.
Figure P2.52
N U M ERICA L AN SW ERS: 10.512, 14.177
Figure P2.54
53.
Consider the circuit o f Figure 2.53a in
which
= 0.5 £^. Suppose each AA-bulb rep­
resents a 12-watt fluorescent bulb at approxi­
mately 12 volts, having an internal resistance o f
12 a
55. Consider the circuit o f Figure P2.55.
(a)
Find
max[
and the average
(b)
value o f Zj(^).
Find ijit), max[ i 2 {t)], and the average
value o f i2 {t).
102
Chapter 2 • K irch h off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
i,(t)
o
58. W ith the car engine turned off, you have
9kn
3kO <
been listening to the car radio. While the radio
6kn<
is on, you turn the ignition to start the engine.
3cost(2t)V
•2kn
You noticed a momentary silence o f the radio.
The following circuit analysis explains this
12V
effect quantitatively. Assume that with the car
Figure P2.55
engine not running, the 12-V car battery is rep­
56. Consider the circuits o f Figure P2.56. In
The load due to the car radio is represented by
resented by the model shown in Figure P2.58.
Figure 2.56,
(a)
Find
(b)
= 120 sin(377?) V and
ou f
out'
an equivalent resistance o f 240 Q. The starter
=5
and the instantaneous
motor draws 150 A o f current when the igni­
power absorbed by 30 D, resistor.
tion is turned on and before the engine starts.
If
Find
is replaced by a current source,
= 120 sin(377?) mA,
up, find
pointing
and the instanta­
neous power absorbed by 30 Q, resis­
tor. Does
affect the current through
at the moment when the ignition
switch is turned on. Compare this to the volt­
age
before the ignition switch is turned on.
W hy do you think the radio goes silent
momentarily?
the other resistors in the circuit?
ignition
model for
car battery
with engine
not running
Chasis
ground
Figure P2.58
C H EC K : /*3oq = 37.97sin^(377r) watts
57. The circuit o f Figure P 2.57 shows a simple
scheme to determine Rq, the internal resistance
o f the battery model. The loading effect due to
the digital voltmeter may be neglected (consid­
er that the meter is represented by an infinite
resistance). W ith the switch open, the meter
reads 12 V. W ith the switch (briefly) closed, the
reading drops to 11.96 V. Find the value o f R^.
CH ECK:
= 900 watts
59. The volume o f a car radio is not much
affected by the on/off state o f the headlights.
The following circuit analysis explains this phe­
nomenon quantitatively. Assume that with the
car engine running, the 12-V car battery is rep­
resented by the model shown in Figure P2.59.
Notice that the effective voltage o f the
car battery increases due to the effect
model for a
12V battery
o f the alternator while the engine is
running. The load due to the car radio
15Q
Figure P2.57
AN SW ER: 0.0376 Q
(a)
represented by an equivalent resist­
ance o f 240 £2. At 12 V dc, each head­
light consumes 35 W on low beam and
65 W on high beam.
Find the equivalent resistance o f each
headlight on low beam.
103
Chapter 2 • KirchhofF’s Current & Voltage Laws and Series-Parallel Resistive Circuits
(b)
Find the equivalent resistance
v(V)
o f each headlight on high
beam.
(c)
Find
(d)
are turned off.
Find
when the low beams
when the headlights
are turned on.
(e)
Find
when the high beams
are turned on.
(f)
v (V)
60 ■
40 -
How much power does each
high beam consume given your
20 -
answer to part (e)? W hy is this
^--------- 1-----------
value different from 65 watts?
(g)
0.5
1
1 V3
>i(A)
i(A)
(b)
How much power must the
Figure P2.61
battery deliver to overcome its internal
losses and operate the high beams and
radio.
C H E C K ; (a) 4.11 Q ; (f) 195.58 watts
DEPENDENT SOURCE PROBLEMS
62. In the circuit o f Figure P 2.62, determine
so that the power delivered to
the 5-kQ load resistor is
lOOPy^, where P^^ is the
mstantaneous
power
con­
sumed by the 8-kQ resistor.
Equivalently,
is the power
delivered by the non-ideal
voltage source.
Figure P2.59
60. A 50-cell lead acid storage battery has an
open-circuit voltage o f 102 V and a total
internal resistance o f 0.2 Q,.
I f the battery delivers 40 A to a load
(a)
Load
resistor, what is the terminal voltage?
(b)
W hat is the terminal voltage when the
battery is being charged at a 50 A rate?
(c)
Figure P2.62
AN SW ER: 6.25 mS
W hat is the power delivered by the
charger in part (b)? How much o f the
power is lost in the battery as heat?
63. Find the equivalent conductance G
and
then the equivalent resistance R “seen” by the
SCRAMBLED ANSWERS: 500, 112, 5600, 94
current source 1^ in the circuit o f Figure P2.63
61. A non-ideal constant voltage source, an
ordinary resistor, and a non-ideal constant cur­
when
in terms o f the literals R\,
rent source have the v-i characteristics given in
Figure P 2.61. Determine the values o f the
source voltage or current, the value o f the
source internal resistance, and, finally, the value
o f the resistance for Figure P2.61.
3
= 1 kQ, R2 =
gm = 0-2 mS.
and g^. Evaluate
104
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
MATLAB PROBLEMS
(a) Find the output voltage,
the output
current (what is its direction), and the
power absorbed by the load (8-Q resis­
tor) for the circuit o f Figure P2.66.
C H EC K :
Figure P2.63
= 10 kO.
64. For the circuit o f Figure P2.64,
write a node equation that allows
you to find
in terms o f
Figure 2.66
Then find
I
(b)
Using MATLAB or equivalent, com­
pute and PLO T with appropriate labels
the power absorbed by the load, denot­
ed by R^, as Rj^ varies from 8 to 64 Q in
increments o f 1 Q. Also plot the current,
again using MATLAB, as a function o f
R^. At what value o f
is the absorbed
power a maximum? Knowing this is
Figure P2.64
important, for example, when matching
C H EC K : 0.75 S.
loudspeaker resistances to the output
65. In the circuit o f Figure P2.65 r^ =
resistance o f your stereo. For this prob­
12.5 kQ and^^ = 12.5 mS:
lem, you should use MATLAB. You will
(a)
Compute the output voltage and out­
need to turn in an original printout (no
put current in terms o f
(b)
Compute the voltage gain, Gy=
copies permitted) o f your code and
plots.
l/-^.
Hint: Begin your program
2kn
©
SmV,
8kQ'
2kn<
with the commands listed
5kO
below. ?? indicates that you
r I,
should
insert
the
proper
number or formula.
8kO<
Figure P2.65
RL = 8:1:64;
% This command generates an array o f
numbers for RL beginning at 8 and
ending at 64 in increments o f 1. I f you
do not end it with a semicolon, it will
list every entry o f the array.
V2 = ??
% This value should be precomputed
I L = ??;
PL = RL .* I L . ^ 2 ;
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
% Note that because IL and RL are
arrays o f numbers . ^ means to square
each number in the array IL and . *
means to multiply each number in IL by
the corresponding number in the array
for RL.
Beginning your MATLAB solution:
% Define element values
R l= 15; R2= 4; R3= 9; R4= 2; R5=8; R6=18;
% To fin d Req start from right side.
Ra= R4 + R5;
Ga= 1/Ra;
plot(RL, PL)
Gb = Ga + 1/Rl;
grid
% Plot IL in mA
Rb = 1/Gb;
% Continue these additions and reciprocals until
plot(RL, IL *]0 0 0 )
obtaining Req.
grid
% typing grid adds a grid to your
% To fin d Vout requires repeated use o f voltage
and current division formulas.
plot. Always add a grid.
% You can put both plots on the
same graph as follows:
plot(RL,IL*WOO,RL,PL)
% The motivated student might
investigate using the “hold”
command instead.
Geq = 1/Req;
IRc = 20*Gc/Geq;
V Rb = IRc*Rb;
% Now write down the MATLAB expression for
finding Vout.
AN SW ERS: (a) 3 Q; (b) 24 V
►67. The analysis o f series-parallel circuits with
numerical element values can be done with
only two types arithmetic operations: adding
two numbers and taking the reciprocal o f a
number. As such, MATLAB is an extremely
convenient tool for finding the equivalent
resistances and the voltages and currents
throughout a series-parallel circuit. This prob­
lem illustrates such a use o f MATLAB.
For the circuit o f Figure P 2.67
(a)
Find R^^_
(b)
Find Vouf
R^= 180
--R =20
R =90
20A
105
©
R,= 150
R =40
Figure P2.67
R =80
106
Chapter 2 • K irchh off’s Current & Voltage Laws and Series-Parallel Resistive Circuits
68. Use MATLAB to find R-^ and
for the
circuit o f Figure P2.68. Turn in your MATLAB
code with your answers. Hint: Label the equiv­
alent seen at each node to facilitate computa-
“on of
K uf
-Cr
200 mA
--- —
2kO
—
-- ---------1,2kO
3kO
^3,2kn
(D
IkO
2.2kO
3kO
^
1.6kn
-
Figure P2.68
A N SW ERS: 591.2 Q, 8.869 V
► 69. Use MATLAB to find R^^,
and /j for
the circuit o f Figure P2.69.
300n
300
2on
ion
eon
i3on
10V
'4
6
4on
i5on
2oon >
135Q
> V
500*
-OFigure P2.69
► 70. Use MATLAB to find Ri„ in the circuit
o f Figure 2P.70.
3on
2on
8000
100
6000
600
1300
1500
100V
4000 <
400
1350
-o
Figure P2.70
A N SW ER: 50.53 II, 133.8 mA
5000
< 500
C H A P
Nodal and Loop Analyses
H ISTO RICAL NOTE
For a network consisting o f resistors and independent voltage sources, one can apply KCL to the
nodes, KVL to the various loops, and Ohm’s law to the elements to construct a large set o f simul­
taneous equations whose solution yields all currents and voltages in the circuit. In theory, this
approach completely solves the basic analysis problem. In practice, this approach proves imprac­
tical because large numbers o f equations are required even for a small network. For example, a 6branch, 4-node network, with each node connected to the other nodes through a single element,
leads to a set o f 12 equations in 12 unknowns: 3 equations from KCL, 3 equations from KVL,
and 6 equations from the element v -i relationships. The 12 unknowns are the 6 branch currents
and 6 branch voltages.
Before the advent o f digital computers, engineers solved simultaneous equations manually, possi­
bly with the aid o f a slide rule, or some primitive mechanical calculating machines. Any technique
or trick that reduced the number o f equations was highly treasured. In such an environment.
Maxwell’s mesh analysis technique (1881) received much acclaim and credit. Through the use o f
a fictitious circulating current, called a mesh current, Maxwell was able to greatly reduce the num­
ber o f equations. For the above-mentioned network, the number o f equations drops from 12 to 3
equations in the unknown mesh currents.
An alternate KCL-based technique (now called nodal analysis) appeared in literature as early as
1901. The method did not gain momentum until the late 1940s, because most problems in the
early days o f electrical engineering could be solved efficiendy using mesh equations in conjunc­
tion with some network theorems. W ith the invention o f multi-element vacuum tubes having
interelectrode capacitances, some compelling reasons to use the node method appeared; primari­
ly, the node method accounts for the presence o f capacitances without introducing more equa­
tions, and secondly, those vacuum tubes that behave very much like current sources are more eas­
ily accommodated with nodal equations. By the late 1950s, almost all circuit texts presented both
the mesh and node methods.
Since the 1960s, many digital computer software programs (SPICE being the most ubiquitous)
have been developed for the simulation o f electronic circuits that otherwise would defy hand cal­
108
Chapter 3 * Nodal and Loop Analyses
culation. These software packages use a node equation method over the mesh equation approach.
One o f several reasons is that a node is easily identifiable, whereas a set o f proper meshes is diffi­
cult for a computer to recognize.
For resistive networks driven by current sources, writing node equations is straightforward.
Certain difficulties arise in writing node equations for circuits containing independent and
dependent voltage sources. During the 1970s, a modification o f the conventional node method
by a research group at IBM resulted in the “modified nodal analysis” (MNA) technique. W ith the
M N A method, the formulation o f network equations, even in the presence o f voltage sources and
all types o f dependent sources, becomes very systematic.
This chapter discusses the writing and solution o f equations to find pertinent voltages and cur­
rents for linear resistive networks.
CHAPTER O U TLIN E
1.
2.
3.
4.
5.
6.
7.
8.
Introduction, Review, and Terminology
The Concepts o f Nodal and Loop Analysis
Nodal Analysis I: Grounded Voltage Sources
Nodal Analysis II: Floating Voltage Sources
Loop Analysis
Summary
Terms and Concepts
Problems
CH APTER O BJECTIVES
1.
2.
Describe and illustrate the method o f node analysis for the computation o f node voltages
in a circuit. Knowledge o f the node voltages o f a circuit allows one to compute all the
branch voltages and, thus, with knowledge o f the element values, all the branch currents.
Define the notion o f a mesh or loop current and describe and illustrate the method o f
mesh or, more generally, loop analysis for the computation o f loop currents in a circuit.
Knowledge o f all the loop currents o f a circuit allows one to compute all the branch cur­
rents. Thus, in conjunction with the knowledge o f the branch element information, one
can compute all the branch voltages.
3.
Formulate the node analysis and loop analysis equations as matrix equations and use
matrix methods in their solution emphasizing the use o f existing software for the gener­
al solution.
4.
Describe and illustrate the modified nodal approach to circuit analysis. This method
underlies the general software algorithms available for computer simulation o f circuits.
Chapter 3 • Nodal and Loop Analyses
109
1. IN TRO D U CTIO N , REVIEW, AND TER M IN O LO G Y
Chapter 1 introduced basic circuit elements, Ohm’s law, and power calculations. Chapter 2 intro­
duced the important laws o f circuit theory, K Y L and KCL, and investigated series, parallel, and
series-parallel circuits. Recall from Chapter 2 that a node voltage is the voltage drop from a given
node to a reference node. As a brief review, consider Figure 3.1, which portrays a circuit labeled
with nodes A through D having associated node voltages, V^, Vg, Vq V^, and eight branches,
one for the current source and one for each o f the seven conductances,
... , Gj. (Since this
chapter deals almost exclusively with dc, the uppercase notation for voltages and currents is com­
monplace.)
FIG U RE 3.1. Diagram of a circuit with labeled node voltages, V^, Vg, Vq V^,
with respect to the given reference node.
KVL states that every branch voltage is the difference o f the node voltages present at the terminals
o f the branch: for circuits in this text and all pairs o f nodes, j and k, the voltage drop
from
n o d ej to node k, is
at every instant o f time, where VJ- is the voltage at node j with respect to the reference and
is
the voltage at node k with respect to reference. Here, j and k stand for arbitrary indices and could
be any o f the nodes. A, B, C, or D , in Figure 3.1. These statements mean that knowledge o f all
node voltages in conjunction with device information paints a rather complete picture o f the cir­
cuit’s behavior. This chapter develops techniques for a systematic construction o f equations that
characterize a circuit’s behavior.
One last introductory point: Throughout this chapter and in many subsequent chapters, software
programs such as MATLAB facilitate calculations. Constructing sets o f equations that character­
110
Chapter 3 • Nodal and Loop Analyses
ize the voltages and currents in a circuit is often a challenge. Solving such sets o f equations with­
out the use o f software tools presents a much greater challenge. Yet facilitated by MATLAB or
equivalent, the calculations reduce to a hit o f the return key. MATLAB and the circuit simulation
program called PSpice or Spice (utilized in Chapter 4) are but two o f the many modern and
important software tools available to engineers.
2. TH E C O N C EPTS OF N O D AL AND LOOP ANALYSIS
Nodal analysis is an organized means for computing ALL node voltages o f a circuit. Nodal analy­
sis builds around KCL, i.e., at each node o f the circuit, the sum o f the currents leaving (entering)
the node is zero. Each current in the sum enters or leaves a node through a branch. Each branch
current generally depends on the branch conductance, a subset o f the circuit node voltages, and
possibly source values. After substituting this branch information for each current in a node’s KCL
equation, one obtains a nodal equation.
As an example, the nodal equation at node A in Figure 3.1 is /^-^ = /j + /y = G j (V ^- V^) + Gy (Vj
- V^). The nodal equation at node C is -/2 + I^ + 1^^- Ij =
Vj- + G j ( V ^ + Gy ( K ( j- V^) = 0. Writing such an equation at each circuit node (except the reference node) pro­
duces a set o f independent equations. O f course, one can substitute a KCL equation at the refer­
ence node for any o f the other equations and still obtain an independent set o f nodal equations.
T he solution o f such a set o f nodal equations yields all circuit node voltages. Knowing all node volt­
ages permits us to compute all branch voltages. Knowing each branch voltage and each branch con­
ductance allows us to compute each branch current using Ohm’s law. The reference node may be
chosen arbitrarily and can sometimes be chosen to greatly simplify the analysis.
A set o f nodal equations has a matrix representation. The matrix representation permits easy solu­
tion for the node voltages using MATLAB or an equivalent software package. A variation o f the
nodal analysis method, termed modified nodal analysis, relies heavily on matrix methods for
constructing and solving the circuit equations. The basic principles o f this widespread analysis
technique are illustrated in Section 4.
Because computer-based circuit analysis packages build on a matrix formulation o f the circuit
equations and because o f the widespread use o f matrices in circuits, systems, and control, we will
stress a matrix formulation o f equations throughout this chapter. The student unfamiliar with
matrix methods might look through a calculus text or a linear algebra text for a good explanation
o f their basic properties and uses.
T he counterpart to nodal analysis is loop analysis. In loop analysis, the counterpart o f a node
voltage is a loop current, which circulates around a closed path in a circuit. A loop or closed path
in a circuit is a contiguous sequence o f branches that begins and ends on the same node and touch­
es no other node more than once. For each loop in the circuit, one defines a loop current, as illus­
trated in Figure 3.2, that depicts three loops or closed paths having corresponding loop currents
/p Ij, and ly O f course, one can draw other closed paths or loops for this circuit and define other
loop currents.
111
Chapter 3 • Nodal and Loop Analyses
90
FIG U RE 3.2. Simple resistive circuit showing three closed paths (dotted lines) that represent
three loop currents, /j, Ij, and 1^; the branch current
which is a difference
o f the two loop currents through the resistor.
Using a fluid flow analogy, one can think o f loop currents as fluid circulating through closed sec­
tions o f pipe. The fluid in different closed paths may share a segment o f pipe. This segment is anal­
ogous to a branch o f a circuit on which two or more loop currents are incident. The net current
in the branch is analogous to the net fluid flow. Note that each branch current can be expressed
as a sum o f loop currents with due regard to direction. For example, in Figure 3.2, the branch cur­
rent 7^3 = ^\ -
Using loop currents, element resistance values, and source values, it is possible
by KVL and Ohm’s law to express the sum o f the voltages around each loop in terms o f the loop
currents. For example, the first loop, labeled
in Figure 3.2, has the loop equation
^ « = 9 / i + 3(/i -/ 2 )+ 6 (/ ^ -/ 3 )
We will explore this concept more thoroughly in Section 5. Here we see that loop analysis builds
on KVL, whereas node analysis builds on KCL.
3. N O D AL ANALYSIS I: G RO U N D ED VO LTAG E SO URCES
As mentioned earlier, nodal analysis is a technique for finding all node voltages in a circuit. W ith
knowledge o f all the node voltages and all the element values, one can compute all branch volt­
ages and currents, and thus the power absorbed or delivered by each branch. This section describes
nodal analysis for circuits containing dependent and independent current sources, resistances, and
independent voltage sources that are grounded to the reference node (see Figure 3.3). Floating
independent or dependent voltage sources (those not directly connected to the reference node) are
covered in Section 4.
For the class o f circuits discussed in this section, it is possible to write a nodal (KCL) equation at
each node not connected to a voltage source. A node connected to a voltage source grounded to
the reference node has a node voltage equal to the source voltage. The other node voltages must
112
Chapter 3 • Nodal and Loop Analyses
be computed from the set o f nodal equations. Each nodal equation will sum the currents leaving
a node. Each current in the sum will be expressed in terms o f dependent or independent current
sources or branch conductances and node voltages. The set o f these equations will have a solution
that yields all the pertinent node voltages o f the circuit. Examples 3.1 and 3.2 illustrate the basic
techniques o f nodal analysis.
EX A M P L E 3 .1 .
The circuit o f Figure 3.3a contains an independent voltage source, an independent current source,
and five resistances whose conductances in S are G j through Gy The nodes other than the refer­
ence are labeled with the node voltages V^,
and V^, which respectively denote nodes a, b, and
c. T he analysis o f this circuit illustrates the process o f nodal analysis to find the node voltages V^,
y,,andK =
FIG U RE 3.3A. Resistive circuit for Example 3.1. Note that node voltage
is specified by the voltage source.
So l u t io n .
Step 1. Consider node c. A voltage source ties node c to the reference node. Hence, the node volt­
age
is fixed at V-^, i.e.,
Because
it is not necessary to apply KCL to this node
unless the current through the voltage source is required, for example, when determining the
power delivered by the source.
Step 2. Sum the currents leaving node a. From KCL, the sum o f the currents leaving node a is
zero. As per the partial circuit in Figure 3.3b, this requires that
Grouping the coefficients o f
and
tion yields our first nodal equation
and moving the source values to the right side o f the equa­
Chapter 3 • Nodal and Loop Analyses
113
G 5J 'V -aV .in')
(3.1)
Step 3. Sum the currents leaving node b. Applying KCL to node b, reproduced in Figure 3.3c,
yields the equation
G 2 ^y b- y a) ^G ,V b^ G, {V ,- VJ = Q
After regrouping terms, one obtains our second nodal equation:
G4(v ,- v
j
V
G 3V,
G .(V ,-V )
FIG U RE 3.3C
-G ^ V ^ + {G ^ + G ,+ G ,)V ,= G,V.„
(3.2)
Step 4 . Write set o f nodal equations in matrix form. Equations 3.1 and 3.2 in matrix form are
Gi + G
2+
G5
-G 2
-G 2
G 2 -I- G 3 -I- G 4
•
(3.3)
Matrix equations organize relevant data into a unified framework. Because many calculators do
matrix arithmetic, because o f the widespread availability o f matrix software packages such as
MATLAB, and because equation solution techniques in circuits, systems, and control heavily uti­
lize matrix methods, the matrix equation formulation has widespread and critical importance.
Step 5 . Solve the matrix equation 3.3: For this part, suppose that the conductance values in S are
Gj = 0.2, G2 = 0.2, G3 = 0.3, G 4 = 0.1, G 5 = 0.4, that = 2.8 A, and that
= 24 V. After sub­
stitution, equation 3.3 simpUfies to
114
Chapter 3 • Nodal and Loop Analyses
■0.8
-0 .2 - ■K,'
- 0 .2
0 .6
■12.4'
(3.4)
2.4
Solving using the inverse matrix method leads to the node voltages (in volts):
■ 0.8
- 0 .2
- 0 .2 '
0.6
-1
T 2 .4 -
1
2.4
0 .6
0.4 4 0.2
0 .2 ' ■12.4'
18-
0.8
10
2.4
V
Alternately, one could have solved equation 3.4 via MATLAB, its equivalent, or the age-old hand
method o f adding and subtracting equations. For example, in MATLAB
»M =[0.8 -0.2;-0.2 0.6];
»b= [12.4 2.4]';
>>NodeV = M\b
NodeV =
1.8000e+01
l.OOOOe+01
»% O R EQU IVA LEN TLY
»NodeV = inv(M )*b
NodeV =
18
10
Step 6. Compute
The branch voltage V^ = V^ -
18 - 10 = 8 V.
Exercises. 1. Utilize the solution o f Example 3.1 to compute the current leaving and the power
delivered by the independent voltage source.
AN SW ER: 3.8 A and 91.2 watts
2. Referring to Figure 3.3a and the values set forth in Step 5 o f Example 3.1, suppose the value o f
is cut in half, the value o f V-^ is 24 V, and the value o f each o f the conductances is also cut in
half W hat are the new values o f the node voltages?
AN SW ER: All node voltages are the same.
3. By what single factor must the values o f
and V-^ be
multiplied so that the node voltages are doubled?
AN SW ER: 2
L
4. Construct a node equation for
A N SW ER: (G, + G ,)V ^ ^ 1 -
in Figure 3.4.
F IG U R E 3.4.
115
Chapter 3 * Nodal and Loop Analyses
EX A M PLE 3.2.
Consider the circuit o f Figure 3.5a. Similar to Example 3.1, the objective is to find the node volt­
ages V^,Vf^, and
. However, in the circuit o f Figure 3.5a, an independent current source has
replaced the independent voltage source o f Figure 3.3a. This change unfreezes the constraint on
the value o f
present in the circuit Figure 3.3a. There will result three nodal equations in the
three unknowns
, and V^.
0.4 U
FIG U RE 3.5A. Circuit containing two independent current sources and
three unknown node voltages
, and V^.
So l u t io n .
Step 1. Sum currents leaving node a. This step is the same as Step 2 o f Example 3.1. By inspec­
tion o f node a,
0 . 2 + 0.2(K^ - V^) + 0.4(V^ - V J - 2 = 0
which upon regrouping terms yields
0.8V ^ -0.2V ^ -0.4V ^ = 2
(3.5)
Step 2. Sum currents leaving node b. This step is the same as Step 3 o f Example 3.1. Again, by
inspection,
0.2
- VJ + 0.3
+ 0.1 (V^ - V;) = 0
Simplification yields
-0 .2 7 ^ + 0.6V ^ ^ -0.1K ^ = 0
(3.6)
Step 3. Sum currents leaving node c. Because a current source
o. 4 ( v - v :
drives node c, the similarity to example 3.1 ends, and we must
write a third node equation. Summing the currents leaving
node c, as shown in Figure 3.5b, yields
0 . 4 ( 1 / - K J + 0.1 ( K ^ - K ^ ) - 1 = 0
Upon simplification, we have
-0.4V ^^-0.1K ^ + 0 .5 V ;= 1
(3.7)
116
Chapter 3 • Nodal and Loop Analyses
Step 4. Write equations 3 .5 -3 .7 as a matrix equation and solve. The matrix form o f our nodal
equations 3 .5 -3 .7 is
0.8
- 0.2
- 0 . 4 ' ■v;'
- 0.2
0.6
- 0.1
- 0 .4
- 0 .1
0.5
'T
(3.8a)
= 0
1
Solving equation 3.8 using MATLAB or equivalent, using a calculator that does matrix operations,
or solving via some form o f row reduction, one obtains the solution (in volts)
■0.8
= - 0.2
Vc
- 0 .4
- 0.2
- 0 .4 '
0.6
- 0.1
-1
■2 ‘
'2 .9
1.4
2.6
■2 ‘
■8.4'
- 0.1
0 = 1.4
2.4
1.6
0 = 4 .4
0.5
1
2.6
1.6
4 .4
1
V
(3.8b)
9.6
Specifically, in MATLAB
» M = [0.8 -0.2 -0.4;-0.2 0.6 -0.1;
-0.4 - 0.1 0.5];
>>b = [2 0 1]';
»NodeV = M\b
NodeV =
8.4000e+00
4.4000e+00
9.6000e+00
Exercises. 1. Suppose the values o f the current sources in Figure 3.5a are doubled. W hat are the
new values o f the node voltages? Hint: Consider the effect on equation 3.8.
AN SW ER: All node voltages are doubled.
2 . Suppose the conductances in the circuit o f Figure 3.5a are cut in half, i.e., the resistances are
doubled. W hat are the new node voltages?
A N SW ER: Node voltages are doubled.
3. Suppose the conductances in the circuit o f Figure 3.5a are cut in half W hat happens to the
magnitudes o f the branch currents? Hint: Express the branch current in terms o f the branch con­
ductance and its terminal node voltages.
AN SW ER: The magnitudes o f the branch currents
stay the same.
I
4. Find two node equations characterizing the cir­
cuit o f Figure 3.6.
AN SW ER: (G j + G^) -V„~G^^Vy =
and
FIG U RE 3.6
Chapter 3 • Nodal and Loop Analyses
117
The matrices in equations 3.3, 3.4, and 3.8a are symmetric. A symmetric matrix, say A, is one
= A; this means that ifA = [a-^ is an n x n matrix whose i-j
entry is a-, then A is symmetric if a-j = a^j^. In words, the off-diagonal entries are mirror images o f
whose transpose equals itself, i.e.,
each other. For example.
■0.8
- 0.2
- 0 .4 '
A = - 0.2
0.6
- 0.1
- 0 .4
- 0.1
0.5
present in the circuit., as in Examples 3.1 and 3.2, the coefficient matrix o f the node equations (as exemplified in equations 3.3, 3.4, and 3.8) is always sym­
metric, provided the equations are written in the natural order.
When only resistances, independent current sources, and grounded independent voltage sources are
present in the circuit, the value o f the entries in the coefficient matrix o f the nodal equations can be
computed by inspection. The 1-1 entry o f the matrix is the sum o f the conductances at node a (or
1); the 2 - 2 entry is the sum o f the conductances at node b (or 2). In general, the i-i entry o f the
coefficient matrix is the sum o f the conductances incident at node i. Further, the 1 -2 entry o f the
matrix is the negative o f the sum o f the conductances between nodes a and b (or between nodes 1
and 2), and the 2 -1 entry has the same value. In Example 3.2, the 1 -2 entry o f - 0 .2 S is the nega­
tive o f the sum o f the conductances between nodes a and b; the 1 -3 entry o f - 0 .4 S is the negative
o f the sum o f the conductances between nodes a and c (or between 1 and 3, if the nodes were so
numbered). Thus, whenever the circuit contains no dependent sources, the node equations can be
written by inspection. Further, if independent voltage sources are absent, then the right-hand side o f
the nodal matrix equation can also be written by inspection: the i-th entry is simply the sum o f the
independent source currents injected into the i—th node at which KCL is applied.
W hen controlled sources are present in the circuit, the resultant nodal matrix is generally not sym­
metric, as illustrated in the following two examples.
EXA M PLE 3.3.
The circuit o f Figure 3.7 represents a small-signal low-frequency equivalent circuit o f an amplifi­
er in which the input signal V-^ is “amplified” at the output,
= V'2. Small-signal means that
the input signal should have a relatively small magnitude so that a LIN EA R circuit will adequately
represent the amplifier. Similarly, low-frequency means that the frequency o f any sinusoidal input
must be relatively low for the (resistive) circuit model o f the amplifier to remain valid.
The amplifier circuit model contains a current-controlled current source (CC CS) and a voltagecontrolled current source (VCCS). These two dependent sources have currents that depend on
other circuit parameters and require some special handling when constructing node equations.
Our objective is to set forth the methodology for writing the node equations when dependent cur­
rent sources are part o f the circuit and to compute the magnitude o f the voltage gain, |
V-^ |
= \y 2 |y^n\■
Note that the source voltage, V-^, specifies the voltage at the node at the bottom o f Gp hence, a
nodal equation at this node is unnecessary. Nodal equations must be written at the remaining
Chapter 3 • Nodal and Loop Analyses
118
nodes, which are labeled with the voltages Vj, Kj (=
and V^. (Numbering and labeling is
often a matter o f personal preference. In this example, we have chosen 1, 2, and 3 as node labels,
in contrast to the previous two examples, where we used a, b, and c.)
G,
FIG U RE 3.7. An equivalent circuit model o f an amplifier.
So l u t io n .
Step 1. Sum the currents leaving node 1. Summing the current leaving node 1 leads to
(^1 -
+ ^2 (^1 - ^ 3) + ^3 (^1 -
+ P
0
or, equivalently, after grouping like terms,
(Gi + G2 + G3)Ki - G3 K2 -
(3.9)
Step 2. Substitute fo r i^ in equation 3 .9 and simplify. In equation 3.9, |3 i^ accounts for the effect
o f the C C C S at node 1 and is not given in terms o f the circuit node voltages. To specify this term
in terms o f the circuit node voltages, observe that in Figure 3.7, i^ is the current from node 2 to
node 3 through G^. Hence,
I3t -I3G^{V2 - K3) = I3G^V^ - I3G^V^
(3.10)
Substituting equation 3.10 into 3.9, again grouping like terms, one obtains the first nodal equation,
{G, + G2 + G3) Vi + il3G^ - G3) K2 - (G 2 + PG^) K3 =
(3.11)
Step 3. Sum the currents leaving node 2. By the usual methods,
G,{V^ - V,) +
+ G,{V^ - K3) +
0
which, after regrouping terms, reduces to
- G 3 K1 + (G 3 + Gg 4. G,)V^ - G4 K3 +
= 0
( 3 . 12)
119
Chapter 3 * Nodal and Loop Analyses
Step 4, Specify
in terms o f node voltages, substitute into equation 3 . 1 2 , an d simplify.
Inspecting the circuit o f Figure 3.7 shows that
is the voltage across
from node 1 to node 3.
Hence,
gm^. = & j y x - y , ) - g n y x - g r r y ,
Substituting
(3-13)
o f equation 3.13 into equation 3.12 leads to our second nodal equation,
^ 3)^ , + (G 3 + G4 + G ,)V ^ -{G , + ^ J K 3 = 0
(3.14)
Step 5. Sum the currents leaving node 3. Applying KCL to node 3 yields,
G 2 ( V ,- V , ) ^ G , i V , - V , ) ^ G , V , - p i ^ - g ^ v ^ = 0
(3.15a)
Using equations 3.10 and 3.13 for , z and g v respectively, we have
0
= G , { V , - V ,) + G , { V , - V ,) + G ^ V ^ - ^ G , { V , - V , ) - g ^ { V , - V , )
Grouping like terms leads to our third equation in the three unknowns V^, V^, and V^:
-(G i + kJ V
x-
+ ^ 4)^2 + (^2 + G4 + I3G^ + G 5 + gJV ^ = 0
(3.15b)
Step 6. Put nodal equations in matrix form. The three nodal equations 3.11, 3.14, and 3.15b
have the matrix form
Gi + G 2 + G 3
Sm - G
PG^ —G 3
~ ^ 2
G 3 + Gg + G 4
3
“ ^2 “ 8 m
~^A -
~
-G 4 -
G 2 + G 4 + fiG^ + G 5 + g„
■y,'
G.V^v/
^2 =
0
^^3
0
Step 7. Substitute values and solve. Suppose that the various circuit conductances have the fol­
lowing values in [xS: G j = 1,000, G j = 2.0, G j = 1.0, G^ = 10, G j = 2 0 ,100, and Gg = 200 .
Suppose further that
= 2.1 V, (3 = 4/1010 and^^ = 21,112 [xS. This allows us to generate the
following M ATLAB code for the solution:
»G 1 = 1000e-6;G 2 = 2e-6; G 3 = le - 6 ; G 4 = lOe-6 ;
»G5 = 20100e-6; G 6 = 200e-6; Vin = 2 . 1;
»beta = 4/1010; gm = 2 1 1 12 e-6 ;
»M =[G1+G 2+G 3
gm-G3
-G2-gm
beta*G 4-G 3
G 3+G 6+G 4
-G 4-beta*G 4
» b = [G l* V in
»NodeV = M\b
NodeV =
2 . 0000 e +00
-l.OOOOe+02
l.OOOOe+00
0
0]';
-G 2-beta*G 4;
-G4-gm ;
G2+G4+beta*G4+G5+gm ];
Chapter 3 • Nodal and Loop Analyses
120
in which case,
■V,-
■ 2
■
V^2 = - 1 0 0
V
1
V3
Step 8. Compute the voltage gain. The voltage gain o f the amplifier is given by
Kut
V2
-1 0 0
Vin
Vin
2.1
= 4 7 .6 2
Exercises. 1. Suppose V-^ in the circuit o f Figure 3 .7 is doubled. W hat are the new node voltages?
Hint: Consider the matrix equation o f Step 6.
A N SW ER: Node voltages are doubled.
2. Suppose all conductances in the circuit o f Figure 3.7 are cut in half (resistances are doubled)
and (3 is held constant. How must^^ change for the node voltages to remain at their same values?
AN SW ER:
must double.
Realistic problems do not permit hand solutions. For hand solutions, the smallest number o f
equations is generally desired. For matrix solutions using software packages such as MATLAB,
more variables with more equations may often be easier to construct and may often result in more
reliable numerical calculations. This can be illustrated using the equations o f Example 3.3. All the
pertinent basic equations o f the circuit o f Figure 3 .7 can be written down as follows: from equa­
tions 3.9 and 3.10 we have
and
However, in contrast to the example, we do not substitute 3.10 into 3.9 to obtain 3.11. Rather,
we just let them be two independent equations. Further, from equations 3.12, 3.13, and 3.15a,
we have
- G 3 K, + (G 3 + G4 + G ,) K2 - G 4 K3 + g^v^ = 0
and
By not substituting for and v^, we avoid unnecessary hand calculation, and if there is an error,
it is easier to find. The resulting equations have the matrix form where i and
now appear as
additional unknowns, easily handled by a software program:
121
Chapter 3 • Nodal and Loop Analyses
G] 4- G 2 + G 3
-G 3
-G 2
13
0
0
G4
-G 4
-I
0
-G 3
G 3 -1- G 4 -HGg
-G 4
0
8m
1
0
-1
0
-1
-G 2
-G 4
G2 + G 4 + G5
-/?
~Sm
1
•V,-
0
^2
=
iy
0
0
0
As a general rule, we would reorder the equations so that rows 1, 3, and 5 came first, as they cor­
respond to the three nodal equations at Kj, V2 , and
equations for i and
. Then we would write the constraint
. Such a reordering leads to certain symmetry properties discussed earlier.
Exercise. Solve the above matrix equation in MATLAB or equivalent, using the numbers o f
A and „ - 1 V.
Example 3.3 to verify that = -1 .0 1 x 10^-3■ A
Matrix methods as used in the above examples and in the ones to follow necessitate the power o f
a calculator or a software program such as MATLAB for easy solution. Such programs permit a
straightforward calculation o f the required answers and are not prone to arithmetic errors.
The next example illustrates how to write node equations for circuits containing a voltage-con­
trolled voltage source (VCVS) grounded to the reference node. The analysis o f CCVSs grounded
to the reference node is similar. The more challenging analysis o f circuits containing floating
dependent or independent voltage sources is taken up in the next section.
EX A M P L E 3 .4 .
The circuit o f Figure 3.8 models a poor operational amplifier circuit' in which the output voltage
1/^^ = V2 approximates
For the analysis, let |i = 70. The adjective “poor” arises because ^
should have a value much larger than 70.
R3= lO k O
V. =
F IG U R E 3.8. A two-node (amplifier) circuit containing a grounded VCVS with jx = 70.
So l u t io n .
The circuit contains two nodes labeled
and
(equivalently nodes 1 and 2) not constrained
by voltage sources. The goal o f our analysis is to find these node voltages by writing two equations
in these voltages and solving. As is commonly the case, resistances are in ohms and will be con­
verted to conductances in S for convenience in writing the node equations.
/~N,
122
Chapter 3 • Nodal and Loop Analyses
Step 1. Compute conductance values in S. Conductances are the reciprocal o f resistances, i.e., G-
= ^IRj- Hence,
= 3.3 3 3 3 3 lQ-5, G^ = 10“^, G^ = 0.01, and G^ = lO'^
G j = 2.0 10-3,
Step 2 . Write a node equation at node 1. Summing the currents leaving node 1 yields
Grouping like terms leads to
{G ,^ G ^ ^ G ,)V ,-G ,V ^ = G,V^n
Inserting numerical quantities yields the first node equation
(3.16)
Step 3. Sum currents leaving node
2
. Summing the currents leaving node 2 yields
^3 (^^2 - ^l) + ^5 ^2 + G4 {I /2 +
Kj) = 0
The dimensionless coefficient |i is placed with the conductance,
obtain
(HG4 -
while grouping like terms to
+ (G 3 + G4+ ^ 5)^2 = 0
Inserting the numerical values produces the second node equation
0.699 9 Vj + 0.0111 ^2 = 0
(3.17)
Step 4. Write equations 3 .1 6 and 3 .1 7 in matrix form and solve. In matrix form
■3.3333
-1
■ -V f
0 .6 9 9 9
O .O Ill
.^2.
■2'
0
Using the formula for the inverse o f a 2 x 2 matrix (interchange the diagonal entries, change the
sign on the o ff diagonals, and divide by the determinant), one obtains
•Vf
1
0.0111
1
0 .0 3 0 1 2 6
.^2.
0.7369
-0 .6 9 9 9
3.3333
-1 .8 9 9 6
In Example 3.4, observe that
=
^^ 2
= -1 -8 9 9 6 , which approximates - 2 V-^ since V- = 1 V.
Exercises. 1. Write MATLAB code to solve the above example. Check that your code works. Hint:
See Example 3.3.
2. If R2 is changed to 100 k£2 in Example 3.4, show that V2 = -1 .9 0 6 3 V.
Chapter 3 • Nodal and Loop Analyses
123
4. N O D AL AN ALYSIS II: FLOATING VO LTAG E SO URCES
A floating voltage source means that neither node o f the source is connected to the reference
node. When a floating dependent or independent voltage source is present with respect to a given
reference node, a direct application o f KCL to either terminal node o f the voltage source is unfruit­
ful. There are several ways to handle this situation. One fruitful method is to enclose the source
and its terminal nodes by a Gaussian surface, i.e., a closed curve, to create what is commonly
called a supernode, as illustrated in Figure 3.9. One would then write KCL for the supernode as
is done in a number o f circuit texts. However, there is a conceptually more straightforward
approach, which is often called the modified nodal analysis, or MNA. In MNA, we add an addi­
tional current label to each floating voltage source. In Figure 3.9, we have added the current label
l^y. This additional current becomes an unknown in a set o f nodal equations generated by apply­
ing KCL to each node. At this point, further explanation is best done by an example, but the con­
cept is similar to the discussion following Example 3.3.
E X A M P L E 3.5.
Find the node voltages V^, Vy,
and the unknown current
in the circuit o f Figure 3.9, when
the bottom node is taken as reference.
FIG U RE 3.9. Resistive circuit containing a floating voltage source for the given reference;
generally, the reference node may be chosen arbitrarily.
So l u t io n .
Step 1. Write a node equation at node a. Summing the currents leaving node a yields
8 + 0.15
+ 3 + 0.2 ( K ^ -K .) = 0
After grouping terms appropriately, we have
(0.15 + 0 .2 )K ^ - 0.15^ -^ - 0 .2 V ;+ 8 + 3 = 0
or, equivalently,
0.351^ ^ -0.15K ^ -0.2V ^ ^ = - 1 1
This provides our first equation in four unknowns.
(3.18)
124
Chapter 3 • Nodal and Loop Analyses
Step 2. Write a nodal equation at node b. Here,
=0
- 3 + 0 .1 5 ( K ^ - K ; + 0.05
or equivalently,
- 0 .1 5
(0 .1 5 + 0.05) n - / ^ ^ = 3
Simplifying this expression leads to
- 0 .1 5 K , + 0.2K^-/^^ = 3
(3.19)
Step 3. Write a nodal equation at node c. Here,
^.^ + 0-25 V ^-25 +
0
2
{V^- K J = 0
or equivalently.
- 0 .2 K ,+ 0.45K^ + /,, = 25
(3.20)
Step 4. Write the node voltage relationship fo r the terminal nodes o f the floating voltage source,
i.e., between the voltages
and V^. The voltages
Mathematically, this constraint is 1/ -
and
are constrained by the voltage source.
= 440, i.e.,
K ^ = 440
(3.21)
Step 5. Write thefour equations 3.18, 3.19, 3.20, and 3.21 in matrixform and solve. In matrix form,
0.35
- 0 .1 5
- 0 .2
0
-1 1
- 0 .1 5
0.2
0
-1
3
- 0 .2
0
0.45
1
25
0
-1
1
0
hb
(3.22)
440
Because o f the extra variable, the equations become too large for hand calculation. Hence, we use
MATLAB as follows:
>>M = [0.35 - 0 .1 5 - 0 .2 0 ;-0 .1 5 0.2 0 - 1 ;
- 0 .2 0 0.45 1 ; 0 - 1 1 0];
»b = [-11 3 25 440]';
»x = M\b
X =
-9.0000e+ 01
-3 .1 0 0 0 e + 0 2
1.3000e+02
-5.1500 e+ 01
Hence,
= - 90 V,
= - 3 1 0 V,
= 130 V,
= -5 1 .5 A
Chapter 3 “ Nodal and Loop Analyses
12 5
In a conventional nodal analysis, all unknowns are node voltages. Here we have the additional
unknown current,
. Because o f this additional unknown current, the method is called a mod­
ified nodal analysis.
Also, in this example, node d was taken as the reference node. However, one could just as easily
take node b as the reference node, in which case, the voltage source would not have been floating.
A home problem investigates this choice o f reference node.
Exercise. 1. For Example 3.5, compute the voltages
and
2. For Example 3.5, compute the power absorbed by the 0.15 S resistor.
3. Compute the power delivered by the floating voltage source.
AN SW ERS in random order: 220 V, 22.6 6 kw, 310 V, 440 V, 7260 watts
The next example investigates a circuit having floating independent and dependent voltage
sources. By convention, the reference node o f this circuit, figure 3.10, and all subsequent circuits,
will be the bottom node o f the circuit unless stated otherwise.
E X A M P L E 3.6.
The circuit o f Figure 3.10 contains a floating independent and a floating dependent voltage
source. Find the node voltages V^, Vy, V^, and the unknown currents
and
Then find the
power delivered by the 30 V source and the dependent source.
500
1
FIG U RE 3.10. Resistive circuit containing a floating dependent voltage source and a floating
independent voltage when node d is chosen as the reference node.
So l u t io n .
Step 1. Sum currents leaving node a. Here,
126
Chapter 3 • Nodal and Loop Analyses
Equivalently,
100
(3.23)
100
“
Step 2. Sum currents leaving node b. Here,
100
500
Equivalently,
(3.24)
Step 3. Sum currents leaving node c. Here,
1
(3.25)
800
Step 4. Write an equation relating the terminal voltages o f the independent voltage source. Here,
K ,-n = 3 0
(3.26)
Step 5. Write an equation relating the terminal voltages o f the dependent voltage source. Here,
40
K ,- K = 4 0 ., = —
Equivalently,
(3.27)
0.6V;, + 0 .4 y ^ - V ^ = 0
Step 6. Write equations 3.23 through 3 .2 7 in matrix form an d solve in MATLAB. Combining
the above equations into a matrix produces
0 .03
-
0.01
-0 .0 1
0
1
0
0.012
0
0
-1
0.0 0 1 2 5
-1
1
0
0
^ac
30
0
0
Icb,
0
0
0
0
-1
0 .6
0.4
________
-1
■2.2'
0
V'i
=
0
Again, this matrix equation is too large for hand computation. Hence in MATLAB,
»M = [0.03 - 0 .0 1 0 1 0;
-
0.01 0 .012 0 0
-
0 0 0.00125 - 1 1;
0 - 1 1 0 0;
0.6 0 . 4 - 1 0 0];
1;
Chapter 3 • Nodal and Loop Analyses
127
>.b= [2.2 0 0 30 0]';
»x = M\b
X =
l.OOOOe+02
5.0000e+01
8.0000e+01
-3 .0 0 0 0 e -0 1
-4 .0 0 0 0 e -0 1
Hence,
100 ■
50
=
80
^ac
- 0 .3
Icb_
- 0 .4
Step 4. Compute the power delivered by the 3 0 Vsource. The power delivered by the 30-V source is
^^./ = - 3 0 / ,^ = 3 0 x 0.4 = 1 2 W
Step 5. Compute the power delivered by the dependent source. The power delivered by the depend­
ent source is
Pdel =
=
- 4 0
X
^ ( - 0 . 3 ) = 0.12(V^ - V^) = 6 W
100
Exercises. 1. For Example 3.6, compute the voltages
and the power absorbed by the 800
Q resistor.
A N SW ERS in random order: 8 watts, 20 V, - 8 0 V
2. Suppose the two independent voltage source values in Example 3.6 are doubled. W hat are the
new node voltages? W hat are the new branch currents?
A N SW ERS: Node voltages are doubled and branch currents are doubled.
3. Suppose all resistances in the circuit o f Figure 3.10 are doubled and the value o f the parameter
on the dependent source is also doubled. W hat are the new branch currents?
AN SW ER: All branch currents are cut in half.
The above example increases the number o f unknowns beyond the node voltages to include the
two currents through the floating voltage sources. However, we could have included additional
currents to the set o f equations making the dimension even higher. W ith a tool like MATLAB,
this poses no difficulty. However, it does make hand computation a challenge. For example, we
128
Chapter 3 • Nodal and Loop Analyses
= l^y as an additional variable with a corresponding increase in the num­
could have included
ber o f equations. By adding addirional unknowns we would simplify the writing o f the individual
node equations but increase the dimension o f the matrix equation. Specifically, the node equation
at “a” becomes
_
t i
I
50
IV
50
and the resulting larger matrix equation is
0.02
0
0
1
0
2.2
0
0.002
0
0
-1
0
0
0
0.00 1 2 5
-1
1
0
0
-1
1
0
0
0
0
-1
0
0
-4 0
-1
0
0
0
-100
0
^ac
30
Ic h
0
0
This completes our discussion o f the standard nodal equation method o f circuit analysis. T he next
section takes up a discussion o f an alternative analysis method entided loop analysis.
5. LOOP ANALYSIS
Loop analysis is a second general analysis technique for computing the voltages and currents in a
circuit. Mesh analysis is a special type o f loop analysis for planar circuits, i.e., circuits that can
be drawn on a plane without branch crossings. For planar circuits, loops can be chosen as mesh­
es, as illustrated in Figure 3.2, or as in 3.11 below. Associated with each loop is a loop current.
Loop currents circulate around closed paths (loops) in the circuit. Similarly, for planar circuits,
the term mesh current is used traditionally for loop current. By KVL, the sum o f the voltages
across each branch in a loop is zero. By expressing each o f these branch voltages in terms o f the
designated loop currents, one can write an equation in the loop currents for each designated loop
in the circuit. For branches that are often common to two or more designated loops, the branch
current equals the net flow o f the loop currents incident on the branch. Writing an equation for
each loop produces a set o f equations called loop equations. If sufficient independent loops are
defined, one can solve the loop equations for the loop currents. Once the loop currents are known,
we can easily compute the branch currents and then the branch voltages in the circuit. Then we
can compute any other quantities o f interest, such as power absorbed, power delivered, voltage
gain, etc.
EXA M PLE 3.7.
Consider the planar circuit o f Figure 3.11 with the three specified loops, which are also called
meshes. Denote the “loop” currents for each loop by /j, 1^, and /j. The objective is to write three
equations in the currents /j,
and /j using KVL and solve these equations for their values. Then
we will compute the power absorbed by the 2 -Q resistor marked with the voltage v. Suppose the
source voltages are
= 4 0 V andV^2 = 20 V.
Chapter 3 * Nodal and Loop Analyses
129
FIG U RE 3.11. Resistive circuit containing only independent voltage sources
for the loop analysis of Example 3.7.
So l u t io n .
Step 1. Write a KVL equation based on loop 1 by summing voltages around this loop. Summing
the voltages around loop 1 using Ohm’s law and the defined loop currents produces
K., = 40 = /i + 4 (/; - /j) + Va + ih - ^3) = 6 /, - 4 /2 - 7 3 + 20
(3.25a)
Here, observe that the 4 Q resistor is incident on two loops; the net current flowing from top to
bottom, i.e., with respect to the direction o f loop 1, is /j - l 2 - The idea is analogous to a pair o f
distinct water pipes that share a common length. The common length is analogous to the 4 -Q
resistor. The flow rate in each pipe is analogous to the currents /] and I 2 , which in fact, are rates
at which charge flows past a cross sectional area o f the conductor. It follows that the net flow
through the common length o f pipe with respect to the direction o f loop 1 is the difference in the
net flow rates o f pipes 1 and 2, respectively. This is precisely the meaning o f /j - /2 . A similar
explanation can be made for the 1 -Q resistor common to loops 1 and 3 for which the net flow
rate with respect to the direction o f loop 1 is 7j - ly
Simplifying equation 3.25a yields
6/1 - AI^ - 73 = 20
(3.25b)
Step 2. Write a KVL equation based on loop 2 by summing the voltages around this loop. Applying
Ohm’s law and KVL to loop 2 produces
0 = 4(72 - 7,) + 272 + 2(^2 -
(^.26)
h'
Notice that with respect to the direction of loop 2, the net flow rate through the 4 Q resistor is
Step 3. Finally, write a KVL equation based on bop 3. Stmiming the voltages around loop 3 yields
V^2 = 20 = 2(73 - 72) + 73 + (73 - 7,) = - 7i - 272 + 473
(3.27)
Chapter 3 • Nodal and Loop Analyses
130
Step 4 . Write eqtiations 3.25b, 3.26, an d 3 .2 7 in matrix form an d solve. Writing the above three
loop equations in matrix form yields
'6
-4
-r
W
-4
8
-2
h
-1
-2
4
■20'
-
0
(3.28)
20
h
Solving Equation 3.28 by the matrix inverse method (by a numerical algorithm or by Cramer’s
rule) yields the loop currents in amps as
h'
6
-4
-r
-1
0 .35
'2 0 '
h
= -4
8
-2
0
h
-1
-2
4
20
= 0 .2 2 5
0 .225
0 .2
■20'
0 .2 8 7 5
0 .2
0
0 .2
0 .4
20
0 .2
11 ■
=
8.5
12
Step 5. Compute the power consumed by the 2 Q resistor. Knowledge o f the loop currents makes
it possible to compute all voltages and currents in the circuit. For our purpose, the voltage
and the power absorbed by the 2 Q resistor is
/ 2 = 24.5 watts.
Exercises. All exercises are for the circuit o f Figure 3.11.
1. Compute the power delivered by the 20 V source.
AN SW ER: 20 watts
2. Compute the power absorbed by the 4 Q resistor.
AN SW ER: 25 watts
3. Suppose the source values are doubled. W hat are the new values o f the loop currents?
AN SW ER: loop currents are doubled
4. Suppose the resistance values are multiplied by 4. W hat are the new loop currents? W hat are
the new node voltages?
AN SW ERS: Loop currents are 0.25 times their original values, and node voltages are unchanged.
Observe that there are no dependent current or voltage sources in the circuit. Similar to the nodal
analysis case, whenever dependent sources are absent and the equations are written in the natural
order, the loop (or mesh) equations are symmetric, as illustrated by the coefficient matrix o f equa­
tion 3.28 where, for example, the 1 -2 and 2 -1 entries coincide, as do the 1 -3 and 3 -1 entries,
etc. Also, the value o f all entries can be computed by inspection. The 1-1 entry o f the matrix is
the sum o f the resistances in loop 1; the 2 - 2 entry is the sum o f the resistances in loop 2, etc. In
general, the i- i entry is the sum o f the resistances in loop i. T he 1 -2 entry o f the matrix is 'L{±R^
(the large sigma means summation), where each
is a resistance common to both loops 1 and 2.
Use the + sign when both loop currents circulate through
in the same direction, and use the -
Chapter 3 • Nodal and Loop Analyses
131
sign otherwise. Further, if independent current sources are absent, then the right-hand side o f the
loop equations can also be written by inspection. The i-th entry is simply the net voltage o f the
sources in the i—th loop that tends to deliver a current in the direction o f the loop current.
Exercises. 1. Use the inspection rules described above to write two mesh equations for the circuit
o f Figure 3.12, when both mesh currents are assigned clockwise direction.
FIG U RE 3.12.
2. Use the inspection rules described above to write two mesh equations for the circuit o f Figure
3.12, when the left mesh current is clockwise and the right mesh current is counterclockwise.
3. Use the inspection rules described above to determine the right-hand side o f the mesh equation
for the circuit o f Figure 3.13.
AN SW ERS: 8, 0, 10
A simplifying reduction to the set o f loop equations occurs if an independent current source coin­
cides with a single loop current. The analysis becomes simpler because that loop current is no
longer an unknown; rather it is equal to the value o f the source current if their directions coin­
cide, or to the negative value if their directions are opposing. Because the associated loop current
is known, there are fewer loop equations to write and solve. One would apply KVL to such a loop
only if it were necessary to compute the voltage across the independent current source, which
might be necessary for determining the power delivered by the source. Th e entire situation is anal­
ogous to an independent voltage source tied between a node and the reference in nodal analysis.
The following example illustrates the details o f this discussion.
Chapter 3 • Nodal and Loop Analyses
132
EXA M PLE 3.8.
The circuit o f Figure 3.1 4 is a modification o f the one o f 3.11 in which (i) a 1 ohm resistor on
the perimeter o f the circuit is replaced by an 8 A independent current source, and (ii) the values
o f the voltage sources are doubled. The currents for each loop are again denoted by /j, I 2 , and ly
Our objective is to find all the loop currents, the voltage V^, and the power delivered by the 8 A
source.
FIG U RE 3.14. A resistive circuit containing an independent current source on the perimeter
o f loop 3 forcing /^ = 8 A.
So l u t io n .
Step 1. So/ve fo r
by inspection. Because
is the only loop current circulating through the
branch containing the independent 8 A current source, /j = 8 A. This phenomena is similar to the
fact that in nodal analysis, the node voltage o f a grounded voltage source is fixed at the voltage
source value.
Step 2. Write a KVL equation fo r loop 1 by summing voltages around this loop. Summing the
voltages around loop 1 using Ohm’s law and the designated loop currents produces
28 = / j + 4 ( /j - /2) + 12 + ( /j - 8) = 6 /, - 4 /2 + 4
Hence,
6 /, - 4 /2 = 24
(3.29)
Step 3. Write a KVL equation fo r loop 2 by summing the voltages around this loop. Applying KVL
and Ohm’s law to loop 2 produces
0 = 4 (/2- /i) + 2 /2 + 2 (/2 - 8 ) = - 4 /, + 8 /2 - 16
Equivalently,
- 4 /1 - 8 /2 = 16
(3.30)
Chapter 3 • Nodal and Loop Analyses
133
Step 4 . Write above loop equations in matrix form and solve. The matrix form o f equations 3.29
and 3.30 is
' 6 -A
'i r
’24‘
-4
l2
16
8
Using the inverse matrix technique to compute the solution, we have
II"
■6 -4
I2
-4
-1
8
24 '
1
16
“ 32
' 24'
4 6 16
'8
4
's'
6
Step 5. Compute V^. By KVL,
v; = 2 (/2 - 8) + 12 + (/j - 8) = 8 V
Step 6. Compute power delivered by 8 A source. Observe that the 8 A current source is labeled
according to the passive sign convention, in which case,
Pdel = -
= - (8
8) = - 64 watts
Hence, the source actually absorbs 64 watts.
Exercise. In the circuit o f Figure 3.15, nvo o f the three mesh currents coincide with independent
source currents. By writing and solving just one mesh equation, find /j.
A N SW ER: 3 A
Not only do independent current sources constrain loop currents, but dependent currents sources
do also. This situation is illustrated in Example 3.9.
E X A M PLE 3 .9 .
This example illustrates the writing o f loop equations for a simplified small signal equivalent cir­
cuit, Figure 3.16, o f a two-stage amplifier that contains a current-controlled current source
(C C C S) and a current-controlled voltage source (CC VS). This process extends the techniques o f
Examples 3.7 and 3.8 to find some important characteristics o f the amplifier. Specifically, find
(a) the input resistance seen by the source, i.e.,
= v-Jij^,
134
Chapter 3 * Nodal and Loop Analyses
(b) the voltage gain, v
j and
(c) the voltage v across the dependent current source.
+
V
FIG U RE 3.16. Small signal equivalent circuit for a two-stage amplifier. Signals in amplifiers are
usually time dependent, so we adopt the lowercase notation for voltages and currents.
So l u t io n .
The circuit o f Figure 3.16 contains three loop or mesh currents. The direction o f the loops is a
user-chosen preference. For convenience, we have chosen mesh current z'2 to be consistent with
the direction o f the arrow in the dependent current source. Because this dependent current source
lies on the perimeter o f the circuit, it constrains the value o f /'2> i-e-) ?2 ^ P ^b-
the control­
ling current,
, z'2 = . ih ~ ^ h'
relationship implies that the mesh current o f loop 2
depends directly on the mesh current o f loop 1. This observation allows us to skip constructing a
mesh equation for loop 2. Only equations for loops 1 and 3 are needed, thereby reducing the
number o f simultaneous equations from three (because there are three loops) to two.
Step 1. Apply KVL to loop/mesh 1. Here, by KVL and the observation that
^in = h h +
('1 + P ' 1) =
+ (1 + p)
12
= P zp
i\
From this equation, we can immediately compute the input resistance
V•
in
V•
in
(3.31)
Step 2. Apply KVL to loop/mesh 3. In this case, observe that z^ = — (z^ + z'g) = - (|3 z'j + Zj). By BCVL,
'3 -
K + K ('3 + ' 2) =
h +
(P h + ' 3) + K ('3 +P ' 1) = 0
Combining like terms, it follows that
K +
ph +
'3 = 0
(3.32)
135
Chapter 3 * Nodal and Loop Analyses
Step 3. Write equations 3.31 and 3.32 in matrix form an d solve. The matrix form o f these equa­
tions
IS
/?^ + (P + l ) ^ ,
0
h
^in
'3
0
(3.33)
)
V.
Because the solution is desired in terms o f the hteral variables, we solve equation 3.33 using
Cramer’s rule, which utilizes determinants. In this task, first define
Using the notation A for the determinant, Cramers rule provides the solution for i^ according to
the formula
det
^in
0
0
R^ + R ^+r„
(3.34)
‘I =
Applying Cramer’s rule for the solution o f Zj, yields
det
Step 4. Compute
/?^ + (|3 + l)7 ? e
V •
m
(3.35)
in terms o f v-^ and then the voltage gain vjv^^. As per the circuit o f Figure
3.11 and equations 3.34 and 3.35,
Vo =
V ;„ = r
ic = '•w(PM + '3) = r,i
After substituting for A, the voltage gain is
o
---V.
in
A
P/?
I— d
r= y
^
—r
Step 4 . Compute v. To compute the voltage across the dependent current source, apply KVL to
mesh 2 to obtain
n
= R^ (z'2 + /'i) + R, («2 + ^3) = [P
h +
h
136
Chapter 3 * Nodal and Loop Analyses
Exercise. Find a simplified loop equation for
in the circuit o f Figure 3.17.
FIG U RE 3.17.
c' '1
To see the importance o f the calculations o f the amplifier circuit o f Example 3.9, suppose two
amplifiers are available for use with a non-ideal voltage source. The non-ideal voltage source is
modeled by an ideal one-volt source in series with a 100 Q source resistance. Suppose amplifier 1
has a voltage gain, vjv^^ = 1 0 and
= 100 k£2. Suppose amplifier 2 has a voltage gain o f 100
and R -^ 2 = 5 £3. If amplifier 1 is attached to the non-ideal source, then by voltage division,
=
100,000/(100,000 + 100) = 0.999 V, whereas in the case o f amplifier 2, v -^ 2 = 5/(100 + 5) =
0 .0 4 7 6 V. In the first case, the gain from
to
is 10, yielding
= 9.99 V. In the second case,
the same gain is 100, yielding v^ 2 =
V. One concludes that amplifier 1 is better suited to this
particular application, although it has a lower voltage gain than amplifier 2. Hence, Example 3.9
illustrates the need to know both the voltage gain and the input resistance to determine the out­
put voltage in practical applications. Further, using the literal solution to the example allows us to
apply the formulas to different sets o f parameter values without repeating the complete analysis.
In the previous two examples, there were current sources on the perimeter o f the circuit. Such cur­
rent sources were incident to only one loop. It often happens that independent and dependent
current sources can be common to two or more loops. When this happens, a situation analogous
to floating voltage sources in nodal analysis occurs. To handle such cases, many texts define some­
thing called a supermesh and write a special loop equation for this supermesh. Supermeshes often
confuse the beginner. There is an easier way.
Example 3.10 below illustrates how to write “loop” equations when current sources are common
to two or more loops. In such cases, we introduce auxiliary voltage variables across current
sources common to two or more loops. The resulting set o f simultaneous equations will contain
not only the loop currents as unknowns, but also the auxiliary voltages as unknowns. Because the
resulting set o f equations contains both loop currents and additional (auxiliary) voltage variables,
the equations are called modified loop equations. T he process o f writing modified loop equa­
tions is extremely systematic and straightforward. Further, it allows us to avoid explaining the very
confusing concept o f a supermesh. On the other hand, the presence o f auxiliary voltage variables
increases the number o f “unknowns,” i.e., the number o f simultaneous equations increases.
Because o f the availability o f software packages such as MATLAB, M ATH EM ATICA, and
MAPLE, this increased dimension is not a hindrance.
137
Chapter 3 * Nodal and Loop Analyses
E X A M P L E 3 .1 0 .
Consider the circuit o f Figure 3-18 in which
= 28 V and 7^2 = 0 .06 A. Note that the inde­
pendent current source is common to loops 1 and 3 and a voltage-controlled current source is
common to loops 1 and 2. Find values for the loop currents 7p Ij, l y and the power delivered by
each independent source.
200Q
FIG U RE 3.18. Circuit containing a current source between loops.
So l u t io n .
To begin the solution, we introduce two auxiliary voltage variables Vj and Vj associated with the
current sources common to two (or multiple) loops. The purpose o f these variables is to facilitate
the application o f KVL for constructing the loop equations. This will require that we obtain three
KVL equations, one for each loop, and two constraint equations, one for each current source.
Step 1. Apply KVL to loop 1. By a clear-cut applicanon o f KVL,
28 = 2007] - V] - V2
(3.36)
Step 2. Apply KVL and Ohm’s law to loop 2. Again applying KVL and Ohm’s law to loop 2, we
obtain 100 7j + 200 (Jr^ —7^) + V2 = 0. After grouping like terms.
3 0 0 7 , - 20 0 /3 + V2 = 0
(3.37)
Step 3. Apply KVL to loop 3. Applying KVL to loop 3 yields 150 7^ + Vj + 200 (/j - Tj) = 0.
Equivalendy,
- 200/2 + 350/3 + V| = 0
Step 4. Write a constraint equation determined by the independent current source. Here, loops 1
and 3 are incident on the independent current source so that
0 .0 6 = / , -/ 3
(3.39)
138
Chapter 3 * Nodal and Loop Analyses
Step 5. Write a constraint equation determined by the dependent current source. In a straightfor­
ward manner, we have
/ , - / , = 0 .0 2 V, = 0 .02 r2 0 0 (/3 -
12
) = 4 7 , - 47,
After simplification,
/i + 3/2 - 4 /3 - 0
(3.40)
Step 6. Write equations 3 .3 6 to 3.40 in matrix form an d solve. The matrix form o f these equations is
0
0
-1
-r
w
300
-2 0 0
0
1
h
-2 0 0
350
1
0
h
0
-1
0
0
''I
0 .0 6
3
-4
0
0
V2
0
■ 28
0
=
0
(3.41)
Solving equation 3.41 by the matrix inverse method or by an available software package yields the
solution (currents in A and voltages in V) given by equation 3 .42 below:
w
200
0
0
-1
h
0
300
-2 0 0
0
0
-2 0 0
350
1
0
-1
I
3
-4
Li
,''2.
=
j
-r -1 ■ 28
0 300 -200 0 1
0 -200 350 1 0
1 0 -1 0 0
1 3 -4 0 0];
»b = [28 0 0 0.06 0 ]’;
»LoopIplus= M\b
Looplplus =
l.OOOOe-01
2,0000e-02
4,0000e-02
-l.OOOOe+01
2.0000e+00
■0.1 ■
1
0
0 .02
0
0
= 0 .0 4
0
0
0 .06
-To
0
0
0
2
1
.
which can be obtained using the following MATLAB code:
M = [ 2 0 0 0 0 -1 -1
■
(3.42)
139
Chapter 3 • Nodal and Loop Analyses
Step 7. Compute the powers delivered by the independent sources. First, the power delivered by the
independent voltage source is
Py-source = 28 /j = 2.8 watts
The power delivered by the independent current source is
Pl-source =
V^ = - 0 .6 watts
This last value indicates that the independent current source actually absorbs power from the cir­
cuit.
Exercise. For the circuit o f Figure 3.19, write the modified loop equations having two unknowns
/j and V, following the procedure described in Example 3.10. Solve the equations and find the
power absorbed by the 2 -Q resistor.
4A
AN SW ER: 18 watts
One final point before closing our discussion o f loop analysis. Loops can be chosen in different
ways. Cleverly choosing loops can sometimes simplify the solution o f the associated equations. For
example, by choosing a loop that passes through a current source so that no other loop is com­
mon to the source, the loop current is automatically specified by that current source.
6. SUM M ARY
This chapter introduced the technique o f nodal analysis. Nodal analysis is a technique for writing a set
o f equations whose solution yields all node voltages in a circuit. With knowledge o f all the node volt­
ages and all the element values, one can compute all branch voltages and currents. As mentioned, when­
ever there are no dependent sources present, the coefficient matrix o f the node equations is always sym­
metric. Hence, whenever dependent sources are absent, it is possible to write the nodal equation coef­
ficient matrix by inspection. Further, if independent voltage sources are absent, then the right-hand side
o f the matrix form o f the nodal equations can also be written by inspection: the entry is simply the sum
o f the independent source currents injected into the node at which KCL is applied. When VCCSs are
present, the steps for writing nodal equations are the same as illustrated in Example 3.3. Generally, in
such cases, the resultant coefficient matrix is not symmetric.
140
Chapter 3 • Nodal and Loop Analyses
W hen floating dependent or independent voltage sources are present with respect to a given ref­
erence node, we introduce new current variable through these floating sources as unknowns. The
node equations then incorporate these additional unknown currents, as was illustrated in
Examples 3.5 and 3.6. This method increases the number o f equations but simplifies the con­
struction o f the individual equations. W ith a tool like MATLAB to compute solutions, there is no
difficulty, although hand computation may become more difficult. This concept is the basis o f the
modified nodal analysis method used in circuit simulation programs like SPICE.
Loop/mesh analysis, an approach dual to nodal analysis, was introduced in Section 5. Mesh analy­
sis is a special case o f loop analysis for planar circuits when the loops are chosen to be the obvious
meshes, similar in geometry to a fish net. In loop analysis, one sums the voltages around a loop or
mesh to zero. Each o f the branch voltages in the loop is expressed as a product o f resistances and
(fictitious) loop currents that circulate through the branch resistance, as illustrated in Figures 3.10,
3.14, and 3.16. The branch current o f the circuit are equal to the net flow o f the loop currents
incident on a particular branch, meaning that each branch current is expressible as a sum o f loop
currents. The desired set o f loop equations is produced by summing the voltages around each loop,
expressing these voltages either as source values or as resistances times loop currents. One solves
the loop equations for the loop currents. Once the loop currents are known, we can then compute
the individual branch currents and then the branch voltages, and thus any other pertinent current,
voltage, or power. Whenever there are no dependent sources present, the coefficient matrix o f the
loop equations is always symmetric. Whenever dependent sources are absent, it is possible to eas­
ily write the loop matrix by inspection.
As the size o f an arbitrary circuit grows larger, there are two good reasons for choosing the nodal
method over the loop method; (i) the number o f nodal equations is usually smaller than the num­
ber o f loop equations, and (ii) the formulation o f nodal equations for computer solution is easier
than methods based on loop equations. Writing nodal equations is particularly easy if the circuit
contains only resistances, independent current sources, and VCCSs — for short, an R—
I—
g„ net­
work. For an
network, one simply applies KVL to every node (except the reference node)
and obtains a set o f node equations directly. For floating independent or dependent voltage
sources, the task is more complex. Examples 3.5 and 3.6 illustrate cases where, besides the node
voltages, additional unknown auxiliary currents are added. By adding additional auxiliary variables
to the formulation o f the nodal equations, we described the concept behind the modified nodal
analysis (MNA) method. The MNA method retains the simplicity o f the nodal method while
removing its limitations and is the most commonly used method in present-day computer-aided
circuit analysis programs.
7. TERM S AND C O N C EPTS
Connected circuit: every pair o f nodes in the circuit is joined by some set o f branches.
Cram er’s rule: a method for solving a linear matrix equation for the unknowns, one by one,
through the use o f determinants; the method has serious numerical problems when
implemented on a computer, but is often convenient for small, 2 x 2 or 3 x 3, hand cal­
culations.
Floating source: neither node o f the source is connected to the reference node.
Chapter 3 ®Nodal and Loop Analyses
141
Gaussian surface: a closed curve or a closed surface surrounding two or more nodes.
Linear matrix equation: an equation of the form Ax =b, where A i s z n x n matrix, x is an n-vector of unknowns, and b is an n-vector of constants.
Loop (closed path): a contiguous sequence of branches that begins and ends on the same node
and touches no node more than once.
Loop analysis: an organized method of circuit analysis for computing loop currents in a circuit.
Knowledge of the loop currents allows one to compute the individual element currents
and, consequently, the element voltages.
Loop current: a (fictitious) current circulating around a closed path in a circuit.
Matrix inverse: the inverse, if it exists, of an n n matrix yl, denoted b y ^ “ ^ satisfies the equation
A A~^ = A~^ ^ = I,, where / is the w x « identity matrix; the solution of the linear matrix
equation, ^
is given hy x = A~^b,
Mesh: After drawing a planar graph without branch crossing, the boundary of any region with
finite area is called a mesh. Intuitively, meshes resemble the openings of a fish net.
Mesh analysis: the special case of loop analysis for planar circuits in which the loops are chosen
to be the meshes.
Mesh current: a fictitious current circulating around a mesh in a planar circuit.
Modified nodal analysis: a modification of the basic nodal analysis method in which the
unknowns are the usual nodal voltages plus some naturally occurring auxiliary currents.
Nodal analysis: an organized method of circuit analysis built around KCL for computing all node
voltages of a circuit.
Node voltage: the voltage drop from a given node to a reference node.
Symmetric matrix: a matrix whose transpose is itself I f ^ =
is a « x « matrix whose i- j entry
is
then A is symmetric if a - = a-.
142
Chapter 3 • Nodal and Loop Analyses
3. For the circuit o f Figure P3.3, suppose
=
1.2 A. Write a single node equation in the volt­
PROBLEMS
age V and solve.
SIN G LE N ODE PROBLEM S
1. For the circuit o f Figure P3.1, write a single
node equation in
, Gj > <^3 >
For a fixed K > 0 , R-^ = R , R2 = 2R ,
Compute Kj
^ 2-
= 2R.
Figure P3.3
in terms o f R and V^j
A N SW ER: - 6 V
4 Ksi-
M ULTIPLE N ODE PROBLEM S
4. The purpose o f this problem is to write the
nodal equations direcdy by inspection o f the cir­
cuit diagram o f Figure P3.4. Recall that when the
Figure P3.1
network has only independent current sources and
resistors, the nodal equation matrix is symmetric
A N SW ER: K, = l.5V^^
and the entries can be written down by inspection
2 . The battery o f your car has been dealt a sud­
as per the discussion following Example 3.2.
den death by the sub-zero North wind and a
Construct the nodal equations in matrix form for
the circuit o f Figure 3.4 by inspection.
faulty alternator. Unable to fight the elements,
you wait a few days hoping for a thaw, which
comes. You replace the alternator. Then, using
your roommate’s car, you attempt a jump-start.
Nothing happens. You let it sit for a while with
your roommates car running juice into your
battery for 20 minutes. Still, nothing happens.
W hy won’t your car start? Consider the circuit
reference node
o f Figure P3.2. Notice that your “dead” batter
Figure P3.4
is labeled as Vq. Your roommate’s battery is
A FEW A N SW ERS: 'Fhe 3-3 entry is G^ + G^
labeled 12 V. Each battery has an internal
+ G^ + Gg, and the 2-1 entry is -G y
resistance o f 0.02 Q and the starter, an internal
resistance o f 0.2 Q. The starter motor requires
5. Consider the circuit o f Figure P3.5 in which
50 A to crank the engine. Find the minimum
/^, = 0.5 A and V^ 2 = 40 V. Furtiier, let G, = 5 mS,
G2 = 2.5 mS, G3 = 2.5 mS, and G^ = 12.5 mS.
(a)
By inspection, what is the
value o f voltage Vq needed before the starter
can draw 50 A and work.
value o f
50 A
>0.020
7 0.20
''load <
v
I
Live
Battery
Write a minimum number o f node
equations and put in matrix form.
0.02Q
V
12V
(b)
(c)
Solve the node equations for the voltages
and
using MATLAB or the for­
mula for the inverse o f a 2 x 2 matrix:
Dead
Battery
Figure P3.2
Starter
Motor
a b
c d
_
1
'd -b'
ad - be
-c a
Chapter 3 “ Nodal and Loop Analyses
143
(d)
Find 1 ^ ,7 ^ and
(e)
Find the power delivered by each
source and the power absorbed by
each resistor. Verify the principle of
conservation of power.
8 . The circuit of Figure P3.8 is an experimental
reference node d
Figure P3.5
measurement circuit for determining tempera­
In the circuit of Figure P3.6,
ture inside a cavern underneath the Polar ice
6 . (a)
= 8
cap. The cavern is heated by a fissure leading to
V* Further,
= 5 kQ,
R l = R^ = R^ = 20 k£2, and R^ = 10
kfl. Find the node voltages,
and
some volcanic activity deep in the earth. The
V^, and also the voltage, V^. Compute
the range -2 5 ° C to +25°C . The nominal tem­
the power absorbed by R^ and the
perature of the cavern is 0°C. In this type of cir­
power delivered by each of the sources.
cuit, the voltage
It is suggested that you write your
the temperature changes. Suppose that
equations in matrix form and solve
V, and in kQ, R-^ = 20, i?2 ” 44, R^ - 20, and
using MATLAB or the formula for a 2
R^ = 12.5. Note that the 4 4
mA, V^2 -
X
(b)
2 inverse given in problem 5.
resistor
changes its value linearly from 15
IdQ to 65 k n as a fiinction of temperature over
- Vg is a measure of how
= 50
resistor is a
result of manufacturing tolerances that often
Repeat part (a) when all resistances are
permit deviations from a nominal of, say, 40
cut in half
kflt, by as much as 2 0 %. As usual, it is cost ver­
sus precision.
(a)
Write a set of nodal equations in the
(b)
Assuming
----------^/S/^-------
variables
and Vq
= 40 kQ at 0°C, put
the nodal equations in matrix form
and solve for the node voltages,
and Vq
(c)
Determine the power delivered by the
source.
Figure P3.6
(d)
Use MATLAB to solve for all the node
7. In the circuit of Figure P3.7,
^2
= 4 V, and
- 1 mA. Further, in mS, G| =
voltages as
0.4, G2 = 2, G3 = 3, and
out. Then find the linear equation
analysis to find
= 5. Use nodal
and V^. Then compute the
power delivered by the independent sources
and the power absorbed by G2 .
varies from 15 to 65
ki2 in 1 kfi increments. Do not print
relating
to temperature. Plot Vq
- y c as a W c tio n of temperature, i.e.,
over the range - 2 5 ° C to +25°C . Over
what range of temperatures about 0
degrees would the sensor be reason­
ably accurate? W hy and why not?
144
Chapter 3 • Nodal and Loop Analyses
II.
Consider the circuit o f Figure P 3 .l l .
Choose node D as the reference node. This
choice eliminates the floating voltage source
and hence the nodal equations can be written
without the need o f a so-called supernode. Let
= 0.0 8 S, Gj = 0.08 S,
0.0 2 S, G 5 = 0 .0 2 S,
= 0.3 A, and
= 0.01 S, G^ =
= 0.3 A, 7^2 = 0 .2 A, 7^3
= 50 V. Write and solve a set
o f nodal equations for the voltages
Figure P3.8
and Vg = Vg0 . Then compute the powers
9. In the circuit o f Figure P3.9, ail resistances
are 1 ItQ, except
= 500 Q. Suppose
delivered by each o f the sources.
=
G,
100 V and 1 ^ 2 = 0-3 A. Compute all tiie node
voltages o f the circuit. You may want to use
MATLAB or a calculator that inverts matrices
G,
to compute the answer. Compute the power
© '
delivered by the independent sources.
Figure P 3 .1 1
12. In the circuit o f Figure P 3.12,
= 20 kQ, T?3 = 20 kQ,
= 5 mA. Find
Figure P3.9
1,7,0.6,7
10. In the circuit o f Figure P3.10,
= 30 V,
= 0.6 A. Use nodal analysis
on the circuit below, as indicated:
(a)
Figure P3.12
Write a nodal equation at node A.
(b)
Write a nodal equation at node B.
(c)
Write a third nodal equation at node C.
(d)
Solve the 3 equations in 3 unknowns
13. Consider Figure P3.13.
(a)
M ATLAB, or using some other soft­
ware program to obtain all the node
this, let G-=\ I R -.
(b)
W ith K ^ = 150V ,7?^ = lk Q ,7?, = 5 k Q ,
(c)
= 15
, Vg the power delivered by
, and the power absorbed by R^ .
Compute I 2 , the current through Rj
^2 = 10 kQ, T?3 = 10 kQ, and
voltages. Show ALL work/procedures.
mS, find
Find the power delivered by the inde­
pendent voltage source.
ion
100
Write the nodal equations and place in
matrix form prior to solving. In doing
by hand, with your calculator, using
(e)
and the power delivered by
the dependent current source.
ANSWERS IN RANDOM O R D E R
7^2 = 1-2 A, and
= 5 kQ,
= 0.55 x 10“3, 7-„
from left to right.
L
ion V.
ion.
©
Reference node
Figure P 3 .1 0
Figure P 3 .13
Chapter 3 “ Nodal and Loop Analyses
145
14. Consider the circuit of Figure P 3.14
(a)
17. Use nodal analysis to find the voltages
Write two node equations in terms of
the literal variables in Figure P 3.14
(c)
(e)
in the circuit of Figure P 3.17.
= 20 Q, i?2 = 10 Q,
= 10 Q, and I^=
= 4 Q,
=
A. Note that in
and put in matrix form.
0.1 S,
Solve the node equations for the volt­
solving this problem, you are to generate three
and Vq when
6
= 0.1 A, 7^2 =
(nodal) equations in which the unknowns are
0-2 A,
= 7 mS;
= 2 mS,
=
500 Q ; i?2 = 333.33 £2; and /?3 = 1 fl.
Determine Kq.
you could eliminate the equation
ages
(d)
, and
Suppose
for
but this problem is to illustrate that
such elimination is not necessary. Finally, deter­
Determine the power delivered by
mine the equivalent resistance seen by the inde­
each source. (Be careful of sign.)
pendent current source.
. 9.iVa
'4
Figure P3.14
Figure P3.17
15. Consider the circuit of Figure P3.15 in which
= 20 V and
(a)
= 0 ; node voltage C is
.
Write the two nodal equations in
terms of the literal variables.
(b)
Suppose 11 = 6 and the following in S
are given:
= 0.5, G2 ”
= 4, 6*5 = 1. Solve for
and
Check: 20 and 10 volts.
(c)
Find
and then find the equivalent
resistance seen by the independent
voltage source.
(d)
"■
18. Consider the circuit of Figure P3.18. By
choosing node C as the reference node, we elim­
inate a floating voltage source. Write an appro­
priate set of nodal equations, with node C as the
reference node. Solve the nodal equations, speci­
fy the voltages V^q Vbo V^Q and V
and the
power delivered by the sources. Finally, find the
equivalent resistance seen by the current source.
Leti?i = 9 k Q ,7 ? 2 = l S k i 2 ’
= 6 kQ,
kQ,
= 3000 Q, and = 20 mA.
= 9
Find the power delivered by the inde­
pendent source and the dependent
source.
(e)
W hat is the power absorbed by the
output resistor?
-------- --------------
Figure P3.18. By choosing node C as the ref­
erence node, it is possible to simplify the con­
struction of the node equations.
19. Consider the circuit of Figure P 3.19 in
Figure P3.15
16. Redo problem 15 with
\
‘ J
= 60V .
0.25 S and
which
60 ^2.
- 20 Q, /?2 ~ 20 Q,
= 2 0 Q,
= 30 Q, R^ =
= 12 V, and
=
146
Chapter 3 * Nodal and Loop Analyses
FLOATING VO LTAG E SO URCE
PROBLEM S
0.6 A. The point o f this problem is to illustrate
how a good choice o f reference node may sim­
plify the calculation o f node voltages, whereas a
poor choice may lead to a complicated formu­
22. For example 3.5 suppose all resistance values
lation o f the node equations.
are doubled, the floaring voltage source remains
Choose a reference node so that there
the same at 440 V, and all current sources are
are no floating voltage sources. Write
scaled down to one-half o f their original values.
three equations in the unknown volt­
(a)
Compute all node voltages and the
ages. Solve for the node voltages.
current
C H EC K : i^ = 2 A k and iy = - 3 0 A.
(b)
Compute the voltages
and
Determine the power delivered by
(b)
(c)
Compute the power absorbed by the
each source.
0.075 S resistor.
(c) Determine the power absorbed by
(d)
Compute the power delivered by the
each resistor.
floating voltage source.
(d) Verify conservation o f power using the
(a)
results o f parts (b) and (c).
23. For the circuit o f Figure 3.10 in Example 3.6,
suppose the 110 V source is changed to 200 V and
the 50 Q resistor is changed to 500 £2. Find the
node voltages
rents
and
Vy,
and the unknown cur­
Then find the powers delivered by
the 30 V source and the dependent source.
24. Consider the circuit o f Figure P 3.24 in
which
= 2 0 0 V, V^2 = 5 0 V, R^ = 5 0 Q, R.^
= 20 Q, R^ = 50 Q, and R^ = 40 Q.
(a)
Identify the floating voltage source and
(b)
W rite
add a current label through the source.
20. The nodal equations for the circuit in
Figure P3.20 are
0.03
0.09
modified
nodal
equations,
which include both node voltages and
- 0 .0 r ■Vf
0.04
unknown currents through any float­
'A-.'
ing voltage sources.
0
.^0.
Compute the values o f R^, Rj ,
(c)
,a n d .
Solve the equations for the node voltages Kg and Vq and the current
through the 50 V source. C H ECK :
Vg = 50 V and
Figure P3.20
= 100 V.
(d)
Find the power consumed by R^ .
(e)
Determine the power delivered by
each o f the sources.
21. Consider the circuit o f Figure P3.15, which
has nodal equations {R^
■ 0 .0 0 8
0 .0 1 9
- 0 .0 0 5 '
-0 .0 0 1
0 .0 0 5 5
- 0 .0 0 2
-0 .0 0 5
- 0 .0 0 2
0 .1 0 7
Compute the values of /?j,
0) given by
■ 0
=
0
O .n/
■
0
=
■
0
G A.
>-^3 >-^4’ -^5 ’ ^nd ,u.
Figure P 3 .2 4
Chapter 3 * Nodal and Loop Analyses
147
25. Consider the circuit o f Figure P3.25 in
27. T he modified nodal equations for the cir­
which l/j = 2 50 V,
cuit o f Figure P 3.27 are
: 20 Q,
(a)
(b)
= 5 0 Q,
= 5 0 V,
= 50 Q ,
= 40 Q , and R^ = 10 Q.
■0.004
-0 .0 0 1
-0 .0 0 2
add a current label through the source.
-0 .0 0 1
0.001
0
-1
W rite
-0 .0 0 2
0
0.004
1
2
-1
-1
0
Identify the floating voltage source and
modified
nodal
equations,
which include both node voltages and
0 •T 4 '
O'
yB
0
^CB
0
unknown currents through any float­
(c)
ing voltage sources.
Compute all four resistor values and (3. Hint:
Solve the equations for the node volt-
Find all the conductances first and then convert
ages Vg and Vq and the current
to resistances.
through the 50 V source. C H EC K :
5 0 V a n d V^= 100 V.
(d)
(e)
Find the power consumed by R^ .
Determine the power delivered by
each o f the sources.
Rc
Figure P3.27
= 0.02
= 10 Q.,
=
28. For the circuit o f Figure P 3.28,
S, ^2 = 0.025 S, Gg = 0.2 S,
0.4 A, and V^ 2 ~ 12 V. Use nodal analysis to
find all node voltages, the current , the power
absorbed by
and the power delivered by
the two sources.
Figure P3.25
26. Consider the circuit o f Figure P3.26. R^ =
10 Q ,
= 100 Q, R^ = 100 Q,
= 50 Q ,
= 100 V, K^2 = 60 V, V^3 = 100 V,
= 14
A. Label appropriate currents
and I j^q
through the floating voltage sources.
(a)
Write the modified nodal equations
for the three unknown node voltages
and two unknown currents.
(b)
Solve for the five unknowns (in MATLAB).
(c)
Find the power delivered by each o f
(a)
Determine Vq
(b)
Label the current /ABUsing V^, Vg,
and 4 as
unknowns, write a 4 x 4 matrix set o f
(c)
nodal equations.
(d)
Solve the nodal equations for V^, Vg,
^AB’ and 4 .
(e)
(0
Determine the power absorbed by G2 .
Determine the power delivered by all
sources.
AN SW ERS (D) IN R A N D O M O R D E R : 12,
0.24, 9.6, 0.16
the sources.
29. Repeat Problem 28, except this time write
Reference node
Figure P 3 .2 6
only three nodal equations in the variables V^,
Vg, and l^g. Notice that you must express
in
148
Chapter 3 • Nodal and Loop Analyses
terms o f
and the appropriate conductance.
One can even reduce the number o f equations
AN SW ERS (R A N D O M IZ ED ): 250, 325, 1 2 5 ,7 5 ,2 5 0 , 25
to two using the so-called supernode approach,
32 Consider the circuit o f Figure P3.32, where
which is the subject o f other texts.
= 4 kQ,
30. For the circuit o f Figure P3.30,
= 100 Q,
mS,fj. = 4 S,
= 1
= 4/3 kQ,
= 0.75
= 160 V, and
= 40 mA.
= 300
(a)
Specify V^.
£2 A,
= 2 A, Vj2 ~
nodal analysis
to find all node voltages, the current
, the
(b)
Write modified nodal equations.
= 20 Q,
= 20 Q, G4 = 0.09 S,
(c)
nodes B and C, and the power delivered by the
(d)
independent sources as follows:
Find the power delivered by each o f
the sources.
(a)
Determine Vq
(b)
Write a set o f modified nodal equa­
(e)
tions that contain extra current vari­
ables including
Solve the modified nodal equations in
MATLAB.
power absorbed by the 20 Q resistor between
and
Compute the power absorbed by each
resistor.
(f)
Verify conservation o f power.
.
(c)
Solve your nodal equations for the
(d)
Compute the power delivered by the
independent sources.
unknowns. C H EC K : Vg= 180 V.
(e)
Compute the power absorbed by the
20
resistor between nodes B and C.
Figure P 3.32
33. Consider the circuit in Figure P 3.33 in
which V-^ = 60 V, G j = 0.1 S, G2 = 0.1 S,
= 0.3 S, G4 = 0.4 S, G 5 = 0.1 S, Gg = 0.1 S, Gj
= 7/480 S, ^ = 3, and (3 = 2 .
(a)
Figure P3.30
using
31. Use nodal analysis on the circuit o f Figure
P3.31 as indicated.
all resistors are 10 Q.
(a)
W rite
= 100 V,
modified
1^2
nodal
= 1 A, and
equations
including the extra variable IgQ
(b)
(c)
Write the modified nodal equations
(b)
(c)
Vg Vq and /^^as unknowns.
Solve the modified nodal equations in
MATLAB.
Find
the equivalent
resistance seen by the independent
voltage source.
Solve the modified nodal equations in
MATLAB.
Find the power delivered by each of
the sources.
Reference node
Figure P3.31
Figure P3.33
AN SW ERS T O (C) IN RA N D O M O R D E R :
40, -2 5 ,3 8 .7 5 ;/ ?
= 12 Q .
Chapter 3 • Nodal and Loop Analyses
149
SIN G LE LOOP-EQUATION
PROBLEM S
34. In the circuit o f Figure P3.34,
= 50 Q, and
= 0.5. If
= 400 Q,
= 50 V, find 4 ,
the power delivered by the independent and
37. Consider the circuit o f Figure P3.37
(a) Suppose
sources.
R,
dependent voltage sources, and the equivalent
resistance,
= 2 0 0 Q, R^ = 300 Q, R2 = 500
Q, /jj = 750 mA and I ^ 2 = 100 ^lAh
the power delivered by each o f the independent
--------- ----------- ---
, seen by the independent source.
Figure P3.37
C H EC K : /j = 100 mA.
(b)
C H EC K S: R^^ = 2 50 Q, and
3 5 . In the circuit o f Figure P 3.35,
= 5 0 Q, and
= 50 V, find
= 8 watts.
= 4 00 £2,
= 50 Q, and jj. = 0.5. If
, the power delivered by the
independent and dependent voltage sources,
and the equivalent resistance, R^^ , seen by the
independent source.
Now suppose /^j = 4 0 0 mA and 1 ^ 2 =
100 mA and the loop equation for /j
written in the standard way directly
yields 2000/|= 60. Find R-^ and R2 if
R^ = 600 fil. Note: If the equations are
not written in the standard way, the
solution is not unique. For example,
multiplying both sides o f the above
equation by 0.5 yields a different
answer in which R-^ = 140 £2 , as
opposed to the correct answer o f R-^ =
400 Q.
38. In the circuit o f Figure P 3.38,
/j2 = 100 mA, 7^3 = 2 0 0 mA, and
mA. Find
Figure P3.35
CH ECKS: R^^ = 200 Q, and
36. In the circuit o f Figure P3.36,
= 56 V,
= 100 mA
and the power delivered by each
independent source.
12.5 watts.
= 200 V
and 1 ^ 2 = 20 mA. Find
Then find the power
delivered by each o f the independent sources.
Finally, find the power absorbed by each resistor
and verify conservation o f power for this circuit.
Figure P3.38
Figure P3.36
C H EC K : 7, = 4 mA.
C H EC K : Sum o f powers delivered by the
sources is 15.68 watts.
150
Chapter 3 • Nodal and Loop Analyses
with internal resistances R
39. Consider the circuit o f Figure P3.39.
(a)
(b)
(c)
Suppose
= 2 50 Q,
= 5 0 0 Q,
■20 Q and R2 = 80
Q (faulty connection) respectively connected in
= 100 V, and /S = 0.5. Use loop analy­
parallel to supply power to a load o f 7?^ = 80 Q.
sis to find /j and R^^ .
Compute the power absorbed by the load R^
Compute the power dehvered by each
and the power delivered by each independent
source and absorbed by each resistor.
source. W hich battery supplies more current to
Verify conservation o f power.
R^ and hence more power to the load? How
Compute R^^ as a function o f R^,
much power is wasted by the internal resistanc­
and 13 . Suppose
5 00 Q , plot R
s 13 s
2
= 250 Q and
=
as a function o f
,0
es o f the battery?
.
Pi,
0 '
C H EC K : P.s\
3.15 watts and P ,2 = 1 .8 watts.
Figure P3.39
C H EC K : Power absorbed by resistors is 15
watts and R^^ > 4 50 Q .
42. Reconsider the circuit o f Problem 3.41,
redrawn with different loop currents in Figure
3.42a and 3.42b. Th e point o f this problem is
40. (a) For the circuit o f Figure P3.40, R^ = 1
kQ,
= 5 kQ, /?3 = 4 kQ,
= 100
mA and g„ = ‘i xlO “'^ S. Find /j and
by writing two equations in the
two unknowns /j and
. The first
to verify that different sets o f independent loop
equations produce the same element currents
and branch voltages.
(a)
/j and
equation is the usual loop equation
(b)
(c)
.
for the circuit o f Figure
3.42a, and then find the voltage across
and the second determines the rela­
tionship o f /j and
Write the new loop equations and find
and the power consumed by R^ .
(b)
Write the new loop equations and find
Given your answer to (a), find the
/j and I 2 for the circuit o f Figure
equivalent resistance, R^^, seen by the
independent source.
3.42b, and then find the voltage across
and the power consumed by R^ .
Find the power delivered by the
dependent source.
Figure 3.42
Figure P3.40
43. The matrix loop equation o f the circuit o f
Figure P3.43 is
M ULTIPLE LOOP PROBLEM S
41. The circuit o f Figure P3.41 represents two
non-ideal batteries
= 21 V and
= 24 V
• 150
-4 0
-1 0 0 '
-4 0
140
0
h
= -2 0
-1 0 0
0
150
h
20
h'
TOO
Chapter 3 ®Nodal and Loop Analyses
151
Find the value of each resistance and each
source in the circuit.
(b)
the current in the locomotive motor and
(c)
repeat parts (a) and (b) when the loco­
the power absorbed by the locomotive
motive is 1/3 distant from either station
CHECKS: V^2 = ^0 V,
= 40 Q.
44. The mesh equations for the circuit of
Figure P 3.44 are
■40
-8 0
- 1 0 ‘ \h'
-3 0
130
-5 0
h
-1 0
-5 0
70
[*3.
■n ■
=
-V2
0
Figure P3.45
Find
*
46. Reconsider Problem 3.45. Let
= 590 V
and R^ = 1.296 Q. This time, suppose there are
two locomotives on the track. One is 1/3 dis­
tant from the East side station, and the other is
1/3 distant from the West side station.
(a)
Determine the resistance R in Figure
P 3.46.
(b)
Using the indicated currents, write a
set of three mesh equations and solve
(c)
45. Figure P 3.45a shows an electric locomotive
(d)
propelled by a dc motor. The locomotive pulls
a train of 12 cars. The motor behaves
like a 590 V battery in series with a
for
, ^2 ’
H•
Determine the two motor currents.
Determine the power delivered by
each of the 660 V sources.
---------------------- -------- ------------- 1--------- -----------------R
__ I
^
R
A
1.296 Q resistor. Suppose the train is
midway between stations, West side 660 V
and East side, where 66 0 V dc
T
sources provide electricity. The
resistance of the rails affects the cur­
rent received by the locomotive. The
equivalent circuit diagram is given by Figure
P3.45b, where R
660VVf
R
R
Figure P3.46
0.15 Q. Using m e ^ analy4 7 . Reconsider the Problem 3.5 and the circuit
sis find
(a)
©
T
l=
the currents
and I 2
of Figure P3.5. Draw two loop currents and
Chapter 3 ® Nodal and Loop Analyses
152
solve for these currents. Then compute the
50. Consider the circuit o f Figure P 3.50 in
node voltages
which
= 40 V and V^ 2 = 20 V. Write a set o f
three loop equations by inspection. Refer to
n\
Example 3.7 and the discussion following the
o
> and
48. In Figure P 3.48, let
= 6 kQ,
= 9 kQ,
.
= 9 kQ,
= 18 kQ,
=
example. Solve the loop equations using matrix
and /2.
software program. Compute the voltage v. Note
= 3000 Q, and
methods via your calculator or an appropriate
20 mA.
(a)
Write two mesh equations in
that /| and I 2 should have values identical with
Put in matrix form and solve.
those in example 3.7. Finally, find the power
(b)
Specify the voltages 1^^,
,
(c)
^BO ^CD’
K iD'
Find the power delivered by each o f
O
delivered by each o f the sources.
IQ
the sources.
49. Consider the circuit o f Figure P 3.49 in
which 1/j = 250 V, V^ 2 = 50, V,
= 50 Q,
= 20 Q,
= 50 Q,
= 40 Q, and
= 10 Q.
(a)
51 . Consider the circuit o f Figure P3.51.
(a)
Write three standard loop equations
matrix form in terms o f the literal
and put in matrix form.
(b)
Write two mesh equations and put in
o
parameters.
Solve the equations for the loop cur­
(b)
rents and determine the node voltages
Solve the mesh equations for the
unknown currents assuming
and
Q,
= 40 Q,
A, and
.
= 20 Q,
V.
= 100
= 80 Q,
(c)
Find the power consumed by
(d)
Determine the power delivered by
(c)
Find
each o f the sources.
(d)
Find the power delivered by the inde­
r^
and V^.
pendent sources.
(e)
Find the power delivered by the
dependent source.
r )
n
Figure P 3 .4 9
Figure P 3.51
Chapter 3 • Nodal and Loop Analyses
153
52. Repeat Problem 51 when 7?, = 100 £3, ^2 =
40 Q,
= 80 Q,
= 60 £2,
= 1 A, and
K2 =
V-
53. Consider the circuit o f Figure P3.53.
(a)
Write two mesh equations in terms o f
the literal parameter values.
(b)
Solve the mesh equations assuming
= 100 Q, ^2 = 40 Q,
= 60 f i,
80 Q,
1^2
= 80 Q,
=
= 1 A, and V^j =
40 V.
.
(c)
Compute
(d)
Find the power delivered by all the
and
sources in the circuit.
CHECK: /, = 0.1 A, K- = 50 V, and V = 20 V.
56. Consider the circuit o f Figure P3.56.
(a)
Write the modified loop equations (using
the indicated loops) in matrix form.
(b)
Figure P3.53
(c)
If I/, = 200 V, 7^2 = 0.3 A, /?! = 7^3 =
100 £2, T?2 = 400 £2, and /i = 0.5, find
7j , 72 , and V^ 2 ■
W hat is the power delivered by the
three sources?
54 . Repeat Problem 53 when
= 40 Q,
30 £3,
0.25 A, and
= 40 Q, 7?2 =
= 20 £2,
= 10 £2, 7^2 =
= 60 V.
>mv,-
M O D IFIED LOOP ANALYSIS
PROBLEM S
55 . Consider the circuit o f Figure P3.55. The
objective o f this example is to illustrate a
numerical approach to loop analysis where the
number o f variables to be found is quite large,
57. Consider the circuit o f Figure P3.57.
(a) Write the modified loop equations in
matrix form.
but the equations are quite easy to write and do
(b)
not require multiple substitutions.
(a)
A, 7?, = 200 £2, 7?2 = 400 Q,
If the loop equation matrix is o f the
Q,
form below, compute the undeter­
(c)
1
1 + ??'
??
-1
0
R3
????
-1
(b)
If
y s -R ih n
A'
=
0
0
= 500 £2, 7^2 = 100 Q,
Q, 7?4 = 100 £2,
= 100
= 500 Q, and TJj = 1150 Q,
compute 7j, 72 , 73 , and
mined entries.
'7?2 + ? ?
If 1/, = 210 V, K^2 = 150 V, 7^3 = 0.1
= 400
= 150 V, /•„ = 0.5
A, find the three unknowns.
Compute the power delivered by the
independent sources.
154
Chapter 3 ° Nodal and Loop Analyses
C H EC K : /j = 0.4 A and
= 50 V.
58. Consider the circuit of Figure P3.58.
(a)
Write the modified loop equations in
matrix form in terms of the literal values.
(b)
If
= 4 0 0 ^ ^ ^ 2 = 2 0 0 V,i?i = 3 0 Q ,
= 20 Q,
= 270 Q,
= 80 Q, T?5
= 140 Q, and compute /p /2 , 73, and
(c)
Compute the power delivered by all
sources.
(d)
Compute the power absorbed by each
resistor and verify conservation of power.
Ri
R3
R.
-VESA­
's ■<-----—N/N^
Rs
Figure P3.58
CHECKS: V, = 265 V and
watts.
= -3 9 7 .5
C H A P
'
‘
"
/
f
L in D lif ie r
The Operational Amplifier
Amplification o f voice allows announcers at sports events to convey their comments on the playby-play action to the crowd. At concerts, high-powered amplifiers project a singer’s voice and the
instrumental music into a crowded auditorium. Electronic amplifiers make this possible. One o f
the simplest and most common amplifiers is the operational amplifier, the subject o f this chapter.
The word “operational,” though, suggests a purpose beyond simple amplification. Often one must
sum signals to produce a new signal, or take the difference o f two signals. Sometimes one must
decide whether one dc signal is larger than another. The operational amplifier is operational pre­
cisely because it can be configured to do these things and many other tasks, as we will see later in
the text.
CH APTER O BjEC TIV ES
1.
Introduce the notion o f an ideal operational amplifier, called an op amp.
2.
3.
4.
Describe and analyze basic op amp circuits.
Describe and illustrate a simple method for designing a general summing amplifier.
Describe and illustrate the phenomenon o f saturation in op amp circuits and describe cir­
cuits that utilize saturation for their operation.
SECTIO N HEADIN GS
1.
2.
3.
4.
5.
Introduction
The Idealized Operational Amplifier: Definition and Circuit Analysis
The Design of General Summing Amplifiers
Saturation and the Active Region of the Op Amp
Summary
Terms and Concepts
Problems
156
Chapter 4 • T h e Operational Amplifier
1. IN TRO D U CTIO N
Chapters 1 and 2 defined and discussed independent and dependent voltage and current sources.
Chapter 3 investigated the nodal and loop analysis o f resistive circuits containing such sources.
Ofi:en, dependent sources supply energy and power to a circuit, making them so-called active ele­
ments. O n the other hand, resistors are passive elements because they only absorb energy. Circuit
models o f real amplifiers (see Examples 3.3 and 3.4 with associated Figures 3 .7 and 3.8, respec­
tively) contain controlled sources that underlie their analysis and performance evaluation. Indeed,
the VCVS is the core component o f the operational amplifier (op amp), the main focus o f this
chapter. Thus, the op amp is an active circuit element whose analysis is done with the techniques
o f Chapters 1 through 3.
A real op amp is a semiconductor device consisting o f nearly two dozen transistors and a dozen
resistors sealed in a package from which a small number o f terminals protrude, as shown in Figure
4.1(a). Despite its apparent internal complexity, advances in integrated circuit manufacturing
technology have made the op amp only slightly more expensive than a single discrete transistor.
Its simplicity, utility, reliability, and low cost have made the op amp an essential basic building
block in communication, control, and the instrumentation circuits that can be found in all under­
graduate EE laboratories.
Top View
Balance 1 [
TO-5
Dual-in-line
Inverting
input
^
Noninverting ^ [
input
m
E- 4 [
3 6 Output
H 5 Balance
(a)
Inverting
E+«-
input
—
V-
Output
Reference
node
Noninverting
E-
input
(c)
(d)
FIG U RE 4.1 (a) Typical op amp packages; (b) typical terminal arrangement of an op amp package;
(c) dual power supply notation; (d) essential terminals for circuit analysis.
Figure 4.1 (b) shows a typical arrangement o f terminals for a dual-in-line op amp package. Th e ter­
minal markings and the symbol shown in Figure 4.1(b) do not appear on the actual device, but
Chapter 4 • T h e Operational AmpUfier
157
are included here for reference. In Figure 4.1(b), the terminal labeled “N C ” (no connection) is not
used. The E+ and E- terminals (Figure 4.1(b)) are connected to a dual power supply, illustrated in
Figure 4.1(c), where
typically ranges between 3 V and 15 V, depending on the application;
adequate voltage is required for proper operation. The three terminals in Figure 4.1(b) marked
“inverting input,” “non-inverting input,” and “output” interact with a surrounding circuit, and
correspond to V,
and Vq in Figure 4.1(d). The two terminals labeled balance or ojf-set have
importance only when the op am is part o f a larger circuit: resistors o f appropriate values are con­
nected to these terminals to make sure the output voltage is zero when the input voltage is zero.
This “balancing process” is best discussed in a laboratory session.
This chapter sketches the basic properties o f the op amp: just enough to understand some o f the
interesting applications. The ideal op amp model and the saturation model are described. Using
these models and the principles o f analysis covered in Chapters 1 through 3, we then analyze the
behavior o f some widely used op amp configurations. These application examples hint at the
importance o f the op amp and furnish motivation for the study o f electronic circuits.
Several o f the examples include a SPIC E simulanon o f the circuit being analyzed. SPIC E is a
sophisticated circuit simulation program. Behind the user-interface, SPIC E uses complex models
o f the real operational amplifier. Our purpose in using SPIC E simulation is to verify or test the
theoretical analysis set forth in the examples. W hat we show is that the simplified theoretical
analysis provides a very good approximation to the actual circuit behavior represented in the
SPIC E simulation results. Industrial circuit designers often use SPIC E to visualize the expected
behavior o f very complex circuits. Later chapters cover some o f the more complex op amp appli­
cations.
2. TH E ID EALIZED O PERATIO N AL AM PLIFIER
This section analyzes resistive circuits containing an operational amplifier. Figure 4 .2 explicitly
shows an op amp embedded in a surrounding resistive circuit.
FIG U RE 4.2
One possibility for analyzing op amp circuits is to represent the op amp by one o f the simplified
models shown in Figure 4.3 that do not account for saturation effects. The first model o f Figure
4.3(a) consists o f an input resistor,
an output resistor,
and a V CV S with finite gain A.
O f practical import is the idealization o f this model (Figure 4.3(a)) to the one o f Figure 4.3(b) by
158
Chapter 4 • T h e Operational Amplifier
(1) letting R-^ become infinite, setting up an open circuit condition at the input terminals; (2) let­
ting
become zero, making the output voltage o f the op amp equal to that o f the V C V S; and
(3) letting the gain A approach infinity. These conditions are idealizations because (1) with R-^
infinite, there is no loading to a circuit attached to the input; (2) with R^^^ = 0, the full output
voltage appears across any circuit connected to the output; and {5) A ^
leads to a simplifica­
tion o f the associated analysis. These conditions, stated below as equation 4.1, define the so-called
ideal operational amplifier;
-> 00 (infinite gain)
R-^
Rout
(4.1a)
00 (infinite input resistance)
(4.1b)
0 (zero output resistance)
(4.1c)
Rest of Circuit
Rout
+
+
V
A(v - V
(a)
—
FIG U RE 4.3
To see how this idealization simplifies op amp circuit analysis, consider an equivalent set o f conditions for the ideal op amp, called the virtual short circuit model:
(4.2a)
(4.2b)
v^ = v_
(4.2c)
From Figure 4.3(b), the conditions that = 0 and i_ = Q follow directly from the open circuit con­
dition at the input terminals. The condition that
= v_ (hence, the term “virtual short circuit”)
will be discussed later, but occurs because A ^ co, forcing {v^ —v j
0.
159
Chapter 4 • T h e Operational AmpUfier
The recommended way to analyze circuits containing op amps is to replace any ideal op amp by the
model o f Figure 4.3{b), the virtual short circuit model o f equation 4.2. Th e following examples illus­
trate the use o f the virtual short circuit model.
EX A M P L E 4 .1 . This example investigates the inverting am plifier o f Figure 4.4, which is used in
a wide range o f commercial circuits. The objective is to compute
Rjs and
in terms o f
+
V
Rf
FIG U RE 4 .4 Inverting amplifier, assuming an ideal op amp in which V„„, = ---- —V:„ .
So l u t io n
Step 1. Compute
and v_. Since the + terminal is grounded,
= 0. From the virtual short prop­
erty o f the ideal op amp, v_= v^ = Q.
Step 2. Compute i^-^. Since v_ = 0, the voltage across R-^ is v-^. From Ohm’s law,
v,„
R
Step 3. Compute iy. Again, since v_ = 0, the voltage across Rf is
From Ohm’s law.
if = R
Step 4. Relate the currents i-^ and ip and substitute the results o f Steps 2 and 3. From KCL, i^^ —i_
+ ir= 0. From the properties o f the ideal op amp, i_ = 0, in which case, ir= -i-^. This imphes that
R*
V
Hence, the voltage gain relationship o f the inverting op am circuit is
Rf
V ou t= --^ ^ in
Rin
(4.3)
Equation 4.3 shows that the input and output voltages are always o f opposite polarity, hence the
name inverting am plifier. One also observes that by choosing proper values for Rjr and R^ a volt­
age gain o f any magnitude is possible, in theory. In practice, other factors limit the range o f obtain­
able gains.
160
Chapter 4 • T h e Operational Amplifier
Exercises. 1. Find
for the circuit o f Figure 4.5.
lOOkO
25kQ
-s/ W '
50 mV
FIG U RE 4.5 Inverting amplifier.
2. Find
for the circuit o f Figure 4.6.
lookn
FIG U RE 4.6 Inverting amplifier with additional resistor.
A N SW ER F O R BO T H :
= - 200 mV.
A few remarks are in order. Op amp configurations in which one o f the input terminals is ground­
ed, as is the non-inverting terminal in Figure 4.4, are said to operate in the single-ended mode. The
input terminal can be grounded directly or through a resistor, as in Exercise 2 above. Also, since
v_ =
or v_-v^ = 0, the terminals are virtually short circuited even though there is no hard-wired
direct connection between them. This condition is called a virtual short circuit. Further, if one
o f the terminals is grounded, then the other terminal is said to be virtually grounded, as is the
case in Figures 4.4, 4.5, and 4.6. Specifically, in Figure 4.4, there is a virtual ground at the invert­
ing input terminal.
The next example continues the investigation o f the ideal inverting amplifier for the two-input,
single-output op amp circuit o f Figure 4.7. The solution again makes use o f the virtual ground
and virtual short circuit properties o f the ideal op amp.
161
Chapter 4 • T h e Operational Amplifier
E X A M P L E 4 .2 . For the circuit o f Figure 4.7 , our objective is to compute
and the two input voltages
in terms o f R^, ^2>
and v^2 Rf
+
V
FIG U RE 4.7 Inverting (ideal) amplifier with rwo inputs for which
Rf
Rj
v„„,=-^v,|V ,2.
So l u t io n . A s in Example 4.1, and by the same reasoning described there, v_ = v^ = 0.
Step 1. Compute ij and
Since v_ = v^ = 0, the voltage across
is
and the voltage across Rj
is v^2 - From Ohm’s law,
R^
Step 2. Compute ij-. Again, since v_ = 0, the voltage across Rj- is
and from Ohm’s law.
'/ = ■R
/
Step 3. Relate the currents /j, ^2
h
and then substitute the results o f Steps 1 and 2. From KCL,
H —i_ + ijr = ^- From the properties o f the ideal op amp, i_ = 0, in which case, ijr = -(/j +
This implies that
Hence,
Rf
Rj
(4.4)
162
Chapter 4 • T h e Operational Amplifier
Exercise. In Figure 4.7, suppose Rjr= 100 IcD. Find
AN SW ER:
= 25 kD and
and ^2 so that
= 50 kD.
E X A M P L E 4 .3 . This example analyzes the non-inverting operational am plifier circuit o f Figure
4.8. As in Examples 4.1 and 4.2, the objective is to compute
in terms o f R^, R2 , and v-^. We
show that
''out
V.
FIG U R E 4.8 A non-inverting op amp circuit.
So l u t io n
Step 1. Compute
and v_. Since the + terminal is connected to the input voltage source,
= v-^.
From the virtual short property o f the ideal op amp, v_ = v^ = Vj^.
Step 2. Compute zj . Since v_ =
the voltage across the resistor
is
Observe that the current,
Zj, has reference direction different from the passive sign convention. Hence, from Ohm’s law,
h=Step 3. Compute ip Again, since v_ = v-^, the voltage across R^-is
lf =
From O hm s law,
Vout - Vin
Step 4 . Relate the currents z'j and ip and substitute the results o f Steps 1 and 2. From KCL, z'j - z_ +
Zyr= 0. From the ideal op amp property o f equation 4.2, z_ = 0, forcing z^= -Zj. This implies that
Rf
/?,
163
Chapter 4 • T h e Operational AmpHfier
Hence, the input-output voltage relationship
Rf\
(4.5)
1+ ^
R i)
/
R \
From equation 4.5, the voltage gain is greater than 1, i.e., i 1 + - i - i > 1, and
I
}
have the same polarity; the circuit is naturally called a non-inverting amplifier.
Exercise. For the non-inverting amplifier o f Figure 4.8, find
and
and
always
so that the gain is 2, and
when v-^ = 5 V, the power absorbed by R^ is 5 mW.
i,n
AN SW ER: Rf = 5\<Q. and /?,^ = 5 kQ
EXA M PLE 4 .4 . This example analyzes the ideal general di^Ference amplifier circuit o f Figure
4.9. We show that
Kf
^’s 2 - - r ^ s \
R\
'’" " ' I
In a basic difference amplifier, the output is the difference o f two input voltages. For the gener­
al difference amplifier o f this example, the output is a difference o f the scaled input voltages,
=
for appropriate positive
and a^.
So l u t io n . From the ideal op amp property o f equations 4.2, v_ =
and no current enters the
inverting and non-inverting op amp input terminals.
Step 1. Write a node equation at the non-inverting input terminal o f the op amp. Summing the cur­
rents leaving the + node o f the op amp yields
G 2 { v + - v , 2 ) + GgV+=Q
164
Chapter 4 • T h e Operational Amplifier
Solving for
leads to
(4.6a)
Step 2. Write a node equation at the inverting input terminal o f the op amp. Recall v_ = v^. T he sum
o f the currents leaving the - node satisfies
G i ( v+ - v, ] ) + G ^ ( v+ - v„„,) = 0
Thus,
Cl
^out =
+
1+
(4.6b)
Step 3. Combine Steps 1 and 2. Substituting equation 4.6a into 4.6b yields
(4.7a)
G t + Go
or, in terms o f resistances,
[R g + R2 j
(4.7b)
^s2
Equations 4 .7 have the desired form:
appropriate positive constants <?2
a ^, which can be obtained by proper choices o f the resistors.
Two special cases o f Example 4 .4 are o f practical importance. First, if
and R^ = /?2> then
equation 4.7b reduces to the classical difference amplifier equation.
out
= K {v , 2 -V si)
with K = I and for an arbitrary K > 0 , Rj- = KR^ and R^ = KR2 fits the bi
bill.
Exercises. 1. In Figure 4.9, if
= 7?2 = 5 kQ and K = 2 , find R p and R^.
ANSW ER: R j - = R^= \0 kQ
2. Using the circuit o f Figure 4.9, design a difference amplifier so that
= 4(t^^2 “ ^ji) ^^d the
feedback resistance R^ = 20 kQ.
A N SW ER: R, = R , ^ 5 kQ and R^ = 20 kQ
3. In Exercise 1, suppose R^ and /?yare scaled by a positive constant A'j, i.e., R^^^ = K^R^y and R2
and R are scaled by a positive constant K^. Determine the new input-output relationship.
o
A N SW ER:
w ith
K
the sam e as in Exercise 1
165
Chapter 4 • T h e Operational AmpUfier
The point o f Exercise 3 is that the group
, Rj] can be independently scaled by
and the group
{Rj, R^ independently scaled by K 2 without affecting the gain o f equation 4.7b.
E X A M P L E 4 .5 . This example analyzes a special case o f the non-inverting amplifier called the
buffer or isolation amplifier, shown in Figure 4.10, where
= v-^. W hen connected between
two circuits, the buffer amplifier prevents one circuit from having a loading effect on the other.
FIG U RE 4.10 The buffer or isolation amplifier for which
So l u t io n . From the connection shown in Figure 4 .1 0 , v-^ =
ties o f the ideal op amp,
= v_, in which case
and
=
= v_. From the proper­
= v-^.
Exercise. Compute the power delivered by the source in Figure 4 .1 0 and the power delivered to
the load R^.
^
AN SW ER: 0 and
R,
The circuit o f Figure 4.10 is called an isolation or buffer amplifier, because no current is drawn
from the source
maintaining
However, the op amp does supply current (and power) directly to the load by
under the condition that
put current rating. Since Vg„f{t) =
not exceed the manufacturer’s maximum out­
the circuit is also called a voltage follower.
Figure 4.11 shows a SPIC E simulation that verifies the behavior arrived at in Example 4.5. Here
a dc voltage sweep, Q <v-^< 12 V, was input to a highly accurate SP IC E model o f a Burr Brown
741 connected to ±10 V power supply. Observe in Figure 4.11 that the output follows the input
up to the 10-volt value, after which, the output remains at 10 V despite increased input values.
This non-ideal phenomenon, called saturation, is due to the power supply voltage level and is dis­
cussed in Section 4.
166
Chapter 4 • T h e Operational Amplifier
buffer -DC Transfer-2
(V)
+0 .0006+000
+2 .000
+4.000
+6.000
V2
+8.000
+10.000
+12.000
FIG U RE 4.11 Spice simulation of voltage-follower circuit.
Exercise. Find
for the circuit o f Figure 4.12, the power supplied by the source
power supplied to the 12 kQ load.
AN SW ERS:
=v,.0.
1 2 x 10-’
FIG U RE 4.12 Isolation of load from source using buffer amplifier.
and the
Chapter 4 • T h e Operational AmpUfier
167
3. TH E DESIGN OF G EN ERAL SUM M IN G AMPLIFIERS^
Often data acquisition equipment and active fdters require multi-input single-output amplifiers
having a more general summing characteristic, such as
^out -
+ « 2 'a 2 ) +
)
(4.8)
where the constants ay>0 and |3>0. The inverting and non-inverting amplifier configurations
(Examples 4.1, 4.2, and 4.3), as well as the difference amplifier configuration o f Example 4.4, are
special cases o f equation 4.8. W ith a little cleverness, it is possible to design by inspection an op
amp circuit whose input-output characteristic is precisely equation 4.8. The op amp circuit o f
Figure 4.13 having the four inputs
V^2 > ^hv ^bl accomplishes this. The circuit looks ordi­
nary except for the presence o f one additional conductance, AG, incident on the inverting termi­
nal o f the op amp. T he dashed lines in Figure 4.13 are present because this conductance may or
may not be needed. Computation o f the values o f AG and
are explained in design Step 2,
below.
b2 0 — s / \ / V
G
FIG U RE 4.13 A general op amp circuit that realizes equation 4.8.
Design Choices for the General Summing Circuit o f Figure 4.13
The first two design steps constitute a preliminary or prototype design, meaning that the feedback
resistor is normalized to 1 Q , or equivalently, 1 S. After completing the prototype design, an engi­
neer would scale the resistances to more practical values without changing the gain characteristics.
T he scaling procedure is explained in Step 3.
D esign Step 1. Prototype design. Set G ^ = l S , G , , = a , S ,G ,2 = « 2 S, G^, = P i S , and CJ^2 = P2
S. For the design to remain simple, the total conductance incident on the inverting terminal must
equal the total conductance incident on the non-inverting terminal. This is achieved by proper
choice o f AG and/or G . T he proper choices are given in Step 2.
168
Chapter 4 • T h e Operational Amplifier
Design Step 2. Prototype design continued: Computation ofG^ andhr tS.G so that the total conduc­
tance incident at the inverting terminal o f the op amp equals the total conductance incident at the non­
inverting terminal.
To achieve this equality, recall that in design Step 1, Cy= 1 S,
S, and
S. Define
a numerical quantity
^ = (1 + a i +
) - (^1 +
)
The sign o f 6 leads to two cases:
Case 1: If 8 > 0, then set (7 = 8 and AG = 0.
.5
Case 2. If 8 < 0, set G to some value, for example, G = 1 S and AG = |8| + G .
o
&
&
Design Step 3. Scaling to achieve practical element values. Multiply all the resistances (divide all
conductances) incident at the inverting input terminal o f the op amp by a constant
Similarly,
multiply all resistances (divide all conductances) incident at the non-inverting terminal o f the op
amp by
It is permissible to choose
but this is not necessary.
EXA M PLE 4 .6 . Design an op amp circuit having the input-output relationship
>^o»/ = - 7 V a l - 3 v ^ 2 + 2 v i , + 4 v^2
(4.9)
So l u t io n
Step 1. Prototype design. Using Figure 4.13, choose Gy- = 1 S, G^j = 7 S, G^2 = 3 S, G^, = 2 S, and
G^2 = 4 S .
Step 2. Equalization o f total conductances at inverting and non-inverting terminals. Since 8 = (1 +
7 + 3) - (2 + 4) = 5 > 0, set AG = 0 and G^ = 8 = 5 S. The circuit in Figure 4.14(a) exemplifies
the prototype design.
Step 3. Scaling. To have practical element values, let us choose
to a design with resistances Rjr= 100 kQ, 7?^, = 14.28 kQ,
25 kD and R„ = 20 kQ.
= 10^. This scaling leads
= 3 3.33 kQ,
= 50 kQ, R^^ =
169
Chapter 4 • T h e Operational Amplifier
14.28 kQ
V3, o------
lOOkO
25 kQ
(a)
(b)
FIGURE 4.14. (a) Prototype design of equation 4.9; (b) final design
after scaling with
= lO^.
EXA M PLE 4 .7 . Design an op amp circuit to have the input-output relationship:
I'ow/ = - 2 v „ i - 4 v^2 + 7 v/,| + 5 v^2
So l u t io n
Step 1. Prototype design. Again, using Figure 4.13, choose
(4.10)
1S,G „=2S,G ,2 = 4S,G^,=7
S, and G^2 = 5 S.
Step 2. Equalization o f total conductances at inverting and non-inverting terminals. 8 = (1 + 2 + 4)
- (7 + 5) = - 5 <
• 0; set C = 1 S, AG = |8| + G = 5 + 1 = 6 S. This prototype design is given in
A
o
Figure 4.15(a).
Step 3. Scaling. To have practical element values, let us again choose
= 10^. This scal­
ing leads to a design with resistances Rjr = 100 kQ, R^j = 50 k fl, R ^2 = 25 kQ,
= 14.28 kQ,
R^j = 20 kQ,
- 100 kQ, AR = 16.67 kQ. The final design is set forth in Figure 4.15(b).
16.67kn
50 kQ
V3, o------v X / X . -
lOOkQ
25 kQ
'^a2 O------- s/\v^v^
14.28 kQ
lOOkQ
'^b2 O--20 kQ
(a)
(b)
FIGURE 4.15 (a) Prototype design of equation 4.10; (b) final design after scaling with
= 10^.
170
Chapter 4 • T h e Operational Amplifier
Exercise. 1. Obtain an alternative design for Example 4 .7 such that
= 0, implying the saving
o f one resistor.
AN SW ER: In prototype design, AG = 5 mho.
2. Design a difference amplifier so that
- v^-^, with
= 10 kD.
AN SW ER: See Figure 4.16.
10 kQ
At this point, the reader may wonder how this simple procedure is derived. The derivation o f this
procedure is beyond the scope o f the Hght edition^ Th e interested reader is directed to the 2nd
edition o f this text.
Exercise. 1. Find
AN SW ER: „
the G- for the circuit in Figure 4.17(a).
in terms o f
_
' out -
G,
G,
Gj
G3
,v2
2. Find t',,
in terms o f for the circuit in Figure 4.17(b).
AN SW ER:
-
- 7z>2 +
6K U
(a)
FIG U RE 4 .17
Chapter 4 • T h e Operational Amplifier
171
4. SATURATION AND TH E A CTIVE REGION OF TH E OP AMP
In the previous sections, we assumed the op am functioned ideally:
= v_ and
= i_=Q. For the
inverting amplifier o f Example 4.1, this led to the very simple gain formula,
'^out
_
Rh,
Thus, as the input voltage increases, the output voltage increases proportionately. For real circuits,
this proportional relationship holds only when
< V^^^for some value o f
that is associat­
ed with the power supply voltage. Intuitively speaking, an op amp cannot generate an output volt­
age beyond that o f its power supply voltage, typically less than or equal to 15 V. W hen the V^^limit is reached, further increases in the magnitude o f v-^ produce no change in the value o f
This behavior is called saturation.
To explain this saturation behavior, we refer to Figure 4 .1 8 . In Figure 4.18,/'(v^ - v j represents
a nonlinear controlled voltage source, as opposed to the linear relationship A(v^ - p_), shown in
Figure 4.4(a). However, because the op amp functions more or less linearly until reaching its sat­
uration limits, we can approximate f(v^ - v_) by the three-segment piecewise linear relationship
shown in Figure 4.19(a), wherein the saturation effects are captured by segments II and III. One
observes that when
A
, the voltage/(f^ - z 'J clamps at
the voltage/(z'^ - v j clamps a
saturation occurs are
A
. If
t
and when
^
A
j
As observed, the critical threshold voltages o f t^^at which
= 15 V and A= 10^, the critical threshold voltages are ±0.15
mV; if A is infinite, as in Figure 4.19(b), then saturation occurs when \v^ > 0.
FIG U RE 4.18 Practical op amp model with a nonlinear controlled voltage source.
The linear r e l a t i o n s h i p , - v^=A{v^ - v J , holds for segment 1 in Figure 4.19(a), which is said
to be the linear region or active region o f the op amp, denoted by
Chapter 4 • T h e Operational Amplifier
172
Typical values o f finite A range from 10'^ to 10*^. The active region is the ordinary region o f oper­
ation. In the active region, the op amp provides a very high (open loop) voltage gain A, the slope
o f segment I. The phrase “open loop” gain means that there is no connection through a wire, a
resistor, or some other device back to the input terminals.
Models o f the three operating regions o f the op amp are summarized in Table 4.1.
A f(v^-v)
V
\
Positive
Saturation
/
Active /
Region
d
= V -V
Negative
Saturation
\
-V
(b)
FIG U RE 4.19 A piecewise linear (three-segment) curve for the op amp that specifies the active
and positive/negative saturation regions of operation: (a) finite gain A, and (b) (ideal) infinite gain A.
TABLE 4.1 Operating Regions o f the Op Amp with Associated Models
C urve Seg m ent
N ame
of
B
R e g io n D efining E quations
Vcl =
Active
f(Vd)
A
and
*^sar
II
Positive
saturation
Vsa,
A
and
sat
III
Negative
saturation
<and
I dealized C ircuit M od el
Chapter 4 • T h e Operational Amphfier
173
The use o f a three-segment curve in Figure 4 .1 9 is different from the techniques o f earlier chap­
ters. The operating point,
determines the proper segment to be used for analysis. If the
input is small, one reasonably assumes the operation is in the active region, segment I. However,
when the input magnitude is large, one must “guess and check” to determine the appropriate oper­
ating region. For example, should the guess be incorrect, then the model for one o f the other
regions must be used and the analysis repeated until a valid solution (and operating region) is
obtained. The following example illustrates the approach.
E X A M P L E 4 .8 . The purpose o f this example is to illustrate that an op amp may operate in any
o f three regions and also to illustrate that the determination o f the region o f operation using the
“guess and check” method. Recall the inverting am plifier o f Figure 4.5. Suppose
= 50 kQ,
V; (b)
= 10 kQ and
= 4 V; and (c)
= 15 V. Find
and
is infinite,
for the following three cases: (a)
= 0.5
= - 5 V. Finally, verify the theoretical analysis using SPICE.
So l u t io n
(a) Assume the op amp operates in its active region. From equation 4.3 in Example 4.1, the out­
put voltage is
vout
„ = - ^ v , m, = - ^JQx 0 . 5 = - 2 . 5 V
Since |-2.5| <
= 15 V the op amp operates in its active region; the answers
= - 2 .5 V and
v^ = 0 are valid.
(b) W ith v-^ = 4 V, assuming operadon in the active region,
However, since |-20| >
^out
~
^
^
50
= -----= -------------- x 4 = - 2 0 V.
10
Rl
= 15 V, the op amp does not operate in its active region. Therefore,
invalid, but does suggest operation in the negative saturation region. The negative
saturation model o f Table 4.1 yields the circuit o f Figure 4 .20 in which
= - 1 5 V.
FIG U RE 4.2 0 Op amp operating in negative saturation region.
By writing and solving a single node equation at the inverting input terminal designated by the
minus sign in Figure 4.20, we obtain Vj = - 0 .8 3 V.
(c) With v-^ = - 5 V, assuming operation in the active region,
R'
~
50
^
= 25 V
This result suggests that the op amp is really operating in the positive saturation region. Using the
positive saturation model o f Table 4.1, Figure 4.21 shows the proper circuit configuration with
174
Chapter 4 • T h e Operational Amplifier
^out ~ 15 V. As in the previous case, by writing and solving a single node equation at the invert­
= 1.667 V. In this case and in case
ing terminal designated by the minus sign in Figure 4 .2 1 ,
(b) above,
0, as we were not in the active region o f operation, and it was necessary to change
the guessed region o f operation to obtain valid results.
FIG U RE 4.21 Op amp operating in positive saturation region.
A SPIC E simulation was used to validate the theoretical analysis^. A D C sweep, - 4 <
< 4 V,
is an adequate input to demonstrate the saturation effects. In the SPIC E simulation, an accurate
model for a 741 op amp manufactured by Burr Brown was used. T he resulting dc transfer curve
is shown below in Figure 4.22.
Lin/Decarlo E xI-D C Transfer-4
Output voltage
(V)
-4.000
- 1.000
+ 0 .0006+000
Vin
+
1.000
+3.000
+4.000
V(IVM)
From this curve, one can see that the op amp saturates for input voltages v-^ such that
> 3,
and the op amp operates in its linear region whenever \v-^ < 3. As hoped, the simplified three-seg­
ment model in Table 4.1 yields very good results in all regions o f operation relative to the realis­
tic SPIC E simulation.
O ne can conclude from the above example that for the purpose o f faithfully amplifying an input
signal, the input should not be so large as to drive the op amp into saturation. Driving an op amp
into saturation distorts the output signal relative to the input. O n the other hand, for some spe­
cial applications, such as the com parator, saturation is precisely the property to be utilized. Figure
Chapter 4 • T h e Operational AmpHfier
17 5
4.23 shows two comparator circuits. A com parator circuit compares tlie input voltage v-^ with a
reference voltage Vygjr{or some multiple o f
Only two different output voltages are produced,
and the other for v-^ <
one for
EXA M PLE 4 .9 . For the com parator circuits shown in Figure 4.23, each op amp has infinite gain
and a saturation voltage
= 1 5 V"^:
v -^ relationship for the comparator o f Figure 4.23(a).
(a)
Find the
(b)
Repeat part (a) for Figure 4.23(b).
Note that in both circuits, there is no connection between the output and inverting input
terminals, a departure from all the previous circuit configurations. Because o f this, for
^ v_, and
almost all voltages,
=
Vin O------s / W 2 0 kQ
''re f O --------
= -2 0 V
80 kn
1 .
1 .
(a)
(b)
FIG U RE 4 .23 Two comparator circuits that are used to determine when an input
voltage is above or below a reference voltage.
S
o l u t io n
(a) For
> - 5 V, the voltage v^=
v_ = -5 - v-^ < 0. Referring to Figure 4.19(b),
^out ^ ~^sat = - 1 5 V. Similarly, for v-^ < - 5 V, the voltage v^ = v^ - v_ = -5 - v-^ > 0, and hence
^out = ^sat = 15 V.
(b) By the fact that no current flows into the input terminals o f Figure 4.23(b), using nodal analy­
sis, we have that
V
- V
20x10^
80x10^
m which case,
- 4 + 0.8v,„
For
> - 5 V, the voltage v^ =
ration curve o f Figure 4.19(b),
- v_ = 0 - {-A +
Here, referring again to the satu­
= - 1 5 V. Similarly, when v-^ < 5 V, the voltage
v^ = v ^ -v _ = Q -{-A + 0.8V.J > 0; hence,
= 15 V.
176
Chapter 4 • T h e Operational Amplifier
To verify this analysis, the circuit o f Figure 4.23(b) was simulated in B2Spice using a Burr Brown
741 op amp model. T he results o f the simulation are given in Figure 4.24. The theoretical analy­
sis based on the simplified models o f Table 4 .1 shows a very good match with the more realistic
SPIC E simulation results.
example4.8-DC Transfer-6
Output voltage
(V )
-1 0 0 °
+2.000
+3.000
V(IVM)
FIG U RE 4.2 4 B2Spice simulation o f the comparator circuit of Figure 4.23(b).
Exercise. For the circuit o f Figure 4.25, suppose
= 12 V, find the range o f v^2 for which the op
amp is in positive saturation. Then find the range o f v^ 2 fof negative saturation.
AN SW ER:
when
< -^ V, and
=-
when
V , C3------- v N / \ - -
75 kO
''s2 o-----s /s y \/-
-o
+
25 kQ
1 .
FIG U RE 4.25
Chapter 4 • T h e Operational Amphfier
177
5. SUM M ARY
This chapter has introduced the operational amplifier and a number o f practical circuits that uti­
lize this new device. These circuits include the inverting and non-inverting amplifiers, the buffer
amplifier, the difference amplifier, and the general summing amplifier. W ith regard to the gener­
al summing amplifier, a simple design algorithm is described and exemplified. The analysis o f
these circuits builds on the definition o f an ideal op amp, meaning that, when properly config­
ured, no current enters the input terminals and the voltage across the input terminals is zero; these
properties are referred to as the virtual short circuit model o f the op amp, i.e., the ideal op amp
has infinite input resistance, zero output resistance, and an infinite internal gain, A. (See equations
4.1 and 4.2.) Practically speaking, the gain A, is not infinite, but ranges between lO'^ and 10*^.
After exploring properties o f the ideal op amp, we discussed the phenomena o f output voltage sat­
uration. By introducing output saturation, the ideal model o f the op amp gives way to a more real­
istic one, characterized by three regions o f operation, each having its own “ideal” model, as set
forth in Table 4.1. In practical design and applications, output saturation is either to be avoided
or utilized to some advantage, as in the case o f the comparator circuit studied in Example 4.8. For
a faithful amplification o f an input signal, saturation is to be avoided.
6. TERM S AND C O N C EPTS
Active element: A circuit element that requires an outside power supply for proper operation and
has the capability o f delivering net power to a circuit such as is the case for an op amp or
negative resistance.
Buffer: A circuit designed to prevent the loading effect in a multistage amplifier. It isolates two
successive amplifier stages. Characteristics o f an ideal buffer are infinite input impedance,
zero output impedance, and constant voltage gain.
Com parator: an op amp circuit that compares the input voltage
(or some multiple o f
with a reference voltage
only two different output voltages are produced, one for
<
v-^, and the other for v-^ < v-^.
Difference amplifier: given two inputs,
and
a difference amplifier produces the output
^out ^
appropriate constant k, often taken as 1.
General summing amplifier: an op amp circuit having the input-output relationship
=
constant a - and p ..
Ideal op amp: An operational amplifier with infinite input resistance and infinite open-loop gain.
-
+ ^ n ^ a r) +
+ - + ^ m ^ bn )
p o s itiy c
Inverting amplifier: An operational amplifier connected to provide a negative voltage gain at dc.
Linear active region: In the op amp output vs. input transfer characteristic, the region where the
curve is essentially a straight line through the origin is called the linear active region.
Non-inverting amplifier: An operational amplifier connected to provide a positive voltage gain
at dc.
Open-loop gain: The ratio o f the output voltage (loaded, but without any feedback connection)
to the voltage across the two input terminals o f an op amp. The slope, p, o f the straight
line in the active region o f an op amp is the open loop gain under no load condition.
When a load
is present, the open loop gain is reduced to
the output resistance o f the op amp.
+ R^, where R^ is
178
Chapter 4 ®The Operational Amplifier
Operational amplifier (abbreviated op amp): A multi-stage amplifier with very high voltage
gain (exceeding 10"^) used as a single circuit element.
Passive elements: a circuit element that cannot deliver net power to a circuit such as a resistor.
Saturation regions: In the op amp output vs. input transfer characteristic, the region where the
curve is essentially a horizontal line is called the saturation region. There are two such
regions: one for positive input voltage, and the other for negative input voltage.
SPICE: Acronym for Simulation Program with Integrated Circuit Emphasis. It is a very sophisti­
cated software tool for simulating electronic circuit behavior.
Virtual ground: When an ideal op amp has one of its input terminals grounded, and is operat­
ing in the active region, then the other input terminal is also held at the ground potential because of the virtual short effect (see below). Such a condition is called a virtual
^
ground (in contrast to a physical ground).
Virtual short circuit: When an ideal op amp is operating in the active region, the voltage across
the two input terminals is zero, even though the two terminals are not hard-wired togeth­
er. Such a condition is called a virtual short circuit (in contrast to a physical short circuit).
Voltage follower: A voltage-controlled voltage source with gain equal to 1, often utilized to sep­
arate stages of amplification in a multi-stage amplifier device.
^The circuit proposed in this section is a modification of one proposed in W. J. Kerwin, L. P. Huesman, and R.
W. Newcomb, “State-Variable Synthesis for Insensitive Integrated Circuit Transfer Functions,” IEEE Jr.
of Solid State Circuits, Vol. SC-2, pp. 87-92, Sept. 1967. The modification consists of an additional
resistor, which greatly simplifies the design calculations and was published by P. M. Lin as “Simple
Design Procedure for a General Summer,” Electron. Eng., vol. 57, no. 708, pp. 37-38, Dec. 1985.
/—
'
^ See Linear Circuit Analysis by DeCarlo and Lin, 2nd edition, New York: Oxford University Press, 2002.
^ Any of the SPICE or PSPICE software programs available by a variety of vendors will suffice to obtain the indi­
cated curve.
^ An op amp and a comparator as seen in a parts catalog are essentially the same, except that the comparator device
has a modified output stage that makes it compatible with digital circuits.
o
o
o
179
Chapter 4 • T h e Operational Amplifier
Problems
ANALYSIS USING IDEAL OP
AM P M O D EL
1. Consider the inverting amplifier circuit o f
Figure P 4 .1, in which
= 4 V.
(a)
If
= 2 kD, find i ?2
dehvered to
(b)
that the power
= 100
Figure 4.3
is 4 W.
Now suppose T(’2 = 12 kQ. Find
so
that the power delivered to Rj^ = 2 Id l
is 450 mW. Then find the power con­
sumed in i?, and Rj-
4. In the circuits o f Figure P4.4, a source is rep­
resented by an ideal voltage source, v-J^t) = 4 V,
in series with R^ = 10 Q resistor. The loadin
both cases is
(a)
= 40 Q.
W ith the load connected directly to
the source, as shown in Figure P4.4(a),
find the load voltage, the load current,
the source current, and the power
delivered to the load.
(b)
Figure P 4.1
Check: (b) 1500 <R^< 2000, and
As in Figure 4.4(b), a buffer amplifier
separates the source and the load.
= 10 m W
Again, find the load voltage, the load
current, the source current, and the
power delivered by the op amp to the
2. Consider the non-inverting circuit o f Figure
P 4.2, in which v- = 4 V.
(a)
If R^ = 2 kQ, find 7?2 so that the
power delivered to
= 100 Q is 4 W.
(b)
Now suppose i ?2 = 13 kO. Find R-^ so
that the power delivered to
load.
+
Rs
= 2 kH
' l >r
is 450 mW. Then find the power con-
■6
sumed in R-^ and i?2(a)
(b)
Figure P4.4
SCRAMBLED ANSWERS: (a) 0.256, 3.2, 0.08
5. Figure P4.5 contains three circuits that
explore loading and the elimination o f loading
effects using either a dependent source or an
equivalent buffering op amp circuit.
(a)
For the circuit o f Figure P4.5(a), com­
pute
and
in terms o f v^. Observe
Figure P4.2
Check: (b) 1500 <
< 3000
3. For the circuit o f Figure P4.3, find the volt­
age gains,
G,
andG , =
^
^ Vin
in terms o f the literal resistor values.
that the 80-Q -240 Q resistor combina­
tion loads down the 320—0 resistor.
(b)
For the circuit o f Figure P4.5(b), com­
pute
and
in terms o f v^. Notice
that
is different from the answer
V,
180
Chapter 4 • T h e Operational Amplifier
computed in part (a) because the 80-
A N SW ER: (a) R, = 5 kQ; (b) 6 .76 m W
£2-240 Q resistor combination is iso­
lated from the 320-Q. resistor.
(c)
For the circuit o f Figiire P4.5(c), again
compute v-^ and
7. (a)
in terms o f v^. Your
answers should be the same as those in
part (b). The buffering op amp circuit
again isolates the 80-£l-240 Q. resistor
comination from the 320 Q resistor.
<,v,
320 n<
Repeat part (a) for the circuit o f Figure
P4.7(b).
(c)
If for Figure 4.7(b),
3 kn ,
12 kD,
= 4 kQ, Rj = 1 k£2,
=
= 1.5
V, and v^ 2 = 2 V, find the power deliv­
ered to the load R^= 100 Q.
-o
+
V r + ^
as a function
and the R^.
(b)
80 n
80 n
For the op amp circuit o f Figure
P4.7(a), find
AN SW ER: (c) 0.04 watts
240
(a)
80 n
son
-o
+
240 n
320 n<
(b)
80 0
son
-v \ ^ -
-o
+
240 n
320 n
Figure P4.5
AN SW ER: (a)-V /
3
0 5V
6. In the circuit below, R^= 10 kQ.
(a)
Find R^ and R^ so that
(b)
Given correct answers to part (a), sup­
= -2t^^j 8. (a)
">^a-
Find the power delivered to the load if
= 20. If
= “ 600 mV.
= 0.6, find the power deliv­
ered to the 8 -Q load.
(b)
Now suppose R^^'^ kQ, and find R^
so that
(c)
out
=
20
Finally, suppose Rj =
and find
their common value so that
F igu re P 4 .6
out
= 6 kQ, and find R^ so that
pose a 1 kO resistor is attached as a load.
= 200 mV and
For the circuit o f Figure P 4.8, suppose
out
=
20
Chapter 4 • T h e Operational Amphfier
181
A N SW ER: (a) 40 k£l; (b) /?, = 15 k£2
2kn
-O-
11. Consider the circuit shown in Figure
P 4 .l l.
-0 -,
sn
and v^2 -
(a)
Find
(b)
If
= 250 m V and v^ 2 - 500 mV,
find the power delivered to the 1 kQ
in terms o f
load resistance.
Figure P4.8
A N SW ERS (in random order): 14 kO, 8 kf2,
10 k n , 18
lOkn
lokn
lokn
5kn
2kQ
9. In the circuit below, Rr= 12 ki2 and ^3 > 1 k£2.
(a)
Find
and ^2 so that
1 kn<
= -lOr^^j -
© '
20^,2(b)
Given correct answers to part (a), find
the power delivered to the load if
=
- 2 0 0 m V and
Figure P 4.11
C H EC K : (b) Pj^ = 12.25 m W
12.
(a)
For
the
circuit
o f Figure
P4.12a, the input voltage
= 2 V and
the input voltage v^2 = 3 V. Find
and
the power delivered to the 1 kI2 load
1 kn
resistor.
(b)
Repeat part (a), when
= 4 V and
the input voltage v^ 2 = 2 V.
C H EC K : (b)
Figure P4.9
= 0.1 watt
(c)
Reconsider part (b). Find the mini­
mum value o f R so that the maximum
amount o f power consumed in either
R-ohm input resistors is 2 mW.
10. Consider the circuit o f Figure P4.10.
(a)
Find the value o f R =
= R2
’^^at
the power delivered to R^= 1.25 kO is
0.5 watt when
(b)
3R
= 1 V.
Suppose 6R^ = i ?2
v^ = 2Y . Find
and R2 so that the power delivered
to
= 1.25 k il is 2 watts.
Figure P4.12
13. (a)
For the circuit o f Figure P4.13(a), the
input voltage
= 1 V, and the input
voltage v^2 ~ 500 mV. Find /?, in terms
Figure P 4.10
o f R so that
= 10 V.
182
(b)
Chapter 4 • T h e Operational Amplifier
Repeat part (a) for the circuit o f Figure
P 4.13(b), given that
= 2.5 V.
12R
r e
R,
-V S/V
2R
5R
-O
+
3R _ d
(a)
Figure P 4 .15
C H EC K : (a)
12R
2R
rO
5R
16. For the circuit o f Figure P 4.16, find
-o
+
terms o f
in
V, v^2
2.5R
(b)
Figure P4.13
AN SW ERS; (a) 6R-, (b) 3R
14. For the circuit in Figure P4.14, the input
voltages are
= 2 V,
= '1-5 V, and
=2
V.
(a)
Find
(b)
If /? = 10 kQ, find the power delivered
17. For the circuit o f Figure P 4.17, find
by each o f the operational amplifiers.
4R
2R
0.5 R
+ Av^2 -
1.5R
-o+
0.75R
0 - .
in terms o f R so that
and
^
Figure P4.14
C H EC K :
i = 0.9 mW,
= 0-20667
watts
15. For the circuit o f Figure P4.15, the input volt­
age
= 5 V, and the input voltage
(a)
(b)
If
= 8 R and Rj - Ry find
If R^ = 8R and R^ = ARy find
(c)
If /?2 =
that
find
= lOV.
in terms o f R so
CH ECK:
Figure 4.17
= 8R
18. For the circuit o f Figure P 4.18, find R^ and
i?2> and Rj in terms o f R so that
= 8f^j +
10v^2~2^s3- Hint: Consider Problem 17 first.
Chapter 4 • T h e Operational Amphfier
183
Time in s
AN SW ERS:
= AR, R^ =
(0
R, = O m
Figure P4.19
NON -IDEAL OP
AM P-SATURATIO N EFFECTS
19. The op amp in Figure 4.19(a) has
V,
= 4 ld2, and /?2 = 20 kQ.
(a)
Plot the
versus
for
20. Repeat Problem 19 for the op amp circuit of
Figure P4.20 when
= 4 ld2, and R2 = 20 kfl.
=15
-O+
given in
Figure 4.19(b).
for 0 < /■< 6 s for v(J) in
(b)
Plot
(c)
Verify your analysis in part (a) using
Figure 4.19(c).
Figure P4.20
SPICE. Assume that the op amp is a
type 741 whose model should be
available within your SPIC E program.
21. For the circuit o f Figure P4.21, each amplifier■saturates
at
= 15 V.
sat
(a)
Suppose the input voltage
= 5 Y
and the input voltage v^2 = “ 2.5 V.
R,
Find
-o (b)
and
If
= -2 -5 V and
= 15 V, find
so that no amplifier saturates.
(a)
4R
1.5R
2R
-O
Figure P4.21
AN SW ER: (a) 15 V; (b)
= 3.75 V
22. For the circuits o f Figure P 4.22, suppose R^
(b)
= 40 k n , and R^ = 120 Q.
(a)
For the circuit o f Figure P4.22(a),
compute
the power delivered by
the source, and power delivered to the
load R^ in terms o f v-^.
184
(b)
Chapter 4 • T h e Operational Amplifier
For the circuit o f Figure P4.22(b),
compute
(b)
the power delivered by
compute
the source, and power delivered to the
load
(c)
For the circuit o f Figure P4.24(b),
the power delivered by
the source, and power delivered to the
in terms o f v-^^.
load
Discuss the differences in your solu­
(c)
in terms o f
Discuss the differences in your solu­
tions to (a) and (b). Specifically, dis­
tions to (a) and (b). Specifically, dis­
cuss the effect o f using a voltage fol­
cuss the effect o f using a voltage fol­
lower to isolate portions o f the circuit.
lower to isolate portions o f the circuit.
(a)
(a)
(b)
Figure P4.24
25. For the circuits o f Figure P4.25, suppose /?, =
20 Q, and /?2 = 160 Q, R^ = 40 Q., and R^ = 120
a.
(a)
For the circuit o f Figure P4.25(a),
compute
the power delivered by
the source, and power delivered to the
Figure P4.22
23. In the circuits o f Figure P4.22, all resist­
ances are 100 Q, and
= 1 V.
(a)
(b)
the source current again when all
resistances are 100 Q.
I f a buffer amplifier separates the
source and the load, as in Figure
For the circuit o f Figure P4.25(b),
compute
the power delivered by
the source, and power delivered to the
For the circuit o f Figure P4.22(a), find
the load voltage, the load current, and
(b)
load Rj^ in terms o f v-^.
load R^ in terms o f v-^.
(c)
Discuss the differences in your solu­
tions to (a) and (b). Specifically, dis­
cuss the effect o f using a voltage fol­
lower to isolate portions o f the circuit.
P4.22(b), find the source current, the
load voltage, the load current, and the
current supplied by the op amp.
AN SW ERS: (a) 0.5V, 5 mA, 5 mA; (b) 0 A, 1
V, 10 mA, 10 mA
24. For the circuits o f Figure P4.24, suppose
= 40 Q , and Rj = Rl = 120 Q.
(a)
For the circuit o f Figure P4.24(a),
compute
the power delivered by
the source, and power delivered to the
load Rj^ in terms o f v-^.
Figure P4.25
Chapter 4 • T h e Operational Amplifier
AN SW ERS: (a)
3.7037 x l O 'V .
185
= 0.003i^y„; (b) Pl =
26. Figure P 4.26 contains three circuits that
explore loading and the elimination o f loading
effects using either a dependent source or an
equivalent buffering op amp circuit.
(a)
27. Two non-ideal voltage sources are each rep­
resented by a connection o f a (grounded) inde­
pendent voltage source and a series resistor.
Denote the parameters o f each connection by
{Vs^,
^st)' Design an op amp cir­
cuit such that the output voltage with respect
to ground is
For the circuit o f Figure P4.26(a),
compute I'j and
Observe that the
8-Q -24 Q. resistor combination loads
down the 3 2 -Q resistor.
(b)
For the circuit o f Figure P4.26(b),
for all values o f
and R^2 tie greater
than or equal to 100 kS2 so that only small
compute
is
amounts o f current are drawn from the buffer
different from the answer computed
amplifiers. Note that the general difference
in part (a) because the 8-Q -24 Q resis­
amplifier circuit o f the chapter will not work
tor combination is isolated from the
here because o f the presence o f the resistances
and
Notice that
3 2 -Q resistor.
(c)
and R^2 - To achieve such a design, it is nec­
For the circuit o f Figure P4.26(c),
essary to isolate the (practical) sources from the
again compute Pj and
Your
difference amplifier inputs using buffer ampli­
answers should be the same as those in
fiers, as shown in Figure P 4.27. Explore your
part (b). The buffering op amp circuit
design for various values o f
again isolates the 8-Q -24 O, resistor
SPICE. Do the SPIC E simulations verify that
combination from the 32 Q resistor.
the output is independent o f the values o f
and R^2 using
Figure P4.27
8Q
80
-o
+
circuit o f Figure P 4.28, suppose the op amp has
24 n
32 0 <
(c)
Figure P4.26
AN SW ERS: (a) 0.6665 V^, 0.5 V^; (b) and (c)
0.8 V^, 0.6
28. Following Example 4.9, for the comparator
infinite gain and a saturation voltage
=15
V. Find the
versus
relationship and plot
as a function o f v-^. Verify your analysis using
SPICE. Assume that the op amp is a type 741
whose model should be available within your
SPIC E program.
186
Chapter 4 • T h e Operational Amplifier
Check your design using SPICE. Hint: How
80 kO
can the circuit o f Figure P4.29 be modified to
achieve the correct polarities?
/?] _
A N SW ER:
32. Find the
comparator
Figure P4.28
29. (a)
Find the
1
1.5
versus v-^ relationship for the
circuit
of
Figure
P 4.32.
Specifically, show that when
versus
relationship for
Vin >
, then
and when
the comparator circuit o f Figure
P4.29. Specifically, show that when
Vin <
=
R,
and when v-^ > ??, then
then
V
= - V^sa f
'^out
(b)
^in<-^^ref , then
Now suppose
^ref ~
characteristic if
= R2 = 100 kQ and
^out versus v-^
= 15 V. Verify
Figure P4.32
your analysis using SPICE. Assume
that the op amp is a type 741 whose
model should be available within your
SP IC E program.
G EN ERA L SUM M IN G
AM PLIFIER (ID EA L OP AM P
M O D EL)
3 3 .(a)
Assuming the op amp in Figure P4.33
is ideal, derive the relationship
Figure P4.29
AN SW ERS: (b)
= 15 V if
< 2 V, and
(b)
Rf
Rf
R,
Ro
Suppose
so that
25 kO. Find
, i?2>
is the negative o f the
average o f
, V2 , and Vy
30. Using a 1.5 V battery, an op amp with
= 10 V, and some resistors, design a comparator
circuit such that
and
= 10 V when
< 1 V,
= - 1 0 V when v-^ > 1 V. Check your
design using SPIC E. Use part (a) o f Problem 29
as a guide.
31. Using a 1.5-V battery, an op amp with
= 10 V, and some resistors, design a comparator
circuit such that
= 10 V when v-^ >
1 V, and
= - 1 0 V when v-^ < I V.
V l-
Figure P4.33
AN SW ER: (b) R. = 75 kO
Chapter 4 • T h e Operational Amplifier
187
34. Using the topology o f Figure 4.13, design
an op amp circuit to have the input-output
38. Using the topology o f Figure 4.13, design
an op amp circuit to have the input-output
relationship
relationship
+H i
- 3^.2 +
Two different designs are to be produced for
+H i
Two different designs are to be produced for
comparison and selection:
comparison and selection:
(a)
Design 1: Rjr= 100 kQ.
(b)
1:
. = 50 kQ.
^
iqq
Design 2: Rj^= 50 kQ. Specify all final
values in terms o f Q..
C H ECK :
= 25 kO, and R^2fiil ^ 33.33 k£2
^
39. Generalizing the topology o f Figure 4.13,
design an op amp circuit to have the input-output relationship
35. Using the topology o f Figure 4.13, design
an op amp circuit to have the input-output
^out ~^a\ ~ '^'^al ~
I„ .he final circm, Rf - 40 k a
" ' “ “ “ '■‘P
^out ~ ~^^a\ ~ '^^al
‘^^h\
^'^bl
Two different designs are to be produced for
comparison and selection:
40. Generalizing the topology o f Figure 4.13,
design an op amp circuit to have the input-output relationship
out
(a)
(b)
Design 1: Rr= 100 k£l.
Design 2: R^= 50 k tl. Specify all final
values in terms o f Q.
,
^
,
= -4v^j + 2z;^j +
n
/ r^
^/=
VARIABLE GAIN AM PLIFIERS
36. Using the topology o f Figure 4.13, design
an op amp circuit to have the input-output
4 1
x h e circuit o f Figure P4.41 is a modificanon o f the basic non-inverting amplifier. In the
relationship
modification, a potentiometer R^ is connected
between the output terminal and Rq, with the
^out ^
~ '^'^al ^
^'^bl
Your design must have Rj-= 10 kX2 in the final
sliding contact between points A and B, as
shown. Show that as the sliding contact o f the
potentiometer is moved between positions A
circuit, and all other resistors should be within
and B, the range o f voltage gain achievable is
the range 2 kQ to 20 kQ.
^out
P
37. Using the topology o f Figure 4.13, design
an op amp circuit to have the input-output
relationship
^out =
- ">^al + ^^b\ +
hi
Your design must have all resistors, including
Rjr, in the range 5 kQ to 25 kQ.
Figure P4.41 Variable gain non-inverting amplifier.
188
Chapter 4 • T h e Operational Amplifier
42. The circuit o f Figure P4.42 is a simple
modification o f the basic inverting amplifier
circuit in which a potentiometer is connected
to the feedback resistor Rp as shown. Show that
the range o f gains achievable by this circuit is
h .
SIM ULATION OF
C O N TR O LLED SO URCES
USING OP AM PS
44. Design an op amp circuit to simulate the
grounded VCVS in Figure P 4.44 when p > 1.
Hint: Consider the non-inverting amplifier o f
Example 4.3.
Figure P4.44 Grounded VCVS.
45. Design an op amp circuit to simulate the
grounded V CV S in Figure P4.44 for any p > 0.
Hint: Try a voltage follower in cascade with two
inverting op amp circuits.
Figure P4.42
43. T he circuit o f Figure P4.43 is another mod­
ification o f the basic inverting amplifier to
obtain a variable gain amplifier. Show that as
the sliding contact o f the potentiometer is
moved between the two extreme positions, the
range o f achievable voltage gain is
R
f
U a = 1, H—
where
:-a -
—H— -
Rq
Rf
46. Reconsider the design o f Problem 45 so
that only two op amps are used. In this case,
one still needs the voltage follower. Why? Hint:
Consider using a voltage divider followed by a
non-inverting amplifier circuit.
R
f
R,
Rp
R■0n // /Ii\f
Rf
Hint: Apply KCL to the non-inverting input
terminal, and make use o f the virtual ground
property o f an ideal op amp.
47. Design an op amp circuit to simulate the
grounded VCVS in Figure P 4.44 when p < 0.
Hint: Consider an inverting amplifier configu­
ration in conjunction with a buffer amplifier.
48. For the circuit o f Figure P4.48, show that
the load current
equals V/R^, which is inde­
pendent o f the load resistance R^. Hence, this
op amp circuit converts a grounded voltage
source into a floating current source. (This is
sometimes called a voltage-to-current con­
verter.)
■o
Figure P4.48 Op amp circuit simulating a
floating current source.
Chapter 4 • T h e Operational AmpHfier
49. In Problem 4.48, since 7^ depends on Vand
only, the load need not be a resistor. For
example, Rj^ may be replaced by an LED (lightemitting diode), as shown in Figure P4.49.
Then by turning the knob o f the 10-kf2 poten­
tiometer, one can control the brightness o f the
LED. The current through the load is supplied
by the op amp. The potentiometer, which con­
trols the brightness o f the LED , uses a low-voltage part o f the circuit. Find the magnitude o f
the LED current if the potentiometer is set at
(a)
= 5 k£^ and (b) 7?, = 8 k tl.
A N SW ERS: 1.32 niA, 2.1 niA
10 kn
50. This problem is a variation o f Problem 4.49
in which the load current flows in the opposite
direction. For the circuit o f Figure P 4.50, show
that the load current
equals v^JR^, which is
independent o f the load resistance R^. Hence,
this op amp circuit converts a grounded voltage
source into a floating current source in which
the current enters the op amp output terminal.
(This is sometimes called a voltage-to-current
converter.)
Figure P4.50 Op amp circuit simulating a float­
ing current source.
189
C
H
A
P
T
E
R
Linearity, Superposition, and
Source Transformation
H ISTO RICAL NOTE
In the mid-nineteenth century, before the introduction o f the alternating current (ac), electricity
was available mainly as direct current (dc). This time period saw the evolution o f basic laws for
the analysis o f electrical circuits composed o f dc voltage sources and resistors: Ohm’s law, KVL,
and KCL. Application o f these laws to the analysis o f circuits led to the development o f the mesh
and nodal techniques requiring the solution o f simultaneous equations. Before the computer age,
manual solution o f a (large) set o f equations was very difficult. To circumvent this difficulty,
researchers developed a number o f network theorems that (i) simplified the aforementioned man­
ual analysis, (ii) reduced the need for repeated solution o f the same set o f equations, and (iii) pro­
vided insight into the behavior o f circuits. These network theorems remain useful even in the pres­
ent day o f high-powered computing.
CH APTER O BjEC TIV ES
1.
2.
3.
Introduce and apply the property o f linearity.
State and explore the two consequences o f linearity called superposition and
proportionality to simplify response computation.
Use superposition and proportionality to simplify manual analysis and to gain better
4.
insight into circuit behavior.
Introduce and apply the source transformation theorem to again simplify manual analysis.
SECTIO N HEADIN GS
1.
2.
3.
4.
5.
6.
Introduction
Linearity
Linearity Revisited: Superposition and Proportionality
Source Transformations
Equivalent Networks
Summary
192
Chapter 5 * Linearity, Superposition, and Source Transformation
7.
8.
Terms and Concepts
Problems
1. IN TRO D U CTIO N
Chapter 3 covered nodal and loop/mesh analyses. Node voltage or loop current calculation pro­
ceeds by constructing a set o f simultaneous node or loop equations and solving them by hand, by
MATLAB, or with some equivalent software package. Few o f us will attempt a paper-and-pencil
solution o f four equations in four unknowns. Yet, MATLAB, Mathematica, or some other com­
putational software program, can easily and reliably crunch numbers, relieving us o f tedious hand
calculations. Nevertheless, manual analysis in some form remains important for a deeper under­
standing or insight into a circuit’s behavior, as well as a way to check the validity o f a program
output.
Experience teaches us that manual analysis is ordinarily practical only for small circuits.
Fortunately, the network theorems studied in this chapter and the next can often reduce seemingly
complex circuits to simpler ones amenable to manual analysis. They also provide shortcuts for
computing outputs and allow us to obtain deeper insights into a circuit’s behavior.
This chapter talks about linearity and superposition, which are motivated by the following ques­
tions: What is the effect on the circuit output (voltage or current) o f a single independent voltage
source, say
acting alone. “Acting alone” means that the independent source, Vj^, has a nonzero
value, while all other independent sources are set to zero. A deactivated voltage source acts as short
circuit (see Chapter 2), and a deactivated current source acts as an open circuit (again, see Chapter
2). Is there a shortcut to computing the response if Vj^ is doubled in value?
To answer the above questions and others, our discussion begins with the important property o f
linearity. Linearity relates the values o f independent sources to a circuit output with a very com­
pact equation. This equation defines the effect o f any independent source on a circuit output.
After studying linearity, we discuss two special consequences called superposition and propor­
tionality. Each o f these concepts helps reduce manual computation o f responses, and each provides
insight into circuit behavior. Next, we state the source transformation theorem and show how this
method can reduce a complex circuit to a more simple form. Finally, we set forth the notion o f an
equivalent two-terminal network and then outline a proof o f the source transformation theorem.
193
Chapter 5 * Linearity, Superposition, and Source Transformation
2. LIN EARITY
This section investigates the circuit property o f linearity, which we introduce with a motivating
example.
E X A M P L E 5 .1 . For the circuit o f Figure 5.1, find the outputs
current /^j, and the source voltage V^2 -
A
3
and Vg in terms o f the source
will derive the relationships Vg = 40/^j + ^ 1^2 and
jgQ .sz
I.
60 0
+
V„
120Q
FIG U RE 5.1. Resistive circuit driven by current and voltage sources.
So l u t io n
Step 1. Find the voltage Vg. A node equation at the top o f the current source is
Vb , V B - V s l ^ r
120
60
Solving for Vg yields
= 4 0 7 ,1 + - V ,2
Here, Vg appears as a constant times /^j, plus another constant times K^2>^ so-called linear com­
bination.
Step 2. Find the current I
From Step 1, we know Vg. The current
40 / , i + - V , 2 - V ,2
Similar to Step 1, the output current
linear combination.
3
satisfies
180 •'
is a constant times /^j plus another constant times V^2>^
194
Chapter 5 ' Linearity, Superposition, and Source Transformation
Exercise. 1. In Example 5.1, suppose the 60 Q resistor is changed to 120 Q. Find the outputs
and Vg in terms o f the sources, /^j and V^2 AN SW ER: Vg = 60/ ,, + 0 .5 V ,2 and
= 0 .5 / ,, -
2. For the circuit o f Figure 5.2, find Vg in terms o f
and V^2-
--- ---------- ---- --------120 Q +
60 Q
> 120Q
V
(j
V..
-
FIG U RE 5.2 Resistive circuit for Exercise 2.
A N SW ER:
= 0.251/, + 0 .5 V^2
In the above example and exercises, the desired output voltage or current was a so-called linear
combination o f the independent source values. This is, in fact, a quite general phenomena, as indi­
cated by the linearity theorem below.
LIN EA R ITY TH EO R EM
For all practical linear resistive circuits, as per Figure 5.3, any output voltage,
or any cur­
rent, ig, can be related linearly to the independent source values, as in the following equa­
tions:
= ^i^s\ + - +
+•
^nfsm
(5.1a)
= “ l Kl + - +
+ ■•• + ^mhm
(5.1b)
or
where the a - and
are properly dimensioned constants.
Chapter 5 • Linearity, Superposition, and Source Transformation
V
195
Linear Circuit
containing
no
independent
sources.
V
+
■
FIG U RE 5.3. A linear circuit driven by n independent voltage sources and m independent current
sources with outputs of
and
A rigorous proof o f the linearity theorem entails solving a set o f modified nodal or loop equations
using matrix algebra and is beyond the scope o f this text.
EXAM PLE 5.2. For the circuit o f Figure 5.4, our objective in this example is to express
ear combination o f /^j, Iq, and
lin­
as per equation 5.1a. In doing this, we review nodal analysis.
g .v .
)v .
FIG U RE 5.4
So l u t io n
Step 1. Write nodal equation at A. For node A,
(5 .2 )
196
Chapter 5 • Linearity, Superposition, and Source Transformation
Step 2. Write nodal equation at output node. At the output node,
or equivalently,
=
(5.3)
Step 3. Write equations 5.2 and 5.3 in matrix form. In matrix form, the nodal equations are
0
G ^ -g^
G
8m
2+
■
G
3
■
/.I
■
•
(5.4)
J s l + G ^ V ,,
Vou t,
Step 4 . Solve equation 5.4. Solving equation 5.4 for
and
yields
-1
-V
a
■
Vou t,
0
'G l- g m
8m
G 2
+
G
h i
3
1
G2 + G 3
{G :-gJ(G 2+ G ,)
-8 m
0
‘ s\
G, - I
fs2+G3V,3
It follows that
Vout--
8m
(G i-g ,)(G 2 + G 3 )
hl+-
Gi-g.
(G ,-g „,)(G
jGl-8m)G3
2+
G3)
(G i-g ,„)(G
2+
G
-V.s3
3)
(5 3 )
as set forth in equation 5.1(a).
Exercise. 1. In equation 5.5, suppose G j = 1 S, G2 = 2 S, G3 = 3 S, and^^ = 5 S. Find the numer­
ical expression for
A N SW ER:
= 0.25/,, + 0.27,2 + 0-6 ^^,3
2. Suppose the dependent current source in Figure 5.4 is changed from^^V^
the new expression for
if G j = 1 S, G2 = 2 S, G3 = 3 S, and
Sm^ouf Compute
= 5 S.
AN SW ER: V ;„,= 0.17,2+ 0 .31/3
3. Suppose the dependent current source in Figure 5.4 is changed from
Compute the new expression for
AN SW ER:
if G j = 1 S, G2 = 2 S, G3 = 3 S, and
to
+ ^0 ^)-
= 5 S.
^ ^.'i + ^ h i + J ^v3
E X A M P L E 5 .3 . A linear resistive circuit has two inputs
and z,2 with output
as shown in
Figure 5.5. Rows 1 and 2 o f Table 5.1 list the results o f two sets o f measurements taken in a lab­
oratory. The measurements are taken in a practical way by first setting the value o f the current
197
ch a p te r 5 " Linearity, Superposition, and Source Transformation
source to zero, i.e., i^ 2 = 0
exciting
with a dc power supply set to 5 V; then the voltage
source is removed and replaced by a short circuit using a jumper cable, i.e.,
rent source is excited by a power supply producing a constant current o f
(a)
Derive the linear relationship
(b)
Find
when
= 0, and the cur­
= 0-2 A.
+ 50/^2 using the data in Table 5.1.
= 10 V and ^'.2 = 0.5 A, i.e., complete the third row o f Table 5.1.
FIG U RE 5.5. Linear resistive circuit driven by two sources.
TABLE 5.1. Two Sets of Measurements o f a Linear Circuit in which
One is Allowed to Set Each Source Value to 0
i^ 2 (amperes)
Vout (volts)
5
0
4
0
0.2
10
10
0.5
(volts)
So l u t io n
From the linearity equation 5.1(a),
^out =
+hhl
for appropriate ttj and ^2 - From the data in rows 1 and 2 o f Table 5.1,
4 = a , x 5 + P2><0 = 5a, ^
U j = 0.8
and
10 = a , X 0 + p2 X 0.2 = O.2 P 2 => P 2 = 50
in which case,
+ 5 0 i,2
So from row 3 o f Table 5.1, if i',] = 10 V and z'^2 = 0.5 A, we have that
0 .8 X 10 + 5 0 X 0 .5 = 33 V.
(5.7)
198
Chapter 5 • Linearity, Superposition, and Source Transformation
Exercise. 1. For Example 5.4, suppose
= 50 V and i^ 2 ~ 0-4 A. Find
AN SW ER: 60 V
2. Suppose the data in row 1, column 3, o f Table 5.1 is changed to 10 V. Find y^^^when
V and
= 50
= 0.4 A.
AN SW ER: 120 V
Comparing the development o f equation 5.7 in Example 5.3 with equation 5.1 suggests that the
coefficients Qj and P 2 can be defined as ratios:
«1 =
''out
and 132= —
I,-,v2
Example 5.3 and these equations suggest the algorithm for finding the coefficients in equation 5.1
by setting all inputs to zero except the input associated with the desired coefficient. This approach
is sometimes impractical. It is not always possible to set an independent source voltage or current
source to zero: imagine turning off a generator for downtown Manhattan to obtain a coefficient.
T he following example illustrates an alternate approach.
E X A M P L E 5 .4 . Consider Figure 5.6, which has two inputs
and
output
Table 5.2
lists measurement data taken in a laboratory. Row 1 ofTable 5.2 lists the nominal operating con­
ditions o f the circuit. Rows 2 and 3 illustrate measurements in which one source has its value only
slightly changed (although the change may be arbitrary) while keeping the other source value the
same. From the linearity theorem, we know
find
to complete row 4 ofTable 5.2.
+ ^2^s2- Compute
FIG U RE 5.6. Linear resistive circuit driven by two sources.
and P j, and then
Chapter 5 • Linearity, Superposition, and Source Transformation
199
TABLE 5.2. Two sets o f measurements of a linear circuit.
i^ 2 (amperes)
hut (amps)
5
0.25
-1
5+0.1
0.25
- 1 .0 3
5
0.25+ 0.05
- 0 .9
15
0.5
????
(volts)
So l u t io n
From rows 1 and 2 o f Table 5.2,
- 1 = ttj X 5 + P2 X 0.25
(5.8a)
- 1 .0 3 = ttj x (5 + 0.1) + P 2 X 0.25
(5.8b)
and
Subtracting equation 5.8(a) from equation 5.8(b), we have
- 0 .0 3 = ttj X 0.1 => ttj = - 0 .3
Similarly, from row 3, we have that
- 0 .9 = ttj X 5 + P 2 X (0.25 + 0.05)
(5.8c)
Again, subtracting equation 5.8(a) from equation 5.8(c), we have
0.1 = p 2 x 0 .0 5 ^
Pa = 2
Equation 5.1 for the given data has the linear form
(5.9)
Hence, for row 4 o f Table 5.2, we have that
2,„, = - 0 . 3 x 1 5 + 2 x 0 . 5 = - 3 .5 V
200
Chapter 3 * Linearity, Superposition, and Source Transformation
Exercise. Find the unknown value in Table 5.3 using linearity.
TABLE 5.3. Two Sets of Measurements of a Linear Circuit
(volts)
i^2 (niA)
20
100
15
22
100
15.9
20
110
15.6
28
80
???
AN SW ER: 17.4 A
As a final comment on linearity, we note that by simply using the data o f rows 1 and 2 o f Table
5.2, one can solve for the coefficients by solving simultaneous equations. Specifically, using the
data o f rows 1 and 2 ofTable 5.2, we have the following matrix equation
■5
0 .2 5 ' ■ «i’
5.1
0.25
.^1.
■ -1
■
-1 .0 3
whose solution yields the proper coefficients o f equation 5.9.
■5
■«r
5.1
.A .
0 .2 5 '
0.25
-1
■ -1
- 1 .0 3
"
= -4 0
'0 .2 5
-5 .1
- 0 .2 5 ' ■ -1
5
■
■-0.3'
-1 .0 3
2
Exercise. Find the unknown entry in Table 5.4 after finding a , and (3j in the equation
+ Pl^.2TABLE 5.4. Two Sets of Measurements of a Linear Circuit
(volts)
AN SW ER:
ia (itiA)
^out
10
100
15
20
100
20
30
150
???
= 0.5z^,i + 100/^2
V
201
Chapter 5 • Linearity, Superposition, and Source Transformation
3. LIN EARITY REVISITED: SUPERPO SITION AND
PRO PO RTIO N A LITY
T he linearity principle o f equation 5.1 has the more simple form
y=
( 5 . 10)
+ ... + a„u„
Here, j denotes an output, whether it be current or voltage, and each
denotes a source input,
whether it be voltage or current. A special consequence o f the linearity principle is the superpo­
sition property. Equation 5.10 says that the total response jy is the sum o f the responses '"aju".
Each “a-u” is the response o f the circuit to u- acting alone, i.e., when all other independent
sources are set to zero. Although implied by linearity, this property is so important that we single
it out.
THE SUPERPOSITION PROPERTY
For almost all linear resistive circuits containing more than one independent source, any out­
put (voltage or current) in the circuit may be calculated by adding together the contributions
due to each independent source acting alone with the remaining independent sources deacti­
vated, i.e., their source values are set to zero.
EXA M PLE 5 .5 . A linear resistive circuit has two inputs
and
with output
as shown in
Figure 5.7, where
= 2 Q,, Rj = 2.5 £2, and R^ = 10 Q. Find
by the principle o f superposi­
tion. Then, compute the power absorbed by the 10
resistor. We show that
=
0.5 V^l + 0.4V^2’ where
is the contribution o f the source
acting alone for k = 1, 2.
FIG U RE 5.7 Linear resistive circuit driven by two voltage sources;
Rj = 2 Q , R 2 = 2.5
So l u t io n
Step 1. Fini/ the contribution to
and R^ = 10 Q.
due only to V^j. Denote this contribution by
W ith V^2
= 0, the equivalent circuit is shown in Figure 5.8(a). Here, the 2.5 Q and 10 Q, resistors are in par­
allel, yielding an equivalent resistance o f 2 = 2.5 x 10/12.5 f i. By voltage division,
V
-
2 +2
202
Chapter 5 • Linearity, Superposition, and Source Transformation
+
Vo^ut
FIG U RE 5.8 (a) Circuit equivalent to Figure 5.7 when
(b) circuit equivalent to Figure 5.7 when
Step 2. Find the contribution to
due to
= 0;
= 0.
Denote this contribution by
W ith
= 0,
the equivalent circuit is shown in Figure 5.8(b). Here, the 2 Q and 10 O resistors are in parallel,
yielding an equivalent resistance o f 5/3 = 2 x 10/12 Q. By voltage division,
2.5 + 3
Step 3. Compute
by superposition. Using superposition,
Vout = ]/’o u t ^+
^ out
= 0 5 K , ^+ 0 4K .,
Step 4 . Compute the power absorbed by the R^= 10 Q resistor.
Pr3 =
(youtT
= 0 . 1(0.5
+ 0.4V,2 f = 0 . 1(o .2 5 v /i + 0 .2 V ,iy ,2 + 0 . 16V / 2 )
Note that the total power, Pj^^, is not the sum o f the powers due to each source acting alone
because o f the presence o f the cross product term. Hence, in general, superposition does not apply
to the calculation o f power.
For dc circuit analysis, the principle o f superposition does NOT apply to power calculations.
Exercise. Reconsider Figure 5.7 in which
principle o f superposition.
AN SW ER;
0.5K^, + 0.25
2
= 2 Q , 7?2 = 4 Q, and R^ = A Q.. Find
by the
203
Chapter 5 • Linearity, Superposition, and Source Transformation
The next example adds a controlled source to the circuit o f Figure 5.7 and repeats the superposi­
tion analysis.
E X A M P L E 5 .6 . For the circuit o f Figure 5.9, suppose i?, = 2 Q,
S. Using superposition, find
in terms o f
and v^2 show that
= 0 . 5 z^^j + 0 .4 t/^2’ where
=5
-^3 = 10
= 0-2
superposition theorem to
is the contribution from the source
acting alone for k = 1, 2 .
q V.A
FIG U RE 5.9 Circuit containing a dependent source for illustrating
the principle o f superposition.
So l u t io n
Step 1. Compute the contribution due only to
Setting v^ 2 ~ ^ leads to the circuit o f Figure 5.10,
where we note that
Applying KCL to the top node yields
0 - 5 ( - L - " . l ) + (0-2 + 0 . 1 + 0 .2 ) .i „ ,= 0
Therefore,
d r = 0 - 5 ^ .i
Step 2. Compute the contributions due only to
o f Figure 5.11, where this time,
= ^out~
Setting
= 0 in Figure 5.9 leads to the circuit
204
Chapter 5 • Linearity, Superposition, and Source Transformation
As in Step 1, we apply KCL to the top node to obtain
(0.5 + 0.1 + 0 .2 ).2 ^ ,- 0.2.^2 + 0 - 2 K i - -. 2) = 0
Therefore,
Step 3. Using superposition, add up the contributions due to each independent source acting
alone.
^out =
Exercise. Repeat Example 5.6 with
AN SW ER:
+ 0-4 ^',2
= /?2 = ^3 = ^ ^
Sm ~
(5.11)
S.
= 0.25i',i + 0.5t^,2
The above examples used voltage division and superposition to compute an output voltage due to
two independent voltage sources.
E X A M PLE 5 .7 . This example illustrates the principle o f superposition for the three-input op
amp circuit o f Figure 5. 12 . Show that
is the contribution o f
acting alone for k = 1 ,2 , 3.
FIG U RE 5.12 Three-input op amp circuit.
+ 2 . 5 K^2 +
Chapter 5 • Linearity, Superposition, and Source Transformation
205
So l u t io n
Step 1. Find the contribution to
due only to
Denote this output by
W ith V^ 2 = ^ 3
= 0, the circuit o f Figure 5.12 reduces to that o f Figure 5.13(a). The properties o f an ideal op amp
ensure that i^ = 0, making
= -0.5i?z^ = 0. Thus, v_ = v^=Q implies
V’i r = - — K . = - 4 V ,
2R
(a)
FIG U RE 5.13
Step 2. Find the contribution to
due only to V^2 - W ith
shown in Figure 5.13(b) where we denote the output as
= 0, the equivalent circuit is
From op amp properties and volt­
age division,
^ -V ,2 = 0 .5 V ,2
R+R
Hence, from Example 4.3,
Step 3. Find the contribution to
due only to
as that o f Figure 5.13(b) with V^ 2 replaced by
T he equivalent circuit in this case is the same
Therefore, the output due to source
acting
alone is
Step 4 . Sum up contributions due to each source. By the principle o f superposition.
Vout =
out
out
Exercise. 1. For Example 5.7, suppose
out
= -^4 sK\ , + 2 .5 Ks2t + 2 .5
K.
^ si
“ ^3 “ ^
(5 . 1 2 )
^ouf
AN SW ER: 2 V
2. Now suppose
A N SW ER: - 1 2 V
= 8 V,
= ^s3 =
°P
saturates at |
= 12 V; compute
206
Chapter 5 • Linearity, Superposition, and Source Transformation
The above examples have generated the linearity formula, equation 5.1, using superposition, i.e., the
response o f a circuit is the sum o f the responses due to each source acting alone. The technique is equiv­
alent to that described in Example 5.3. However, superposition alone is not equivalent to linearity.
Linearity is equivalent to the properties o f superposirion AND proportionality, which is now stated.
T H E P R O P O R TIO N A LITY PRO PERTY
For almost all linear resistive circuits, when any one o f the independent sources is acting
alone, say « j, with output 7 , then y =
for some constant a^. Proportionality says that if
is multiplied by a constant K, then the output is multiplied by K, i.e.,
However, for dc analysis, the proportionality property does N O T apply for power calculations.
T he proportionality property is easily illustrated by equation 5.12 o f Example 5.7:
- V'.., • V I , * V i , ■
. K,
If
and
* 2-5^2 ^ 2-5^3
. V„ . 0, then V ^ . - 4 (« V „ ) . « - 4 K „ ) -
Exercises. For certain nonlinear circuits, the principle o f superposition may be satisfied, but pro­
portionality not satisfied, or vice versa. This exercise explores these distinctions.
1.
sV ^^^ow that the principle o f
If a circuit has input-output relationship
superposition is satisfied, but proportionality is not satisfied.
2.
If a circuit has input-output relationship
= a,Wjj +
^^ow that the
principle o f proportionality is satisfied, but superposition is not satisfied.
A very interesting and significant application o f the proportionality property occurs in the analy­
sis o f a resistive ladder network. A resistive ladder netw ork is one having the patterned structure
shown in Figure 5.14, where each box represents a resistor.
-H
v.Q
V
FIG U RE 5 .14 A ladder network.
A typical analysis problem follows:
Given
and all resistances in Figure 5.14, find all node voltages. O ne can, o f course, solve the
problem by writing and solving a set o f mesh equations or node equations. A simple trick using
the proportionality property allows us to solve arbitrarily long ladder networks without simulta­
207
Chapter 5 • Linearity, Superposition, and Source Transformation
neous equations, as follows: assume
= 1 V. We can sequentially compute currents and voltages
= \ V. Suppose we call this
in a backwards fashion to obtain the required source value to yield
voltage
Define K =
to be the proportionality constant, where is the actual source
voltage. Then the correct output voltage is
E X A M P L E 5 .8 . Find all the voltages V^-, i = 1, ..., 6 in the resistive ladder network o f Figure 5.15.
----------- —
— ^
L
"
R =100
'4 W R4 = 6 Q
R =5Q
-
—
+
r
1.
tS
V =50V
-------- — '—
^
V.
V,
R3 = 5 0
R j= 1 0 Q
t .
V,
R, = 1 0 Q
FIG U RE 5.15 A simple resistive ladder network.
So l u t io n
Assume Vj = 1 V. Repeatedly apply Ohm’s law, KCL, and KVL as follows: (Q, V and A are used
throughout):
(Ohm’s law)
/2 = / i =
V2 = /?2 / 2 =
V3 =
/3
=
/2
1 0
(KCL)
0 .1
x
0 .1
+ V2 =
=
(Ohm’s law)
1
(KVL)
2
(Ohm’s law)
= ^ = 0.4
^3
+ / 3 = 0 .1 + 0.4 = 0.5
V4 = R4 / 4 =
Vg =
\/3
6
X 0.5 = 3
(Ohm’s law)
(KVL)
+ y4 = 5
(Ohm’s law)
7, = "^ = 0 .5
^5
^6 = 7 4 + 7 5 = 0.5 + 0.5 =
V6 = 7?67e=5
y, = y5 + y6 = io
(KCL)
1
(KCL)
(Ohm’s law)
(KVL)
We conclude that if Vj = 1 V, the source voltage must be
= 10 V. But the actual source voltage is
50
50 V. Define K =
= — = 5 . By the proportionality property, if
= 50 V, then
- I^x 1 =
5 V Similarly,
- 5 V,
= W V,
= 15 V, and V5 = 25 V.
208
Chapter 5 • Linearity, Superposition, and Source Transformation
In the solution given above, we have separated the expressions into calculation blocks to empha­
size the repetitive pattern. For example, the expressions in block #3 are simply obtained from
block #2 by increasing all subscripts by 2. When the ladder network has more elements, the
sequence o f expressions contains more blocks, each o f which entails two additions and two mul­
tiplications. This method then allows us to straightforwardly solve ladder networks o f any size
without writing or solving simultaneous equations.
Exercise. In Example 5.7, change all resistances to 2 Q and find V^. Would it make any differ­
ence in the voltage Kj if all the resistors were changed to R ohms?
50
AN SW ER: Vi = — = 3.85 V, and no difference.
'
13
4. SO U RCE TRAN SFO RM ATIO N S
The words “source transformation” refer to the conversion o f a voltage source in series with an Rohm resistor to a current source in parallel with an R-ohm resistor, and/or vice versa. This section
explains the details o f such transformations and how they can simplify analysis. But first we must
recall from Chapter 2 that voltage sources in series add together (such as batteries added to a flash­
light) and that current sources in parallel combine into an equivalent single current source. This
is illustrated for multiple voltage sources in series in Figure 5.16. Similarly, Figure 5.17 shows how
multiple current sources combine into a single source.
4.5 V
FIG U RE 5.16 (a) Three voltage sources in series; (b) equivalent single voltage source.
s,eq
FIG U RE 5.17 (a) Three independent current sources in parallel;
(b) equivalent single source circuit.
Chapter 5 * Linearity, Superposition, and Source Transformation
209
SOURCE TRANSFORM ATION THEOREM FOR INDEPEN DEN T SOURCES
A 2-terminal network consisting o f a series connection o f an independent voltage source
and a nonzero finite resistance R is equivalent to a 2-terminaI network consisting o f an
independent current source, /^ = VJR'm parallel with R.
Conversely, a 2-terminal network consisting o f a parallel connection o f an independent cur­
rent source
and a nonzero finite resistance R, is equivalent to a 2-terminai network con­
sisting o f an independent voltage source,
= RIy, in series with R. T he reference directions
for voltages and currents are as shown in Figure 5.18.
V =RL
FIG U RE 5.18 Illustration of source transformation theorem for independent sources.
A justification for the source transformation theorem will be given in the next section. Practically
speaking, it can save significant computational effort. For example, in the circuit o f Figure 5.19 in
Example 5.9 below, a solution approach using mesh analysis requires writing and solving three
simultaneous equations. Nodal analysis at A and B requires writing and solving two simultaneous
equations. Applying the source transformation theorem is a third avenue that avoids all simultane­
ous equations.
E X A M P L E 5 .9 . Find
in Figure 5.19 by repeated applications o f the source transformation
theorem. Then find the power absorbed by the 4 kD resistor.
5 kO
6kQ
50 V
FIG U RE 5.19 Circuit for Example 5.9.
210
Chapter 5 * Linearity, Superposition, and Source Transformation
So l u t io n
Step 1. Substitute all series V^- R combinations by their parallel
- R equivalents, where in each
case, /^ = — . Applying the source transformation theorem four times results in Figure 5.20.
R
10 mA
10 mA
FIG U RE 5.20 Circuit equivalent to that o f Figure 5.19 by source transformation theorem.
Step 2. Combine the parallel resistances and the parallel current sources.
To the left o f point A are two independent current sources and two resistors, all in parallel.
Similarly, to the right o f B are two current sources and two resistors in parallel.
Combining current sources and resistors to the left o f A results in a single current source o f 5 mA
directed upward and an equivalent resistance o f 4 kD. To the right o f B, the current sources can­
cel each other out, and the equivalent resistance is 2 kXl. T he resulting simplified circuit is shown
in Figure 5.21.
5 mA
10 mA
FIG U RE 5.21 Simplification o f the circuit in Figure 5.20.
Step 3. Apply the source transformation theorem a second time to each o f the
allel I ; - R pairs become series V^- R pairs, as illustrated in Figure 5.22.
4kO
--A
20 V
4kQ
B
20 V
FIG U R E 5 .2 2 Further simplification of Figure 5.21.
- R pairs. These par­
Chapter 5 • Linearity, Superposition, and Source Transformation
Step 4 . Find
and P^j^- From Ohm’s law,
^AB -
Thus,
211
20 + 20
4+4+2
= 4 mA
= 4 0 0 0 X (0 .0 0 4 )2 ^ <54
Exercises. 1. For the circuit o f Figure 5.23(a), /^j = 50 mA and
R combination to a series V^- R combination, where
= 500 Q. Convert the parallel
= ? and
= ? in Figure 5.23(b).
AN SW ER: 25 V, 500 Q
f^series
A
-o Circuit
V.
Circuit
r +
B
■O
B
-O
(b)
(a)
FIG U RE 5.23
2. For the circuit o f Figure 5.24(a), /^j = 50 mA and R^ = 500 Q , while I ^ 2 = 1 5 0 mA and R^ =
300 Q . Convert the two parallel 1^—R combination to a single series V^—R combination, where
= ? and
= ? in Figure 5.24(b).
AN SW ER: - 2 0 V, 800 Q
A
A
ho-
-o Circuit
Circuit
V
B
-O -
B
-O
(b)
(a)
FIG U RE 5.24
3. Consider the circuit in Figure 5.25(a). Using a source transformation and resistance combina­
tions, determine the values o f
and Rp^^^ in Figure 5.25(b).
Chapter 5 * Linearity, Superposition, and Source Transformation
212
100Q
100Q
FIGURE 5.25
AN SW ER: 150 LX 50 niA
5. EQ U IVALEN T N ETW ORKS
The source transformation theorem above is based on the notion o f equivalent networks, as is the
material o f the next chapter. So we now explore a precise understanding o f equivalent 2-tenninal net­
works. Figure 5.26 illustrates four 2-terminal networks, all enclosed by dashed boxes, labeled N j,
N 2, Ng, and N^. Their characteristic is that there are only two accessible nodes for connection to other
circuits. Note however that any controlling voltage or current must be contained within the dashedline box. Such dashed-line boxes are often omitted to avoid cluttering in circuit diagrams.
FIGURE. 5.26 Examples of 2-terminal networks, i.e., networks in which only two terminals are
available for connection to other networks.
Observe that networks N j and N 2 in Figure 5.26(a) and (b) have the same terminal characteris­
tics: at the terminals o f N j, the v - i characteristic is
v = 2i+ 10
Chapter 5 • Linearity, Superposition, and Source Transformation
213
At the terminals o f N 2, the characteristic is
v = 2( + 10
i= - - 5
2
The two equations are identical. We then say that a pair o f 2-terminal networks are equivalent if
they have the same terminal characteristics. Therefore,
and N 2 are equivalent.
Now, observe that networks
and
are also equivalent to A^j and A^2- To see this, note that
for N y
V
v-1 5
6/ = 3 v - 3 0
v = 2i + 10
And for N^, first observe that i = \0i^ from KCL, in which case
10 + 20/'a ^
= O.lz; further, from KVL,
v = 2 i + 10
as was to be shown.
Because equivalent 2-terminal networks have the same terminal v —i characteristic, if one network
is interchanged with its equivalent, all currents and voltages outside the box remain the same as
illustrated in Figure 5.27; i.e., all voltages and currents in the “rest o f the circuit” are the same as
before.
A
■o
N
N.1
+
V
J
Rest of
Circuit
-O B
FIG U RE 5.27 The networks denoted N^, i = 1,2, are equivalent when the v-i values at the terminals
are identical; logically then, all voltages and currents inside the “Rest o f Circuit” remain the same.
These examples allow us to justify the source transformation theorem as follows. Both 2-terminal
networks in Figure 5.18 have the same v - i relationship: v = Ri +
v = Ri + Rlj, = Ri +
and
.
Therefore, the two networks o f Figure 5.18 are equivalent, and the source transformation is a valid
analysis technique.
214
Chapter 5 * Linearity, Superposition, and Source Transformation
6. SUM M ARY
This chapter covers the notions o f Unearity, superposition, proportionaHty, and source transfor­
mations. Linearity states that for any hnear resistive circuit, any output voltage or current, denot­
ed as y, is related to the independent sources by the formula j =
+ ... +
are the voltage and current values o f the independent sources, and
ate constants. O nce values for the
where
through
through
are appropri­
are known, one can compute the output for any (new) set o f
input values without having to resolve the circuit equations, a tremendous savings in time and
effort. A special consequence o f linearity is the widely used principle o f superposition.
Superposition means that in any linear resistive circuit containing more than one independent
source, any output (voltage or current) can be calculated by adding together the contributions due
to each independent source acting alone with the remaining independent source values set to zero.
Practically speaking, this is the customary path to computing the coefficients, a^, in the linearity
formula.
Proportionality, another consequence o f linearity, means that if a single input is scaled by a constant,
with the other inputs set to zero, then the output is scaled by the same constant. This property led
to a clever technique for analyzing ladder networks without writing simultaneous equations.
Since
power
is
proportional
to
the
square
o f a voltage
or
current, P = — =
R
for dc resistive circuits, the principle o f linearity and its consequences, superposition and propor­
tionality, D O N O T APPLY for power calculations.
Using the notion o f an equivalent 2-terminal network, the chapter set forth the theorems on
source transformations for source-resistor combinations: a 2-terminal network consisting o f a
series connection o f an independent voltage source
and a nonzero finite resistance R is equiv­
alent to a 2-terminal network consisting o f an independent current source,
= V^IR, in parallel
with R, as illustrated in Figure 5.18. These transformations, applied multiple times to a circuit,
often simplify the analysis o f a circuit.
Chapter 5 • Linearity, Superposition, and Source Transformation
215
7. TERM S AND C O N C EPTS
2-term inal network: an interconnection o f circuit elements inside a box having only 2 accessible
terminals for connection to other networks. T he concept is extendible to n-terminal net­
works.
Equivalent 2-term inal networks: two 2-terminal networks having the same terminal voltage-cur­
rent relationship. I f two 2-terminal networks
and N 2 are equivalent, then one can be
substituted for the other without affecting the voltages and currents in any attached net­
work.
and Uj, each acting alone, be y-^ and y 2 . U2 are applied simultaneously, the response is =
Linearity property: let the responses due to inputs
W hen the scaled inputs
and
0 2
+ ®2^2- Linearity implies both superposition and proportionality, and vice versa.
Linear resistive element: a 2-terminal circuit element whose terminal voltage and current rela­
tionships is described by Ohm’s law.
Linear resistive circuit/network: a network consisting o f linear resistive elements, independent
voltage and current sources, op amps, and controlled sources.
Proportionality property: when an input to a linear resistive network is acting alone, multiply­
ing the input by a constant, K, implies that the response is multiplied by K.
Source transformation: a 2-terminal network consisting o f an independent voltage source in
series with a resistance is equivalent to another 2-terminal network consisting o f an inde­
pendent current source in parallel with a resistance o f the same value.
Superposition property: when a number o f inputs are applied to a linear resistive network simul­
taneously, the response is the sum o f the responses due to each input acting alone.
Chapter 5 • Linearity, Superposition, and Source Transformation
216
4 . Consider the circuit shown in Figure P5.4.
Problems
(a)
Find the coefficients a , and (3j in the
linear relationship
LIN EARITY
(b)
and |3j in the
linear relationship v^ 2 =
1. Consider the circuit o f Figure P5.1 in which
= 5 £2 and
= 20 Q.
(a) Using linearity,
= a j V^j +
Find the coefficients
^l^sT
rnay be expressed
Compute
ttj and
(b)
If v^{t) = 10 cos(10?) V and I 2 = 2 A,
(c)
find v^Jt).
Redo part (a), but this time express
ttj and P 2 in terms o f the hterals
G] = — and G t = — .
Figure P5.4
5. Consider the circuit o f Figure P5.5.
(a)
Find the linear relationship between
(b)
If
(c)
(Challenge) W hat is the effect o f dou­
and the input sources V^j and l^j= 20 V and ^ = 0.5 A, find 1/
bling all resistance values on the coef­
ficients o f the linear relationship
’Q
'^(t)
found in part (a)?
Figure P5.1
A N SW ER: (b) 8 cos(lO^) - 8 V
2. For the circuit o f Figure P5-2,
(a)
(b)
find Vg in terms o f
and Gy and
*^2’
find Ig in terms o f
and Gy
V^2 > ^i> ^ 2’
CH ECK:
= 0 .2 5 ^ 1 +????/,2
6. For the circuit o f Figure P5.6, find the linear
relationship between
and the independent
sources. Hint: Write a single loop equation.
Figure P5.2
3. For the circuit o f Figure P5.3,
Figure P5.6
(a)
find Ig in terms o f 7^,, /^2>^1 >-^2’
Ry and
(b)
find Vg in terms o f /^,, /^2>
+
-
. ^2>
7. Consider the circuit shown in Figure P 5.7 in
^3-
which /?, = 80 Q. R2 = 20 Q, R^ = 80 Q. and
r^ = 20.
(a)
^oia
the two independent sources.
Hint: Write two loop equations.
-t" ’
(b)
Figure P5.3
Find the linear relationship between
If
= 20 V and 7^2 = 0.125 A, com­
pute the power delivered by the
dependent source.
Chapter 5 • Linearity, Superposition, and Source Transformation
217
G,
— <+ i;
I .Q
>R,
© '
Figure P5.7
C H EC K : (a)
Figure P5.9
????? I/, + 16/^2’ 39 m W >
10. A linear resistive circuit has two independent
sources, as shown in Figure P5.10. If
with v^2 ^t) = 10cos(2 ?) V, then
8. Consider the circuit o f Figure P5.8. in which
V. On the other hand, if
Ry = 18 Q.
=9
and R^= 18 D.
with v^^{) = 0, then
(a)
^3 = 18
Ra = 36 Q,
it) and the four independent sources.
out
(b)
= 20 cos(2 i)
= 10cos(2^) mA
= 2 cos(2z-) V. Find the
linear relationship between
Find the linear relationship between
and
= 0
and the inputs,
Now compute
when
= 20cos(2/) mA and v^2 (^) = 20 V.
(Challenge) If each o f the resistances
is doubled, what is the new linear rela­
tionship. (Reason your way to the
answer without having to resolve the
circuit. Hints: Investigate the effect o f
Linear
resistive circuit
with
dependent sources
changing the resistance in Ohm’s law
for fixed current. Investigate the effect
o f equal changes in all resistances on a
V________________^
voltage divider formula.)
Figure P5.10
11. Again, consider the configuration o f Figure
= 0 with v^2 ^t) = 10 V, then
P5.10. If
= 55 V. On the other hand, if z^j(r) = 4cos(2^) A
with v^2 ^t) = 0, then
= -2cos(2?) V.
(a)
If z^j(?) = 2cos(2z-) A and v^2 ^t) = -
(b)
If /^j(z-) = -4cos(5^) A and v^2 ^t) =
10cos(2z-) V, find
20cos(5^) V, find
C H EC K : (b) v^Jt) = 108cos(5^) V
Figure P5.8
CH ECK:
27
- - Vi + ?? *4 - —
12. Consider again Figure P5.10. Suppose the
+ ?? V2
measured data are as follows: (i)
when
9. For the circuit o f Figure P5.9, express
a linear combination o f /^j, /^2>
equation 5.1(a). Assume
= 0.4 S, ^2 = G j =
0.05 S, and^^ = 0.1 S.
= 15 V
= 2 A and v^ 2 = 10 V, and (ii)
=10
V when
= 3 A and v^ 2 = 5 V.
(a)
Determine the linear relationship
(b)
Find
AN SW ER: 7.5 V
when z^j = 1 A and
=
Chapter 5 * Linearity, Superposition, and Source Transformation
218
13. Consider the linear network o f Figure
(a)
P 5.13, which contains, at most, resistors and
linear dependent sources. Measurement data is
given in Table P5.13.
Compute the coeiTicients o f a linear
relationship among the output and
three inputs.
(b)
(a)
Find the linear relationship
(b)
Find the power consumed by the 10
resistor when
= 20 V and
=
If 7,1 = - 1 A, 1/2 = 4 0 V, and I/3 = 10
V, find the power absorbed by R^.
AN SW ER: 16 W
= 500 mA
Resistive Circuit
witli
Dependent
Sources
+
V. . -
Figure P 5.I3
Figure P5.15
Table P5.13
I',, (volts)
z'^2 (amperes)
5
0.4
-1
16. Again consider Figure P5.15. Suppose the
data measurements are given in Table P5.16.
10
1
2
Table P5.16
Km
14. Reconsider Figure P5.13 Two separate dc
/j, (mA) ^.2 (V)
measurements are taken. In the first experi­
ment,
= 7 V and hi - 3 A, yielding
=1
A. In the second experiment,
= 9 V and z'^2
= 1 A, yielding
= 3 A.
(a)
Find the coefficients o f the linear rela­
(b)
tionship
+ p 2i,2 Given the equation found in part (a),
compute
when
= 5 A.
AN SW ER: (b) 90 watts
= 15 V and z^2
15 . The box in the circuit o f Figure P 5.15 con­
Case 1
30
2
-1
11.5
Case 2
40
2
-1
13
Case 3
30
2.2
-1
11.6
Case 4
30
2
- 0 .9
11.9
Case 5
40
8
10
(a)
ments.
Case 1
50
-2
Case 2
0
Case 3
0
^^.3 (V)
1
(b)
Find the power consumed by Rj^ for
the data in Case 5.
A N SW ER: 25 watts
Table P5.15
(mA) 1^.2 (V)
Find the coefficients in the linear rela­
tionship
+ a2^^2 +
without any matrix inversions.
tains resistors and dependent sources. 7?^ = 100
Q. Table P5.15 contains various data measure­
(V)
17. Again consider Figure P 5.15. Suppose the
5
-1 3
3
5
2
2
4
0
data measurements are given in Table P5.17.
219
Chapter 5 • Linearity, Superposition, and Source Transformation
Table P5.17
h\
Table P5.18
^ 2 (V)
^",3 (V)
Kut (V)
i,4 (mA)
-out (V)
Case 1
30
2
-1
11.5
1
6
Case 2
-2 0
4
2
27
2
10
Case 3
-1 0
-3
1
-1 4
5
?
Case 4
40
10
10
???
>
0
(a)
Find the coefficients in the linear rela­
tionship
using a matrix inversion.
(b)
Find the power consumed by 7?^ for
the data in Case 4.
A N SW ER: 102.01 watts
SUPERPO SITIO N AND
PRO PO RTIO N ALITY
19. For the circuit o f Figure P5.19,
R2 = 50 Q, ‘'sX = 12 V, and
(a)
=
1^2
using
Find
= 200 Q,
mA.
superposition.
18. The linear resistive circuit o f Figure P5.18 has
Specifically, first find
due to
four independent sources. Three o f these sources
acting alone, and
to z,2 act-
have ftxed values. Only one,
is adjustable. In
a laboratory, the data set forth in rows 1 and 2 o f
out
ing alone.
AN SW ER: 2.4 V, 2.4 V, 4.8 V
Table P5.18 were taken. Complete the last two
rows o f Table P5.18 using linearity and the data
(b)
Find
and i^ 2
from the first two rows. For the data in row 3,
find the power delivered by the current source z^.
Hint: To solve this problem, recall from the lin­
in terms o f the literals
compute the spe­
cific numerical relationship.
(c)
If
= 10
determine
earity equation 5.1
, R^-,
X
12 V and «j2 = ^ x 60 mA,
using the proportional­
ity theorem by first computing
^out =
+ ^2^.2 +
\‘'s\
due to the modified
and
'ih-i
acting alone,
due to the modified z'^2 act­
ing alone.
We have used the fact here that the term (a,v^
+ ^2^j2
constant because the associat­
ed source values are constant. Thus,
+
^2^i2
^^
some K. Hence, one can
use the data from the first two rows o f Table
P5.18 to solve for
and K.
Figure P5.19
20. Consider the circuit o f Figure P5.20 in
which
= 20 £2 , R2 = 60 Q, R^ = 20 Q.
(a)
Find the coefficients o f the linear rela­
Linear resistive
network with
dependent
sources
tionship
+ a 2 i, 2 + ^3^,3 by
superposition. Specifically, first find
—a
due to z/^j acting alone, due to z^2
acting alone, and
ing alone.
Figure P5.18
g^f due to z^g act­
220
Chapter 5 * Linearity, Superposition, and Source Transformation
(b)
(c)
Repeat part (a), but express your answers
Using superposition, find
acting alone, and then find
Find
power delivered to
v^2 acting alone. What is
V,
(d)
(a)
in terms o f the literals R-,i= 1, 2, 3.
= 2 A, and
and
when
(b)
due to
^
Redo part (a) using the literals G/ = —
Ri
= 4 A.
Repeat part (c) when
is tripled, and
and the
= 100
due to
is doubled, i^2
is halved.
1
AN SW ER: (a)
5
2
7
= -''.v i- '' m » = - ‘',v2
AN SW ER: (c) 100 V, 500 watts
Figure P5.22
Figure P5.20
23. For the circuit o f Figure P5.23, suppose R^ =
21. In the circuit shown in Figure P5.21, R-^ =
180 Q, R^ = 360 a , T?3 = 90 Q, an R^ = 720 Q.
(a)
Find the coefficients o f the linear rela­
tionship
+
fi by superposition.
Repeat part (a), but express your
0 -2 1
(b)
answers in terms o f the literals G- =
20 Cl, /?2 = 50 Q,
(a)
= 100 Q and
= 0.02 S.
Using superposition, find
due to
acting alone, and then find
due to v^ 2 acting alone. W hat is
A N SW ER:
= 0.5j',| + 0.9i^,,
(b)
Redo part (a) using the literals G,- = —
l lR- ,i = 1 , 2 , 3 , 4.
(c)
Find
and the power absorbed by
R^ when
= 100 V and v^ 2 = 50 V.
= 60 V;
= 5 watts
AN SW ER:
(d)
Repeat part (c) when
= 0.5 x 100 V
and v^ 2 = “ 10 x 2 V.
Figure P5.21
Figure P5.23
24. For the circuit o f Figure P 5.24, find the
contribution to
from each independent
source acting alone, and then compute
'.,© .,.> U
- 0
power absorbed by the 900 Q resistor.
AN SW ER: 38 V and 1.6 watts
22. For the circuit o f Figure P5.22, suppose R^
= 20 Q., /?2 = 50 Q and R^ = 100 Q.
by
the principle o f superposition. Finally, find the
221
Chapter 5 • Linearity, Superposition, and Source Transformation
i8on
225 n
0
20V
Figure P5.27
0.1 A
28. (a)
■^900 0
MATLAB program.
Figure P5.24
25. For the circuit shown in Figure P5.25,
160 V. Find
=
(b)
If it is known that
(c)
Find the equivalent resistance seen by
Then find the intermediate
node voltages. Hint: Assume
For the circuit shown in Figure P 5.28,
If /j = 1 A, find /j by writing a
= 200 mA, find /j.
the current source.
= 1V
and use proportionality, as per Example
5.11.
C H E C K : Answer is an integer.
+
Figure P5.28
LIN EARITY AND OP AM P
CIRCU ITS
26. For the circuit shown in Figure P5.26,
64 mA. Find
Hint: Assume
=
29. Consider the circuit in Figure P5.29.
(a)
= 1 A and
then use proportionality.
1n
1n
1n
1n
III
©
2n
1n
2n
2n
2n
Find the contribution to
due only
to
(b) Find the contribution to
due only
“ '^^2o>^i (c) Find
by superposition.
r2n
Figure 5.26
2 7 .(a)
For the circuit shown in Figure P5.27,
If Vj = 1 V, find V; by writing a
MATLAB program to solve the prob­
lem, given that R-^ = 10 Q ,
= 10
= 5 Q, ^4 = 6 Q ,
= 10 Q ,
Figure P5.29
=
(b)
5 Q,
= 20 Q, and
= 5 Q..
If it is known that
= 175 volts, find
(c)
^1Find the equivalent resistance seen by
(d)
the voltage source.
Suppose
is changed from 1 to 10
Q in steps o f 0.25 O.. Obtain a plot o f
vs
by modifying the MATLAB
code o f (a). Assume Vj = 1 V.
30. Consider the circuit in Figure P5.30.
(a)
Find the contribution to
due only
(b)
to I/j.
Find the contribution to
(c)
to
Find
by superposition.
due only
222
Chapter 5 ®Linearity, Superposition, and Source Transformation
O
Figure P5.30
Figure P5.32
31. (a) For the circuit of Figure P 5.31, find
V^ut^ the voltage due to each source
acting alone in terms of the literal values,
(b)
Find
(c)
Now suppose that
= 2Rq,
= 3i?o»
= 4 i^ , Rr= URq, and
= 100 Q,
33. (a)
in terms of the literals.
(b)
If the input voltages are
= - 2 .5 V and
Vout.
Suppose each voltage source has value 2
V: (i) Find the power absorbed by the
load to each source acting alone, and
For the circuit in Figure P 5.33, find
the linear relationship between
(c)
= 5 V, V^2
= 2 V, determine
If the voltages are all halved, what is
the new
(ii) the actual power delivered to the
load when all sources are active.
r^
34. (a)
For the circuit in Figure
P 5.34, find the linear relationship
Figure P5.31
Kd-
between V^^^^and
32. Consider the circuit in Figure P5.32.
(a)
Find the contribution to
due only
(b)
to Ki
Find the contribution to
due only
to
(c)
Find
(d)
If
= ^3 = 0 . 5 ^ = 5 kQ
and V;i = 2V;2 = 4 V, find the
(b)
If the input voltages are
V^2 = ” 0-5 V and
Vour
(c)
n
= 0.25 V,
= 2 V, determine
If the voltages are all halved, what is
the new
by superposition.
power delivered to the 1
load.
n
Figure P 5 .3 4
n
Chapter 5 • Linearity, Superposition, and Source Transformation
37. Consider the circuit in Figure P 5.37 in
35. Consider the circuit in Figure P5.35.
(a)
Find the contribution to K^^^due only
(b)
Find the contribution to
(c)
to 1/2.
Find
(d)
223
(a)
= Q.25R^= R2 = Ry
Find the contribution to
due only
(b)
to
Find the contribution to
due only
(c)
to
Find
which
due only
by superposition.
by superposition.
Find the power delivered to Rj^ when
is acting alone, i.e.,
®
then find the power delivered to R^
(e)
when V^2 acting alone when Rj =
R and R^ = R.
Find the total power delivered to R^
when R 2 = 4 R and R^ = R.
Figure P5.37
38. Consider the circuit in Figure P5.38.
Find the contribution to Vout due only
(a)
to l/i(b)
Figure P5.35
(c)
36. Consider the circuit in Figure P 5.36 in
(b)
= Q.25Rjr.
“Find■ the
■ contribution to
only to
Find the contribution to
(c)
only to V^2 Find
by superposition.
(d)
Find the power delivered to Rj^ when
which
(a)
(d)
Find the contribution to Vout due only
to K2Find the contribution to Vout due only
to V^,3.
Find
by superposition.
due
due
is acting alone, i.e., V^2 =
and
then find the power delivered to
(e)
when V^ 2 is acting alone when R ^ ARy
Find the total power delivered to
Figure P5.38
when Rjr=
SO U RCE TRAN SFO RM ATIO N S
39. Use a series o f source transformations to
simphfy the circuit o f Figure P 5.39 into one
consisting o f a single voltage source in series
with a single resistance.
Chapter 5 * Linearity, Superposition, and Source Transformation
224
200 n
0
Figure P5.39
AN SW ER: 6 V source in series with 12 Q resis­
40 0
80
40 n
20 n
son
Figure P5.41
AN SW ER: 13.5 and 9.1125
42. Use source transformations on the circuit o f
tor
Figure P5.42, to compute the value o f
40. Consider the circuit o f Figure P 5.40 in
which /,i = 10 mA, 1^2 = 20 V, and I/3 = 80 V.
(a)
need­
ed to deliver a current o f I = 0.25 A.
Use a series o f source transformations
to find a single voltage source in series
with a resistance that is in series with
the 9.6 kQ resistor.
(b)
Then find the power absorbed by the
9.6 kQ resistor.
Figure P5.42
A N SW ER: 28 V
Figure P5.40
AN SW ER: (a) 48 V in series with 3.2 kQ; (b)
135 mW
43. For Figure P5.43, use a series o f source
transformations to find the value o f
so that
the power delivered to
is 16 watts.
40 A
41. In the circuit o f Figure 5.41, 1/j = 240 V
and 1^2 = 0-25 A.
(a)
Use a series o f source transformations
to reduce the circuit o f figure P5.41 to
©
sn
2A
<
50<
= 1n
a current source in parallel with a sin­
gle resistor in parallel with the 20
resistor across which
V ^ appears.
(b)
Find
Find the power dissipated in the 20 Q
resistor.
(c)
I f both sources have their values
increased by a factor o f two, compute
the new value o f
Can you do this
by inspection? Explain.
—
is n
Figure P5.43
44. Apply source transformations to the circuit
shown in Figure P 5.44. Then write two nodal
equations to find
and V2 .
Chapter 5 • Linearity, Superposition, and Source Transformation
Figure P5.44 Source transformations simplify
writing node equations.
AN SW ER: 25 V, 20 V
45. Apply source transformations to the circuit
o f Figure P5.45. Then write two nodal equa­
tions to find K| and Vj-
Figure P5.45
AN SW ER: 2.8 V, - 0 .4 V
225
C
H
A
P
T
E
R
Thevenin, Norton, and
Maximum Power Transfer Theorems
H ISTO RICAL NOTE
In the early days o f electricity, engineers wanted to know how much voltage or current could be
delivered to a load, such as a set o f street lamps, through a complex transmission network. Before
the days o f computer-aided circuit simulation, simplification o f complex circuits allowed engi­
neers to analyze these very complex circuits manually. In 1883, a French telegraph engineer, M.
L. Thevenin, first stated that a complex (passive) network could be replaced by an equivalent cir­
cuit consisting o f an independent voltage source in series with a resistor. Although stated only for
passive networks, the idea o f a Thevenin equivalent evolved to include active networks. Its wide­
spread use has simplified the homework o f students for many years now and probably will con­
tinue to do so for many years to come.
A more recent but quite similar idea is the Norton equivalent circuit consisting o f an independ­
ent current source in parallel with a resistance. At the time o f E. L. Norton (a scientist with Bell
Laboratories), the invention o f vacuum tubes made independent current sources a realistic possi­
bility. Many electronic circuits were modeled with independent and dependent current sources.
The appearance o f Norton’s equivalent circuit was a natural outcome o f advances in technology.
CHAPTER OBJECTIVES
1.
2.
3.
4.
Define and construct the Thevenin and Norton equivalent circuits for passive networks.
Define and construct the Thevenin and Norton equivalent circuits for active networks
containing dependent sources or op amps.
Illustrate several different techniques for constructing the Thevenin and Norton equiva­
lent circuits.
Investigate maximum power transfer to a load using Thevenin or Norton equivalents.
228
Chapter 6 • Thevenin, N orton, and M axim um Power Transfer Theorem s
SECTIO N H EADIN GS
1. Introduction
2. Thevenin and N orton Equivalent Circuits
3. A General Approach to Finding Thevenin
4. Thevenin and N orton Equivalent Circuits
5. Thevenin and N orton Equivalent Circuits
6 . Thevenin and N orton Equivalent Circuits
7. M aximum Power Transfer Theorem
8 . Summary
9. Terms and Concepts
10. Problems
for Linear Passive Networks
and N orton Equivalents
for Active Networks
for Op Amp Circuits
from Measured Data
1. IN TRO D U CTIO N
Practicing electrical engineers often want to know the power absorbed by one particular load. The
load may be a large machine in a factory or a lighting network in the electrical engineering build­
ing. Simple resistances often represent such loads. Usually the load varies over time in which dif­
ferent resistances are used at different times to represent the load. W hat is the effect o f this load
variation on the absorbed power and on the current drawn by the load? To simplify analysis, the
rest o f the linear network (exclusive o f the load) is replaced by a simple equivalent circuit consist­
ing o f just one resistance and one independent source.
For our purposes, a (resistive) load is a two-terminal network defined in Chapter 1, meaning that
the current entering one o f the terminals equals the current leaving the other. More generally, a
two-terminal network is any circuit for which there are only two terminals available for connec­
tion to other networks. (See Figure 6.1.) T he important question for our work in this chapter is:
How does one characterize a two-terminal networks As is shown in Figure 6.1(a), there is a voltage
v{t) across the terminals and a current i(t) entering one terminal and leaving the other. T he rela­
tionship between the voltage v{t) and the current i{t) characterizes the two-terminal network. For
example, if v{t) = Ri{t), we would recognize the terminal network as an equivalent resistance R.
Or, if v{t) = Ri{t) + Vq, we might recognized this equation as that o f a resistance in series with a
voltage source. In fact, this equation could be represented as graph, e.g. Figure 6.1(b).
This leads to our next question: When are two 2-terminal networks equivalent'^ As developed in
Chapter 5, two 2 -terminal networks are said to be equivalent when their terminal v-i characteris­
tics are the same. O f particular interest for this chapter is an equivalent network consisting o f a
voltage source in series with a resistance, called the Thevenin equivalent network, and a current
source in parallel with a resistance, called a Norton equivalent network. Figure 6.1c shows a
Thevenin equivalent for a linear resistive circuit.
Chapter 6 • Thevenin, N orton, and M axim um Power Transfer Theorems
N-
229
i(t)
2-terminal
Linear
—o
+
Resistive
Network
V(t)
FIG U RE 6.1. (a) a 2-terminal linear network with terminal voltage v{t) and current i{t)\
(b) graphical representation of the equation v{t) =
+ Vq,
(c) Thevenin equivalent network
having the same terminal v{t) and i{t) relationship as (b).
This chapter investigates the replacement o f a network N by its Thevenin equivalent or its Norton
equivalent. The first section describes the Thevenin and Norton equivalent theorems for passive net­
works, those containing only independent sources and resistors. Following that, we generalize the
statements to include active networks. However, because op amps have peculiar properties, Thevenin
and Norton equivalents o f circuits with op amps are explored exclusively in Section 4. Following
this, in Section 5, we describe how to obtain a Thevenin or Norton equivalent from measured data
without having to know anything about the internal circuit structure. This is particularly useful
when one has equipment such as a power supply but no schematic diagram o f the internal circuit­
ry. Unfortunately, not all linear devices have a well-defined Thevenin or Norton equivalent. The
homework exercises illustrate a few cases. Section 6 explores the problem o f maximum power trans­
fer to a load in the context o f the Thevenin equivalent circuit, which ends the chapter.
2. TH EVEN IN AND NORTON EQ UIVALEN T CIRCU ITS FOR
LINEAR PASSIVE N ETW ORKS
Our first objective is to develop and illustrate the celebrated Thevenin theorem for passive net­
works. Then we will state and illustrate Norton’s theorem, dual to Thevenin’s theorem.
To develop Thevenin’s theorem, consider Figure 6.2(a) consisting o f two 2-terminal networks, N
and Nj^ , joined at A and B. Only resistors and independent sources make up N, while
con­
Chapter 6 • Thevenin, N orton, and M axim um Power Transfer Theorem s
230
tains arbitrary even nonlinear elements. Suppose
undergoes various changes as part o f an
experiment, while N, complicated in its own right, remains unchanged. To simplify repeated cal­
culations, N is replaced by its Thevenin equivalent, as illustrated in Figure 6.2(b). The more sim­
ple Thevenin equivalent consists o f a single voltage source,
in series with a single resistance.
Rthf
------ o-----+
Resistances
and
independent
Sources
B
J
V
Arbitrary
Networi<
r-\
J
V
(a)
i, = 0
-A
NResistances
and
independent
Sources
-o
-I-
-o
NReslstances
with independent
Sources
-o
R..
Deactivated
B
(c)
(d)
FIG U RE 6.2 (a) Network TVattached to an arbitrary network load, N^,
(b) N replaced by its so-called Thevenin equivalent,
(c) circuit for computing
still attached to 7V^;
(d) circuit for computing
in which all independent sources inside N are deactivated.
This brings us to a formal statement o f Thevenin s theorem for passive networks.
Chapter 6 • Thevenin, N orton, and M axim um Power Transfer Theorem s
231
TH EVEN IN 'S TH EO REM FOR PASSIVE N ETW ORKS
Given an arbitrary 2-terminal linear network, N, consisting o f resistances and independent
sources, then, for almost all such N, there exists an equivalent 2-terminal network consisting
o f a resistance,
in series with an independent voltage source,
Th e voltage,
called the open-circuit voltage, is what appears across the 2 terminals o f N. R^j^, called the
Thevenin equivalent resistance, is the equivalent resistance o f N when all independent sources
are deactivated. Figure 6.2(c) shows the appropriate polarity for v^J^t),
In the above theorem, “for almost all” means there are exceptions. For example, an independent
current source does not have a Thevenin equivalent. More generally, any two-terminal network
characterized by i{t) = constant does not have a Thevenin equivalent. This leads us to suggest that
there ought to be an equivalent current source formulation o f an equivalent network. From
Chapter 5, the source transformation theorem tells us that the Thevenin equivalent o f Figure
6.2(b) when
0 is equivalent to a current source in parallel with R^j^, as in Figure 6.3(b).
Figure 6.3 leads us to a formal statement o f the so-called Norton theorem.
■NResistances
and
Independent
Sources
(c)
FIG U RE 6.3 (a) Arbitrary 2-terminal linear network o f resistors and independent sources;
(b) Norton equivalent circuit; (c) circuit for computing
with
computed, as per Figure 6.2(d).
N O RTO N 'S TH EO REM FOR PASSIVE NETW ORKS
Given an arbitrary 2-terminal linear network, N, consisting o f resistances and independent
sources, then for almost all such N, there exists an equivalent 2-terminal network consisting
o f a resistance, R^^, in parallel with an independent current source, z'^^(r). Th e current,
called the short circuit current, is what flows through a short circuit o f the 2 terminals o f N,
as per Figure 6.3(c). R^^ as before, is the Thevenin equivalent resistance o f N computed
when all independent sources are deactivated.
A single voltage source does not have a Norton equivalent, and— as mentioned— a single current
source does not have a Thevenin equivalent. Both Thevenin and Norton equivalents exist for a 2terminal linear circuit when R^^
0 and is finite. W hen both the Thevenin and Norton equiva­
lents exist for the same network, the source transformation theorem and Ohm’s law imply that
thhc'kt)
(6.1a)
232
Chapter 6 • Thevenin, Norton, and M axim um Power Transfer Theorem s
4^ 0, then
and when
r,
th -
(6.1b)
■
This formula turns out to be useful in calculating
for a variety o f circuits, especially op amp
circuits.
E X A M P L E 6 .1 . For the circuit o f Figure 6.4, using literals, find the open circuit voltage,
short circuit current,
200 Q ,
and the Thevenin equivalent resistance,
the
Then, if R^ = 50 Q,
=
= 100 V, and i^ 2 = 2 A, construct the Thevenin and Norton equivalent circuits.
o
A
A
-O
-O
40 n
160V
40 0
4A
-o
-O
(c)
(b)
FIG U RE 6.4. (a) Resistive 2-terminal network; (b) Thevenin equivalent; (c) Norton equivalent.
S
o l u t io n
Step 1. Find
Using superposition, we have by voltage division and Ohm’s law,
/?2
^1^2
~ R\+R2
^1 + ^2
Substituting the given values into this formula yields
= 0.8
X
100 + 4 0 x 2 = 160 V
Step 2. Find i^^. As per Figure 6.4, with terminals A and B shorted ^together, all the current from
flows through the short circuit. From superposition, isc = 's2'^----- ■ Substituting numbers
into this formula yields
i^^ = 2 + 0.02
X
100 = 4 A
233
Chapter 6 • Thevenin, N orton, and M axim um Power Transfer Theorem s
Step 3. Find
Replacing
by a short circuit and z'^2 t>y an open circuit implies that
^
^ 40
q
R, +R2
Step 4 . Determine the Thevenin and Norton equivalent circuits. The Thevenin equivalent circuit fol­
lows from Steps 1 and 3 and is illustrated in Figure 6.4(b). The Norton equivalent circuit follows
from Steps 2 and 3 and is illustrated in Figure 6.4(c). We also note that
R ,, =
V =155 =
he
40 Q
4
as expected.
It is important to note that for many circuits, especially when the deactivated circuit is a seriesparallel connection o f resistances, one can obtain the Thevenin equivalent by a series o f source
transformations.
Exercises. 1. Redo Example 6.1 using a series o f source transformations.
2. In Example 6.1, suppose
= 100 Q,
= 400 Q, v^^ = 100 V, and i^ 2 = 2 A. Find
v^^, and i^^.
AN SW ER; 80 Q, 240 V, 3 A
Among the three quantities, R^j^, v^^, and
if two have been calculated, then the remaining one
follows easily from Equation 6.1. In some cases, the choice o f which two to find first either
increases or decreases the amount o f calculation. The following exercises illustrate this point.
Exercises. 1. For the circuit o f Figure 6.5, R^^= 200 Q, R2 = 50 O., R^= 10 Q,
= 50 V. Find R^f^, i^^ and v^^ in this order.
= 100 V, and v^2
A N SW ERS: 8 Q, 1.5 A, 12 V
2. For the circuit o f Figure 6.5 with the same values as in Exercise 1, find v^^, i^^, and
this order.
AN SW ER: Same as in 1, but v^^^. is harder to find.
3. For the circuit o f Figure 6.5, find the Thevenin equivalent circuit using a series o f source trans­
formations.
The next example illustrates the computation o f the Thevenin and Norton equivalent circuits
using loop analysis.
Chapter 6 • Thevenin, Norton, and M axim um Power Transfer Theorem s
234
E X A M PLE 6 .2 . Find the Thevenin and Norton equivalent circuits seen at the terminals A-B for
the circuit depicted in Figure 6.6, where
= 100 V and
= 3.2 A. We show that
= 4 0 0 Q,
= 200 V, and
= 0.5 A.
1500
500 Q
A
FIGURE 6.6 Two-source circuit for Example 6.2 with loop currents shown;
So l u t io n
Step 1. Compute
= 100 V and i^ 2 = 3.2 A.
To compute R^^^, we set all source values to zero. Each voltage source becomes
a short, and each current source becomes an open. This leads to the circuit o f Figure 6.7. Here, we
have a 500 Q in series with 100 Q, yielding 600 Q. Since this 600 Q resistance is in parallel with
400 £2, the resulting equivalent resistance is 240 Q. Hence,
500 0
= (150 + 240 + 10) = 400 Q..
1500
FIG U RE 6.7 The circuit of Figure 6.6 with all independent sources deactivated.
Step 2. Compute an expression for
A-B is N O T present. Hence,
Because we are computing
the short across the terminals
= 0 and no current flows through the 150 Q resistor. This means
its voltage drop is zero. (One ofi:en says that the 150 Q resistor is dangling.) Thus, from KVL we
have
= '^oozj + ioz;2
(6 .2)
Chapter 6 •Thevenin, Norton, and Maximum Power Transfer Theorems
23^
Step 3. Compute iy The only unknown in Equation 6.2 is /j, since
~ 3-2 A. Hence, around
loop 1,
^s\ =
+ 100/^2= 1000/jand
in which case,
/| = 0.0 0 ly^j + 0.1/^2
Thus, from Equation 6.2,
= 400(0.001
+ 0.1/^2) + lO/^, = 0.4r/^, + 50/^2 =
+ 160 = 200 V
Step 4 . Construct the Thevenin and Norton equivalent ciraiits. Equation 6.3 with
(6.3)
= 400 Q yields
the Thevenin equivalent o f Figure 6.8(a). Further, from the source transformation theorem,
(6.4)
Equation 6.4 leads to the Norton equivalent circuit o f Figure 6.8(b).
400 Q
A
FIGURE 6.8 (a) Thevenin equivalent o f circuit of Figure 6.6;
(b) Norton equivalent o f Figure 6.6.
Step 5. Compute i^^ directly so as to verify the above calailation. This step is merely given to illus­
trate the direct calculation o f i^^ and is unnecessary at this point to the solution o f the problem.
Referring again to Figure 6.6 and assuming that the short acro.ss A-B is present, then /2 = i^^.
Hence, around loop 1,
^S\
= 500/, + 400 (/j - / J
+ 100(/j - /^2)
in which case.
v^\ + 100/^2 = 1000/j - 400/^^ = 420 V
Around loop 2 we have
560/;^-400/, = 10/^2 = 32 V
In matrix form, the pertinent equations are
1000
-4 0 0 '
-4 0 0
560
h ■
Jsc
■420'
32
2M^
Chapter 6 • Thevcnin, Norton, and Maximum Power Transfer Theorems
Thus,
' i\ ■
1000
-4 0 0 '
j.sc
-4 0 0
560
-1
■420‘
■0.62'
32
0.5
A
Consequently, / = 0.5 A as was found earlier using the easier method o f
=
Vo,.
th
Exercises. 1. Suppose all source values in the circuit o f Figure 6.6 are doubled. What is the new
v j Does
change?
A N S W E R : /•, =-)()() V. no
2. Suppose all resistances in the circuit o f Figure 6.6 are multiplied by 4 and the independent cur­
rent source is changed to 0.6 A. Find
and
Hint: For
in equation 6.3, the value “5 0 ”
is in ohms, so if the resistances are multiplied by four, what is the new value?
AN SW ER: r
= 160 V. A',
//» 4 x 400
1600 Q. and /SC
. = 0.1 A
3. A 400 £L resistor is connected in series with terminal A o f the circuit o f Figure 6.6. Find the
V
“ ‘I V
ANSWLR:
im
V, A',,,
=»K) ti. ,nd
- «.2S A
4. A 400 Q resistor is connected across terminals A and B o f the circuit o f Figure 6.6. Find the
V
“" ‘I V
ANSV('-UR:
O.SA, /(•, titJi - 2 0 0 U . a n d r oC- lOl) V
In the above two examples, deactivation o f all independent sources led to a series-parallel network.
Calculation o f
w^as then straightfor\vard. In fact, we can state a corollary to Thevenin and
Nortons theorems.
CO RO LLA RY TO TH EVEN IN AND N ORTO N'S TH EO REM S
FOR PASSIVE N ETW ORKS
When a network contains no independent sources,
= 0 , and the Thevenin or
Norton equivalent consists o f a single resistance R^f^. For a series-parallel net%vork, R^f^ can be
computed by straightforward resistance combinations.
3. A GEN ERAL APPROACH TO FINDING THEVEN IN AND
NORTON EQUIVALENTS
Consider Figure 6.10(a) where we have a network N connected to the remainder o f a larger cir­
cuit. Our goal is to replace the net\vork N by its Thevenin equivalent, as shown in Figure 6.10(b).
Chapter 6 •The\’enin, Norton, and Maximum Power Transfer Theorems
23'
The terminal v-i characteristics o f the network N and its Thevenin equivalent must be the same.
Consider that the v-i characteristic at A-B o f the Thevenin equivalent o f N is
(6.5)
while the Norton equivalent o f N as per Figure 6.10(c) has the v-i relationship
•
'a
- 1
^th
(6 .6 )
- ^sc -
These relationships tell us that if we have a linear net\vork and assume there is a voltage
across
its terminals and a current /^j entering the network, as shown in Figure 6.10(a), then obtaining an
equation o f the form
(6.7)
or o f the form
(6.8)
Vj n —
allows us to match the coefficients o f equations 6.7 and 6.5 to determine
the coefficients o f equations 6.8 and 6.6 to determine ^th = — - and
and
or to march
This sometimes proves
^th
an easier approach for non-simple circuits, as the next two examples illustrate.
Remaining
Network
^
A
------oLinear
Network
A8
- o .......+ —
B
(a)
Remaining
Remaining
Network
(c)
I'lG U R M 6.10 (a) Nervvork N attached to an unknown network;
(b) theTheveinin equivalent of N attached to the unknown nervvork; (c) the
Norton equivalent of N attached to the unknown network.
23.S
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
E X A M PLE 6 .3 . This example revisits Example 6.2 using the new approach. Again, we find the
Thevenin and Norton equivalent circuits seen at the terminals A-B for the circuit depicted in
Figure 6. I I , where
= 100 V and
= 3.2 A. Our goal is to find the v-i characteristic at the ter­
minals A-B.
FIGURE 6.11 Two-source circuit for Example 6.2 with loop currents
shown;
= 100 V and ip = 3.2 A.
So l u t io n
Step 1, Consider i^ loop. Around the loop for i^, we have
^AB ~ 560/^ + 400/j + 10/^2 = 560/^ + 400/j + 32
(6.9)
Step 2. Consider loop 1. From Example 6.2, around loop 1 we have,
= 100/^2 = 420 = 1000/, + 400;;^ V
Thus,
=
42 0 - 400/^
(6 . 10)
1000
Step 3. Substitute. Substituting equation 6.10 into 6.9 yields
4 2 0 - 400/4
+ 32 = 400/.A + 200 V
V.
o = 560/.A + 4 0 0 ----------------^
AU
jQQQ
Step 4. Match coefficients. Matching coefficients o f equations 6.11 and 6.5 implies that
R,h = 4 0 0 Q,\-
= 200 V and /,, = ^
= 0.5 A.
th
(6.11)
Chapter 6 • Thcvenin, Norton, and Maximum Power Transfer Theorems
ly )
EXAM PLE 6.4. For the circuit o f Figure 6. ] 2, find theThevenin equivalent o f the 2-terminaJ Network
N defined by the dashed line box. We show that
= 9.6 V,
= 4.4 Q., and = 2.1818 A.
FIGURE 6 . 12 A current source is is attached to N for computing
and
So l u t io n
Our objective is to compute the relationship o f the form o f equation 6.7 using Nodal analysis and
then match coefficients with equation 6.5 to obtain R^j^ and
Assume /^, = 2 A and
~^
Step 1. Write nodal equations. For writing the equations o f this circuit, the reader might first
review Example 3.2. Alternately, using the inspection method, the matrix nodal equations are
- 0 .2
■0.8
- 0 .4 ■ ■ ^’1 ■
- 0 .2
0.6
-0.1
- 0 .4
-0 .1
0.5
Step 2. Solve equation 6.12 fo r
h\
■ ' ’1 ■
=
\'2
\>2
( 6. 12)
0
js l + U .
^'ab
using Crammer's rule. First, we note that
0.8
det(A/) = det
- 0 .2
0.2
0.6
- 0 .4
-0 .1
-
- 0 .4 '
-
0.1
= 0.:
0.5
From Crammers rule,
■0.8
- 0 .2
det - 0 .2
0 .6
h\
0
- 0 .4
-0 .1
‘s2 + U
^’ab -
V
d et(M )
which from the properties o f determinants becomes
■0.8
- 0 .2
O'
det - 0 .2
0.6
0
- 0 .4
-0 .1
1
d et(M )
=
4 . 4/4
■0.8
- 0 .2
r
■ 0.8
- 0 .2
O'
det - 0 .2
0.6
0
det - 0 .2
0.6
0
- 0 .4
-0 .1
0
- 0 .4
-0 .1
1
. /. -1------"
d et(M )
+ 2 . 6/^1 + 4 .4 /^ 2 = 4.4/,^ + 9 .6
• / . 4- -----
d et(M )
( 6 .1 3 )
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
240
Equation 6.13 shows that
is calculated finding four determinants numerically using MATLAB
or equivalent.
Step 3. Match coefficients o f equations 6.13 and 6.5. Matching coefficients o f equation 6.13 with
equation 6.5, we obtain
V,
= 4.4 a and
2.1818 A
Exercises. 1. If the independent current sources in the circuit o f Figure 6.11 arc set to zero, find
the Thevenin equivalent circuit.
AN.SW'j-^R: The I hcvcnin ci|uiv.ilcnt «.onsist.s oi a single rc.sisn)r.
2. Find
= 4.4 12.
when i^^= 10 A and i^2 = 5 A.
A N S W I- R ; -48 \’
3. A 4.4 d resistor is connected in series with terminal A o f the circuit o f Figure 6.12. Find the
V
V
•ANSW'l-.R:
. •).(, \’.
= 8.8 12. and
=1
A
4. A 4.4 Q resistor is connected across terminals A and B o f the circuit o f Figure 6.12. Find the
ANS\V1;R:
- 2 .1SIS A.
= 2.2 f l and
M .S V
At this point, we end our development in this section with an example that shows how to com­
pute a Thevenin equivalent from measured, e.g., in a laboratory setting where there is a power sup­
ply with an adjustable voltage.
EX A M PLE 6 .5 . Consider Figure 6.13, which show^s the Thevenin equivalent o f an unknow'n netw’ork N attached to a variable voltage,
power supply, which also shows the current delivered
to the unknown network N, i.e.,
Two measurements o f the unknown network N are taken, and
the data is displayed in Table 6.1. Find the Thevenin equivalent o f N.
N
r
R..
Variable
Voltage
Power
Supply
oc
B
MCURI-. 6.13 Thevenin equivalent of an unknown network N
connected to a variable voltage power supply.
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
2/f
TABLE 6.1
/^I (mA)
-
S
(V)
10
24
20
40
o l u t io n
Substituting the measured data in Table 6.1 into equation 6.5
yields
24 = 0.01«,;,+ V
from row 1 o f Table 6.1, and
40 = 0 . 0 2 « ,^ * V
from row 2 o f Table 6.1. In matrix form,
0.01
r
0.02
I
^th
•24‘
40
Solving produces
^r/i
'V>(Hence, R,,, = 1600
■ 1
-1 ■■24‘
r -I ■24‘
= -100
-0.02 0.01 40
40
0.02 1
0.01
—
and
=
1600'
8
= 8 V.
Thus, one can use the technique o f Example 6.5 to determine Thevenin equivalent circuits in the
laboratory.
4. TH EVEN IN AND NORTON EQ U IVALEN T CIRCU ITS FOR
A CTIVE N ETW ORKS
Constructing Thevenin and Norton equivalents for active networks, those containing dependent
sources and op amps, presents us with some unique challenges. Except with one extra condition,
Thevenin and Norton’s theorems and their corollary are valid for active networks. Because active
networks contain dependent sources, the extra condition is that all controlling voltages or currents
be within the 2-terminal network whose Thevenin/Norton equivalent are being sought.
Chapter 6 *Thevenin, Norton, and Maximum Power Transfer Theorems
242
TH EVEN IN AND N ORTO N'S TH EO REM S FOR A CTIVE
N ETW O RKS
For almost every 2-terminal linear network, N, as in Figure 6.14(a), consisting o f resistances,
independent sources, and dependent sources whose controlling voltages and currents are con­
tained within N ‘, there is an equivalent 2-terminal network consisting o f either (i) a resist­
ance,
in series with an independent voltage source,
(Figure 6.14(b)), or (ii) a resistance,
called the Thevenin equivalent
in parallel with an independent current source, /y^(^),
called the N orton equivalent (Figure 6.14(c)). In most cases, both the Thevenin and Norton
equivalent circuits exist. Computation o f y^^is characterized by Figure 6.14(a), computation
of
by Figure 6.14(d), and computation o f b y Figure 6.14(e).
/^Neq
NResistances,
independent
and
+
R..
V
dependent
sources
(a)
(b)
N-
(c)
NResistances,
Independent
sources
deactivated
independent
and
(d)
(e)
-o—
dependent
sources
M GURE 6.14 (a) Arbitrary linear nerwork N; (b) Thevenin equivalent of N;
(c) Norton equivalent of N; (d) N with independent sources deactivated for calculating R^j',
(e) N with short circuited terminals for calculating
As in the previous section, a corollary to Thevenin and Nortons theorems is that if the nerwork
N has no internal independent sources, then the Thevenin and Norton equivalent circuit consists
o f a single resistance /?^yr. However, in contrast to passive networks, R^f^ can be negative. As a first
example illustrating the above theorems, we consider an active nervvork containing no internal
independent sources.
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
EXA M PLE 6.6 . Find the Thevenin equivalent circuit for the 2-terminal network (marked by
dashed line box) in Figure 6.15(a) using the method o f Section 3. (The dependent source acts as
a voltage amplifier.)
B
O(b)
FIGURE 6.15. (a) circuit with terminal voltage i/^^and input current /^;
(b)
/?„, = /?,//■
SO L u n o s
Step 1. Since there are no independent internal sources, the Thevenin equivalent consists o f a single
resistance,
i.e., v^^ = i^^= 0.
Step 2. Write a nodal equation. Writing a single node equation we have
Step 3. Match coefficients with equation 6.6. Matching coefficients implies that
= ((7j + (p +
1)G^) in which case,
^2
R^R
j A2
1
-
{ 1
J_ \
R2+(M + ^)R\
d 4.
R^
We recognize equation 6.14 as the parallel combination o f the resistance
To illustrate a typical calculation, suppose p = 199, /?j = 100
(6.14)
'(A^ + 1)
and _.^2—
and Rj = 4 kf2.
Then
R.u = 500 / /20 = 19.23 a 20 =
iJ + \
Exercises. 1. For the above example, suppose p = 99, R^ = 500 Q, and /?2 = 1
A N SW ER: 10 Q
P*rid R^jy
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
244
2. For the circuit o f Figure 6.16, find the Thevenin equivalent resistance by obtaining
of
in terms
V
'AB-
FIG U RF 6.16 A circuit having no independent sources, in which case
=
0 and
the Thevenin equivalent consists only of a single resistance,
^ih -
('jth
E X A M PLE 6 .7 . Find the Thevenin and Norton equivalent circuits seen at the terminals A-B in
Figure 6.17 when = 50 mA. Our computations will proceed using loop analysis to find the ter­
minal v-i characteristic A-B.
5001
MCiURF 6.17 Arbitrary network for finding Thevenin equivalent.
S
o l u t io n
We first note that \.=
and
= 0.05 A.
Step 1. Write a set o f loop equations for the circuit o f Figure 6.17.
For loop 1, vvc obtain
0 = 1000(/j + i^) + 1000(/, - i^) - m i ^
which simplifies to
- 5 = 2000/, - 1500/;^
For loop A, we have
Chapter 6 * Thevcnin, Norton, and Maximum Power Transfer Theorems
Jn
Step 2. Write the loop equations in matrix form and solve. Writing the loop equations in matrix
form yields
■ 2000
- 1 5 0 0 ' 'i\
-1 0 0 0
■-50'
■
1100
Solving for i^ using for example Crammers rule produces
■ 2000
del
IA=
det
-5 0 '
det
-1000
[2 0 0 0
-1 0 0 0
-1 5 0 0
-= -5 0 -
■ 2000
1•
-1 0 0 0
0
det
AB
700x10-
■ 2000
()■
-1 0 0 0
1
700x10-
1100
50
2
------- 1* AR-----700
700
(6.15)
or equivalently,
(6.16)
‘'/IS = 350/;^ + 25
Step 3. Match coejficietits o f equation 6.15 with equation 6.6 or equation 6.16 with equation 6.5 to
obtain
50
1
1
= — = — A, G,i, = — = —
S. R„, = 350 Q, and v,,,. = 25 V
700
14
700
350
Exercises. I. In Example 6.7, if /j= 5 n-L\, find
I
140
A.
=
350
S.
R^f^, and
Hint: Use proportionalir)^
= 350 a a n d i„ ^ = 2.5 V
2. In Example 6.7, if /^= 5 mA and the 100 Q resistor is replaced by a short circuit, find
and
ANSWKR:
Hint: We have removed the dangling resistor in this case.
= 2.S \'. K,i, = 250 Q.G,,, =
S, and
= - ^ = 0 .0 1 A
3. Find the Norton equivalent at the terminals A-B o f the circuit o f P'igure 6.18 when mA.
AN SW ER:
= 200 LI and
= 0.1 25 A
8 0 0 I.
200 0
--- O
<—
I
<
^ 800 0
800 Q
----- (
B
FIG U R H 6 .1 8 Modification o f the circuit o f Figure 6 .1 7 .
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
2-i6
5. TH EVEN IN AND NORTON EQ UIVALEN T CIRCU ITS FOR OP
AM P CIRCU ITS
Op am circuits arc active circuits. However, because the op amp is a device with special proper­
ties, such as the virtual short circuit in the ideal case and such as output saturation in the non­
ideal case, their discussion warrants special consideration. Our discussion begins with a Thevenin
equivalent o f a non-inverting amplifier with a dangling resistor at the output terminal.
EXA M PLE 6 ,8 . Find the Thevenin equivalent seen at the terminals A-B for the op amp circuit
o f Figure 6.19.
S o lution
Step 1. Find
2 0 -h 5
' ’Cfi Step 2. Find v^g. By inspection.
Step 3. Match coefficients with equation 6.5. Matching coefficients we obsen'e that
R.i. = 16 Q. V’ r =
, and
=
16
Our next example illustrates how to construct a negative resistance using an ideal op amp.
Chapter 6 *Thevenin, Norton, and Maximum Power Transfer Theorems
EXA M PLE 6 .9 . Find the Thevenin equivalent seen at the terminals A-B for the (ideal) op amp
circuit o f Figure 6.20.
lO k O
S o lution
By V-division and the properties o f the op amp,
VcB -
5 ,
Thus, computing /^j we have
.
^’AB - Vc
_
10x10^
lOxlO-"' ■■
Matching coefficients with equation 6.6 we have
-4
S, R„, = - 2 .5 kQ, and VV;,. = 0
10x10^
Exercise. For the circuit o f Figure 6.21, find the Thevenin equivalent circuit at A-B.
AN SW ER:
= 0 and R^,, = -R
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer 1 hcorems
:-i«
Our third example constructs a Thevenin equivalent of the standard inverting op amp configura­
tion with a terminal resistance. However, we will consider both the ideal and non-ideal cases.
EXA M PLE 6 .1 0 . Find the Thevenin equivalent seen at the terminals A-B for the op amp circuit
o f figure 6.22 when
(a) w'hen the op amp is assumed ideal, and
(b)
when the op amp has a saturation voltage,
= 15 V.
S o lution
Step 1. Fiuci the Theveniti equivalent seen at the terminals A-B assuming an ideal op amp. I'h e prop­
erties of an ideal op amp imply that
On the other hand,
with set to zero,
ed into node A. Hence,
x
= 0 for all possible currents,
inject­
-5V^ implies
= 0 and
= -5K^. The
= 0. In flict,
Thevenin equivalent seen at the terminals A-B consists only o f a voltage source o f value
=“5
for the ideal op amp case.
Step 2. Find the Thevenin equivaloit seen at the termitials A-B assuming an op amp ivith output sat­
uration. When the non-ideal op amp operates in its linear region, the Thevenin equivalent by Step
I is a voltage source having value v^^^ = - 5 V^. When, |-5 V} >
= 15 V , or equivalently, when
I
> 3 V, then the op amp saturates at ± 15 V. Specifically, when K > 3 V, then v^^ = - \5V and
when K, < - 3 V, then v^^ = 15 V. The Thevenin equivalent for an op amp with output saturation
is summarized in Figure 6.23, where v^^. takes on three separate values depending on the region o f
operation o f the amplifier.
{
1 5 V fo rV ^ < -3 V
- 5 V f o r |V j< 3 V
1 5 V fo rv / > 3 V
FIGURE 6.23 Thevenin equivalent at output terminals of an inverting amplifier (Figure 6.22)
with non-ideal op amp.
Chapter 6 • Thcvcnin, Norton, and Maximum Power Transfer Theorems
249
This cnd-s our investigation of I hevenin equivalents o f op amp circuits. There arc many more
interesting examples that are beyond the scope o f this text.
7. M AXIM UM POW ER TRANSFER THEOREM
Figure 6.24 shows the Thevenin equivalent o f a network N connected to a variable load desig­
nated Rj. 1'he load voltage, Vj, the load current, ij, and the power,
delivered to the load arc
all functions o f Rj. The main objective o f this section is to show that for fixed R^j^, maximum
power is transferred to the load when R^ = R^f^. We illustrate this assertion with an example that
shows the power delivered to Rj as a function o f Rj. Throughout this section, it is assumed that
all resistances are non-nesative.
FIG URE 6.24 Thcvcnin equivalent o f network N connectcd to a variable load, Rj.
E XA M PLE 6 .1 1 . For the circuit o f Figure 6.24, suppose that R^j^ = 20 Q and
= 20 V, Plot the
power delivered to the load R^ as a function o f
SO L U T IO N
The power delivered to the load Rj is
L
(6.17)
Plugging in the known values yields
I’L = —
^ (2 0 )'
To obtain the plot w'e use the following MA'FLAB code, resulting in the plot o f Figure 6.25.
»voc = 20; Rth = 20;
»RL = 0 :0.25:100;
»PL = RL .* voc^2 ./ ((RL + Rth) .^2);
>>plot(RL,PL)
»grid
Chapter 6 •Thevenin, Norton, and Maximum Power Transfer 1 heorems
250
RL in Ohms
FIG URE 6.25 Plot of power delivered to the load in Figure 6.24 as a function of Rj.
= R^j^ = 20
From the curve, maximum power is transferred at
In a neighborhood o f R^ =
20 ^2, the curve remains fairly flat. At R^ = 40 Q. and Rj = \0 Q, the curve shows that about 88%
o f maximum power is transferred.
This experimentally observed fact, that maximum power transfer occurs when R^ =
plays an
important role when matching speaker “resistances” to the output “resistance” o f a stereo ampli­
fier or when trying to get as much power as possible out o f an antenna and into a receiver.
M AXIM UM POW ER TRAN SFER THEO REM
Let a two-terminal linear network, N, represented by its Thevenin equivalent, as in Figure
6.24, be connected to a variable load, R^. For fixed R^j^, maximum instantaneous power is
transferred to the load when
=«,/,
and the maximum instantaneous power is given by
rL.max
. „
In the dc case, the instantaneous power is a constant for all t.
A verification o f the maximum power transfer theorem proceeds using differential calculus. From
equation 6.17, the power absorbed by the load is
2S1
Chapter 6 • Thevcnin, Norton, and Maximum Power Transfer Theorems
R
Pl =
L____
VV? o c
[^L + ^th)‘
Following the standard procedure o f calculus for determining a maximum/minimum, we com­
pute the derivative o f w i t h respect to Rj, set to zero, and solve for R^.
dPL ^ d
cJRl
(IRl
Rr
2
-9 .
oc
Rl v I c
= y2
/
\3
from which R^ = R^^^ and R^ = ^ are the only possible solutions. But, if R^ = oc, then
= 0.
Hence, because equation 6.17 is positive for /?^ > 0, /?^ = R^j^ produces maximum power,/>^, deliv­
ered to the load.
Further, substituting
= R^f^ into equation 6.1 7 yields
Rfll
P L ,m a x
=
^n/(6.18)
This completes the verification o f the maximum pow'er transfer theorem.
E X A M PLE 6 .1 2 . Consider the circuit o f Figure 6.26a. Find (i) the value o f 7?^ for maximum
power transfer and (ii) the corresponding
■.......N ................................
600 O
<
V -/
300 0
Thevenin
equivalent
(b)
FIGURE 6.26 (a) A network N connected to a load /?/-, (b) Thevenin equivalent of N connected to
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
2^2
SO L U T IO N
■urrent
the independent voltage source becomes a short and the independent cur,
Step 1. To compute
= 200
source becomes an open. Finding the equivalent: resistance seen at the terminals produces
Hence, maximum power is transferred when
Step 2.
= 200
may be computed by any o f the methods discussed throi4ghoi4t this chapter. For example, by
repeated source transformations, the network N reduces to its Thevenin equivalent shown in Figure
6.26(b) with
Plugging
16 V. In fact, this approach would have found R^j^ and
16 V and
at the same time.
= R^^ = 200 Q into equation 6.18 yields
PU
vi
(1 6 )“
4R,th
800
= 320 mW
Exercise. Suppose the 400 Q resistor in Figure 6.26(a) is changed to 100 Q. Find
R^j^, and pj
A N S W l'R S : 24 V. ISO 12, 0.96 wat t s
EXA M PLE 6 .1 3 . This example shows that the Thevenin equivalent cannot be used to calculate
power consumption within the network N it represents. For this demonstration, consider the net­
work N given in Figure 6.27(a) with its Thevenin equivalent given in Figure 6.27(b). Compute
the power loss within the actual network N and within its Thevenin equivalent. We show that
these are different.
•N.
-O1n
1A
20
3V
--Thevenin equivalent -(a)
(b)
FIGURE 6.27 (a) A network N; (b) Thevenin equivalent of (a).
S O L U T IO N
W ithin the network N, the power loss is
;
watts
PN, actual = 2 x 2 “ + 2 x 1 - = 10 w
Within
the power loss is
Pm
- 1 X 1- -
1
Chapter 6 • Thevenin, Norton, and iMaximum Power Transfer Theorems
2^3
This means that theTIicvenin equivalent is not, in general, representative o f pov/er relationships
within the network, i.e., the losses that are dissipated as heat, for example.
When a network N is a voltage source in series with a resistance R^, and hence is its own Thevenin
equivalent, one may ask about maximum power transfer when
fixed, assuming
is variable and the load Rf is
is also fixed. The following example is an experiment for investigating this sit­
uation.
E X A M P L E 6 .1 3 . For the circuit o f Figure 6.28, suppose /?^ = 20 Q and
delivered to the load as a function o f R^ along with the power loss,
FIGURE 6.28 A network N in which R^ can be adjusted with
= 20 V. Plot the power
in R^.
and /?^ fixed.
SO L U T IO N
The power delivered to the load R^ is
R,
(R
l
+ Rs )
0
•vr =
20 X 400
(20+ / ?,)
To obtain the plots, we use the following MATLAB code, resulting in the plot o f Figure 6.29.
«vs = 20; RL = 20;
..Rs = 0:.25:50;
>>PL = RL .* vsA2 ./ ((RL + Rs) .^2);
»plot(Rs,PL)
»grid
»hold
»Ploss = Rs .* vs'^2 ./ ((RL + Rs) .'^2);
»plot(Rs,Ploss,’b’)
Chapter 6 • Thcvcnin, Norton, and Maximum Power 1 ransfer Theorems
25-1
FIGURE 6.29 Plot of power delivered to load as a function of
for circuit o f Figure 6.28.
According to Figure 6.29, the maximum power o f 20 warts is delivered when
if R^< R^ (the usual case), then mi n i m i z i n g m a x i m i z e s
Ploss
= 0. Observe that
However, if R^ > Rj, minimizing
niaximizep^.
The proof for the maximum power transfer theorem given earlier considers Rj as the independ­
ent variable and sets
dRi
to zero, standard practice in calculus. There is, however, an alternate
approach whose derivation is simpler mathematically, but is more meaningful for applications in
the sense that the load can be a general 2-terminal linear network, N^, instead o f a single resistor.
For this alternate derivation, refer to Figure 6.30. We ask the question. What v-i characteristic
should the load network
have so that maximum power is transferredfrom N to N J
FIGURE 6.30 The Thcvcnin equivalent of a network N connected to a loading ncrvvork N^.
Chapter 6 • Thevcnin, Norton, and Maximum Power Transfer Theorems
To find the value o f v, we note that the power
= y i= y l^ - V
R,th
p
To find the value o f v that maximizes
transferred from N lo Nj^ is
we differentiate
with respect to v and set the result to zero:
lIPL
dv
0
^oc R,i,
Solving for v yields
V = 0.5v„
(6.19a)
at which value
(6.19b)
ih
th
which are the conditions on i/and i for maximum power transfer to
resistor
If
consists o f a single
it has the v-i characteristic o f equations 6.19. Then from O hm s law,
^ L - ~ - ^ih
I
At V= 0.5v , the corresponding maximum power is
( 6 .20 )
PLmax = V X /= -
4/?,th
EX A M PLE 6 .1 4 . In the circuit o f Figure 6 .31,
and R^f^ = 2 0.. Find the value o f that
maximizes power transfer to the network N^.
S o lution
According to equation 6.19(a), maximum power transfer occurs when v =
equation 6.19(b), i =
= 0.5 A. Thus,
^th
V= \ = 0 .2 / +
= 0.\ + Vj
=o
Vf = 0 .9
V
= 1 V and from
2S 6
Chapter 6 • Thcvcniii, Norton, and Maximutn Power Transfer'I’hcorcnis
Exercises. 1 If the 0.2 £2 resistor is changed 4 Q, find the value o f
that maximizes power trans­
fer to the network N^.
ANSWHR; -1 V
2. If the 0.2 Q resistor is changed 2 Q, find the value o f Vj that maximizes power transfer to the
network N^.
AN SW ER: 0 V
3. IF the 0.2 Q. resistor is variable and Vj = 0.5 V, find the new' value of the 0.2
resistor that
maximizes power transfer to the Nj.
AN SW ER: 1 Q
8. SUM M ARY
This chapter has set forth a powerful strateg)' for analyzing complex networks by replacing portions o f the nerwork by their simpler Thevenin and Norton equivalents. The Thevenin and
Norton theorems assure us that almost any 2-terminal linear nerwork, no matter the number o f
internal elements, is equivalent to a simple nerwork consisting of an independent source either in
series with or in parallel with a resistance. O f course, an independent current source does not have
a Thevenin equivalent, and an independent voltage source does not have a Norton equivalent.
\4ore generally, there are some circuits that have one but not the other. Further, some circuits have
neither.
'
rhe chapter has illustrated various techniques for constructing the Thevenin and Norton equiv-
^
alents. For passive networks, the ordinary approach is to find
first by deactivating all internal
independent sources. If the resultant circuit is series-parallel, then
can be found by combin-
ing series and parallel resistances as learned in Chapter 2. If the resultant nerwork is not seriesparallel, then one should use the main technique set forth in this chapter, which is to find the vi characteristic o f the terminals. This technique is valid for all circuit t\'pes.
With the ideas o f a Thevenin and Norton equivalent circuit, we then investigated the problem of
transferring power to a load. When R^f^ is fixed, maximum power is transferred when R^ is adjust­
ed to be R^j^. If
is adjustable and R^ is fixed, then maximum power is transferred when R^j^ =
0. It is important to imderstand that a practical dc voltage source (such as a battery in an auto­
mobile) is designed to provide nearly constant output voltage for the intended load current.
Accordingly, it has a rather small source resistance R^. Any attempt to transfer the maximum
power from such a source continuously will overload the source and may cause damage to its
internal structure. For example, in a lead acid battery, the plates may warp or the solution bub­
ble. Hence, maximum power transfer is not o f critical importance for power transmission net­
works, whereas for communication networks, maximum power transfer is important.
^
Chapter 6 •Thcvcnin. Norton, anil iMaximum Power Transfer Theorems
25'
9. TERM S AND C O N CEPTS
2-term inaI network: an interconnection o f circuit elements inside a box having only 2 accessible
terminals for connection to other nervvorks.
D eactivating an independent current source: replacing the source by an open circuit.
D eactivating an independent voltage source: replacing the source by a short circuit.
Equivalent n-term inal networks: t\vo n-terminal networks having the same terminal voltagecurrent relationships. Alternately, two n-terminal networks N j and N-, are equivalent
when substituting one for the other in every possible network N; the voltages and cur­
rents in N are unaffected,
the current through a short circuit placed across the output terminals of a 2-terminal network.
M aximum Power Theorem : let an adjustable load resistor
be connected to the Thevenin
equivalent o f a 2-terminal linear network. Maximum power is absorbed by the resistor
when Rj = R^i^.
N orton’s equivalent circuit: any 2-terminal net\vork consisting o f independent sources and lin­
ear resistive elements is equivalent to an independent current source in parallel with a
resistance.
R(h (Thevenin’s equivalent resistance): the resistance that appears in the 7'hevenin equivalent
circuit o f a 2-terminal linear network. It is also the equivalent resistance of the 2-terminal net\vork w'hen all internal independent sources are deactivated.
Thevenin’s equivalent circuit: any 2-terminal network consisting o f independent sources and lin­
ear resistive elements is equivalent to an independent voltage source in series with a resist­
V
ance.
: the open circuit voltage o f a 2-terminal network N when no load is connected.
' For a generalization o f this condition to the case where the controlling voltage or current is out­
side o f N, see the article by Peter Aronheim entitled “Frequenc)' Domain Methods” in The Circuits
and Filters Handbook, BocaRaton, FI.: C R C Press, 1995, pp. 682-691.
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
258
4. In the circuit o f Figure P6.4,
Prob ems
= 12 V,
=
= 60 Q, /?2 = 60 Q, and R^ = 40 Q.
0.4 A,
(a) Find the Thevenin and Norton equiva­
TH EVEN IN /N O R TO N FOR
PASSIVE CIRCU ITS
lents seen at the terminals A-B.
(b) If a load resistor o f 90 Q is connected to AB, find the power absorbed by this resistor.
1. For the circuit o f Figure P6.1, find R^i^,
and
in terms o f the literals. Hint: Consider
(c) Repeat (b) for a 30 Q resistor. Which resis­
tor, 30 Q. or 90
absorbs die most power?
using C/ = — .
R.
Figure P 6.1
Figure P6.4
2. Find the Thevenin and Norton equivalent
circuits seen at the terminals A-B for the circuit
depicted in Figure P6.2.
5. In the circuit o f Figure P6.5, R^ = 2 kH, Rj
= 8 k n , Ri = 6 kQ,
= 60 V, i^ 2 - ^9 mA and
= 5 mA. Find the Thevenin and Norton
equivalents o f the circuit in the dashed box.
Then find ij and the power absorbed by R^.
5kQ
-o-r
Figure P6.2
-O -v
3. For the circuit o f Figure P6.3, /?, = 3 kH, Rj
= 6 kD,
= 30 V, and /p = 10 mA.
(a) Find the Norton and Thevenin equiva­
lents.
(b) Suppose a variable load resistor
attached
across
M ATLAB
A-B.
Plot
is
using
or equivalent the power
Figure P6.5
C H E C K : 200 V, 10 kQ
6. For the circuit o f Figure P6.6,
R.
^«(l)
'■ 6
«
Figure P 6 .6
Figure P6.3
ANSWF.R: /?;/; =2 kil, isr = 20 mA, // . = 40 V
= 48
V. Find the Norton and Thevenin equivalents.
absorbed by R^ when \00 < R^< 4 kH.
- 6
= 18 kH,
/?2 = 9 kQ, ^3 = 3 kD, R^ = 6 kQ, and
CH ECK:
= 2 mA
Chapter 6 • Thcvcnin, Norton, and Maximum Power Transfer Theorems
2S9
7. Find the Thevcnin equivalent seen at A-B o f
(c) Find the value o f Rj for maximum
the circuit o f Figure P6.7, where R, = 18 kH, R-,
power transfer and the resultant power
= 9 kD, /?3 = 3 kD. /?4 = 6 kD.
= 3.6 kQ.
= 32 k n .
= 48 V, and s2 = 8 mA. Hint: Use
the result o f Problem 6 to find the Thevenin
equivalent of the network between C and D.
delivered to the load.
(d) If the value o f
is doubled, what is the
power delivered to the load under the
condition o f maximum power transfer?
R,
'■ 6
8R
6R
Figure P6.9
C H ECK S: (a) 900 Q, 30 V; (c) 250 mW, (d) 1
Figure P6.7
C H EC K : R.th = 10 kD
10. Find the Thevenin equivalent seen at A-B
o f the circuit o f Figure P6.10. Hint: For this
t)'pe o f problem, the more natural solution
8. (a) Find the Thevenin equivalents for the
technique is source transformations. Why?
circuit o f Figure P6.8 in terms o f the lit­
erals
and i^2 (b) If the A-B is terminated in a 15 kl^ load,
and
= 30 V, and 1^2 =
power delivered to the load.
(c) What is the proper resistance across the
terminals A-B for maximum power
transfer and what is the resultant power
Figure P6.10
delivered to the load?
ANSW'ER:
= 2R,
V()
= — +
V
11. Find the Thevenin equivalent o f the circuit
o f Figure P 6 .ll enclosed in the dashed-line
box. Then compute
and the power absorbed
by the 2 kQ resistor. Hint: W hat resistances are
extraneous to the solution?
Figure P6.8
A N SW FR: R^f, = 10 k ii,
= 204 V
9. Consider the circuit Figure P6.9, in
which v^= 120 V and R = 300 Q.
(a) Find the Thevenin equivalent circuit
to the left o f the terminals A-B.
(b) For Rj^ = 300 Q, 600 iX and 1200 Q.,
find die power absorbed by R^. Does
the use o f a Thevenin equivalent reduce the
effort needed to obtain these answers?
Figure P 6 .11
A N S W E R : 5 2 V. 2 4 k £ X 2 mA
Chapter 6 • Thev enin, Norton, and Maximum Power Transfer Theorems
260
12.(a) Find theThevenin equivalent circuit for
the 2-terminal non-series-parallel net­
work shown in Figure P6.12. Use the
1 kn
2kO
general method.
(b) If a load resistance
is connected to
terminals A-B, use MATLAB to calcu­
late and plot the power absorbed by the
Figure P6.14
load for 1/ = 30 V, and 10 < /?^ < 200 ^2
in 5
steps. At what value o f
is max­
15. (a)
imum power achieved?
= 3 kH for the circuit
Find a so that
o f Figure P6.15.
(b)
100 o
100 Q
Repeat part (a) so that
R^^^ = -1 kQ.
Hint: Do problem 14 first, and then
100 Q
modify the Thevenin resistance appro­
200 Q
priately.
C H EC K : (b) a = 4000 Q
i.
Figure P 6.12
ANSW FR: For (a),
------O
= 1(H) LI
1 kO
1 kO
A
2kn
TH EVEN IN /N O R TO N FOR
A CTIVE CIRCU ITS
13.(a) Find the value o f
so that the
Figure P6.15
Thevenin equivalent resistance o f the
16.(a)
circuit shown in Figure P 6 .13 is 5
(b) Repeat part (a) for the case when
=
- 2 5 0 Q. C H EC K : (b) 6.25 mS
Find a so that R^^^ = 5 kQ for the cir­
cuit of Figure P6.16.
(b)
Repeat part (a) so that
200 n
1 kn
2k O
»800 n
A N SW FR:
= -1 kl^ .
Figure P 6.13
= 1000 uS
Figure P 6 .16
17. For the circuit shown in Figure P6.17, find
the Norton equivalent circuit.
14.(a) Find a so that G,/, = —
S for the circuit
^ ih
o f Figure P6.14.
(b) Repeat part (a) so that G,/, =
200 n
= -0 .0 0 1
A
7/l
18kQ
Figure P 6 .17
ANSWFR: - 6 0 0
/„ = 0
261
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
18, Use loop analysis to compute the Thevenin
equivalent for the circuit shown in Figure
2 1 .(a) Find the Thevenin and Norton equiva­
lent circuits for the network shown in
Figure P6.21, assuming that k = 0,025 S
P6.18, What is the Norton equivalent?
and
= 20 V,
(b) For what value o f k is the open circuit
voltage zero. For this value o f k, deter­
100 0
-----1----- -
loon
mine R^fj.
'300 0
' 0.01
0>
800 n
ANSW I-R;
Figure P 6,18
= 0,
= 250 12
19.(a) Find the Norton and Thevenin equiva­
lents o f the circuit o f Figure P 6 ,19,
(b) If a load resistor
output
is attached across the
terminals,
plot
the
AN SW ER:
Figure P6.21
and
= 60
= 18 V
power
absorbed by the load for 1 <^Rj < 24 Q
usingMATLAB or equivalent. For what
22. For the circuit shown in Figure P6.22, b =
- 0 .0 2 S and a = 25
Find the Norton equiv­
value o f
alent circuit.
does the load absorb maxi­
mum power? Determine the power
delivered to the load at maximum power
transfer.
i.j
50 0
V. <v 50 O
'■©
50 0
bv
Figure P 6.19
A X 'S W I- .R :/'DC=6/<. / w' = U Mf, P,//
Figure P6.22
•W SW I'R; R^,, = 100 12.
= 50/,
20. Find the Thevenin equivalent o f the circuit
23. Consider the circuit shown in Figure
in Figure P6.20 where
P6.23.
(a) Find the Thevenin equivalent.
= 0.2 A.
(b) If a load resistor R^ is connected across
400 Q
—►
'• 0
1 kO
'400 0
200i
terminals A-B, determine R^^ for maxi­
mum power transfer and determine the
maximum power delivered to R^.
(c) If a resistor /?, were added in series with
terminal A o f figure P6.23, what is the
Figure P6.20
AN SW FR: r^. = 60 \‘. A’,;, = SOO LI
Thevenin equivalent resistance o f the
augmented network.
Chapter 6 •Thevenin, Norton, and Maximum Power Transfer Theorems
2(>:
ANSW ER: (a) R,i, = R^
R
'■ 6
Figure P6.23
27. Find the Thevenin and Norton equivalents of
2 4 .(a) Find the Thevenin equivalent for the
network shown in Figure P6.24.
(b) If the values of each source are cut in
the circuit o f Figure P6.27 when
= 20 mA.
AA
half, what is the new
ANSWER: (a) 1.6 kLl - 260 V: (b) /• = - 130 V
400 n
2000i
io o v (^ ^
0
0.1 A
ANSWI-:R:
= 12.S V,
= 650 Q
Figure P6.24
25. Find the Norton equivalent for the circuit
shown in Figure P6.25 when = 30 niA,
=
0.04 S, /?j = 100 Q and
= 400 n .
28. Find the Thevenin equivalent seen at A-B
for the circuit o f Figure P6.28 when /^= 10 mA,
= 24 0 0 Q and /?2 = HOOO Q, R^ = 5600 Q,
R^ = 1500 n , R^ = 1000 Q, and
= 0.25 x
10-^ S.
C H EC K :
Figure P6.25
0.09 A
26. Consider the circuit o f Figure P6.26, where
= 32 V, /?, = 80 Q, /?2 = 240 Q,
and
(a) Replace the circuit to the left o f nodes A
and B with its Thevenin equivalent.
(h) Given your answer to (a), assume that Rf
= 150 n , and find and the power con­
C H EC K : R^,^ = iQ Q.
Figure P6.28
= 8 mA
= 60 Q,
= 2.
sumed by Rj.
CH ECK:
OP AM P PROBLEM S
2 9 .(a) Find the Thevenin equivalent seen at the
terminals A-B for the op amp circuit o f
Figure P6.29.
Chapter 6 • Thevenin, Norton, and Maximum Power Transfer Theorems
(b) What is the value o f a load resistor /?^
263
C H EC K :
= 5000 Q
3 2 .(a) Find
the
attached across the terminals A-B for
maximum power transfer. What is the
power absorbed by this Rj}
Thevenin
and
Norton
Equivalent circuits o f the op amp con­
figuration o f Figure P6.32 seen at A-B.
(b) Repeat (a) for the terminals C-B.
Figure P6.29
C H EC K : /?,/,=/?3
30. (a) Find the Thevenin and Norton equiva­
lents seen at the terminals A-B for the op amp
circuit o f Figure P6.30.
Figure P6.32
AN SW ER: ( b ) = A*,,
=0
3 3 .(a) Find
the
Thevenin
and
Norton
Equivalent circuits o f the op amp con­
figuration o f Figure P6.33 seen at A-B.
(b) Determine the value o f a load resistor R^
connected across the terminals A-B for
maximum power transfer. If
= 4 V
and ryp = 5 V, determine the maximum
power transferred to this R^.
C H E C K : (b)
1 ( /?! + /?2
rA
«i
\
20 kQ
= 0.9 watts
50 kO
‘I
3 1.(a) Find the Thevenin equivalent to the
right o f the terminals A-B for the (ideal)
op amp circuit o f Figure P6.31.
(b) If the practical source indicated in the
figure is attached to A-B, find the current
in terms o f
15 kn
34. Find the Thevenin equivalent seen at the
terminals A-C for the op amp circuit o f Figure
P 6.34 when the op amp has output saturation,
>5 V.
Chapter 6 • Thcvcniii, Norton, and Maximiun Power Transfer Theorems
2 64
N
Linear resistive
network with
dependent sources
and fixed
independent
sources
Power
Supply
Figure P6.36
Tabic P6.36
3 5 .(a) Find the Thevenin equivalent seen at the
terminals A-C for the op amp circuit of
/^l (mA)
Figure P6.35 when the op amp has out­
put saturation,
1
6
4
12
= 12 V.
(b) Find the Thevenin equivalent seen to the
left o f the terminals B-C and the maxi­
C H E C K : ( b )P „ ,,,= 2 m W
mum power that will be absorbed by the
24 k n resistor for all variations in V..
37. Repeat Problem 36 with the data given in
■Rible P6.37.
Table P6.37
r j (mA)
Figure P6.35
C H EC K :
/W =
144
2800
10
54
40
66
C H E C K : (b)
.5625 W
= 6 mW
THEVEN IN AND NORTON
EQUIVALENTS FROM
M EASURED DATA
36. In a laboratory, the data set forth in rows I
and 2 o f Table P6.36 were taken.
(a) Compute the Thevenin and Norton
equivalents o f N.
38. The data listed in Table P6.38 was taken for
the network N o f Figure P6.38.
(a) Fill in the values for the third column o f
Table P6.38 and find the Thevenin and
Norton equivalents o f the linear resistive
nervvork N.
(b) To what resistance should
be changed
to achieve maximum power transfer?
What is P..
(b) After the power supply is removed, what
resistance, Rj, should be connected
Table P6.38
across A-B for maximum power transfer?
(niA )
What i s />„„„?
2
4
10
10
j>
Chapter 6 •lhc\xnin, Norton, and Maximum Power Transfer Theorems
26S
41. This problem is the first of t%vo problems
N
that outline a laboratory measurement proce­
dure for finding the 'Fhevenin equivalent o f a
Linear resistive
network with
dependent sources
and fixed
independent
sources
linear resistive 2-terminal nerwork. For this
problem, consider Figure P6.41 in which the
circuit under test contains no independent
sources. I'he experimental apparatus includes a
C H EC K :
Figure P6.38
= 10.667 mW
resistance decade box, denoted R, a dc volt­
meter with internal resistance
and a signal
generator having known internal resistance, R_.
39. Repeat Problem 38 using the data in Tible
P6.39.
To begin the procedure, one sets R = R^ = 0 and
adjusts the dc level,
o f the signal generator
to obtain a reasonable meter reading, say
Table P6.39
R, (£2)
=
£q, where the subscript “ 1” indicates our first
(niA)
meter reading. (For an analog meter, the read­
ing should be almost full scale.) Leaving the sig­
200
2
>
nal generator set at this value o f V^, increase R
1200
6
>
until the meter reading drops to
Record this value o f R as R^.
^''o-
(a) Suppose R^^^ = x and R^ = 0. Show that
CH ECK:
31.25 m\V
«,/, = «2(b) Now suppose R^^^ = x> and R^ ^ 0. Show
40. The data listed in Table P6.40 was taken for
the network N o f Figure P6.40 with a volt­
meter (VM ) whose internal resistance is 10
M n . Fill in the values for the third column o f
Table 6.3 6 and find theThevenin equivalent o f
the linear resistive network N.
that/^,/^=y?2(c) F'inally, suppose R^^^and R^are nonzero and
finite. Show that
and then solve
^ th
for R.th-
= R j - R^
^m
Table P6.40
(pA)
R, (M Q)
2
0.4
>
10
1
>
N
Linear resistive
network with
dependent sources
and fixed
independent
sources
42. 'Fhis problem is the second of rwo problems
that outline a laboratory measurement procedure
for finding the The\'enin equivalent of a linear
resistive 2-terminal net\vork. For this problem,
consider the new configuration o f Figure P6.42 in
which the circuit under test contains independent
Figure P6.40
=4 V
sources and has a non-zero
'Fhe experimental
apparatus includes a resistance decade box, denot­
ed R, and a dc voltmeter with internal resistance
Chapter 6 • Thcvenin, Norton, and Maximum Power Transfer Theorems
266
All devices are connected in parallel. Because
?= 0, a sign;il generator is not needed, as in
Problem 41. l b begin the procedure, open circuit
AN SW ER: 4 WLl and 20 V
45. The linear circuit shown in Figure P6.45 is
= co,
found experimentally to have the voltage and
and set the scale on the voltmeter to obtain a rea­
current relationship shown. Find its Thevenin
Norton equivalent.
the decade box, or equivalently set /? =
sonable meter reading, say
= £q, where the
subscript “ 1” indicates our first meter reading.
(For an analog meter, the reading should be
almost full scale.) Next, reconnect the decade and
decrease R until the meter reading drops to V^p =
0.5
Record this value o f R as /?2.
(a) Suppose R^^^ = co. Show that R^f^ = R-,.
(b) Suppose R^^^ is nonzero and finite. Show
R„,R
that
= /? 2 ^ind then solve for
R ,l,+ R ,
R^!^. Then show that
1+
Rm/
'0
ANSWT.R: 0.5 Q, 4 A
46. Repeat Problem 45 for the measurement
curve shown in Figure P6,46, Then determine
the value of a load resistor for maximum power
transfer and compute
.
Figure P6.42
43. The Thevenin equivalent o f a linear resistive
network containing no independent sources is to
be found experimentally using the method o f
Problem 41. fhe voltmeter has an internal resist­
ance R^^^= 20 k n . The dc signal generator has an
internal resistance, R^ = 2 kI2. The following
measurements are taken: (i) with R = 0,
is
adjusted until the voltmeter reads 4 V; (ii) keeping
fixed, the decade box is adjusted until the
voltmeter reads 2 V. For this voltage, the decade
box shows R = 6 kf2. Find R^/^.
AN SW FR: 5 kLl
47. The i-v curve o f the network N in Figure
P6.47a is measured in a laboratory, and is
approximated by the straight-line segments
shown in Figure P6.47b. The meter readings
are shown in Table P6.47.
Table P6.47
44. The Thevenin equivalent of a linear resistive
network containing independent sources is to be
found experimentally by the procedure o f
Problem 42. The voltmeter has an input resistance
R^j = 1 M ti. The following me;isurements are
taken: (i) when R is opcn-circuited, the voltmeter
reads 4 V, and (ii) when R is decreased to 800 kl^,
the voltmeter reads 2 V. Find R^j^and
A
0.2 V
0.1 mA
B
0.7 V
10.1 mA
(a) Find the Thevenin equivalent for the
range 0 < i < 0.1 mA.
(b) Find the Thevenin equivalent for the
range 0.1 < / < 10.1 mA.
Chapter 6 ♦ rhcvcnin, Norton, and Maximum Power Transfer Theorems
(c) If R = 500 Q, V^it) = 50 sintdOOO t) mV,
100 mV, find i{t). Hint: Use a
and
dated voltage
267
and the power delivered to
the load.
suitable Thevenin equivalent for N.
(d) If R = 50
and
v^{t) = 200 sintdOOO t) mV,
= 500 mV, find /(/).
R
scale)
'■ 6
Figure P6.50
(a)
Figure P6.47
ANSWl-.R: (a)
= 0.
= 2 kl2; (h)
=
51 (a) For the circuits o f Figure P6.51, find the
load resistance R^ needed for maximum
0.195V, R,/, = 5 o 'h ; (c) 0.04 + 0.02 sin (1000
power transfer, the associated voltage
and the power delivered to the load.
/) mA: (d) .^.05 + 2 sindOOO t) mA
(b) If the load resistance is constrained as 5
k n < R^ < 10 kD, repeat part (a).
M AXIM UM POW ER TRANSFER
(c) If the load resistance is constrained as 15
48. For the circuit o f Figure P6.48, /?, = 160 Q.,
R-, = 480
and
k n < R^ < 20 k n , repeat part (a).
= 80 V. Find the value o f
for maximum power transfer and
e
2 mA
12 kO
8kn
6
24 V
6kfi
Figure P6.48
C H EC K : 7.5 watts
Figure P 6.51
49. For the circuit o f Figure P6.49, R^ = 900 Q,
R j= 180
/?3 = 50 Q,
21 V. Find the value o f
= 60 mA, and v^ 2 =
for maximum power
52. Consider the circuit o f Figure P6.52.
(a) Find the value o f
for maximum
power transfer to the three-resistor load.
transfer and
(b) Find the power delivered to each load
resistor, i.e., to R^, R J2 , and /?^/3.
40 V
Figure P6.49
lon
50. For the circuits o f Figure P6.50,
= 10 V
and v^ 2 - ^5 V. Find the load resistance R^^
needed for maximum power transfer, the asso-
Figure P 6 .5 2
Chapter 6 • I'hevcnin, Norton, and Maximum Power transfer Theorems
268
53. For the circuits o f Figure P6.53, /?j = 200
= 1000 Q, /?3 = 400 n ,
>^ v.,(V)
= 8 mS, and /;, =
0.4 A. Find the load resistance
80 -
i
needed for
<—
maximum power transfer, the associated voltage,
--- 0 +
a
and the power delivered to the load.
— o b
40 -
/
0
(a)
>
1
0.2
1
1
0.4
i(A)
^
r
(b)
Figure P6.56
57. (a)
Figure P6.53
For the circuit o f Figure P6.57 com­
pute, (i) the value o f R which leads to
maximum power transfer to the load,
54. T he circuits o f Figure P6.54 have the load
(ii) the voltage across the load, and
resistor
(iii) the power absorbed by the load.
connected in different ways. For
each circuit, (i) compute the value o f
which
Hint: In MATLAB, the roots o f a
leads to maximum power transfer, (ii) the volt­
quadratic, aQ x- + a, x + a2> are given
age across the load, and (iii) the power
absorbed by the load. Which configuration
absorbs more power?
by “roots([aO al a2])”.
(b)
To verify the results o f (a) write a
MATLAB program to calculate and
plot the power absorbed by the load as
son
+
R varies from 0 Q to 400
15Q
30 V
15
in 2 Q
increments.
^ (b
Load
(a)
R.
30 Q
(b
+ V
15Q
I
.s v Q
30 V
Figure P6.57
(b)
C H EC K : (a)
Figure P6.54
10 watts
55. Suppose the polarity o f the 15-V-source in
Problem 54 is reversed. Repeat Problem 54 and
determine which configuration transfers more
power to the load.
58. The i-v relationship o f certain type o f LED
(light emitting diode) in its operating range o f
1-7 V -3 V is represented by a 2 V voltage-source
in series with a 50 H resistance. The load con­
sists o f a network o f n such diodes connected in
parallel. The source network is represented by a
5 V voltage-source in series with a 50 ^2 resist­
ance. Assume that the power delivered to each
diode is totally converted into light. Determine
56. T he linear resistive circuit o f Figure
P6.56(a) is found experimentally to have the
voltage-current relationship plotted in Figure
P6.56(b). Find the maximum power that can
be absorbed by placing a load resistor across ter­
minals a-b?
how' many LEDs should be connected in paral­
lel for maximum brightness. W hat is power dis­
sipated by each diode?
ANSW ER: M - 5,
= 2S mW
C H A P
Inductors and Caoacitors
CAPACITIVE SM O O TH IN G IN POW ER SUPPLIES
Every non-portable personal computer contains a power supply that converts the sinusoidal volt­
age o f the ordinary household outlet to a regulated dc voltage. “Regulated” means that the output
voltage stays within very tight limits o f its nominal value (e.g., 12 ± 0.1 V) over a wide range o f
power requirements. Engineers design power supply circuits with regulators that produce voltages
with a small oscillation because to generate a truly dc voltage is practically impossible.
REGULATION
RECTIFICATION
O
O
SMOOTHING
This process o f converting ac to dc has three stages: First, the ac waveform is rectified into its
absolute value. Then a smoothing operation takes place that reduces the variation in the voltage
to a reasonable but still unacceptable level. This first level o f smoothing is nccessar)’ becausc the
voltage regulator is a precision subcircuit that requires a fairly constant voltage for its proper oper­
270
Chapter 7 • Inductors and Capacitors
ation. The partially smoothed waveform is fed into a voltage regulator, which limits the voltage
oscillation between critical levels even when the load drawn by any connected device (e.g., your
computer) varies in the course o f its operation.
As mentioned, the rectified sine wave is smoothed before entering the voltage regulator. A crude
smoothing can be accomplished with a capacitor, a device studied in this chapter. Intuitively,
capacitors resist voltage changes and are designed to steady the voltage at a constant level. In this
chapter, we will study the capacitor and investigate a simplified smoothing operation for a power
supply.
CH APTER O BJECTIVES
1.
Define the notion o f inductance and introduce the inductor, whose terminal voltage is
proportional to the time derivative o f the current through it.
2.
3.
Investigate the ability o f an inductor to store energy and the computation o f the equiva­
lent inductance o f series-parallel connections.
Define the notion o f capacitance and introduce the capacitor, whose current is propor­
tional to the time derivative o f its terminal voltage.
4.
Investigate the ability o f a capacitor to store energy and the computation o f the equiva­
lent capacitance o f series-parallel connections.
5.
Define and illustrate the principle o f conservation o f charge.
CHAPTER O U TLIN E
1.
2.
Introduction
The Inductor
3.
The Capacitor
4.
5.
6.
Series and Parallel Inductors and Capacitors
Smoothing Property o f a Capacitor in a Power Supply
Summary
7.
8.
Terms and Concepts
Problems
1. IN TRO D U CTIO N
This chapter introduces two new circuit elements, the linear inductor and the linear capacitor,
hereafter referred to as an inductor and a capacitor. The inductor, shown in Figure 7.3, is a device
whose voltage is proportional to the time rate o f change o f its current with a constant o f propor­
tionality I , called the inductance o f the device, i.e.
as set forth in equation 7.1. The unit o f the inductance Z., is the henry, denoted by H.
Macroscopically, inductance measures the magnitude o f the voltage induced by a change in the
current through the inductor.
Chapter 7 • Inductors and Capacitors
The capacitor, shown in Figure 7.15, is a device whose current is proportional to the time rate o f
change o f its voltage, i.e.,
ic(0= C
dvcO)
(it
as set forth in equation 7.5. Here, the constant o f proportionality, C, is the capacitance o f the
device with unit farad, denoted by F. Capacitance measures the devices ability to produce a cur­
rent from changes in the voltage across it.
By adding the inductor and the capacitor to the previously studied devices (the resistor, inde­
pendent and dependent sources, etc.), one discovers an entire panorama o f possible circuit
responses, to be explored in the next four chapters. Together, these devices allow one to design
radios, transmitters, televisions, stereos, tape decks, and other electronic equipment. In this chap­
ter, our goal is to understand the basic operation o f inductors and capacitors.
2. TH E IN DU CTO R
SomePhysics
In Figure 7.1, a changing current flowing from point A to point B through an ideal conductor
induces a voltage
between points A and B according to Faradays law. Joseph Henr}' inde­
pendently observed the same phenomenon at about 1831. The induced voltage,
to be proportional to the rate o f change o f current, i.e.,
was found
=— .
dt
FIG URE 7.1 A time-varying current flowing through an ideal conductor.
The following experiment illustrates the idea. Suppose the conductor in Figure 7.1 is 6 feet o f #22
copper with resistance 16.5 Q/1,000 ft. The 6-foot length has a resistance o f about 0.1 U. Using
a current generator, we apply a pair of ramp currents (shown in Figure 7.2a) to the conductor, as
per Figure 7.2b. The measured responses are shown in Figure 7.2c and, as expected, satisfy Ohms
law.
Chapter 7 * Inductors and Capacitors
(a)
(b)
(c)
(d)
(e)
FIGURK 7.2 (a) Ramp currcnt inputs to iincoilcd and coilcd wire, (b) Six feet of #22 wire attached
to a current generator, (c) Voltage responses to ramp current inputs of uncoiled wire.
(d) Six feet o f #22 wire coiled into 45 turns 1” long and 1” in diameter.
(e) Voltage responses to ramp current inputs o f coiled wire.
Chapter 7 • Inductors and Capacitors
273
Now suppose the wire is coiled into a qrlinder 1” in diameter and 1” long, as in Figure 7.2d. Apply
the same ramp currents o f Figure 7.2a to the coiled wire. This time, the measured responses are as
shown in Figure 7.2e. These responses have the same shape as those of Figure 7.2c, except for the
offsets o f 30 mV and 60 mV, respectively. These offiet voltages are proportional to the derivatives
of the input currents, i.e.,
Offset -
for k = 1 , 2 , where L is the proportionality constant, called the inductance o f the coil. Since the
derivative o f i s lO'^ A/sec, and the derivative o f
the coil can be computed as
^
3x10
Offset
0.03
= —j:— = — ^
10^
ini^t)
is 2 x lO'^ A/sec, the inductance
0.06
L
of
,
.
hennes
2x10^*
dt
As mentioned earlier, the heniy, equal to 1 volt-sec/amp and abbreviated H, is the unit of induc­
tance. Also, from the above experiment, one concludes that the inductance o f a cylindrical coil o f
wire is much greater than the inductance of a straight piece of wire, which in the above experi­
ment was not measurable by our apparatus.
The physics of the preceding interaction is governed by Maxwell’s equations, which describe the
interaction between electric and magnetic fields. A time-varying current flow through a wire creates a time-varying magnetic field around the wire. The magnetic field in turn sets up a time-varying electric field, i.e., an electric potential or voltage. One can verify the presence o f this magnetic field by bringing a compass close to a wire carrying a current. The magnetic field surrounding
the wire will cause the compass needle to deflect. Physically speaking, a changing current causes a
change in the storage o f energy in the magnetic field surrounding the conductor. The energy trans­
ferred to the magnetic field requires work and, hence, power. Because power is the product of volt­
age and current, it follows that there is an induced voltage between the ends of the conductor.
W hat is even more interesting is that if a second wire is immersed in the changing magnetic field
of the first wire, a voltage will be induced between the ends of the second wire. A proper (mathematical) explanation of this phenomenon is left to a fields course. For our purposes, three fects
are important: (1) energy storage occurs, (2) the induced voltage is proportional to the derivative
o f the current, and (3) the constant of proportionality is called the inductance of the coil and is
denoted by L.
As mentioned, a straight wire has a very small inductance, whereas a cylindrical coil o f the same
length o f wire has a much greater inductance. This inductance can be increased many times over,
possibly several thousand times, simply by putting an iron bar in the center o f a cylindrical coil.
Alas, the calculation o f inductance is the proper subject of more advanced texts, e.g., on field the­
ory or transmission line theory. Nevertheless, there are empirical formulas for estimating the
inductance of a single-layer air-core coil as described in the homework exercises.
Chapter 7 • Inductors and Capacitors
274
BASIC DEFIN ITIO N AND EXAM PLES
D EFIN ITIO N OF TH E LINEAR IN D U CTO R
The linear inductor, symbolized by a coiled wire as shown in Figure 7.3, is a two-terminal
energy storage device whose voltage is proportional to the derivative o f the current passing
through it. The constant o f proportionality, denoted by Z,, has the unit o f H enry (H), equal
to 1 volt-sec/amp. L is said to be the inductance o f the coil. Th e specific voltage-current rela­
tionship o f the linear inductor is given by
(7.1)
dt
i,(t)
h/Y Y V
+ V jt) FIG U RE 7.3 The inductor and its differential voltage-current relationship
as per the passive sign convention.
EXA M PLE 7.1
Compute Vi{t) for the inductor circuit o f Figure 7.4 when ij{t) = e'‘~.
0.5H
FIG URE 7.4 A 0.5 H inductor driven by a current source.
S o lution
From equation 7.1, direct differentiation o f the inductor current /^(/) leads to
y^(t) = 0 . 5 - " '
dt
^ = 0 .5 (-2 f)e-''
V
Exercises. 1. In Example 7.1, suppose ijit) = 0.5sin(20r + 7t/3) A. Compute v^{t).
AN SW FR: 5 cos(20t + tt/3) V.
2. In Example 7.1, suppose ii{t) = (1 AN SW ER: ///(/) = lOOe--”'*'V.
V for / > 0 and 0 otherwise. Find
t> 0 .
The differential equation 7.1 has a dual integral relationship. Safely supposing that at / =
inductor had not yet been manufactured, one can take
= 0, in which case
the
Chapter 7 • Inductors and Capacitors
275
(7.2)
L-fh
The time
represents an initial time tiiat is o f interest or significance, e.g., the rime when a switch
is thrown or a source excitation is activated. The quantity
specifies the initial current flowing through the inductor at ^q. This quantity,
sums up the
entire past history o f the voltage excitation across the inductor. Because o f this, the inductor is said
to have memory.
EXA M PLE 7 .2
For the circuit o f Figure 7.5a, determine /^(O) and ij{t) for / > 0 when Vj{t) =
V as plotted in
Figure 7.5b.
\(t)
ijt)
,(t,
L = 0.5H
(b)
(a)
Inductor Current (A)
FICJURE 7.5 (a) Simple inductor driven by a voltage source, (b) Source waveform Vj{t).
(c) Resulting inductor current
Chapter 7 • Inductors and Capacitors
1~ G
S olution
A direct application o f equation 7.2 leads to
I
T,
= z j l ’.
1
i / o " " ' ' ’’ = z + i ’
It follows that
il^(0) = — = 2 A and //^(0 = — 2 - e '
-
The graph o f ij\t) for all t is given in Figure 7.5c.
Exercises. 1. In Example 7.2, compute an expression for i^{t) for / < 0.
A N S V V K R : i j U ) - 2<-'' A tor t < 0.
2. Repeat Example 7.2 with L = — H and with v,{t) = cos(27ir) V for r > -0 .2 5 sec and zero
I
.
4jt
otherwise.
ANSWl'.R: //(()) = 2 A, /y(/) = 2 + .sinUni) A for r > 0.
E X A M PLE 7 .3
Consider the circuit o f Figure 7.6a with voltage excitation v^{t) shown in Figure 7.6b. Find the
inductor current /^(r) for f > 0, assuming that /^(O) = 0.
i,(t)
,(t )
Q
L = 0.5H
(a)
FIGURE 7.6 (a) Voltage source driving inductor, (h) Square wave excitation !ȣ(/).
S o lution
It is necessary to apply equation 7.2 to each interval, [0, 1], [1, 2], ... , [;;, n + 1], .... For this we
need to first specify the initial conditions for each interval.
Step 1. Compute i^\). From equation 7.2,
Chapter 7 • Inductors atul Capacitors
Step 2. Cotnpute ii{2).
// (2 ) = /^(0) + — f \'i^{T)dT = — X Net Area = 0
L
0
Step 3. Compute the iuitial condition for the interval [n,u + 1] for u even. Again from equation 7.2,
with t = n and n even, we liave
1
Hence i^{n) = 0 for ail even values o f n.
Step 4. Compute the initial condition for the interval [n,n + 1] for n odd. From equation 7.2, with
t = n and n odd, we have, utilizing steps 1 and 3,
I
+“
/I-1
f
j
+j
/j
f
ri-1
j
«
y i(r)d T = — J
v7 (tV /t = 2
n-l
since n - 1 is even.
Step 5. Compute ij^t) over [n,n + 1] ivith n even. If n is even, then the value o f the inductor cur­
rent over the interval [;/,;/ + 1] is
ilSt) = iL{n) + — ^ cIt = i i{ n) + l { t - n) = 2{t - n) A
l^Jn
Observe that i^it) = 2 t - In A is the equation of a straight line having slope +2 and^-intercept -In .
Step 6. Compute i^(t) over [n,n + 1] with n odd. If n is odd, then for the inter\'al [n,n + 1], the
inductor current is
/^(/) = //(/2) + — ( dT = i i i n ) - 2{t - n ) = 2 - 2{t - n) A
[^J It
Here, i^{t) = 2 + 2n - 2t is the equation o f a straight line, with slope - 2 and )'-intercept 2+2«.
Step 7. Piece segfnentsfivm steps 5 and 6 together. Thus the segments computed in steps 5 and 6 inter­
cept the /-axis at the same points. Figure 7.7 sketches the resulting triangular response for /> 0.
FIGURE 7.7 Triangular shape o f inductor current for the square wave voltage excitation of
Figure 7.6b applied to the circuit of Figure 7.6a.
Chapter 7 * Inductors and Capacitors
27H
Exercises. (All time is in seconds.) 1. Again consider the circuit o f Figure 7.6a. Compute iyr (?) for
(i) 0 < r < 1, (ii) 1 < t < 3 , and (iii) 3 <t For the waveform o f Figure 7.8a, assuming i^(0) = 0.
2. Again consider the circuit o f Figure 7.6a. Compute i^(^) for (i) 0 < r < 1, (ii) 1 < ^ < 3, (iii) 3 <
( < 4, and (iv) 4 < /, for the waveform o f Figure 7.8b, assuming i^(0) = 0.
F IG U R U 7.8 Voltage excitations for Exercises 1 and 2.
It is important to recognize that the square wave voltage input o f Figure 7.6b is discontinuous but
the current waveform o f Figure 7.7 is continuous. Integration (computation o f “area”) is a smooth­
ing operation: it smoothes simple discontinuities. This means that the inductor current is a con­
tinuous function o f t, even for discontinuous inductor voltages, provided that the voltages are
bounded. A voltage or current is bounded if the absolute value o f the excitation remains smaller
than some fixed finite constant for all time. Thus, equation 7.2 leads to the continuity property
o f the inductor: if the voltage Vf{t) across an inductor is bounded over the time interval /] < t <
tj, then the current through the inductor is continuous for
< t < tj. In particular, if
then /^(^o”) =
^
The notation
and “+” on /q is used to dis­
tinguish the moments immediately before and after /q- For example, in Figure 7.9, t = 2 shows a
discontinuity o f
The value o f *^^(2“) is 1 and the value o f
is - 1 . The value Vj{2*) can
be seen as the limiting value o f z^^(r) when approaching r -» 2 from the right, whereas Vf{2~) can
be seen as the limiting value o f v^{t) when approaching t
2 from the left.
Chapter 7 • Inductors and Capacitors
279
\ {2 )
/
c
(U
Lu
T3
C
rtJ
01
CT>
TO
*->
o
>
u
■O
c
Time (seconds)
F IG U R E 7.9 A possible discontinuous voltage v^{t) appearing across an inductor of 1 H,
and the resulting continuous inductor current.
PotverandEnergy
Rccall that the instantaneous power absorbed by a devicc is the product o f the voltage across and
the current through the device assuming the passive sign convention. For an inductor,
dt
Plit)=\'L{t)ilU) =
where
w atts.
is in volts, i^{t) in amps, and L in henries.
Since energy (absorbed or delivered) is the integral o f the instantaneous power over a given time
interval, it follows that the net energy
stored' over the inter\'al [/q, /■,] in the magnetic
field around the inductor is
f'o V
(h
/
(7.3)
=
/£(/,)-//^(/q ) joules.
for L in henries and
in amps. From equation 7.3, whenever the current waveform is bounded,
the net energy stored in the inductor over the interval [/q, rj] depends only on the value o f the
inductor current at times r, and /q, i.e., on //(^j) and //(/q)’ respectively. This means that the
stored energy is independent o f the particular current waveform between
and
If the current waveform is periodic, i.e., if ij{t) =
+ T) for some constant 7'> 0, then over any
time interval o f length T, the net stored energ)' in the inductor is zero because
= /^(/-q + 7)
Cliapter 7 • Iiuluctors and Capacitors
280
forces equation 7.3 ro zero. To further illustrate this propert)', consider Figure 7.10a, which shows
a 0.1 H inductor driven by a periodic current /^(^) = sin(27tr) V. This current signal has a funda­
mental period T = I, i.e., the smallest 7 'over which the signal repeats itselh From equation 7.3,
VV^(OJ) =
pi iOdt = I L/7(l)
L/7(0) = 0
However, we can interpret this result in terms o f the waveform o f pi{t). First note that the volt­
age across the inductor in Figure 7.10a is Vjit) = 0.27Tcos(27if) V. Hence, the instantaneous power
is pf{t) =
= 0.2jtcos(2Tt^)sin(27tr) watts, as plotted in Figure 7.10b. Observe the shaded
regions o f Figure 7.10b in which the area under the power curve has equal parts ot positive and
negative area. This means that all the energ)' stored by the inductor over the part o f the cycle o f
positive power is delivered back to the circuit over the portion of the cycle when the power is neg­
ative. Fhis is true for all periodic signals over any period. Because no energ)' is dissipated, and
because energy is only stored and returned to the circuit, the (ideal) inductor is said to be a loss­
less device.
IlW
v jt)
sln(27T) A
0.1 H
= 0.2n cos(2n) V
(a)
P lW
(b)
FIGUllE 7.10 (a) Inductor excited by periodic current, (b) Plot of the power absorbed by the inductor.
It is convenient to define the instantaneous stored energy in an inductor as
=
(7 .4 )
Chapter 7 • Inductors and Capacitors
281
for all t. Equation 7.4 can be viewed as a special case o f equation 7.3 in which r,) = -oo and /^(-oo)
= 0. Thus, equation 7.4 can be interpreted as the change in stored energ)' in the inductor over the
inten'al (^x>, t].
E XA M PLE 7 .4
Find the instantaneous energy stored in each inductor o f the circuit o f Figure 7.11 a for the source
waveform given in Figure 7.1 lb. In Figure 7.1 lb , note that ij^t) = 0 for r < 0.
FIGURE 7.11 (a) Series inductors excitcd by a source current, (b) Graph o f the source current.
S olution
From KCL, i^{t) =
for all t. Since i^{t) = 2r A for 0 < r < 1 and i^{t) = 2 A for /> 1, equa­
tion 7.3 or 7.4 immediately yields the instantaneous stored energies (in J) as plotted in Figure 7.12:
r
0^/<l
1
4r
0:sr<l
1s
> t
> t
(a)
(b)
FIGURE 7.12 (a) Encrg)' (in J) stored in inductor Z.,. (b) Energy (in J) stored in inductor Ly
Chapter 7 • Inductors and Capacitors
282
Exercises. 1. For the circuit o f Figure 7.1 la, find analytic expressions for the instantaneous stored
energ)' for the current excitation in Figure 7.13a for r > 0.
2. Repeat Exercise 1 for Figure 7.13b.
FIG U RF 7.13 Current excitation for Exercise 2.
ANSWUPvS:
1.
0 .2 5 r
UV/) =
{) ^ / < 2
t-
()s/<l
0 .2 5 (9 - 6/+ /■ )
i s / < 3 and U '2(0 =
3s I
0
r
0 s /< 2
4
2s/
4/“
0 s /< 1
(9-fv + r )
I s / <3
0
3</
EXA M PLE 7.5
For the circuit o f Figure 7.14 in which v^{t) = cos(t) V for r > 0 and 0 otherwise, find the input
current i^{t) for r > 0 and the energy stored in each o f the inductors for the intervals [0, t] for 0 <
t < 1 and [0, t] for 1 < t.
i.(t)
L^ = 1H
FIGURE 7.14 Parallel inductive circuit with switch in which v^{t) = cos(t) V for /> 0 and 0 otherwise.
Chapter 7 • Inductors and Capacitors
283
S olution
Step 1. Since no voltage is applied to either inductor for / < 0, /,(0) = 0. Further, no voltage
appears across the second inductor until r > 1. Hence, /^(l) = 0.
Step 2. Equation 7.2 implies that, for 0 < r < 1,
( s ( 0 = /i(/) =
L|
v ,(T )r/ T =
-jL|
/ ^ c o s (t)Jt =
sin(/) A
Step 3. At ^ = 1, the switch closes. T he r\vo inductors are then in parallel, and the source voltage
appears across each. Hence, by equation 7.2,
co s (t
W t = sin( I ) + sin(/) - sin( I ) = sin(/) A
Also, equation 7.2 applied to L-, implies
i2{t)= /2(l)+ Jj^
cos(T)r/T = sin(/) - s i n ( l ) A
From the KCL, the input current ij^t) = /,(/) + ijit) = 2sin(/) - sin(I) A for / > 1.
Step 4. Compute the energy stored in the inductors over the interval [0, t]. From equation 7.3, it fol­
lows that for 0 < r < 1,
t) - 0.5 sin^(r) joules, whereas
->((), t) = 0.
Step 5. Compute the energy stored in the inductors over the interval [0, f] for 1 < t. Again from equa­
tion 7.3, for 1 < t,
sin^(l)] joules.
t) = 0.5 sin^(^) joules and
t) = 0.5[sin~(r) - 2 sin (l) sin(/) +
Exercise. Repeat the calculations o f Example 7.5 for
= 2 sin(r) V for / > 0 and 0 otherwise.
A N SW ERS: For 0 < / < 1. U'} ,(0, f) = [2 - 2 cos(r)]“ J, whereas W) ,(0, t) = 0; for 1 < U'^^,(0.
/) = [2 - 2 cos(/)]- J and U'} ,(0, ;) = [ 1.0806 - 2 cos(/)]- J.
Chapter 7 • Inductors and Capacitors
28-»
3. THE CAPACITO R
DefinitiojisandProperties
D EFIN ITIO N OF T H E C A PA CITO R
Like the inductor, the capacitor, denoted by Figure 7.15a, is an energy storage device.
Physically, one can think o f a capacitor as two metal plates separated by some insulating mate­
rial (called a dielectric) such as air, as illustrated in Figure 7.15b. Placing a voltage across the
plates o f the capacitor will cause positive charge to accumulate on the top plate and an equal
amount o f negative charge on the bottom plate. This generates an electric field between the
plates that stores energy. Hence, for a capacitor.
(7.5)
clt
d\
where q{() is the accumulated charge on the top plate, which is proportional to the voltage
V({t) across the plates; thus q{i) = Cv^t), with proportionality constant C denoting capaci­
tance and having the unit o f Farad (F). One Farad equals 1 amp-sec/volt. The capacitance C
is a measure o f the capacitor’s potential to store energ)' in an electric field.
ic(t) >r
^
V ,(t)
(a)
+ -h
,
-i- + 4-
4- + + -I--1- +
A- + + + + + +.
(b)
FIG URE 7.15 (a) The symbol for the capacitor with conventional voltage and current direc­
tions. (b) Illustration o f electric field between plates of a parallel-plate capacitor.
Modern-day capacitors take on all sorts o f shapes and sizes and materials. In keeping with craditio n , the parallel-platc concept remains the customar)' perspective. Calculating the capacitance o f
t^vo arbitrarily shaped conducting surfaces separated by a dielectric is, in general, ver)- difficult
Fortunately, the ordinary capacitor o f a practical circuit is o f the parallel-plate variety, with the
plates separated by a thin dielectric. The two plates are often rolled into a tubula, ruu„, .,„>1
complete structure is sealed.
EXA M PLE 7.6
For the capacitor circuit o f Figure 7 . 16a, compute i^t) when v j t ) = r5""'sin(1000r) V for / > 0.
28S
Chapter 7 • Inductors and Capacitors
ic(t)
v Jt )
2mF
(a)
Time in milli-seconds
(b)
FIG URE 7.16 (a) A 2 mF capacitor connectcd to a voltage source,
(b) Plots of capacitor voltage and current waveforms.
S o lution
A direct application o f equation 7.5 yields
i^{t) =
sin(1000/) +
dt
Exercises. 1. In Figure 7.16, suppose
=e
(preferably in MATLAB) i^^t) for 0 < r < 0.5 sec.
cos(l()00/) A
V for r > 0. Compute
for r > 0. Sketch
ANS\V1-:R: - 0 . 0 5 , A.
2. Repeat Exercise 1 with
= e~-^^ cos( 100/) V for f > 0 but plot over die time interval [0, 0.15 sec].
A.\’S\V1-:K: -.- -^ q 0 .0 5 COS. KJOr) r 0.2 sin(lOOr)] A.
The differential relationship o f equation 7.5 has the equivalent integral form
' t < '> =
^ f ° J c W d T + I j ; " ic W d T
=
(7.6)
C •'^0
where
is in volts, /^r) is in amps, and C is in farads, and where we have taken
=0
because the capacitor was not manufactured at t = -oo. The time /q represents an initial time o f
interest or significance, e.g., the time when the capacitor is first used in a circuit. The quantit)'
286
Chapter 7 • Inductors and Capacitors
specifies the initial voltage across the capacitor at ^q. This initial voltage,
sums up the entire
past history o f the current excitation into the capacitor. Because o f this, the capacitor, like the
inductor, is said to have memory.
EXA M PLE 7 .7
Suppose a current source with sawtooth waveform
shown in Figure 7.17h, drives a relaxed
0.5 F capacitor (zero initial voltage) as in the circuit o f Figure 7.17a. Compute and plot the volt­
age across the capacitor.
i.(t) A/v^(t)V
(a)
(b)
FIG U RE 7.17 (a) Current source driving a capacitor.
(b) Sawtooth current waveform and voltage response of a 0.5 F capacitor.
S o l u t io n
The input waveform is periodic in that it repeats itself every 2 sec. Therefore, the solution will pro­
ceed on a segment-by-segment basis.
Step 1. Consider the interval 0 < f < 2. For this interval ij^t) = {It - 2) A. With
= 0, it fol­
lows from equation 7.6 that
v^(/) =
(2 t - 2)ilT = 2 ( r - 2/) V for 0 < t <2
Step 2. Consider the interval 2 < t <A. Observe that at / = 2,
= 0; hence, the capacitor volt­
age over the interval 2 < /“< 4 is simply a right-shifted version o f the voltage over the first inter­
val. Right-shifting is achieved by replacing t with t - 2 . In other words,
v^{t) = 2[{t - 2)2 - 2{t - 2)] V for 2 < r < 4
Step 3. Consider the general interval 2k < t < 2{k + 1). For interval 2k < t < 2{k + 1),
v^t) = 2{{t - 2k)^ - 2{t - 2k)], /^= 0 , 1, 2 ,
...
Lastly, obser\^e that the voltage across the capacitor, as illustrated in Figure 7.17b, is continuous
despite the discontinuity o f the capacitor current. Again, this follows because the capacitor volt­
age is the integral (a smoothing operation) o f the capacitor current supplied by the source.
Chapter 7 * Inductors and Capacitors
287
Exercise. Consider the capacitor circuit o f Figure 7.18. Suppose the current source is i^{t) = e~‘ A
for /■> 0 and
= 1 V. Compute the capacitor voltage
the resistor voltage
and the
voltage vj^t) across the current source for r > 0.
-H V„(t) -
+
I (t) = e-'u(t) 0
20
M O
;
.
0.5 F
v,(t)
FIGURE 7.18 Scries RC circuit driven by a currcnt source for accompanying exercise.
AN SW ERS: v^^t) = 3 -
for t> 0,
= lc~' for t > 0. and, by K\'l., v^{t) = 3 V for t > 0.
It is important to emphasize that the sawtooth current input depicted in Figure 7.17b is a dis­
continuous function, but the associated voltage waveform is continuous because integration
(equation 7.6) is a smoothing operation. This means that the capacitor voltage is a continuous
function o f t even for discontinuous capacitor currents, provided they are bounded. This obser­
vation leads to the continuity property o f the capacitor: if the current i(^t) through a capacitor
is bounded over the time interv'al
< ^ < ^2> then the voltage across the capacitor is continuous
for fj < r < tj. In particular, for bounded currents, if fj <
< tj, then
= V(- (tQ"^), even when
At the macroscopic level, there appear to be some exceptions to the continuit}' propert}' o f the
capacitor voltage, e.g., when two charged capacitors or one charged and one uncharged capacitor
are instantaneously connected in parallel. In such cases, KVL takes precedence and will force an
“instantaneous” equality in the capacitor voltages, subject to the principle o f conservation o f
charge, to be discussed shortly. Another example is w'hen capacitors and some independent volt­
age sources form a loop. When any o f the voltage sources has an instantaneous jump, so will the
other capacitor voltages. Upon closer examination, however, we see that there is really no excep­
tion to the stated continuity rule: it can be shown that in all o f the cases where the capacitor volt­
age jumps instantaneously, an “impulse” current flows in the circuit. Physically, an impulse cur­
rent is one that is ver)' large (infinite from an ideal viewpoint) and o f very short duration. The cur­
rent is not bounded, and consequently, the capacitor voltage may jump instantaneously. This
jump does not violate the rule, which presumes that the currcnt is bounded.
Relatio7ishipofChargetoCapacitorVoltageandCurrent
We have defined the capacitance o f a two-terminal device strictly from its terminal voltage-current
relationship— the differential equation 7.5 and the integral equation 7.6, which is now repeated:
v 'c ( 0 = V c(fo) + ^ f ^
C •'M)
Physically speaking, the integral o f i(^t) over [/q, t\ represents the amount o f charge passing
through the top wire in Figure 7.19 over [rQ, r].
288
Chapter 7 • Inductors and Capacitors
ic(t)
/ ''+ + +q + + /
/+ + +
^
/ ++++++
A- + + + + + +y
-q
FIGURE 7.19 Capacitor cxcitcd by a currciu.
Bccausc o f the insulating' material (the dielectric), this charge cannot pass through to the other
plate. Instead, a charge o f +q{t) is stored on the top plate, as shown in Figure 7.19. By KCL, if
i(it) flows into the top plate, then
of
must flow into the bottom plate. This causes a charge
to be deposited on the bottom plate. The positive and negative charges on these two
plates, separated by the dielectric, produce a voltage drop
from the top plate to the bottom
plate. For a linear capacitor, the only t)'pe studied in this text, the value o f V(^t) is proportional
to the charge
The proportionalit)' constant is the capacitance o f the device. Specifically,
qit) = C\U)
(7.7)
where q{t) is in coulombs, Cis in farads, and t^i^) is in volts. Thus, equation 7.6 has the following phys­
ical interpretation: the first term,
is the capacitor voltage at /q; the integral in the second term,
ic(T )d r.
represents the additional charge transferred to the capacitor during the interval [r,j, /]. Dividing
this integral by Cgives the additional voltage attained by the capacitor during [^q, ^]. Therefore,
the sum o f these rwo terms, i.e., equation 7.6, is the voltage o f the capacitor at r. Since q(/) =
it follows direcdy that
=
(7.8)
(It
cl!
ThePj'hicipleofConservationofCharge
It is important in terms of modern trends in circuit applications to further investigate the rela­
tionship o f charge to capacitor voltages and currents. The principle o f conservation o f charge
requires that the total charge tramferred into a junction {or out o f a junctiori) be zero.~ This is a direct
consequence of KCL. To exemplify, consider the junction o f four capacitors shown in Figure 7.20.
Chapter 7 • Inductors and Capacitors
289
v,(t)
+ V ,(t) -
i,(t)
i3(t)
- V3(t) +
v,(t)
+
l4(t)
FIGURE 7.20 Junction of four capacitors.
By KCL
/,(/) + ijit) + i^{t) + i^{t) = 0
Since the integral o f current with respect to time is charge, the integral o f this equation over
( - 00, t] is
/ ^ (m
'2
+'3
M^
=^/1 ) +^/2(0 +qj, (/) +f/4(/) =0
(7.9)
where qj^{t) is the charge transferred to capacitor k. By equation 7.6, at ever}' instant of time,
qi{t) = C-v.{t)
(7.10)
which defines the relationship between transported charge, capacitance, and the voltage across the
capacitor. Hence, from equations 7.9 and 7.10, at every instant of time,
C^v^it) + C2 V2 U) +
+ C^v^{t) = 0
This simple equation relates voltages, capacitances, and charge transport. The following example
provides an application o f these ideas.
EXA M PLE 7 .8
This example shows that under idealized conditions, capacitor voltages can change instanta­
neously. Consider the circuit o f Figure 7.21 , in which £^q(0~) = 1 V and
and V(^{t) for f > 0.
= 0 V. Find
Chapter 7 • Inductors and Capacitors
290
t=0
C l = 1F
C2
Cl
C2 = 1F
FIGURE 7.21 Two parallel capacitors connectcd by a switch.
S o lution
At t = 0“ , the charge stored on C, is C, ^ ^ (0 “ ) and that o f C j is
requires that
For t > 0, KVL
Therefore, after the switch is closed at r = 0, some charge must be
transferred between the capacitors to equalize the voltages. According to the principle o f conser­
vation o f charge, the total charge before and after the transfer is the same. Thus, conservation o f
charge requires that
F,quivalently, <7i(0'^) - <7i(0 ) + q^iO*) - ^2(0 ) = 0- From equation 7.10,
C,
=
Since ^ ^ (0 ) = 1 V, V(^{0 ) = 0, and from K \T
C,
= V(y{0*), it follows that
v - c , ( 0 ^ ) - l l + C 2 [ i ’c i ( 0 ^ ) - 0
Hence, (Cj + C ,)y Q (0 ‘^) = I implies that
0
=
0
= Vq^{0*) = 0.5 V.
Exercises. 1. In Example 7.8, make C, = 0.75 F and C2 = 0.25 F, and compute ^/^(O'^).
AN SW ER: /Y-,(0") = 0.75 V.
2. In Example 7.8, sufipose ^’q ( 0 " ) =10 V and Vqj,{Q~) = - 8 V. Also let C, = 0.75 F and C-, = 0.25
F. Compute
ANSW ER: /.v.,(0^) = 5.5 V.
Example 7.8 is illustrative o f a charge transport that is germane to switched capacitor circuits,
which are o f fundamental importance in the industrial world.
EnergyStorageinaCapacitor
As with all devices, the energy stored or utilized in a capacitor is the integral o f the power absorbed
by the capacitor. The net energ}' entering the capacitor over the interval [/q, /J is
Jff,
Pci'^)dT = f'' Vc(T)/c(T)i/T
Chapter 7 • Inductors and Capacitors
2 ‘)1
ch’ciT )]
= c r '' (
(It
Vcih)
Oo)
dT =
(7.11)
= -C
1
for C in farads,
in volts, and energy in joules (J). From equation 7. 11, the change in energy
stored in the capacitor over the inter\'al [rQ, r j depends only on the values o f the capacitor volt­
ages at times /q and
i.e., on
and v^t^. This means that the change in stored energy is
independent o f the particular voltage v/aveform between
odic, i.e., if V(\t) =
and r,. If the voltage waveform is peri­
+ T) for some r > 0, then over any time interval [t, / + 7], the change in
the stored energy in the capacitor is zero because
+ 7) =
forces equation 7.11
to zero. Analogous to the inductor, for all periodic voltages, the capacitor stores energy and then
returns it to the circuit and is thus called a lossless device.
As with the inductor, it is convenient to define the instantaneous stored energy in a capacitor as
Wc{t) = l^Cvc{t)
(7.12)
which is really the integral o f power over the interval (—x , /], assuming that all voltages and cur­
rents are zero at r = - x .
E X A M PLE 7 .9
Consider the circuit o f Figure 7.22, in which
= 0. It is known that
for f > 0, the source current is i^{t) and the voltage across the capacitor is
V for ^ > 0. Compute (i) the energ)', in joules, stored in the capacitor for
/ > 0, (ii)
and (iii)
FIGURE 7.22 Parallel RC circuit.
S o lution
(i) Since
= 0, from equation 7.11 (or 7.12),
VV^^(0,/) = lc v J ( / ) = 8 C / ?
\-e
RC
(ii) To find the capacitor current, recall
= C ^ ^ ^ = 4—
<li
RC
A
) - 4/? \ - e
Chapter 7 * Inductors and Capacitors
2 ‘)2
(iii) To find
we first compute
flowing fi-om top to bottom:
\ -e
A
Thus
1- e-
+ 4^
E X A M PLE 7 .1 0
= 4 A
/
\
ijt )
— ►
For the circuit o f Figure 7.23a, it is known that the voltage across
the capacitor is
= 20sin(2f + rr/6) V for r > 0. Compute and
plot the instantaneous power absorbed by the capacitor and the
energy stored by the capacitor during the time interval [0, f].
+
N
5mF
k
-
T im e t in seconds
(b)
F IG U R E
7.23 (a) Capacitor with known voltage v^^t) connectcd to a network N.
and the net cncrg)',
t), stored over the interval lO, ^].
(b) Plot of power,
Chapter 7 * Inductors and Capacitors
S o lution
Step 1. Compute
From equation 7.5, for / > 0
,-^(,) = C ^ ^
dt
= 0 .2 co s 2t + -
6}
Step 2. Computep(\t). By direct multiplication and a standard trig identit}',
/
k\
/
k\
{A
watts
P c(t)= V(-(/)/(-(/)= 20 sin 2 / + - X 0. 2 cos 2/ + - = 2 sin 4t + 6j
\
3!
l
6j
Step 3. Compute
t). From equation 7.11 with
= 20 sin(7r/6) = 10 V, we obtain
\V^(0.0 = 0.5 C \’c {t) - 0.5 C v c (0 ) = sin-
2t + - - 0 . 2 5 J
6}
Plots o f pf4f) i^nd V\^^0, t) arc given in Figure 7.23b. Notice that WT^O, /) can be negative,
because W^—oo, 0) = 0.25 joules, meaning that at r = 0, there is an initial stored energy that can
be returned to the circuit at a later time . Figure 7.23b substantiates this.
4. SERIES AND PARALLEL IN D UCTO RS AND CAPACITO RS
Sei'iesInductors
Just as resistors in series combine to form an equivalent resistance, inductors in series combine to
form an equivalent inductance. As it turns out, series inductances combine in the same way as
series resistances.
E X A M PLE 7 .1 1 .
Compute the equivalent inductance o f the series connection o f three inductors illustrated in
Figure 7.24. Then find the voltages
as a fraction o f the applied voltage
Leq
o
+
V ..
+
Leq
Vl2
+
V .,
Q-
o+
^eq
o -
(a)
FICIJRH
Leq
(b)
(a) Scries connection of three inductors, (b) Equivalent inductance.
29-4
Chapter 7 • Inductors and Capacitors
S o lution
First we must answer the question o f what it means to be an equivalent inductance. Earlier, we
defined the inductor in terms o f its terminal voltage-current relationship. Two 2-terminal induc­
tor circuits have the same inductance if each circuit has the same terminal voltage-current rela­
tionship as defined in equation 7.1.
Step 1. The voltage labeled
il
appears across the series connection, and, by KCL, the current
= ‘l y
flows through each o f the inductors, i.e.,
equivalent inductance,
is defined by the relationship
=
in terms o f Z j, L-,, and Ly
O ur goal is to express
Step 2. Find
(7.13)
dt
in terms ofi^^^. To obtain such an expression, observe that, by KVL,
^Leq = ^/.l + ^12 +
Since each inductor satisfies the v-i relationship
dt
it follow's that
^'U’q - (^1 + ^2 + ^ 3 )
dt
Hence, the series inductors o f figure 7.24a can be replaced by a single inductor with inductance
= -^-1 + ^2 +
Finally, since
= Lj
dt
= Lj —^
dt
and
^
_
dt
= { L, +L^ +
~
'
dt
, it follows that
Lj
(L| 4- Zy-> -l- L-^)
^
which is analogous to the voltage divider formula for resistances.
Exercises. 1. If, in Example 7. 11, Z,, = 2 mH,
AN SW ER:
= 8 mH.
2. Find
in terms o f
3
ANSW ER: '■/.:=
S
= 5 mH, and
“
= 1 mH, find
.
295
Chapter 7 • Inductors and Capacitors
Extension o f the formulas in the above example to n inductors is fairly clear, and we state the
results without rigorous proof: the formula for series inductances is
(7.14a)
and the formula for voltage division o f series inductances is
^'IJ =
(7.14b)
L\ + Lo + ... + Lfj
InductorsinParallel
The same basic question as with inductors in series arises with a parallel connection o f inductors:
what is the equivalent inductance? Rather than derive the general formula, let us consider the case
o f three inductors in parallel, as illustrated in Figure 7.25a.
E X A M PLE 7 .1 2
For this example our goal is to show that the equivalent inductance o f the circuit o f Figure 7.25a
is given by the reciprocal o f the sum-of-reciprocals formula,
- ~\
i
T
— +— +•
U
(7.15)
Ly
We then show a formula for current division.
Leq
Leq
O +
L3
L2,
^eq
Leq
L,
L,
o-
o (a)
(b)
FIG URE 7.25 (a) Parallel connection of three inductors, (b) Equivalent inductance.
So l u t io n
Once again, equation 7.13 defines the relationship for the equivalent inductance:
^U-q
The goal is to construct
in terms o f Z ,,
^eq
and
in a way that satisfies equation 7.13. This
will produce equation 7.15.
Step 1. Write KCL for the parallel connection shown in Figure 7.25a. Here, by KCL,
‘U q = 'Z.1 +
+ 'L 3
Chapter 7 • Inductors and Capacitors
2%
DifFerentiating both sides with rcspect co time yields
(Hl\ ^
dt
dt
_
ill
I
dt
di,
Step 2. Find — Ul in terms o f a n d L^,. From equation 7.1, For each inductor
dt
dt
Li
~ ^i\ ~ ^L1 ~ ^L5
Substituting into the result o f step 1 and noting that
di,
dt
( \
Ln
1
n
— +— +—
Li-q
L]
L t^)
L'S
This has the form o f equation 7.13, which implies equation 7.15, i.e.,
Le, =
1
1
1
— + ---- + —
Z/j
L~i L,'^
lb generate a currcnt division formula we first note that
= ^/1 =
1 '
1 '
= — J y i J c i T ) d T = — J \ ' , ^ { T ) d T and
^ _-TT
'
= / v^,^(tV/t
^ _nr.
Thus
(/)
Exercises. 1. If, in Flxample 7.12, Z.j = 2.5 mH, Z., = 5 mFl, and
ANSWq-R:
= 1 mH, find
= 0.625 mH.
The above arguments easily generalize. Suppose there are u inductors, /,,, Z-,, ... , Z.,^, connected in
parallel. Then the equivalent inductance is given by the reciprocal o f the sum-of-reciprocals formula.
(7.16a)
and the current division formula.
Chapter 7 * Inductors aiul Capacitors
U
ii.jn
~i-------- [
(7.16b)
Exercise. For two inductors Zj and L-, in parallel, show chat the equivalent inductancc satisfies the
formula
(7.17)
Series-ParallelCombinations
This subsection examines series-parallel connections o f inductors. This allows us to use the for­
mulas developed above in an iterative way.
EXA M PLE 7 .1 3
Find the equivalent inductance,
o f the circuit o f Figure 7.26.
S o l u t io n
Step 1. In the circuit o f Figure 7.26, several inductors are enclosed by an ellipse. Let
denote
the equivalent inductance o f this combination. Observe that the series inductance o f the 5/6 H
and 0.5 H inductors equals 4/3 FI. This inductance is in parallel with a 1 H and a 4 Fi induc­
tance. Hence,
4
, =
I
1 I
3= 1
—!----- !---1
4
H
4
Step 2. The equivalent circuit at this point is given by Figure 7.27. This figure consists o f a series
combination o f a 1.5 FI and a 0.5 FI inductor connectcd in parallel with a 6 H inductor. It fol­
lows that
29S
Chapter 7 • Inductors and Capacitors
=
1
I
-------------+ 0 .5 + 1.5 6
= -= 1 .5
4
H
Exercise. In Example 7.13, suppose the 5/6 H and 0.5 H inductors are both changed to 0.4 H
inductors. Find L o f the circuit.
AN SW ER: 1.443 H.
CapacitorsinSeines
Capacitors in series have capacitances that combine according to the same formula for combining
resistances or inductances in parallel. Similarly, capacitances in parallel combine in the same way
that resistances or inductances in series combine. This means that the equivalent capacitance o f a
parallel combination o f capacitors is the sum o f the individual capacitances, and the equivalent
capacitance o f a series combination o f capacitances satisfies the reciprocal o f the sum-of-reciprocals rule. These ideas are illustrated in the examples to follow.
EXAM PLE 7 .1 4
Compute the equivalent capacitance,
o f the series connection o f capacitors in Figure 7.28a.
2‘)9
Chapter 7 • Inductors and Capacitors
o-
Cl
'C2
o+
C3
eq
Q-
o
(a)
(b)
FIGURI^ 7.28 (a) Series combination of three capacitors, (b) Equivalent capacitance,
So l u t io n
The equivalent capacitance denoted in Figure 7.28b is defined implicitly by the current-voltage
terminal conditions according to equation 7.5, i.e.,
^
; - r
^
dt
Our goal is to express this same terminal v-i relationship in terms o f the capacitances, C j, C 2, and
Cy After this we set forth a formula for voltage division.
Step 1. Set forth the i-v relationship fo r each capacitor. For each capacitor, k = 1, 2, 3,
‘Ck - Q
But, by KCL, i^ =
dt
Hence,
dt
C,
Step 2. Apply K V L From KVL,
Differentiating this expression with respect to time and using the result o f step 1 yields
dv/^
d v f^ ]
d v 'c '')
d v (--i
(
\
1
1
\ .
— ^ = —— + —— + —— = \— + — + — \ir
dt
dt
dt
dt
l,C|
Cj
c J
3D0
Chapter 7 • Inductors and Capacitors
Step 3. Compute
. From the result o f step 2, solve for Iq to obtain
\
iQ -
[C l
dt
H--------h --
Cl
dt
C^j
It follows that
Q ,=
C,
G
C3
To set forth a formula for voltage division, we first note that
= /q =
1 '
1 '
'
''CA-(0 = — J ick^T)dr = — J ic{T)dT and C,^ Vci 0 = f ic(r)dT
^ -rr-
^ —rri
Thus
''a - ( 0 = — f ic(r)dT = -—
v c(f) = -^------- j------- p v c (0
C,
C2
Exercises. 1. In Example 7.14, suppose Cj = 5 pF, Cj = 20 |.iF, and
C3
= 16 pF. Compute
ANSW ER: 3.2 uE
2. Find
in terms of
.ANSWER: /V;: = 0 .1 6 /y ;
Generalizing the result o f Example 7.14, we may say that capacitors in series satisfy the reciprocal
o f the sum-of-reciprocals rule. Thus, for n capacitors C j, C2, ... , C^, connected in series, the
equivalent capacitance is
C
1
- 1 -------------- r
— + — + ... + —
C,
G
C„
=■
(7.18a)
and the general voltage division formula is
1
'• c ( 0
------- i---------
— + --- +...
C,
Co
C.,
( 7 .1 8 b )
Chapter 7 • Inductors and Capacitors
3 01
Exercise. Show that if two capacitors Cj and C-y are connected in series, then
c
(7.19)
c ,+ c .
CapacitorsinParallel
If rvvo capacitors are connected in parallel as in Figure 7.29a, there results an equivalent capaci­
tance
= Cj + C2 and a simple current division formula to be derived.
I.
a
+
'c
‘Cl
o
+
'C2
eq
a-
a(a)
(b)
FIGURE 7.29 (a) Parallel combination of two capacitors, (b) Equivalent capacitance,
Since the voltage
appears across each capacitor, and since /^= /q +
by KCL it follows that
Hence,
^eq -
^2
One surmises from the above example that, in general, capacitors in parallel have capacitances that
add. And, indeed, this is the case: if there are n capacitors C j, C2, ... ,
in parallel, the equiva­
lent capacitance is
C
= C, + C2 + ... + C„
(7.20a)
Exercise. Show that in the above derivation
Ck .
Cf,
Ceq
.
Q + Q
and that for n capacitors in parallel.
'CA- -
Q
C] + C j +••• + C„
'c
( 7 .2 0 b )
Chapter 7 • Inductors and Capacitors
302
Series-ParallelCombinations
Wc round out our discussion o f capacitance by considering a simple series-parallel interconnection.
EX A M PLE 7 .1 5
Consider the circuit o f Figure 7.30. Compute the equivalent capacitance,
0.45 mF
0.6 mF
FIG U RE 7.30 Series-parallel combination o f capacitors.
S
o l u t io n
Step 1. Combine series capacitances. Observe that the rwo series capacitances o f 0.5 mF and 0.5
mF combine to make a 0.25 mF capacitance.
Step 2. Combine parallel capacitances. First, as a result o f step 1, the three capacitances, 0.3 mF,
0.25 mF, and 0.45 mF, add to an equivalent capacitance o f 1 mF. Further, the two parallel capac­
itances, 0.3 mF and 0.6 mF, at the bottom o f the circuit, add to make a 0.9 mF capacitance. The
new equivalent circuit is shown in Figure 7.31.
a-
II mF
1.125 mF
0.9 mF
a-
FIGURE 7.31 Circuit equivalent to that o f Figure 7.30.
Step 3. Combine series capacitances. From equation 7.18,
^ eq-
\
1
1 1
-------- + - + ----1.125
1 0.9
1
Exercise. Suppose the two 0.5 mF capacitors in Figure 7.30 are changed to 2.5 mF capacitors.
Find the new
A N SW ER: 0.4 mF.
Chapter 7 * Inductors and Capacitors
303
5. SM O O TH IN G PROPERTY OF A CAPACITO R IN A POW ER
SUPPLY
As mentioned in the chapter opener, a power supply converts a sinusoidal input voltage to an
almost constant dc output voltage. Sucii devices are present in televisions, transistor radios, stere­
os, computers, and a host o f other household electronic gadgets. Producing a truly constant dc
voltage from a sinusoidal source is virtually impossible, so engineers design special circuits called
voltage regulators that generate a voltage with only a small variation between set limits for a given
range o f variation in load. The voltage regulator is a precision device whose input must be fairly
smooth for proper operation. A capacitor can provide a crude, inexpensive sm oothing function
that is often sufficient for the task. This section explores the design o f a capacitive smoothing cir­
cuit. In practice, such a circuit is used only for low-power applications.
i.(t)
FIGURE 7.32 Simple power supply with capacitive smoothing for low- power applications.
Consider, for example, the circuit shown in Figure 7.32. The four (ideal) diodes are arranged in a
configuration called a fidl-wave bridge rectifier circuit. An ideal diode allows current to pass only
in the direction o f the arrow. The diode configuration ensures that i^{t) remains positive, regard­
less o f the sign o f the source current. Specifically, the diodes ensure that /j(f) =
Using the
integral relationship (equation 7.6) o f the capacitor voltage and current, it follows that
V cit)= yc(h)) + ^ f
ici-^)ch = VcUo) + ^ f
[M t)|-/o(t)]^/t
(7.21)
Because o f the difference |/j(r)| - i^it) inside the integrand o f the integral, i^{t) tends to increase
the capacitor voltage, whereas i^^{t) tends to decrease the capacitor voltage. Further, because the
diodes are assumed ideal, it follows that
v^t) > I v^{t)
(7.22)
To see this, suppose the opposite were true; i.e., suppose
One o f the diodes would
then have a positive voltage across it in the direction o f the arrow. The diode is said to be forivarcl
biased. But this is impossible, because an ideal diode behaves like a short circuit when forward
biased. The consequence is that V(4,t) will be 12 V whenever |?>'^(^)| is 12 V. This occurs every 1/120
o f a second. Thus, the rectifier output will recharge the capacitor every 1/120 o f a second. Between
charging times, the current, i(){t), will tend to discharge the capacitor and diminish its voltage.
Chapter 7 • Inductors aiui Capacitors
304
The design problem for the capacitive smoothing circuit is to select a value for C that guarantees
that v^t) is sufficiently smooth to ensure proper operation of the voltage regulator. Here, “suffi­
ciently smooth” means that the maximum and minimum voltages differ by less than a prescribed
amount. To be specific, suppose that i>(\t) must remain between 8 V and 12 V. Recall that i^{t)
tends to increase the capacitor voltage, while
tends to decrease it. The design requires select­
ing a value for C to ensure that i^{t) can keep up with
so that the capacitor voltage remains
fairly constant. The value for /(,(^) is obtained from the specification sheet o f the voltage regulator.
Suppose this value is a constant 1 A. It remains to select C so as to ensure that V({t) remains above
8 V between charging times. From equation 7.21, it is necessary that
CJio
Now we need consider only values for t between 0 and 1/120, because the capacitor will recharge
and the process will repeat itself every 1/120 o f a second. Thus, because i^{t) will only increase the
capacitor voltage, to ensure that v^{t) remains above 8 V, it is sufficient to require that
With i^{t) = 1 and
= 12, it follows that
1 A X 120 sec
= 2.083 mF
4 V
A 2,100 |.iF capacitor satisfies this requirement. A method for computing the capacitor voltage
waveform is described in Chapter 22 o f o f 2"^ edition. However, using SPIC E or one o f the other
available circuit simulation programs, one can generate a plot o f the time-varying capacitor volt­
age produced by this circuit, as shown in Figure 7.33. In the figure, it is seen that the capacitor
voltage varies between 12 and 9.02 V, which is smaller than the allowed variation o f (12 - 8) V.
Two factors contribute to this conservative design: (1) we used C = 2,100 uF instead o f the cal­
culated value, C = 2,083 uF, and (2) the increase in the capacitor voltage due the charging current
is is not included in the calculation.
volts
FIGURE 7.33 Time-varying capacitor voltage generated by the circuit in
Figure 7.32 when C = 2,100 |.iK
Chapter 7 • Inductors and Capacitors
The preceding brief introduction made several simplifying assumptions to clarify the basic use o f
a capacitor as a smoothing or filtering device. Practical power supply design is a challenging field.
A complete design would need to consider many other issues, some o f which are the nonzero
resistance o f the source, the non-ideal nature o f the diodes, the current-handling abilit)' o f the
components, protection o f the components from high-voltage transients, and heat-sinking o f the
components.
6. SUM M ARY
This chapter has introduced the notions o f a capacitor and an inductor, each o f which is a lossless
energ}' storage device whose voltage and current satisfy a differential equation. The inductor has a
voltage proportional to the derivative o f the current through it; the constant o f proportionalit}^ is
the inductance L. T he capacitor has a current proportional to the derivative o f the voltage across
it; the constant o f proportionalit}' is the capacitance C. It is interesting to observe that the roles of
voltage and current in the capacitor are the reverse o f their roles in the inductor. Because o f this
reversal, the capacitor and the inductor are said to be dual devices.
That the (ideal) inductor and the (ideal) capacitor are lossless energy storage devices means that
they can store energ)- and deliver it back to the circuit, but they can never dissipate energ)^ as does
a resistor. The inductor stores energy in a surrounding magnetic field, whereas the capacitor stores
energy in an electric field between its conducting surfaces. Unlike energ)' in a resistor, the energy
stored in an inductor over an interval [r^, fj] is dependent only on the inductance L and the val­
ues o f the inductor current //(/^()) and //(/^j). Likewise, the energ)' stored in a capacitor over an
interval [r^, /,] is dependent only on the capacitance C and the values o f the capacitor voltage
and
Both the inductor and the capacitor have memor)'. The inductor has memory because at a partic­
ular time Tq, the inductor current depends on the past histor}' o f the voltage across the inductor.
The capacitor has a voltage at, say, time
that depends on the past current excitation to the capac­
itor. The concept o f memory stems from the fact that the inductor current is proportional to the
integral o f the voltage across the inductor and the capacitor voltage is proportional to the integral
o f the current through the capacitor. This integral relationship gives rise to the important proper­
ties o f the continuity of the inductor current and the continuit)' o f the capacitor voltage under
bounded excitations.
rhe dual capacitor and inductor relationships are set forth in Table 7.1.
Finally, we investigated the smoothing action o f a capacitor in a power supply.
306
Chapter 7 • Inductors and Capacitors
lABLE 7.1. Summary of the Dual Relationships ot the Capacitor and Inductor
ic « )
icit) = C
d\'c )
dt
V[{t) = L
dilit)
dt
/■/(0 =/■/(^o)+ t /^ V;(T)^/T
LJ k>
7. TERM S AND CO N CEPTS
Bounded voltage or current; voltage or current signal whose absolute value remains below some
fixed finite constant for all time.
Capacitance o f a pair o f conductors: a propert)' o f conductors separated by a dielectric that per­
mits the storage o f electrically separated charge when a potential difference exists
between the conductors. Capacitance is measured in stored charge per unit o f potential
difference between the conductors.
Capacitor (linear): a two-terminal device whose current is proportional to the time derivative o f
the voltage across it.
C oil: another name for an inductor.
Conservation-of-charge principle: principle that the total charge transferred into a junction (or
out of a junction) is /.ero.
C ontinuity property o f the capacitor: property such that if the current i(\t) through a capaci­
tor is bounded over the time interval
< t < t-,, then the voltage across the capacitor is
< tj, then t^(^tQ~) =
when
continuous for /, < r < tj. In particular, if fj <
C ontinuity property o f the inductor: propert}' such that if the voltage
is bounded over the time interval r, < r <
continuous for r, < r <
In particular, if r, <
across an inductor
then the current through the inductor is
then /^(/‘o~) =
when
Coulom b: quantit)' o f charge that, in 1 second, passes through any cross section o f a conductor
maintaining a constant 1 A current flow.
Dielectric: an insulating material often used between two conducting surfaces to form a capacitor.
Farad: a me;Lsure o f capacitance in which a charge o f 1 coulomb produces a 1 V potential difference.
Faradays law o f induction: law' asserting that, for a coil of wire sufficiently distant from any mag­
netic material, such as iron, the voltage induced across the coil by a time-varying current
is proportional to the time derivative o f the current; the constant of proportionality,
Chapter 7 • Inductors and Capacitors
307
denoted Z, is die inductance o f the coil. Faradays law is usually stated in terms of flux
and flux linkages, which are discussed in physics texts.
H enry: the unit o f inductance; equal to 1 V-sec/amp.
Inductance: property of a conductor and its local environment (a coil with an air core or iron
core) that relates the time derivative o f a current through the conductor to an induced
voltage across the ends o f the conductor.
Inductor (linear): a two-terminal device whose voltage is proportional to the time derivative of
the current through it.
Instantaneous powen p(t) = */(/)/(<), in watts when v{t) is in volts and i{t) in amps.
Lossless device: device in which energy can only be stored and retrieved and never dissipated.
Lossy device: a device, such as a resistor (with positive R), that dissipates energy as some form o f
heat or as work.
Maxwell’s equations: a set o f mathematical equations governing the properties o f electric and
magnetic Beids and their interaction.
M emory: property o f a device whose voltage or current at a particular time depends on the past
operational history o f the device; e.g., the current through an inductor at time /q depends
on the history o f the voltage excitation across the inductor for t< /q.
Unbounded voltage or current: a voltage or current whose value approaches infinity as it nears
some instant o f time, possibly r = oo.
Voltage r^ u la to r: circuit that produces a voltage having only a small variation between set lim­
its for a given range o f load variation from a fairly smooth input signal.
^The word “stored” emphasizes that the energy in the inductor is not dissipated as heat and can be recovered by
the circuit, whereas the word “absorbed” is used to mean that the energy cannot be returned to the circuit. In a
resistor, energy absorbed is dissipated as heat.
^ More generally, conservation of charge says that the total charge transferred into a Gaussian sur&ce (or out of a
Gaussian surface) is zero.
31)8
Chapter 7 • Induccors and Capacitors
Problems
4. (a) For i^{t) = 10sin(2000r) mA in Figure
P7.4, calculate and sketch
for 0 <
t < \5 ms assuming both inductor cur­
TH E IN D U CTO R AND ITS
PROPERTIES
(b) What is the instantaneous power deliv­
1. If the length o f a single-layer air coil is
(c) Compute and sketch the energy stored in
rents are zero at f = 0.
ered by the dependent source?
greater than or equal to 0.4 times its diameter,
the 2 niH inductor for 0 < f < 15 msec.
then its inductance is approximately given by
the formula
L =
lOv ft)
o f tiirns)~
4 X
18 {dkimeter)+ 40 (lengths)
length o f the coil are in meters. A 2 cm diam­
Figure P7.4
eter coil has 48 turns wound at 12 turns/cm.
Compute the approximate value o f the induc­
tance.
C H EC K : 18 pH < I < 2 0 pH.
2mH
0.2 mH
where L is in henries, and the diameter and
5. For Vsii) sketched in Figure P7.5a, compute
and sketch
for the circuit o f Figure
P7.5b. What is the instantaneous power deliv­
ered by the dependent source?
2. (a) Find and plot for 0 < r < 5 sec the
V (t) (mV)
inductor voltage Vj{t) for the circuit o f
Figure P7.2a driven by the current
2•
waveform of Figure P7.2b.
1
1”
(b) Find and plot the instantaneous stored
energ)^
t
I
(c) Find and plot the stored energy U^(l,r)
1
-1
2
3
4
5
1
6
^
as a function o f time for 5 > / > 1.
-1 -
L(A)
-2 L(t)
©
t(s)
(a)
Figure P7.2
-2
(a)
+
, 0.8 mH
(b)
„(t)
3. Repeat Problem 2 for:
(a)
and
(b) the waveform sketched in Figure P7.3.
6 ijt)
v,{t)
0.6 mH
0.75 mH
1.5 mH
i.JA )
(b)
Figure P7.5
6. (a) Find and plot for 0 < r < 6 sec the induc­
Figure P 7.3
tor current /^(f) for the circuit o f Figure
3 0 ')
Chapter 7 • Inductors and Capacitors
(a) the voltage waveform sketched in Figure
P7.6a driven by the voltage waveform of
P7.8b, and
Figure P7 .6 b.
(b) the voltage waveform sketched in Figure
(b) Find and plot the instantaneous stored
energy.
P7.8c.
(c) Find and plot the stored energy
as a function o f time for 5 > ^ > 1.
(d) Find and plot t/j (/)
i,(t)
vJt)
(a)
(b)
Figure P7.6
7. Repeat Problem 6 for
(a)
u(t), and
(b) the voltage waveform in Figure P7 .7 .
Figure P7.8
9. Consider the circuit in Figure P7.9 in which
Z,
= 0 .2 H , Z2 = 0 .5 H , and
=
100sin(0.257t/) mV for / > 0 and zero other­
wise.
(a) Find the current /■^(/) for r > 0.
(b) Compute the energy stored in each
inductor over the intervals 0 < r < 2 sec
and 2 <
8 . For the circuit in Figure P7.8a, suppose Z, =
0.8 H and L2 = 0.2 H. Compute and plot the
waveforms / j (/) and /^J f ) for
0
310
Chapter 7 • Inductors and Capacitors
+
v,(t)
-
% iQ °
ic(t)
Figure P7.12
Figure P7.9
13. In Figure P7.13a, the capacitors C, = 4 mF
10. The circuit o f Figure P 7 .10 has two induc­
tors, Z., = 20 mH and Lj = 50 mH, in parallel.
The input is v^{t) = 200cos(5007rr) mV for t >
and C j = 12 mF are driven by the voltage
specified in Figure P7.13b. Plot /q(/),
and iA i).
0 and zero otherwise. The switch between the
two inductors moves down at r = 4 ms.
Compute the currents
and
for 0 <
r < 4 ms and 4 ms < /. Also find the energy
stored in each inductor as a function o f t for the
same time intervals.
(a)
Figure P 7.10
THE CAPACITOR AND ITS
PROPERTIES
11.(a) Suppose that a 20 pF capacii
charged to 100 V. Find the charge that
(b)
resides on each plate o f the capacitor.
Figure P7.13
(b) If the same charge (as in part (a)) resides
on a 5 pF capacitor, what is the voltage
14. T he C = 2 pF capacitor o f Figure P 7 .l4 a
across the capacitor?
(c) What is the voltage required to store 50
pC on a 2 pF capacitor?
(d) Find the energy required to charge a 20
has current ;^ r) shown in Figure P 7 .l4 b . If
= 4 V, compute
at ^ = 1, 2, 3, 4 ms.
Now compute the energy stored in the capaci­
tor over the intervals, [0, 2 ms], [2 ms, 3 ms],
pF capacitor to 100 V.
and [0, 4 ms].
12.(a) The C = 2 pF capacitor o f Figure P7.12
has a terminal voltage o f
= 100[1 +
cos(lOOOTtr)] V. Find the current i(^t)
through the capacitor.
(b) Now suppose the voltage is v^^t) =
10sin(2000r) V and Iq = 10cos(2000r)
mA. Find the capacitance C.
31
Chapter 7 • Induaors and Capacitors
15. Suppose
as specified for all time in
Figure P7.15a, excites the circuit of Figure
P7.15b, in which Cj = 0.2 pF and C2 = 0.1 pF.
(a) Plot
for 0 < ^< 8 msec.
and
(b) Compute and plot the energy stored in
the 0.2 pF and 0.1 pF capacitors for
0 < r < 8 . Hint: use MATLAB to plot the
answers.
(c) Find
and
as t
00.
O '
w
O '
0
Figure P7.16
w
(b)
0
Figure P7.15
0
16. For the circuit in Figure P7.16a, C = 0.25
mF. Compute and plot the waveforms of the
0
voltage,
given
as sketched in Figures
P7.16B and c.
17.(a) Consider the circuit sketched in Figure
P 7.17 in which Cj = 20 pF and C2 = 0.1
mF. Suppose v^{t) = 5sin(2000f) V for t
> 0 and suppose
= 10 V. Find
for r > 0. Is the output voltage
independent of the initial voltage on Q ?
Why?
0
(b) W hat is the instantaneous power deliv­
ered by the dependent source?
w
(c) Find the energy stored in Cj over the
interval [0 , t].
0
i jt )
(a)
0
0
0
Figure P 7 .1 7
312
Chapter 7 • Inductors and Capacitors
18. Repeat Problem 17 when v^{t) =
V for / > 0 and 0 otherwise.
19. Reconsider the circuit o f Figure P 7 .17.
Suppose, however, that
is given by the plot
in Figure P 7 .19.
(a) Find and sketch
Figure P7.21
for 0 < / < 6
msec.
(b) What is the instantaneous power deliv­
22. In the circuit o f Figure P7.22 v^{t) = 25 V
and the
ered by the dependent source?
(c) Compute and plot the energ)' stored in
C,.
= 100 mF capacitor is uncharged at
f = 0 . Compute V({t) for 0 < r < 2 sec and 2 sec
< t when C, = 4 0 0 mF.
t = 2s
Figure P7.22
23 . Fhe circuit ol Figure P7.23 has two capac­
itors in parallel, C| = 30 mF, C-, = 50 mF. Fhe
20. Repeat Problem 17 for the waveform of
input current is i^i) = 360f’~'®^^ mA for r > 0
Figure P7.20.
and 0 otherwise. Suppose each capacitor is
imcharged at /= 0 . The switch between the t%vo
capacitors opens at r = 2 msec.
(a) Find the voltage,
for 0 < r < 2 ms
and 2 ms < t.
(b) Compute the energy stored in each
capacitor as a function o f t for the same
time intervals.
Figure P7.20
(c) Com pute the current through each
capacitor over each time interval.
21. For the circuit o f Figure P7.21, suppose C,
= 0.6 ml', Ct = 1.2 mF,
t = 2 ms
= 0.4 mF, Q = 1.6
mF, ijj) = 120sin(100r) niA for / > 0 and 0 for
r < 0.
(a) Find the equivalent capacitances
the series combination and
for
for the
parallel combination.
Figure P7.23
(b) Find and sketch
(c) What is the instantaneous power deliv­
ered by the dependent source?
(d) What is the instantaneous energ)' stored
in C4?
M IX ED C A PA CITO R AN D
IN D U C TO R PRO BLEM S
24. Consider the circuit o f Figure P7.24,
=
2.5 H, Cj = 1 mF, which is excited by the cur-
313
Chapter 7 • Inductors and Capacitors
rent waveform
= 200fr“ ’^' mA for ^ > 0
27. For the circuit o f Figure P7.27,
as a function o f i^{t) and
(a) Compute
and 0 otherwise.
(a) Compute and sketch I'lit),
the capacitances Cj and C j.
and
in terms o f i^{t) and the
(b) Now find
circuit parameter values.
(b) Compute and sketch the energy stored
in the inductor for r > 0.
(c) Compute and sketch the energy stored
L.
in the capacitor for t> 0.
+
Figure P7.27
28. For the circuit o f Figure P7.28, compute
and
Figure P7.24
as a function of
and the circuit parameters.
25. In the circuit o f P'igure P7.25, suppose Z., =
0.25 H, Cj = 2.5 mF, ij^t) = 20sin(400/‘) mA for
r > 0 and 0 otherwise. Ail initial conditions are
zero at r = 0.
(a) Find Vjit).
+
“" ’0
c:
\|>
-5
L ,2
L,
(b) Find V(^t).
(c) Find the instantaneous stored cnerg)' in
the capacitor.
Figure P7.28
SERIES-PA RALLEL IN D U C TO R S
29. In the circuit o f Figure P7.29, all inductors
are initially relaxed at /^= 0 and /.j = 6 mH, L-,
Figure P7.25
26.(a) In the circuit o f Figure P7.26, (i = 10, C,
= 38.5 mH,
= 22 mH. A voltage
200re~' mV is applied for r > 0. Find,
Vjj(t), and
Challenge: Find
= 20 pF, C , = 80 mF, Z., = Z , = 20 mH
are
initially
uncharged.
If
vj^t) =
10/sin(20^) V for r > 0 and 0 otherwise,
(a) Find i^{t) for r > 0.
(b) Now find
for / > 0.
(c) Compute the energ)- stored in the 20
mH inductor for r > 0.
i.(t)
C H E C K S: 20 mH and \AOte~' mV.
Figure P 7 .2 6
=
314
Chapter 7 • Inductors and Capacitors
30. Consider the circuit o f Figure P7.30.
Suppose Z., = 3 mH, L j= \2 mH,
(mA)
= 36
= 120cos(1000r) mA.
mH, and
(a) Find L
and
(b) Find
(c) Plot the instantaneous power deliv­
ered by the source for 0 < r < 14
msec.
/
Y
Y
V
L.
-200
--
Figure P7.32
(!)
33. In Figure P7.33, Z-j = 5 mH, L-, = 20 mH,
= 20 mH,
= 80 mH, and ij^t) =
lOsin(lOOOr) mA for r> 0 and 0 otherwise.
(a) With the switch in position C, find the
Figure P7.30
C H EC K :
12 mH, - 1 .4 4 sin(lOOOr) V, 90
equivalent inductance,
Vj^, and
(b) Repeat part (a) with the switch in posi­
cos(lOOOr) mA.
tion D.
31. For the circuit o f Figure P7.31, Z., = 260
mH,
= 26 mH, L-^ = 39 mH, and
=
10^"^“ tiiA.
(a) Find
(b) Compute
(c) Com pute
stored in
and i[^2 and
the
instantaneous
energy
as a function o f t.
C H EC K S: 52 mH. 0.2
0.8 i j t ) ,
=
-0.104^6’" '“ V, Vj2 i^) =
34. Consider the circuit o f Figure P7.34a with
voltage source excitation given in Figure P7.34b.
Let the inductor values be those given in
Problem 33. Suppose the switch is in position C.
Note that each inductor is relaxed at r = 0.
(a) Find Z^^^.
(b) Compute and sketch
for r > 0.
(c) Find the instantaneous (total) energ)'
stored in the set o f four inductors as a
function o f time.
Figure P7.31
32. Repeat Problem 31 for the waveform of
Figure P7.32.
(d) Compute and sketch //2(0CH ECK: L
= 20 mH; /.„(1) = /.„(3) = 0.8 A
while /y,/2) =
=0
31
Chapter 7 * Inductors and Capacitors
ijt )
____rvY V
v X
L.
L,
1-6
L,
i„(t)
f r r \ ____ T Y Y \
(a)
i-N/YYA___ TYYV
v„(t)(V)
>k
16'
0
1
2
3
4
16-
L.
(b)
rOA___ TYYV
Figure P7.34
35. Find
L,
(b)
for the circuit o f Figure P7.35, (a)
when the s\vitch is open, and (b) when the
switch is closed. The unit o f L is henries.
Figure P7.36
SCRA M BLED ANSWERS: 0.1, 0.08, 0.6 (in 11)
36. Find
for each o f the circuits in Figure
P 7.36, where Z., = 5 niH,
= 20 mH,
niH, I 4 = 150 mH, Z.5 = “50 mH,
mH, L j = 120 mH,
= 40
= 180
= 35 mH. Notice that
the circuit o f (b) is a modification o f (a) and
that o f (c) is a modification o f (b). Connections
can create interesting behaviors.
37 . Three 60 mH inductors are available for
interconnection. List all equivalent inductances
obtainable over all possible interconnections o f
these elements.
C H EC K : There should be seven different val­
ues.
316
Chapter 7 • Inductors and Capacitors
38. Find L
P7.38.
for each o f the circuits in Figure
/
4 mH
Y
Y
AN.SWER: C//I
for
. all / values.
40. Like Problem 39, this is a conceptual prob­
lem and requires no calculations for the answer.
Y
lOmH
1 mH
I
Consider circuits 1 and 2 o f Figure P7.40. All
5 mH'
wish to determine the relationship bet\veen
inductors are 1 H except the one labeled L. We
3mH'
36 mH
and L^^-, in the presence o f the finite posi­
Bo(a)
tive inductor o f L henries between points a and
b. W hich of the following statements is true?
/YYV
7mH
/
Y
Y
V
/
2.4 mH
Y
Y
(h)
V
< /v,2-
1.2 mH
(d) There is no general relationship between
^eq\
0.6 mH
relationship depends
on the value o f L
Explain your reasoning.
(b)
Figure P7.38
ANSWHR: (a) 13 m il: (b) 2 in 11
39. This is a conceptual problem
and requires no calculations for
the answer. Consider circuits 1
and 2 o f Figure P7.39. All induc­
tors are 1 H except the one labeled
L. We wish to determine the rela­
tionship between
and
presence
o f the finite positive inductor o f L henries
between points a and b. Which of the following
statements is true?
^eq\ ^ ^eql(b)
< K ,l(d) There is no general relationship between
and
Any relationship depends
on the value o f L.
Explain your reasoning.
Circuit 1
Circuit 2
Figure P7.40
ANSW ER: /.
SERIES-PARALLEL CAPACITORS
41. (a) Find the indicated equivalent capacitance
for the circuit o f Figure P 7.4la where C,
= 4 pF, C 2 = 3 pF, C3 = 2 pF, Q = 4 pE
Then find vj^t) when ii^t) = lOcos(lO'^r)
mA for r > 0 and 0 otherwise.
(b) Repeat for Figure P 7 .4 lb in which C, =
60 pF, C , = 18 pF, C 3 = 18 pF,
Q = 36 pF, and C 5 = 10.8 pF.
Then find vj^t) when i^{t) =
1 Osin( 1O^f) mA for r > 0 and 0
otherwise.
31'
Chapter 7 • Induaors and Capacitors
-C,-
Vw/'
C. c,
<!>
^ 1
Figure P7.43
44. For the circuit of Figure P 7.44, Cj = 8 mF,
C2 = 6 mF, C3 = 12 mF, and
=
240sin(200r) mA for ^ > 0 and 0 otherwise.
(a) Find C,^.
(b) Find v- {i). Note: All capacitors are ini­
(b)
tially uncharged. Why?
Figure P7.41
CHECKS: 6 pF, 66 ^F
If
if if
if
c,
c.
+
4 2 .(a) Find the indicated equivalent capaci­
tance for the circuit o f Figure P7.42a
assuming Cj = 48 pF, Cj = 16 |jF,
c,
c
=
20 )jF, Q = 80 pF, and C5 = 8 pF.
(b) Repeat for Figure P7.42b assuming Cj =
Figure P7.44
3 pF, C2 = 6 pF, C3 = 3.6 pF, Q = 6 pF,
C5 = 4.5 pF, Q = 48 pF,
= 48 pF, Cg
= 24 pF, Cg = 24 pF.
4 5 . Three 12 pF capacitors are available for
interconnection. List all equivalent capaci­
tances obtainable over all possible interconnec­
tions of these capacitors.
O ’
oc
C -L
c.
C,
(a)
4 6 . This is a conceptual problem and requires
no calculations for the answer. Consider cir­
cuits 1 and 2 o f Figure P7.46. All capacitors are
1 F except the one that is labeled C. We wish to
determine the relationship between S ' and
^eql
presence o f the finite positive C F
capacitor between points a and b. Which of the
following statements is true?
(b)
Figure P7.42
^ ^eq2 '
(W
^eq\ - ^eql43. Find
for the circuit of Figure P 7.43, (a)
when the switch is open, and (b) when the
(d) There is no general relationship between
and
Any relationship depends
switch is closed, assuming that Cj = C4 = 12 pF,
on the value of C.
C2 = C5 = 40 pF, C3 = Cg = 2 0 pF.
(c) Repeat parts (a) and (b) for Cj = 12 pF, Cj
Explain your reasoning.
= 40 pF, C3 = 20 pF, C4 = 4 0 pF, C5 = 20 pF,
Cg = 100 pF.
318
Chapter 7 • Inductors and Capacitors
ages at f = 0 “ are zero, find all three voltages,
I’Cjit), i = 1,2,3, for r > 0. Then find the instan­
taneous stored energy at / = 0.05 sec.
v jt)
c,
©
V,
Figure P7.48
C H EC K ;
= 6(1 -
V
49. In the circuit o f Figure P7.49, suppose Cj
= 3 pF, C3 = 0.5 pF, C2 = 1.5 pF, and
=
10sin(400r) mV for f > 0 and 0 otherwise.
(a) Find v^^{t) and v^^it) for ^ > 0.
(b) Find an expression for the energy stored
in C] and C , over the interval [0 , /].
Circuit 2
Figure P7.46
ANSWMR: C ;„ <
v jt)
©
47. In Figures P7.47a and b, the charge on Cp
and C3 is Q = 72 X
C. hi Figure
P7.47a, the voltages on C ,, C , and
Figure P7.49
are 2 V,
3 V, and 4 V, respectively; and in Figure 7.47b
Vq = A \ while the charges on Cp C , and
50. In the circuit o f Figure P7.50, suppose C,
are Qj = 48 x 10“^ C, Q 2 = ^0 ^ ^0“^ C, and
Q 3 = 72 X
C, respectively.
= 5 mF, Cj = 20 mF,
(a) Find
for the circuit o f Figure P7.47a.
(b) Find
for the circuit o f Figure P7.47b.
= 4 mF, Q = 80 mF,
= 10 0 e~^^ V for f > 0 and 0 otherwise.
(a) Find
and
for r > 0 .
(b) Compute the energy stored in the C-, over
the interval [0 , t].
/
(b)
(a)
ANSWHK: (a)
Figure P7.47
= 8 mF
48. In the circuit o f Figure P7.48, C| = 6 mF,
C2 = 12 mF, C3 = 36 mF, v J t ) = 20(1 V for t> 0 and 0 otherwise. If all capacitor volt-
v,„(t)
Figure P7.50
51 . In the circuit o f Figure P7.51, suppose Cj
= 4 mF, C-) = 80 mF, C3 = 20 mF, and
1OO^*”^^ mA for t > 0 and 0 otherwise.
=
(a) Find /q(^) and
for t > i).
(c) Compute the energy stored in C-, over
the interval [0 , t].
31')
Chapter 7 • Inductors and Capacitors
the green left turn signal. The interesting variation
is that the v-i inductor relationship is different for
time-var\'ing inductances:
WYiLit)
v^(/) = -
at
The following highly simplified circuit illus­
Figure P7.51
trates the principle o f operation, although the
M ISCELLAN EO U S
52. Find and sketch
configuration and values may not be what are
for 0 < r < 4 sec for
actually used. Consider the circuit o f Figure
the circuit o f Figure P7.52a, assuming all
P7.54a consisting o f an inductor driven by a
capacitors are initially at rest for the excitation
current source. When the car with its steel
o f Figure P7.52b. Are any o f the capacitors
frame moves over the coil o f wire, the induc­
redundant as far as
tance o f the coil changes from I , to some larg-
is concerned?
er v^alue
0.5 F
0.2
+
pkv,
y+\
__
^ . 4 f^ . 2
-_ y
f
3L^ as illustrated in Figure
P7.54b, where the time
depends on the
speed at which the car is slowing down and
i/y
0.4 F
1-^4
2 T f)
(a)
Plot v^it) for t >0 assuming ij{t) =
V ,(t)
a con­
stant value, and that the front edge o f the car
begins to cross the first edge o f the coil at t = 0 .
Explain how this voltage signal might be u.sed
to control the traffic light.
6
Figure P7.52
0.5H
(a)
C H EC K : v,„,,(/) = :^ v ,(/ )
53. Using the circuit given in Figure 7.32,
select a capacitor value to filter the voltage
for a regulator requiring 14 V <
Use
< 20 V.
= 20 cos(2007tf) and /^(f) = 2 A.
C H E C K : C > 1.667 mR
54. When driving a car into a left-hand turn
lane, one often sees a large circular or hexago­
nal cut in the concrete. Embedded in these
cuts is a coil o f wire. When your car (contain­
ing a large percentage o f iron) passes over this
coil, its inductance changes. This change o f
Time (in sec)
inductance can be used as a sensor to activate a
(b)
circuit that stops oncoming traffic and lights
Figure P7.54
L(t)
C
H
A
P
T
E
R
First Order RL and RC Circuits
When watching a manufocturing process, a visitor
might see a pair of robotic arms assemble an engine or
machine a block o f metal with perfectly timed
maneuvers. Timing is a critical aspect of a manufac­
turing process. In T V transmitters there is a signal
called the raster, which is critical to the generation of
the screen image. In an oscilloscope a timing signal
called a horizontal sweep acts as a time base, which
allows one to view measured input signals as a func­
tion o f time. All these applications utilize a signal hav­
ing sawtooth shape and called a linear voltage sweep.
The linear voltage sweep is nicknamed the sawtooth,
rhis sawtooth is pictured here together with an
approximating exponential curv'e for comparison.
Linear Sweep or Sawtooth Waveform
Exponential Approximation
Ideally, the sawtooth voltage increases linearly with time until reaching a threshold where it imme­
diately drops to zero, which reinitiates the process. The threshold voltage corresponds to a fixed
unit o f time. The linear voltage increase then acts as an electronic second hand, ticking o ff the
322
Chapter 8 • First Order RL and RC Circuits
smaller units o f time. In practice, the linear increase in voltage is approximated by the “linear” part
o f an exponential response o f an RC circuit. ^X1^en the voltage across the capacitor reaches a cer­
tain threshold, an electronic switch changes the equivalent circuit seen by the capacitor, allowing
the capacitor to discharge ver)^ quickly, i.e., the capacitor voltage drops to zero almost instanta­
neously. Once the voltage nears zero, the electronic switch reinstates the earlier circuit structure,
causing the capacitor to charge up again. The process repeats itself indefinitely.
CHAPTER O BJECTIVES
1.
Explore the use o f a constant-coefficient first-order linear differential equation as a model
2.
for first-order RL and RC circuits.
Derive from the differennal equation model, the exponential response form (voltage or
4.
current) o f first-order RL and RC circuits without sources and with constant excitations.
Interpret the solution form o f the differential equation model in terms o f the circuit time
constant and the initial and final values o f the capacitor voltage or inductor current.
Develop techniques to handle s\vitching and piecewise constant excitations within first-
5.
order RL and RC circuits.
Investigate waveform generation and RC op amp circuits.
3.
SECTIO N HEADIN GS
1.
2.
Introduction
Some Mathematical Preliminaries
3.
Source-Free or 2^ro-Input Response
4. D C or Step Response o f First-Order Circuits
5. Superposition and Linearity
6 Response Classifications
7. Further Points o f Analysis and Theory
8 . First-Order RC Op Amp Circuits
9. Summary
10. Terms and Concepts
11.
Problems
1. IN TRO D U CTIO N
Our study prior to Chapter 7 focused exclusively on resistive circuits. Recall that all nodal equa­
tions and loop equations for resistive circuits lead to (algebraic) matrix equations whose solution
yields node voltages and loop currents, respectively. Chapter 7 then introduced the capacitor and
the inductor. Interconnections o f sources, resistors, capacitors, and inductors lead to new and fas­
cinating circuit behaviors. How? Inductors and capacitors have differential or integral voltage-cur­
rent relationships. Interconnecting resistors and capacitors or resistors and inductors leads to cir­
cuits that must satisfy both algebraic (KVL, KCL, and Ohm’s law) and differential or integral rela­
tionships for L and C values. When only one inductor or one capacitor is present along with resis­
tors and sources, these relationships lead to first-order RL and RC circuits. When the sources are
Chapter 8 • First Order RL and RC Circuits
dc, such circuits have vohages and currents o f the form A + Be~^ for constants A, B, and X. The
main purpose o f this chapter is to develop techniques for computing the exponential responses o f
first-order RC and RL circuits driven by dc sources. A simple example serves to explain some of
these points.
In the series RC circuit o f Figure 8.1, suppose an initial voltage
is present on the capacitor,
where 0~ designates the instant immediately before zero. Often vve distinguish among 0“, 0, and
O'*' when switching occurs or when discontinuities o f excitation functions occur at r = 0.
R
v(t)
+
©
v,(t)
FIG URE 8.1 Series/?Ccircuit.
A loop equation for the series RC circuit leads to
vp) = Ri(ir) +
Since iciO = C —
(8.1)
equation 8.1 becomes
v,(/) =
at
+Vc(/)
Dividing through by RC yields the constant-coefficient first-order linear diflferential equation
( 8 .2 )
dt
RC
RC
subject to the initial condition Vf^Qr). This equation says that the derivative o f the capacitor volt­
age plus MRC times the capacitor voltage equals MRC times the source voltage. The equation
enforces constraints on the capacitor voltage, its derivative, and the source voltage, and is differ­
ent from the algebraic node or loop equations studied earlier. The terminology first-order differ­
ential equation applies because only the first derivative appears. Equation 8.2 is linear because it
comes from a linear circuit. Our goal is to find capacitor voltage waveforms that satisfy the con­
straints imposed by the differential equation 8.2.
Exercise. For the circuit o f Figure 8.1, show that the capacitor current i({t) satisfies a differential
equation o f the form
cti.it)
dt
1 .
RC
1 dv,{t)
R
dt
32 4
Chapter 8 • First Order RL and RC Circuits
Our scope in this ciiaptcr is limited to circuits containing one inductor or one capacitor— equiv­
alently, first-order RL or RC circuits. W ithin this category we further constrain our investigation
to circuits with no sources but nonzero initial conditions, circuits driven by constant (dc) sources,
circuits driven by piecewise constant sources, and circuits containing switches. First-order circuits
driven by arbitrary source excitations are covered in later chapters using the Laplace transform
method.
2. SOM E M ATHEM ATICAL PRELIM INARIES
Ver)' often our interest is in source excitations such as v^{t) = 2e~^’ V for /> 0 and 0 otherwise. To
conveniently represent such time-restricted waveforms, we define a signal called the unit step
function, denoted by u{t), as
«(/) =
1
0
/ < ()
The unit step function is a universally used function and will appear many times in the remain­
der o f this text. MATLAB code for specifying the step function is
function f = ustep(t)
t = t + le-12;
f = (sign(t)+l)*0.5;
With the unit step so defined, v^{t) - 2e~"'u(t) V, and both relations are plotted in Figure 8.2.
F IG U R E 8 .2 Unit step function and v^{t) = 2 e
V.
32S
Chapter 8 * First Order RI. and liC Circuits
Further, if v<^t) = l e
for t>t^ and 0 for r < /q, then v^{t) = lOf’
r^) would be the prop­
er representation because the shifted unit step function, //(/- /q), means
Plots o f v^{t) = 2e~^‘u{t - ^q) and u{t - t^) are given in Figure 8.3 for
= 0.5.
FIGURH 8 . 3 . Plots of u{t - 0.5) and v^{t) = 2e ^'u{t - 0.5).
Exercise. Plot //(—/) and «(/q — t). Hint: For what values o f t are the functions zero and for what
values are they 1?
A working model ot a physical system underlies an engineer’s ability to methodically anai)'ze,
design, or modify its behavior. Linear circuits are physical systems that have differential equation
models. The RL and RC circuits investigated in this chapter have differential equation models o f
the form
dx{t)
dt
.
= > ..V (0 + /(/ ).
-Y(/(,) = .Vo
(8.3a)
or, equivalently.
dxU)
dt
(8.3b)
valid for t> /q, where a-(/q) =
is the initial condition on the differential equations 8.3. T he term
J{t) denotes a forcing function. Usually, y(r) is a linear function o f the input excitations to the cir­
cuit.
326
Chapter 8 • First Order RI. and RC Circuits
Before proceeding, it is appropriate to explore the intuitive nature o f a differential equation.
Equations 8.3 are first-order constant-coefficient linear differential equations. They are first order
because o f the presence o f only the first derivative o f some unknown function x{t). For example,
in equation 8.3a the derivative o f x{t) equals a constant X times x{t) plus a known forcing func­
tion y(r), w h e r e in c o r p o r a te s the effect o f all the circuit excitations. Rigorously speaking, “lin­
ear” means that under the assumption o f zero initial conditions, if the pairs o f voltage waveforms
(/j(f), A'j(r)) and (fjit), X2 it)) each satisf)' equations 8.3, then for any scalars a^ and a^, the pair
{a/^it) + ajfjU),
+ ajXjU))
also satisfies equations 8.3.
The parameter X, denotes a riaturalfrequency o f the circuit. Natural frequencies are natural modes
o f oscillation such as, for example, in the ringing o f a bell. For physical objects natural frequen­
cies are called natural modes o f vibration. All physical objects have a vibrational motion even
though it may be imperceptible. Knowledge o f these modes is important for the safety and reliabilit)' o f large buildings and bridges. For example, the Tacoma Narrows Bridge had natural modes
o f vibration that the wind excited. Undulations in the wind intensit}' resonated with the natural
vibrations of the bridge, causing a swaying motion to increase without bound until the bridge col­
lapsed. In circuits, the natural modes o f oscillation are reflected in the shapes o f the voltage and
current w'aveforms the circuit produces. A more thorough and mathematical discussion o f the
notion o f natural frequency will take place in the next chapter, when we study second-order {RLQ
circuits.
Let us return to the goal of finding a solution to the differential equations 8.3. The solution to
equations 8.3 (a derivation will appear shortly) for t >tQ has the form
( c ^ ' - " /( x U k
Jk)
(8.4)
This means that the expression on the right-hand side o f the equal sign (1) satisfies the differen­
tial equations 8.3 [its derivative equals K times itself plus/r)], and (2) it satisfies the correct ini­
tial condition, xit^) = x^^. A simple example illustrates this point.
E XA M PLE 8.1.
Compute and verify the solution o f equation 8.3a using equation 8.4.
SO L U T IO N
Suppose in equation 8.3a., J{t) = u{t - 1), a shifted unit step function, X = - 1 ,
10, in which case
cit
From equation 8.4, for ^ > 1,
,v(/) =
\)(k=
+ I
= 1, and at(1) =
32'
Chapter 8 • First Order Rl. and RC Circuits
To verify that [9^
+ 1] does indeed satisfy the differential equation, observe that for ? > 1,
dx{t) _ d
dt
+ I = -,v(/) + 1
= -9 e
~~dt
1] = 10, which is the mandatory initial condition. Thus, x{t) = 9^
Further, at ^ = 1, [9f’
+ 1 is a valid solution for r > 1.
Example 8.1 spells out the application o f the solution (equation 8.4) to the differential equation
8.3a. It also verifies that the computed solution satisfies the differential equation and the proper
initial condition. Although not shown, equation 8.4 also satisfies equation 8.3b. A formal deriva­
tion o f the solution o f equation 8.4 requires the use o f the integrating factor method, the sub­
ject o f a differential equations course. Briefly, the first step o f this method entails multiplying both
sides o f equation 8.3a or 8.3b by a so-called integrating factor e~^. For equation 8.3b, this results
in
e
dx(t)
_>j
— ------ h e x i l ) = e
dt
f/,\
f{ t )
(8.5)
By the product rule for differentiation, the sum on the left equals
d_
dt
e""x{t)
in which case equation 8.5 becomes
( 8 .6)
dt
One can integrate both sides o f equation
f -dr
J'o
e
8 .6
from
Tq
to t as follows:
dr = e
(8.7)
JI q
Bringing the term e'^‘Ox{tQ) to the right-hand side o f equation 8.7 and multiplying through by
results in the solution to the differential equation 8.3a or 8.3b, given by equation 8.4. This
completes the derivation o f the very powerful formula o f equation 8.4.
There are four points to remember about the preceding discussion: (1) circuits have behaviors
modeled by differential equations such as equations 8.3; (2) the solution to a first-order differen­
tial equation is a waveform (also called a signal or response) satisfying equation 8.4; (3) the for­
mula o f equation 8.4 works for all continuous and piecewise continuous time functionsy(/); and
(4) a solution to a differential equation means that the waveform satisfies the given differential
equation with the proper initial condition.
328
Chapter 8 • Hirst Order RL and RC Circuits
Exercise. Show that the hinction .v(/) = (1 — 0. ^or r > 0, is a solution to the difFerential equation
^
dt
= -.v(/) + //(/) with initial condition a-(0) = 0 by showing that x{t) satisfies the difFerential
equation and has the proper initial condition at r= 0.
3. SOURCE-FREE OR ZERO-IN PUT RESPONSE
Figure 8.4 depicts the most basic (undriven or source-free) RL or RC circuit: a parallel connection of
a resistor with an inductor or a capacitor without a source. In these circuits, one assumes the pres­
ence of an initial inductor current or initial capacitor voltage. The complication introduced by a volt­
age or current source is taken up later. Once the source-free or zero-input behavior is understood,
one can understand more easily the responses resulting from constant source excitations.
i,(t)
>r
+
^
yf
+
v^(t)
R ^
ijt )
\ fi)
+
^R(t) <
-
L
S v jt )
*
■
(a)
(b)
FIGURK 8.4
Our first goal is to derive differential equation models for the RL and RC circuits o f Figures 8.4a
and 8.4b, respectively. We do this in parallel.
(1) At the top node o f Figure 8.4a, KCL implies
(1) Similarly, for Figure 8.4b, KVL implies
= v^t)
ifiU) = -iiit)
(2) However,
(2) However,
Vi it)
.d\'c{t)
n it ) — R icit) ——RCdt
L d ii U)
(3) Making the obvious substitution and
(3) Making the obvious substitution and
multiplying by R/L yields the differential
dividing by RC yields the differential
equation model
equation model
dii it)
,
with //(/()) a given initial condition.
dV( {t) _
(8.8a)
dr
with
1
RC
v'c(/)
a given initial condition.
(8-81’ )
Chapter 8 • First Order RL and RC Circuits
Both differential equation models have the same general form,
^
dt
= hc{t) = - - m
(8.9)
X
i.e., the derivative of x(t) is a constant, X = -1/x, times itself. Applying equation 8.4 to equation
8.9 implies that both equations 8 .8a and 8 .8 b have solutions given by
where x is a special constant called the time constant o f the circuit. Equation 8.10 means that the
responses for
W
W
(q o f the undriven JiL and /?C circuits are, respectively, given by
1^(0 =e
-^ '-'0 )
^
vc(0 = e
- ■ ^ ' - ‘0)
(8.11)
vc(/q)
L
where the time constant o f the RL circuit is T = — and the time constant o f the RC circuit is x
= RC.
^
The time constant of the circuit is the time it takes for the source-free circuit response to drop to
e~^ = 0.368 o f its initial value. Roughly speaking, the response value must drop to a little over onethird of its initial value. This is a good rule of thumb for approximate calculations involving decaying exponentials.
The mathematics that underlie the solution to the differential equation 8.9 given in equation 8.10
is nothing more than elementary calculus. To see this, consider the exponential solution form
(8. 12)
where K 'ls an arbitrary constant. The fiinaion
has the property that its derivative is----- e~' ^
This is precisely what equation 8.9 requires. Therefore equation 8.12 satisfies the differ-
^
ential equation 8.9 and is said to be a solution. To completely specify x{t) it only remains to iden­
tify the proper value of K from the initial condition. Evaluating x(r) at ^
yields
Mt„) = /Cf-Vr
in which case
Substituting this value o f i n t o equation 8.12 produces the solution given in equation 8.10,
which is adapted to specific RL and R C circuits in equations 8 . 11. Figure 8.5 plots equation 8.12
for arbitrary K and x > 0 . This plot proves instructive for understanding how the response decays
as a function o f the time constant.
330
Chapter 8 • First O rder RL and RC Circuits
Time
FIGURI: 8.5 Plot o f equation 8.12. For f = x, one time constant,
decays to 0.368
of its maximum value.
In summary, the circuits o f Figure 8.4 motivate the development o f the rudimentary machinery
for constructing solutions to undriven RL and RC circuits. For more general circuits, those con­
taining multiple resistors and dependent sources, it is necessary to use the Thevcnin equivalent
resistance seen by the inductor or capacitor in placc o f the R in equation 8.11. Figure 8.6 illus­
trates this idea.
f
Linear
\
>f
Resistive
Circuit
i,(t)
f
\
Linear
Resistive
Circuit
+
L
p
No
Sources
-N V jt)
No
Sources
C
i,(t)
+
v,(t)
F I G U R I : 8 .6 R ep la c em en t o f “resistive” part o f c irc u it by its T h e v e n in eq u iv alen t.
Chapter 8 • First Order RL and RC Circuits
331
These facts imply that the general formulas for computing the responses o f undriven RL and RC
circuits have the structures
(8.13)
VrU) = e
The difference between equations 8.11 and 8.13 is that in equations 8.13 R^f^ is the Thevenin
equivalent resistance seen by the inductor or capacitor.
EXA M PLE 8.2
For the circuit o f Figure 8.7, find i^{t) and v^(t) for r > 0 given that
S closes at r = 0.4 sec. Then compute the energy dissipated in the 5
val [0.4, co).
= 10 A and the switch
resistor over the time inter­
t = 0.4sec
S
20Q
5Q
+
v,(t)
8H
FIGURE 8.7 Parallel RL circuit containing a switch.
S
o l u t io n
Step 1. With switch S open, compute the response for 0 < /< 0.4 sec. From the continuity property
o f the inductor current, /^(O'*’) = ^^(0“) = 10 A. Using equation 8.13,
'1 ( 0 = c
A
We note that //(0.4) = 3.679 A.
Step 2. With switch S closed, compute the response for t > 0.4 sec. For this time interval the Thevenin
equivalent resistance seen by the inductor is
= 20||5 = 4 Q, i.e., the equivalent resistance o f a par­
allel 20 Q and 5 ^ combination. According to equation 8.13, the response for t>tQ = 0.4 sec is
,- ,( 0 = /
t
Step 3. Write the complete response as a single expression using step fitnctions:
i^{t) = 10^>-2-5^[«(^) - «(/- 0.4)] +
0.4)A
(8.14)
Chapter 8 * First Order RL and R C Circuits
332
Step 4. Plot the complete response. To plot this using MATLAB, we use the following m-file along
with the code given earlier for the unit step function:
»t = 0:0.005:1.4;
»iL= 10*exp(-2.5*t) .* (ustep(t).* ustep(t - 0.4)) + 3.679*exp(-0.5*(t-0.4)) .* ustep(t - 0.4);
»plot(t,iL)
»grid
Using this code, Figure 8.8 illustrates the complete response, showing the two different time con­
stants. The 0.4 sec time constant has a much faster rate o f decay than the lengthy 2 sec time con­
stant.
a
E
<
Time (seconds)
FIGURE 8.8 Sketch of response i^{t) for Example 8.2.
Step 5. Compute i^^(r). It is a simple matter now to compute v^^t) since
vi^t) =
In particular, t'^(0'*') = -200 V. Hence for 0 < t< 0.4
.IL l
For t> 0.4, however, the circuit structure changes and
= 14.716 V. Thus,
yi(f) = e
(0 .4 "') =
4
in which case i'^(0.4'^) = 4 x 3 .679
-
1
4
.
7
1
V
Chapter 8 • First Order RL and RC Circuits
Step 6. Compute the energy dissipated in the 5 O. resistor over the interval [0.4, oo). The power
absorbed by the 5
resistor for 0.4 < / is
v’i ( 0
5
5
= 43.31
The energ)' dissipated over [0.4, oo) is given by
IVjn ( 0 .4 ,cc) =
PfaU/)A/ = 4 3 .3 12 j;^ ^
= 4 3 .3 12 J
Exercises. I. Plot v^it) using the above m-file, ustep, and the appropriate .code.
2. Repeat the calculations o f Example 8.2 with the 8 H inductor changed to 8 mH and a switch
closing time o f 0.4 ms.
A N SW ER: ij{t) = l{)r-“^'^"'//(f)//(0.4 x lO"-^ - r) +
- 0.4 x Ur-^) A
E X A M PLE 8.3
Find V(^t) for r > 0 for the circuit o f Figure 8.9 given that y^^^O) = 9 V.
So l u t io n
Because there is a switch that changes position at r = 1 sec, there are two time intervals to consider.
Step 1. Compute the response forO < t < 1. Over this time interv\il, the equivalent circuit is a par­
allel /?C circuit, as shown in Figure 8.10a.
Chapter 8 • First Order RL and RC Circuits
334
0.1 F
0.1 F
80
3Q
V jt)
+
V jt)
(a)
(b)
FIG URE 8.10 Equivalent circuits for Figure 8.9: (a) 0 :s r < 1 and (b) 1 s t.
) = 9 V. Therefore from equation 8.11,
By the continuity o f the capacitor voltage,
1
------- 1
v c(t)= e
Vc(O^) =9^’- ’ “5^V
Step 2. Compute the response for r > 1. Figure 8.10b depicts the pertinent equivalent circuit.
Observe that
= 2.58 V and
=3
Again by equation 8.11, for r >
r-I
-St-to)
V c (0 = f
= 1,
v .c (^ ) = 2 .5 8 e
V
Step 3. Use step functions to specify the complete response. By using the shifted unit step function,
the two expressions obtained previously can be combined into a single expression:
t-\
V cir)=9e~^-'^'[u{t)-u{f-\)] + 2.5Se
- \) V
Step 4. Obtain a plot o f the response. Using MATLAB and code similar to that used in Example
8.2, the plot in Figure 8 . 1 1 w'as obtained. Here the part o f the response with the 0.3 sec time con­
stant shows a greater rate o f decay than the longer 0.8 sec time constant.
Time (seconds)
F IG U R E 8 .1 1 Response,
for the circuit o f Figure 8.9 .
Chapter 8 • First Order RI, and RC Circuits
Exercises. 1. Show that ti{t) - u{t - 1) =
- t)-
2. Suppose that in Example 8.3 the switch moves to the 4.5 ^ resistor at r = 0.5 sec instead oF 1
sec. Compute the vakie V(\t) at f = 1.2 sec.
ANSW ER: 0.4671 V
For all o f these examples x > 0 and the response is a decaying exponential. Intuitively, the response
decays because the resistor dissipates as heat the energy initially stored in the inductor or capaci­
tor. One o f the homework exercises will ask the student to show that the total energ)' dissipated
in the resistor from
to oo equals the decrease in energ)^ initially stored in the inductor or capac­
itor at ^Q. When controlled sources are present,
may be negative, in which case x < 0. Here the
negative resistance supplies energ)' to the circuit and the source-free response will grow exponen­
tially. This is illustrated in the next example.
E X A M PLE 8 .4
Find Vf^t) for the circuit o f Figure 8.12, assuming that^^^^ = 0.75 S and i^(;(0“) = 10 V.
—
------ o -----q V
0.25F
N
:
^
4Q
+
v,(t)
<
0 -----0.25F
-2 Q
v,(t)
-o-
-o-
b
b
FIGURE 8.12 Parallel /?Ccircuit with dependent currcnt sourcc.
So l u t io n
It is straightforward to show that theThevenin equivalent seen by the capacitor is a negative resist­
ance,
- -2 Q, as shown in Figure 8.12b. Again, by equation 8.11,
v cit) = e
vc{0^)=\0e^'u{t) V
Because o f the negative resistance, this response grows exponentially, as shown in Figure 8.13. A
circuit having a response that increases without bound is said to be unstable. Practically speaking,
an unstable circuit will destroy itself or exhibit a nonlinear phenomenon that clamps the voltage
at a finite value, as in the case o f saturation in an op amp.
Chapter 8 • First Order RL and RC Circuits
336
Tim e (seconds)
FIGURE 8.13 Plot of unbounded voltage response due to presence of negative resistance.
Circuits with such responses arc said to be unstable.
Exercises. 1. For Example 8.3 show that
icU ) = e
2. in Example 8.3,
f > 0.
ANSWI-.l^S: 8 £2.
ic{0-^) = 5e-'ii{t)A
= 0.125 S. Find the equivalent resistance seen by rhc capacitor and
V
3. Show that in general, for / >
the form o f the capacitor current is similar to the voltage form.
Hint: Apply the capacitor v-i relationship to equation 8.11.
4. DC OR STEP RESPONSE OF FIRST-ORDER CIRCU ITS
The circuits o f the previous section had no source excitations. This section takes up the calcula­
tion o f voltage and current responses when constant-voltage or constant-current sources are pres­
ent. It is instructive to start with the basic series RL and RC circuits as shown in Figure 8.14.
Chapter 8 • First Order RL and RC Circuits
O
Linear
Resistive
Circuit
with
i^Ct)
Linear
Resistive
Circuit
with
Constant
Sources
Constant
Sources
O
(a)
(b)
+
\{t)
FIGURE 8.14 (a) Driven first-order RL circuit, (b) Driven first-order RC circuit, (c) Thevenin
equivalent representation oF (a), (d) Thevenin equivalent representation of (b).
Given these basic circuit representations and initial conditions at Tq, what is the structure o f a dif­
ferential equation mode! that governs their voltage and current behavior for t>tQ' The first objec­
tive is to derive the “differential equation” models characterizing each circuits voltage and current
responses. It is convenient to use ij{t) as the desired response for constructing the differential
equation for the series RL circuit (Figure 8 .l4 c ), whereas for the series RC circuit (Figure 8 .l4 d ),
is the more convenient variable.
(1) The circuit mode! for the inductor is
vl O )= L
(i) The circuit mode! for the capacitor is
dv(^(t)
dilit)
ic(t)= C
dt
dt
(ii) By KCL and Ohms law,
(ii) By KVL and O hm s law,
ic (t) -
(iii) Substituting for ;^(^) leads to the
differential equation model
d iijt)
dt
^
R.
(iii) Substituting for /^r) leads to the differentia!
equation model
dvc(r)
+
(8-> 5a)
^’oc R,ill
dt
_
-^ ^ 'C (0 + -
—
v o c (8 .1 5 b )
Chapter 8 • First Order RL and RC Circuits
338
initial condition V(\t^ ) =
with initial condition //(^o") =
is constant (not impulsive).
since
is constant (not impulsive).
Exercise. Construct differential equation models for the parallel RL and RC circuits o f Figure
8 . 15 . Note that these circuits are Norton equivalents o f those in Figure 8.14a and Figure 8.14b.
Again choose /^(r) as the response for the RL circuit and v^^t) as the response for the RC circuit.
(constant)
(constant)
FIG URE 8.15 Driven RL and RC parallel circuits.
dll (/)
.\NSWI-RS: — V di
V/(
I.
R
//(/)+ —
L
and ^ -■ ■ = ------^— ''r ( n + —
d!
R,i,C ^
C
A simple application o f basic circuit principles has led to the two differential equation models of
equations 8.15. The next important question is: What do these t%vo differential equation models
tell us about the behavior o f each circuit? Equivalently, how do we find a solution to the equa­
tions? Observe that both differential equations 8.15 have the same structure:'
dx(t)
1
dt
X
(8.16a)
for RL circuits and x = R^j^C for RC circuits, and F= v J L for RL
where the time constant T =
circuits and F = vJiR^i^j for the RC case. This equation is valid for t>
Equation 8.4, rewritten
here with/^y) = F, presents the general formula for solving the differential equation 8.16a:
) + ( e^^^'-‘f^Fdcj
A-(/)=
where A = —
(8.16b)
•'h)
j
is a natural frequenc)' o f the circuit, and where we have emphasized the use o f
T
the initial condition at
Note that as long as x{t) is a capacitor voltage or inductor current, the
initial condition is continuous, i.e., x{t^p = x(fQ+), because F h a . constant (non-impulsive) forc­
ing function. A straightforward evaluation o f the integral o f equation 8.16b yields
t
x(/) = e
^ \v(fo)+ F e
( I-In
f e
Jif)
dq = e
^ 'x (t^ )+ F T
Some rearranging o f terms in equation 8.16c produces the desired formula
- e
(8.16c)
Chapter 8 • First Order RL and RC Circuits
339
(t-tn
x(
0=
f t
+ U 4 )-
ft
”
(8.17)
= -r
which is valid for t > /q. After some interpretation, this formula will serve as a basis for comput­
ing the response to RL and RC circuits driven by constant sources. A homework exercise will ask
for a different and direct derivation of this formula.
At this point it is helpful to interpret the quantity i r in equation 8.17. For RL circuits, when x(r)
= /*£(/), equation 8.15a implies that
v J L , x = UR^f^ and hence F l = vJR^f^ =
For RC cir­
cuits when ;c(/) = V({t), equation 8.15b implies that F = vJR ^ C , x = Rf^,C, and hence Fz =
This interpretation is valid for both positive and negative values of x. If x > 0, then
t-tQ y
jr(oo) = lim x(t) = lim
/—*00
t~ * 0 0
Fx + \ x {t^ )-F x y
T
isc =
= Ft =
for RL case
Rih
(8.18)
for R C case
This means that for the RL case, /^(oo) =
= vJR^f^ and for the RC case, V({<x>) =
O f course,
is computed by replacing the inductor with a short circuit, and
is computed by replacing
the capacitor with an open circuit. See Chapter 6 for details. Mathematically, any constant, such
as x(/) = constant, that satisfies a differential equation is called an equilibrium state of that dif­
o
ferential equation. Since the constant x(/) = F i satisfies the differential equation 8.16, / r is an
equilibrium state of the differential equation 8.16.
Whenever X > 0, equation 8.18 implies that the formula (equation 8.17) for the solution o f equa­
tion 8.1 6 given constant or dc excitation becomes
.
a: ( 0
_ ^-'o
(8.19a)
= j :(«>) + U ( ^ ^ ) - J c( oo) e
and when x(^) = /£(/),
(t-to )
iL (0 = / z .(“ ) + k ( 4 ) - ' L ( “ )
o
and when x(/) = V({t)y
(8.19b)
t-t
vc(0 = vc(00) + Vc(^o )-^ c (°°)
K/.C
(8.19c)
Note that x > 0 is true whenever R^j^ > 0, C > 0, and Z, > 0, i.e., the circuit is said to be passive.
This allows us to state a nice physical interpretation of equation 8.19a:
elapsed time
w
w
's . ;
x { t )= [ F in a l value^ + i^Initial v a lu e ]-[F in a l value\)e tt^e constant
Graphically, equation 8.19a is depicted in Figure 8 .1 6 for xipo) >
x (/ q ).
Chapter 8 • First Order RL and RC Circuits
340
Elapsed Time
I'iG U RE 8.16 Graphical interpretation o f equation 8.19a for the case xico) > x{tQ).
Exercise. Redo the curve o f Figure 8.16 for the case x{<x)) <
The initial value
computed from initial conditions and possibly the value o f the source
excitation, or it can be computed from past excitations up to
trate the use o f equation 8.19.
Several examples will now illus­
EXA M PLE 8.5
For the circuit o f Figure 8.17, suppose a 10 V unit step excitation is applied at r = 1 when it is
found that the inductor current is
= 1 A. The 10 V excitation is represented mathematically
as
= 1 0 « (/ - 1) V for r> 1. Find ij{t) for r > I.
R = 5Q
v^(t)
1^(1-)=1 A
F IG U R L 8 .1 7 Driven series R L circuit for Example 8.5 with /^(1“) = 1 A.
Chapter 8 • First Order RL and RC Circuits
.Vtl
S o l u t io n
Step 1. Determine the circuit’s differential equation model. Since the circuit o f Figure 8.17 is a driv­
en series RL circuit, equation 8 .1 5a implies that the differential equation model o f the circuit valid
for r > 1 is
cl'iAt)
R
1
1
10
w here the tim e c o n sta n t t = 0,4 sec.
Step 2. Determine the form o f the response. Since /^(1“ ) =
equation 8.19b implies that
/^(/)=/^(oo)+^/^(r)
^
Here the presence o f u{t - 1) emphasizes that the response is valid only for / > 1.
Step 3. Compute i^i^X)) and set forth the fin al expression for /^(r). Since x = 0.4 > 0, we replace the
inductor in the circuit o f Figure 8.17 with a short circuit to compute i^^. = //(oo) = 2 A. It follows
that
[2 + (1 -
- 1) = (2 -
- 1) A
Step 4 . Plot i^it). One cannot presume that the response is zero for r < 1. Hence, using MATLAB
or the equivalent, one can construct the graph o f i^{t) for / > 1 as given in Figure 8.18.
c
0;
3
u
O
tj
3
■o
c.w
Time (sec)
FIGURE 8.18 Plot of /^(r) for Example 8.5.
Step 5. Compute v^it). Given the expression for the inductor current in step 3, it follows that for
t> 1,
cliLit)
Vi{t) = Ldt
/-I
u{t -\^ ) = 5 e
* ^ / (r -l'* ')V
Chapter 8 • First Order RL and RC Circuits
342
Exercises. 1. Verify that in Example 8.5 v^it) can be obtained without differentiation by
=
V -V iW 2. In Example 8.5, suppose R is changed to 4 Q. Find i^it) at r = 2 sec.
ANSW ER: 1.8647 A
Note that we have used the differential equation 8.16 (or equations 8.15) to obtain the solution
form o f equation 8.19. However, when using equations 8.19, it is not necessary to reconstruct the
differential equation of the RL or RC circuit. Specifically, we need only compute xit^), x{cc), and
the time constant x = LIR^i^ or
The method described for computing final values can also be used to find the initial values o f
and i^ at f =
if dc excitations have been applied to the circuit for a long time before t = ?q. The
next example illustrates this technique and extends the preceding discussion.
EXA M PLE 8.6
The source in the circuit o f Figure 8.19 furnishes a 12 V excitation for / < 0 and a 24 V excita­
tion for r < 0, denoted by yy^(r) = [\2u{-t) + lAu{t)] V. The switch in the circuit closes at / = 10
sec. First determine the value o f the capacitor voltage at r = 0", which by continuity equals
Next determine
for all ^ > 0.
R = 6 kn
+
v,(t)
FIGURE 8.19 Switched driven
circuit for Example 8.6.
So l u t io n
Step 1. Compute initial capacitor voltage. For r < 0, the 12 V excitation has been applied for a long
time. Therefore, at r = 0“ , the capacitor has reached its final value and looks like an open circuit
to the source. Hence the entire source voltage o f 12 V appears across the capacitor at r = 0~, i.e.,
^(40~) =
= 12 V by the continuity property o f the capacitor voltage.
Step 2. Use equation 8.19c to obtain v^it) fb r Q < t< 10 sec. Equation 8 .1 9c requires only that we
know V(iQ*) (step 1), x, and Vf^co). For 0 < t< 10 sec, r = R^C= 3 sec. It is important to realize
here that for 0 < r < 10 sec the circuit behaves as if the switch were not present. Hence, the com­
putation o f v^(co) proceeds as if no switching would take place at r = 10 sec. Here v^^ = v^^oo) =
24 V. Hence, for 0 < ^ < 10, equation 8.19c implies
Chapter 8 • First Order RL and RC Circuits
.vi3
v'cW = ''c(= o ) + [wc(0^)
= 2 4 + (1 2 -2 4 )c^ - ^ = 2 4 - 1 2 ^
Step 3. Compute the initial condition fo r the interval 10 <
V
(8.20)
i.e., t'^^lO'^). Plugging into equation
8.20 and using the continuity property o f the capacitor voltage yields
= 2 4 - 12f>-io/3 = 2 3 .57 V
Step 4. Find V(^t) for / > 10. For r > 10, the resistive part o f the circuit can be replaced by its
Thevenin equivalent, which yields Figure 8.20.
Rtn = 2 k O
+
V(t)
FIGURE 8.20 Circuit equivalent to that of Figure 8.19 for / > 10.
Here, equation 8.19c applies again. The value for y^^co), however, is now 8 V and the new time
constant is
= 1 sec. Hence, for ; > 10,
/-lo
v c (/ )= V c (“ ) + [v c (IO *)-V (-(= c )]e
= 8 + ( 2 3 . 5 7 - 8 ) t '- * '- '" ’ = 8 + I5 .5 7 e ‘ * ' '" ” V
Step 5. Set forth the complete response using step functions. Using step functions, the response V(\()
for f > 0 is
v ^ {t)= 2 A -\ le
+ S + \5.51e^'
Step 6. Plot V(it). Plotting i^(^t) yields the graph of Figure 8.21.
Chapter 8 • First Order RL and RC Circuits
3-m
o
OJ
O)
Q.
u
Time (seconds)
FIGUllK 8.21 Capacitor voltage
for r > 0 .
Exercise. Suppose the switch in Example 8.6 opens again at r = 20 sec. Find v^it) at r = 25 sec.
ANSW ER: 20.98 \’
EXA M PLE 8 .7 The circuit o f Figure 8.22a has a capacitor voltage given by the cur\'e in Figure
8.22b. We note that ^f^O.l) = 7.057 V. Find, Kq, y^^O), the time constant r = RC, the exact value
of j/(j(Q.25), and the value o f C i f /? = 100^2.
------ O-
v.(t)= V „u (t)
v,(t)
-o(a)
345
Chapter 8 • First Order lU. and RC Circuits
01
IB
I
Q.
fO
U
T im e (s)
(b)
FIGURE 8.22 (a) Scries /?Ccircuit, (b) Capacitor voltage,
So
lu t io n
. A simple inspection o f the graph indicates that
V(^{Q) = 2 V. One recalls that
/
) - \’(-(0C) e ^
Hencc as / -> oo, v^{t) -> v^co) = 10 V. Since the capacitor looks like an open circuit at / = oo, Vq
=
= 10 V. From the given problem data,
y ^ 0 .1 ) = 7 .0 5 7 = 10
Simplifying yields
-
T =
0.1
/ 1 0 - 7 .0 5 7
= 0.
8
Therefore C = 0.01 F and M 0 .2 5 ) = 9..34.33 V.
When switching occurs frequently, or the excitation changes its constant level frequently, then
hand analysis, as in lixample 8.6, becomes very tedious. For such problems a SPIC E simulation
(or the equivalent) proves useful and saves time. The next example u.ses SPIC E to compute the
waveform o f a simple RC circuit whose input excitation is a square wave. Like the previous exam­
ple, the solution is broken down into time intervals such that during each time interval inputs are
constant. Because no switching occurs, the time constants for all time intervals are the same. In
applying equation 8.19c, the quantities that vary from one time interval to the next are the initial
values and final values.
346
Chapter 8 • First Order RL and RC Circuits
EXA M PLE 8.8
The first-order RC circuit o f Figure 8.23a is excited by the 50 Hz square wave input voltage o f
Figure 8.23b given that the capacitor is initially relaxed.
(a)
(b)
Plot
for 0 < r < 60 ms, using SPIC E or equivalent software.
Find the initial value and the final value in equation 8.19 when t is very large, for exam­
ple, at the beginning and end o f the interval 1 < t< 1.01 sec. Plot the v^t) wave for this
interval using MATLAB or the equivalent.
V (t) (V)
■> t (msec)
10
20
30
(b)
FIG U RE 8.23 (a) Series RC circuit excited by the 50 Hz square wave o f (b).
So
lu t io n
Part (a)
Doing a SPICE or equivalent simulation gives rise to the response curve shown in Figure 8.24, over
which the square wave input is superimposed. Observe that the response V(\t) has an approximate tri­
angular shape. What is happening is that from zero to 10 msec, the circuit sees a step and hence the
capacitor voltage rises toward one volt. At 10 n:isec, the square goes to zero for 10 msec. The capacitor
then discharges its stored energ)' through the resistor, causing a decrease in its voltage value. The decrease
does not go to zero, however. So when the square wave again is at 1 volt the capacitor voltage begins to
rise again and achieves a slighdy higher value at f = 30 msec compared to f = 10 msec. In fact, one notices
in Figure 8.24 that the peak and minimum values are increasing slighdy as time increases. Eventually the
peak and minimum values will reach their respective fixed viilues, c;illed steady-state values. To find these
values, a simulation program could require a very lengdiy simulation interv'al, which often proves
impraaical. The steady-state values can be computed analytically as in part (b).
10
20
30
40
50
t(m se c)
FIG URE 8.24 Response of circuit of Figure 8.23a calculated using SPICE. For reference, the input
square wave excitation is superimposed on the plot.
Chapter 8 • First Order Rl. and RC Circuits
_____________________________________________________
Part (b)
Let Tq = mT, where 7'= 20 msec is the period o f the square wave and m is some large integer. Then,
0.57-
+ 0.571 = 1 + {v^t^) - 1) /
Further, in steady state, V(it^ + T) =
= 1+
- 1]^’- ’
(8.21)
which implies that
0.5T
+ 71 =
+ 0 .5 7 ) ^ ' KC .
+ 0.571^'-' =
(8.22a)
Equivalently, equation 8.22a implies that
+ 0 .5 7 ) = V(itQ)e^
(8.22b)
Substituting equation 8.22b into equation 8.21 yields
= 1 + [i/c(ro)the solution o f which is
^ "
1+e
= 0-2689 V
It follows that
r/J/o + 0 . 5 7 ) = i^(3<ro)e> = 0 .7 3 1 1 V
An examination o f the response in Figure 8.24 shows that the minimum and peak values are
approaching the steady-state values o f 0.26 8 9 V and 0.7311 V, respectively.
Exercise. Based on the response in Figure 8.24, roughly sketch the capacitor current,
At
what time instants is the capacitor current discontinuous?
5. SUPERPO SITION AND LINEARITY
Superposition, a special case o f linearity, helps simplify the analysis o f resistive circuits, as discussed
in Chapter 5. Recall that linear resistive circuits are interconnections o f resistors and sources, both
dependent and independent. Does superposition still apply when capacitors and inductors are
added to the circuit? The answer is yes, provided one properly accounts for initial conditions.
In order to justify the use of superposition for RC and RL circuits, consider that resistors satisfy
O hm s law’, a linear algebraic equation. Capacitors satisfy the differential relationship
; - r -----ir
^
dt
r\
348
Chapter 8 • First Order RL and RC Circuits
which is also a linear equation. To see linearity in this i-v relationship, suppose voltages
and
V(^ individually excite a relaxed capacitor producing the respective currents
:
,■
Let
_^ dvc2
, >c2 - c —
‘a - c
be the current induced by a voltage equal to the sum o f
and
i.e.,
‘C3 = (^-^(^Cl+^C2)
However, the linearity o f the derivative implies the property of superposition;
.
ic=C-
_ dvf
^
+ C — — = ic i + t 2
dt
dt
By the same arguments, the current due to the input excitation t/Q - ^\^c\
+
On the other hand, suppose two separate currents / q and
^2^Cl
^C3 “ ^l^Cl
individually excite a relaxed capac­
itor C Each produces a voltage given by the integral relationship
^Ci (0 =
(”<:) d t , vc2 (0 =
ic2 (T^) ^
By the distributive property of integrals, the combined effect of the input,
+ <?2^C2’ would
be a voltage,
vc3 (0 = ^ f_ Ja \ ic \ (T ) + a2ic2('^)] d r
= ai
\ 7 S ' - J c 2 W ‘‘^
Thus linearity and, hence, superposition hold.
Arguments analogous to the preceding imply that a relaxed inductor satisfies a linear relationship,
and thus superposition is valid, whether the inductor is excited by currents or by voltages.
The interconnection of linear capacitors and linear inductors with linear resistors and sources sat­
isfying KVL and KCL produces linear circuits because KVL and KCL are linear algebraic con­
straints on the linear element equations. Hence, the property of linearity is maintained, and as a
consequence superposition holds for the interconnected circuit.
To cap off this discussion we must account for the presence of initial conditions on the capacitors
and induaors o f the circuit. For first-order RC and RL circuits, this need is clearly indicated by
the first term o f equation 8.4. For a general linear circuit, one can view each initial condition as
being set up by an input that shuts off the moment the initial condition is established. Hence the
r-
Chapter 8 • First Order RL and RC Circuits
efFect o f the initial condition can be viewed as the effect o f some input that turns o ff at the time
the initial condition is specified. This means that when using superpositioti on a circuit, one first
looks at the effect o f each independent source on a circuit having no initial conditions. Then one
sets all independent sources to zero and computes the response due to each initial condition with
all other initial conditions set to zero. The sum o f all the responses to each o f the independent
sources plus the individual initial condition responses yields the complete circuit response, by the
principle o f superposition. A rigorous justification o f this principle is given in a later chapter using
the Laplace transform method.
The following example illustrates the application o f these ideas.
E XA M PLE 8.9
The linear circuit o f Figure 8.25 has two source excitations applied at r = 0, as indicated by the
presence o f the step functions. The initial condition on the inductor current is
= -1 A.
Compute the response /^(r) for r > 0 using superposition.
So
lu t io n
Because the circuit is linear, having a linear differential equation, superposition is but one o f sev­
eral methods for obtaining the solution. An alternative approach is to find the Thevenin equiva­
lent circuit seen by the inductor. As we will see, the superposition approach sometimes has an
advantage over the Thevenin approach.
Superposition must be carefully applied, however. First one computes the response due only to the
initial condition with the sources set to zero. Second, one computes the response due to Vj with
all initial conditions and all other sources set to zero. Third, one computes the response due to /j
with all initial conditions and all other sources set to zero. Finally, one adds these three responses
together to obtain the complete circuit response.
Step 1. Compute the part o f the circuit response due only to the initial condition, with all independ­
ent sources set to zero. With both sources set to zero, there results the equivalent circuit given by
Figure 8.26. The Thevenin equivalent resistance is
= 4 Q, resulting from the parallel combi­
nation o f
and R->. Figure 8.26 depicts the equivalent undriven RL circuit having response
352
Oliapccr 8 • First Order RI. and RC Circuits
An approach based on the Thevcnin equivalent circuit seen by the inductor would allow one to
quickly compute the complete response, but not in a way that identifies the contributions due to
each o f the individual sources. Answ^ers to the preceding three questions would have required
repeated solutions to the circuit equations. However, if one keeps the source values in literal form,
then the Thevenin equivalent approach would be as efficient.
6. RESPONSE CLA SSIFICA TIO N S
Having gained some understanding of the form o f the behavior of RL and RC circuits, it is instruc­
tive to classify the responses into categories. The zero-input response o f a circuit is the response
to the initial conditions when all the inputs are set to zero. The zero-state response o f a circuit is
the response to a specified input signal or set of input signals given that the initial conditions are
all set to zero. By linearit)', the sum o f the zero-input and zero-state responses is the com plete
response o f the circuit. This categorization is the convention in advanced linear systems and lin­
ear control texts.
I'Vequently circuits texts include two other notions o f response, the natural response and the
forced response. However, decomposition o f the complete response into the sum of a natural and
a forced response applies only when the input excitation is (1) dc, (2) real exponential, (3) sinu­
soidal, or (4) exponentially modulated sinusoidal. Further, the exponent of the input excitation,
for example, a'm j{t) =
must be different from that appearing in the zero-input response.
Under these conditions it is possible to define the natural and forced responses as follows: (1) the
natural response is that portion of the complete response that has the same exponents as the zeroinput response, and (2) the forced response is that portion of the complete response that has the
same “exponent” as the input excitation provided the input excitation has exponents different
from that of the zero-input response.
This decomposition is important for rwo reasons. First, it agrees with the classical method o f solv­
ing linear ordinary differential equations with constant coefficients where the natural response cor­
responds to the com plem entary function and the forced response correspontls to the particular
integral. Students fresh from a course in linear differential equations will feel quite at home with
these concepts. The second reason is that the forced response is easily calculated for dc inputs. For
general systems this type o f decomposition is not used.
7. FURTHER POINTS OF ANALYSIS AND TH EO RY
In deriving equations 8.17 and 8 .1 9a, the quantin' x{t) was thought o f as a capacitor voltage or an
inductor current. It turns out that any voltage or current in an RC or RL first-order linear circuit
with constant input has the form
{I- to)
x{l) = X^,+
~ r~
(8.23)
For T negative or positive equation 8.23 is identical to equation 8.17 with Pi = X^. Further, T is
the circuit time constant and X is that voltage or current of interest computed under the condi-
Chapter 8 • First Order Rl. ami RC Circuits
3 S3
tion that the inductor is replaced by a short circuit for the RL case or the capacitor is replaced by
an open circuit for the RC case.
How do we justify the form o f equation 8.23 for all variables? We invoke the linearity theorem o f
Chapter 5 and the source substitution theorem o f Chapter 6 o f 2"^ edition. Suppose in a firstorder RC circuit we have found V(\t). By the source substitution theorem, the capacitor can then
be replaced by a voltage source whose voltage is the computed V(it). This new circuit consists o f
constant independent sources, one independent source o f value
and resistors and depend­
ent sources. By linearity, any voltage or current in the circuit has the form
x(r) =
for appropriate ATj and Kj- By equation 8.17,
which implies that
Mr) = K , + ^6^
for appropriate
"
This is the same form as equation 8.23 for proper choices o f
Exercise. Show that, for t > 0,
and
-X^.
^rid
Note that equation 8.23 requires that the initial value be evaluated at ^
instead o f r =
This
is because only the inductor currents and the capacitor voltages are guaranteed to be continuous
from one instant to the next for constant input excitations. The capacitor current and the induc­
tor voltage as well as other circuit voltages and currents may not behave continuously.
E XA M PLE 8 .1 0
This example illustrates the application o f equation 8.23. For the circuit o f Figure 8.29,
-18//(-r) +
V. Find i- it) for f > 0.
6kQ
3kQ
2kQ
v jt )
0.5 mP
FIGUFIK 8 .2 9 R C circuit with /■ (/) as the desired response.
=
Chapter 8 • First Order RL and RC Circuits
354
SO L U T IO N
Step 1. Compute /y„(0^). To obtain /y„(0*), we first compute V(\Q~) =
Since for r < 0, - 1 8
V has excited the circuit for a long time, the capacitor looks like an open circuit. By voltage divi­
sion,
j = —( - 1 8 ) = - 6 V
Thus at r = 0*, the equivalent circuit is as shown in Figure 8.30.
rigurc K.30 Circuit equivalent to that of Figure 8.29 at t = 0"^.
Application o f superposition to Figure 8.30 shows that Kj = - 1 .5 V. Flence,
9 - ( - 1 .5 )
6x10-^
= 1.75 mA
Step 2. Find the circuit time constant and the equilibrium value o f ijj,t). From Figure 8.29, the
equivalent resistance seen by the capacitor is
= 4 kl^. Hence, the time constant is t = 2 sec.
Further, since t><d,X^ = ij„{'^), which is computed when the capacitor is replaced by an open cir­
cuit. In this case.
=
9 x 1 0 -'
= 1 mA
Step 3. Apply equation 8.23. Using equation 8.23, we have, for r > 0,
= 1 + 0.75^'
-0.5/
mA
Exercise. In Flxample 8.10, find /^O ), /(^O"^), and i^^t) for r > 0 using equation 8.23 directly.
AN SW ERS: 0, 2.25 mA. 2.25^’-** "^ mA
Note that in Example 8 .1 0 we used
instead o f /,„(0~) to obtain the correct answer. Some
straightforward arithmetic shows that /y„(0“) = - 2 mA. Since /y„(0'^) = 1.75 niA, the input current
is discontinuous at f = 0, unlike the capacitor voltage, which is continuous at f = 0. This empha­
sizes the need to use x{t^) in equation 8 .23.
Chapter 8 • First Order RL and RC Circuits
3S5
In several o f the examples o f sections 3 and 4, the circuits contain switches that operate at pre­
scribed time instants. In some electronic circuits, the switch is a semiconductor device whose
on/off state is determined by the value o f a controlling voltage somewhere else in the circuit. If the
controlling voltage is below a certain level, the electronic switch is off; if the voltage moves above
a fixed level, the electronic switch is on. The time it takes for a controlling voltage to rise (or fall)
from one level to another is very important because timing is as critical in electronic circuits as is
scheduling for large organizations. For first-order linear networks with constant excitations, cal­
culation o f the time for a voltage or current to rise (or fall) from one level to another is straight­
forward because all waveforms are exponential functions, as per equation 8.19. The situation is
illustrated in Figure 8.31.
c
0)
k_
I—
D
u
<U
cn
(Z
4->
o
>
FIGURE 8.31 First-order response showing a rise from the voltage or current level A'j to the voltage
or current level A'2, for which the elapsed time is h - ty
In equation 8.19a, let .v(rj) =
and .v(^^) = X , be the two levels o f interest. A straightforward
manipulation o f equation 8.19a leads to the elapsed tim e formula for first-order circuits.
h -ri
[^1 --V(oo)
(8.24)
^2 - -v(x)
E X A M PLE 8.11
This example uses the elapsed time formula o f equation 8,24 for the circuit o f Figure 8.32. The
switch in this circuit is used to produce r^vo different “final” capacitor voltages. When the switch
is open, the final capacitor voltage is 12 V. When the switch closes, the final capacitor voltage, by
V-division, changes to 4 V. Thus the switch causes the capacitor to charge and discharge repeat­
edly. For our purposes we show that the choice o f resistances produces an approximate triangular
waveform.
For the purposes o f this example, suppose the switch in Figure 8.32 is controlled electronically so
that it closes when
SNvitchings.
rises to 9 V and opens when V(^ falls to 5 V. Find and plot v^^t) for several
Chapter 8 • First Order RL and RC Circuits
FIGURE 8.32 Switched driven RC circuit used to generate an approximate triangular waveform.
So
lu t io n
subsequently opening at ^ = tj^ and closing again at t^tc,
Suppose the switch first closes at f =
and so on. For 0 < r <
the time constant T = 3 sec, V(\Q) = 0, and
= 12 V. From equa­
tion 8 .1 9c,
t
VcU) = 12 1 - e 3
V
From the elapsed time formula o f equation 8.24,
- 0 = 3//?
Now, for
19-12^
= 3 X 1.386 = 4 .1 5 9 s
t < tj^, there is a new' time constant x= 1 sec,
= 9, and
= 4. .^gain using
equation 8.19c,
From the elapsed time formula, equation 8.24,
//9-4\
- t „ =/n
Finally, for the time interval
f
- /n{5) = 1.61 s
\5-4/
= 5, and t'^co) = 12. Using equation 8 .1 9c,
, x= 3 sec,
JJz Itl
v c(/) = 1 2 - 7 e
^
and from the elapsed time formula,
^ ///5- 12\
- th =3/n
rh e w aveform o f
v^t)
for 0 < / <
9-12/
^ / ; (1\
\3
= 2.54 s
is plotted in Figure 8 .3 3 .
Chapter 8 • 1-irst Order RL and RC Circuits
Capacitor Voltage (V)
From the preceding solurion
= 4.16 sec,
= 4.1 6 + 1.61 = 5.77 sec, and t^ = 5.77 + 2.54 = 8.31
sec. If we proceed to calculate the waveform for t >
the waveform begins to repeat itself, as is
evident Irom Figure 8.33. Practically speaking, the first c)'cie o f a periodic, approximately trian­
gular waveform occurs in the time interv^al [t^, r j , and the period is
= 8.31 - 4.1 6 = 4.15
sec. Note that the triangular waveform has a frequency
I
/ =
period
1
1
2.5 4 + 1.61
4. 15
= 0.241 Hz
The waveform in figure 8.33 is approximately triangular. This is due the fact that two time con­
stants, 1 and 3 s, have the same order o f magnitude. If we select the resistances so that the charg­
ing time constant is much larger than the discharging time constant, then the capacitance voltage
waveform will look more like a sawtooth waveform. Sawtooth waveforms arc used to drive the
horizontal sweep of the electronic beam in an oscilloscope or a T V picture rube.
8. FIRST-ORDER RC OP AM P CIRCU ITS
RC op amp circuits have some singular characteristics that set them apart from standard passive
RC and RL t)^pes o f circuits. Specifically, because ol the nature o f the operational amplifier, the
time constant o f the circuit will often depend only on some o f the resistances. We present four
important examples to illustrate the behavior o f RC op amp circuits.
3^H
Chapter 8 • First Order RL and RC Circuits
EXA M PLE 8 .1 2
Compute the response
for the ideal op amp circuit o f Figure 8.34.
1.
FIGURE 8.34 Differentiating op amp circuit.
So
lu t io n
Observe that
Chapter 4. Also,
by the virtual short-circuit propert)' o f the ideal op amp, as set forth in
Hence, from these equalities and the definition o f a
capacitor,
= « ,,( ,) = -« / c (0 = - R C ^
=- R C ^
(8.25)
Since the output is a negative constant (user chosen) times the derivative o f the input, the circuit
is called a differentiator.
Exercise. Suppose y/„(^) = cos(250^). Find R for the circuit o f Figure 8.34 so that
= sin(250/)
V a n d C = 1 pF.
AN SW FR: 4 kH
EXA M PLE 8 .1 3
Compute the response
for the ideal op amp circuit o f Figure 8.35 assuming V(^0~) =
= 0.
1.
FIG U R K 8 .3 5 Integrating op amp circuit.
Chapter 8 • First Order RI. and RC Circuits
So
lu t io n
= v-^{t)IR by the virtual short circuit property o f the ideal op amp. Also, i^^t)
Observe rhat
=
3S9
Hence, from these equalities and the integral i>-i relationship o f a capacitor,
(8.26)
Since the output is a negative constant (user chosen) times the integral o f the input, the circuit is
called an integrator.
Exercise. Suppose
= cos(250r) V. Find R for the circuit o f Figure 8.35 so that
sin(250r) V and C = 1 pF.
=
A N SW ER: 4 kD
EXA M PLE 8 .1 4
This example considers the so-called leaky integrator circuit o f Figure 8.36, which contains an
ideal op amp. The input for all time is v^{t) =
V. Rj represents the leakage resistance o f the
capacitor. Given C and Rj, the resistance /?| is chosen to achieve a dc gain o f 10. The objective is
to compute the response
assuming
= 0 and compare it to a pure integrator having a
gain o f 1. This problem is reconsidered in Chapter 13.
+
v jt )
1
FIG U RE 8.36 Leaky integrator op amp circuit in which v^{t) = -5u{t) V.
So
lu t io n
Because there is only one capacitor, the circuit o f Figure 8.36 is a first-order linear circuit. Because
the inverting terminal o f the op amp is at virtual ground,
and the capacitor sees
an equivalent resistance R^i^ = /?2- Hence x = /?,C = 10 sec > 0, Equation 8.19 implies that
Chapter 8 • First Order RL and RC Circuits
360
(8.27)
Because the voltage source v^{t) = 0 for r < 0,
) = -V(i^ ) = 0. For f > 0, v^{t) = - 5 V. Since
the source voltage is constant, the capacitor looks like an open circuit at r = co having final value
..„ „ ( co) = - A ( . 5 ) = 5() V
Entering numbers into equation 8.27 yields
= 50 + (0
= 5 0 (l
ii{t) V
A plot o f the op amp output voltage appears in Figure 8.37 along with that o f an ideal integrator.
O ne observes that the more realistic leaky integrator circuit approximates an ideal integrator only
for 0 < r < 0 .1 5 t before the error induced by the feedback resistor R-^ becomes noticeable. Such
integrators need to be reinitialized periodically by resetting the capacitor voltage to zero.
OJ
Ol
ra
>
4-'
D
CL
4-1
D
o
Q.
E
<
Q.
O
Time (s)
FIGURE 8.37 Outpiu voltage of leaky integrator that
approximates an ideal integrator.
So h r we have assumed an ideal op amp. In practice, the output voltage will saturate at a level
determined by the power supply voltage and the specs o f the particular amplifier used. Further,
practical op amps have complex models. To evaluate the preceding analysis, Figure 8.38 shows a
SPIC E simulation using the standard 741 op amp.
361
Chapter 8 * First Order RL and RC Circuits
Leaky lntegrator-Transient-0
Time (s)
Observe rhar the response approximaces the ideal up to about 0.1 5t = 1.5 sec, which corroborates
our analysis using the ideal op amp. Note, however, that the simulation accounts for saturation
present in practical op amps but absent from the ideal.
EXA M PLE 8.1 5
In trying to build an inverting amplifier in a laboratory, a student inadvertently reverses the con­
nection o f the two input terminals, which results in the circuit of Figure 8.39a. Assume a practi­
cal op amp model with
output o f
= 15 V and a finite gain of/I = 10“^. Instead o f seeing the expected
= - 4 V, the student observes a 15 V output. Explain how this 15 V output could
possibly exist.
4kO
Capacitance
(a)
(b)
F IG U R E 8.39 (a) Incorrectly connectcd Inverting amplifier, (b) Circuit model, including a stray
capacitancc C = 1 pF and a finite gain A = lO"^.
362
Chapter 8 • First Order RL and RC Circuits
So l u t io n
C'hapter 4 op amp models contain only resistors and controlled sources. One way to explain the
situation described in this example is to postulate a small stray capacitance, C = 1 pF, across the
input terminals. In fact, this is a more accurate circuit model and is shown in Figure 8.38b. This
means that the response will be o f the form o f equation 8.23.
The first quantit)' to compute is the circuit time constant t =
C. The equivalent resistance
looking to the right of C, is obtained from the circuit o f Figure 8.40.
FIG URE 8.40 Circuit for computing
note the artificial 1 V excitation.
From Figure 8.40 and our knowledge o f constructing Thevenin equivalems,
'
4000
Hence,
R
= 1 =
/,
1 -1 0 ^
Observe that in the actual circuit,
Q
is in parallel with 1 k il. Hence, the Thevenin equivalent
resistance seen by C is
R^,, = 1000
= lOOOll ( - 2.5) = - 2 .5 0 6 Q.
The time constant o f the first-order circuit is
T = R ^^C =
- 2 .5 0 6
X
10"'^ sec, or -2 .5 0 6 picosecond (psec)
The negative time constant spells instabilit)'. The complete response may be written directly with
the use o f equation 8.23, where x{t) =
Suppose a very small noise voltage,
To use equation 8.23, we need a^O"^) =
=E V, appears across C Then
= lO-le V
and
Chapter 8 • First Order Rl. and RC Circuits
363
To compute the equilibrium output voltage
Kg =
we open-circuit the capacitor and compute
with C open-circuited
/
-2 5
\ j
= 1 X ------ — 10'’ = -2 5 .0 6 3 V
V
1000-2.5/
From equation 8.23, the complete response is
= -2 5 .0 6 3 + (lO'^E + 2 5 .0 6 3 )e
0.339xl0‘^r
(8.28)
For any small positive initial capacitor voltage E, equation 8.28 implies that the output would
increase exponentially, had the op amp been ideal. Because this particular real op amp saturates at
15 V, the output more or less instantaneously saturates at 15 V, the phenomenon observed by the
student. Had the initial capacitance voltage been sufficiently negative,
-2 5 .0 6 3
£ < ------- j ---- V,
10-^
equation 8.28 implies that v
would saturate at - 1 5 V.
9. SUMMARY
This chapter has explored the behavior o f first-order RL and RC circuits (1) without sources for
given ICs, (2) for constant excitations (dc), (3) for piccewise constant excitations, and (4) with
switching under constant excitations. In general, first-order RL and RC circuits have only one
capacitor or one inductor present, although there are special conditions when more than one
inductor or capacitor can be present. Our discussion has presumed only one capacitor or one
inductor is present in the circuit.
Using a first-order constant-coefficient linear differential equation model o f the circuit, the chap­
ter sets forth rwo t}'pes o f exponential responses, the source-free response and the response when
constant independent sources are present. The source-free responses for the RL and RC circuits
have the exponential forms
ii(0= e
where
^
iiito)
vcit)=e
is the initial condition for the inductor and
itor. For an RC circuit, the time constant x=
the initial condition on the capac­
where R^^^ is theThevenin equivalent resistance
seen by the capacitor. For an RL circuit, the time constant T= L/R^i^, where R^^^now is theThevenin
equivalent resistance seen by the inductor.
When independent sources are present in the circuit, the response o f a first-order RC or RL cir­
cuit has the general form
I-10
x{t) = ,v(3c) + [,y(/o ) -
lie-
36-i
Chapter 8 • First Order RL and RC Circuits
forT > 0. Stated in words, this Formula is
elapsed lime
x{t) =[Final value] +{[Initial value]- [Final value])e
provided the time constant
t
> 0. When the time constant
t
<^<>nsvdn{
< 0, then it is necessary to modify
the interpretation as discussed in section 7, with equation 8.23 identifying the form o f any volt­
age or current in the circuit:
The time constants of- a circuit can be changed by switching within the circuit. By changing time
constants in a circuit, one can generate different t)'pes o f waveforms such as the triangular wave­
form o f Figure 8.32. As mentioned at the beginning o f the chapter, wave shaping is an important
application o f circuit design. When inductors, resistors, and capacitors are present in the same cir­
cuit, many other wave shapes can be generated. RLC circuits arc the topic o f the next chapter and
allow even greater freedom in waveform construction.
As a final application o f the concepts o f this chapter, we looked at the leaky integrator op amp cir­
cuit. Integrators are present in a host o f signal processing and control applications. Unfortunately,
ideal integrators do not exist in practice. The leaky integrator circuit o f Figure 8.35 provides a rea­
sonable model o f an ideal integrator.
10. TERM S AND CO N CEPTS
Com plete response: sum o f zero-input and zero-state responses.
D ifferential equation o f a circuit: equation in which a weighted sum o f derivatives o f an impor­
tant circuit variable (e.g., a voltage or current) is equated to a weighted sum o f derivatives
o f the source excitations to the circuit.
D ifferentiator circuit: op amp circuit whose output is a constant times the derivative o f the
input.
Equilibrium state o f a differential equation: c o n stan t, say x{t) = X^, th at satisfies the differential
eq u atio n in the variable A.*(r).
First-order differential equation o f a circuit: difterential equation whose highest derivative is
first order.
Forced response: that portion o f a complete response that has the same “exponent” as the input
excitation, provided the input excitation has exponents difl^erent from that o f the zeroinput response, under the condition that the input excitation is either (1) dc, (2) real
exponential, (3) sinusoidal, or (4) exponentially modulated sinusoidal.
Integrating factor method: mathematical technique for finding the solution o f a differential
equation in which multiplication by the integrating factor e~^’‘ on both sides o f the dif­
ferential equation leads to a new equation that can be explicitly integrated for a solution.
Integrator circuit: op amp circuit whose output is a constant times the integral o f the input.
Leaky integrator circuit: op amp circuit having a response approximating an ideal integrator, as
described in Example 8.13.
Chapter 8 • First Order RL and RC Circuits
365
Natural iirequency of a circuit: natural mode of “oscillation” of the circuit. For a first-order circuit having a response proportional to
it is the coefficient X in the exponent.
Natural response: that portion of the complete response that has the same exponents as the zeroinput response.
Passive RLC circuit: circuit consisting of resistors, inductors, and capacitors that can only store
and/or dissipate energy.
Sawtooth waveform: triangular waveform resembling the teeth on a saw blade and typically used
to drive the horizontal sweep of the electronic beam in an oscilloscope or a T V picture
tube.
Source-free response: response of a circuit in which sources are either absent or set to zero.
Step response: response, for ^> 0, of a relaxed single-input circuit to a unit step, i.e., a constant
excitation of unit amplitude.
Stray capacitance: small capacitance always present between a conductor and ground. It usually
can be ignored, but as Example 8.14 shows, it can critically affect the response of a cir­
cuit.
Superposition: in linear RC and RL circuits, the complete response is the superposition of the
relaxed circuit responses due to each source with all other sources set to zero, plus the
responses to each initial condition when all other initial conditions are set to zero and all
independent sources are set to zero.
Time constant: in a source-free first-order circuit, the time it takes for the circuit response to drop
to e~^ = 0.368 of its initial value. Roughly speaking, the response value must drop to a
litde over one-third of its initial value or rise to within one-third of its final value. For RL
circuits x = LIR^f^ and for RC circuits x =R^/jC.
Unit step (unction: function denoted «(/) whose value is 1 for f > 0 and 0 for ^< 0.
Unstable response: response whose magnitude increases without bound as t increases. The time
constant for first-order circuits is negative for an unstable response.
Zero-input response: response in which all sources are set to zero.
Zero-state response: response to a specified input signal or set of input signals given that the initial conditions are all set to zero.
w
w
^ It happens that all variables in a first-order RL or R C circuit satisfy a differential equation o f the same form. The
interpretation o f the solution is somewhat different fi-om what follows. A detailed explanation o f the general solu­
tion is presented in section 7.
Chapter 8 • First Order RL and RC Circuits
366
Problems
i,(t)
UNDRIVEN RESPONSE WITH
GIVEN INITIAL C O N D ITIO N S
1. For the RC circuit o f Figure P8.1, R = 2.5
kiQ and C = 50 |.iF.
(a) If V(^Q) = 10 V, find
Plot your
answer for 0 < / < 5x, where x is the cir­
cuit time constant.
(b) If V(\Q) = 10 V, find
with­
out differentiating your answer to part
(a). Plot your answer for 0 < f < 5x,
where x is the circuit time constant. At
what time is the energy stored in the
capacitor about 1% o f its initial value?
(c) Compute ;^(r) for «^^0) = 5 V and y<;^0)
= 20 V without doing any further calcu­
lations, i.e., by using the principle o f lin­
earity.
Figure P8.2
♦ 3 . In Figure P8.1, suppose R = 25 kH and
v^O) = 20 V.
(a) Find C so that v^O.25) = 2.7 0 6 7 V.
(b) Given your answer to part (a), find
C H EC K ; C = 5
pF
♦ 4 . In Figure P8.2, suppose R = 2.5 kl^ and
/^(O) = 20 mA.
(a) Find L so that at /, = 1 msec,
=
2.7 0 6 7 mA.
(b) Given your answer to part (a), find
C H E C K ; Z.= 1.25 H
5. The response o f an undriven parallel RC cir­
cuit is plotted in Figure P8.5. Find the time
ic(t)
constant of the circuit, at least approximately. If
+
C= 0.25 mF, find R.
. v,(t)
Figure P8.1
2. Consider the RL circuit o f Figure P8.2 in
which R = 50 Q and Z, = 0.1 mH.
(a) If the energy stored in the inductor at t
= 0 is 2 [ij, find /^(O) and
Plot your
answer for 0 < r < 5x, where x is the cir­
cuit time constant.
(b) If the energy stored in the inductor at t
= 0 is 2 pj, find Vjit) without differenti­
Figure P8.5
ating your answer to part (a). Plot your
answer for
0
< t < 5x, where x is the cir­
cuit time constant.
(c) Repeat part (b) for /j^(0 ) = 50 mA and
/^(O) = 250 mA. Hint; What principle
makes this a straightforward calculation
given your answer to part (a)?
6. In the circuit o f Figure P8.6, suppose /?j = 50
Q., Rj = 200 Q., L = 2 H, /^(O) = 100 mA, and
the switch opens at t = 50 msec.
(a) Find ij {t) for r ^ 0. Plot your answer in
MATLAB for 0 s / ^ 0.1 sec.
(b) Find y^(0) and
t > 0. Plot your
answer in MATLAB for 0 < / < 0.1 sec.
Chapter 8 * First Order RL and RC Circuits
36"
where T is the circuit time constant.
(b) Let a = - 1 1 . Compute the equivalent
resistance seen by the capacitor. Find
y^ f) for
> 0. Plot
for 0 < f < 2 t
where T is the circuit time constant.
(c) Find the range o f
(X
for which the time
constant is positive.
Figure P8.6
R.
7. In the circuit o f Figure P8.7, suppose /?j = 5
= 20 V,
kQ, /?2 = 20 k n , C = 50 pF H,
and the switch opens at r = 0.4 sec.
R,
(a) Find v^^t) for f > 0 . Plot your answer in
0.25 mF
MATLAB for 0 < f < 4 sec.
(b) Find
Figure P8.9
and /^(/), t > 0. Plot your
answer in MATLAB for 0 < r < 4 sec.
t=0.4s
C H E C K : (c) a > - 3
10. In the circuit o f Figure P8.10, /?, = 100 Q,
= 20
P = 2 0 0 ,1 = 0.5 H, and /'^(O) = 250
i„{t)
+
mA. The switch opens at / = 0.03 sec. Find the
,Vc(t)
Thevenin equivalent resistance seen by the
inductor before the switch opens, and then
compute /^(r) and
for r > 0.
Figure P8.7
Consider the circuit o f Figure P8.8.
(a) Find the value o f
and the initial con­
dition /^(O) so that /^(0.05 msec) =
R,
9.197 niA and /^(0.15 msec) = 1.2447
mA.
ijt)
1
(b) Given your answer to parr (a), compute
and plot /^(r) for 0 < r < 5x, where I is
the circuit time constant.
-
Figure P8.10
C H EC K :
= -2 5 n
ijt )
11. Consider the circuit o f Figure P 8 .ll in
1 kf)
which /?j = 25 ^2,
“ 50 Q, and L = 2.5 H.
Suppose /^(O) = 2 A.
80 mH
(a) With
CX
= 0.1, compute the Thevenin
equivalent resistance seen by the induc­
Figure P8.8
tor; then compute /^(f) for f > 0. Plot in
AN SW ER: (a) 600 Q, 25 mA
M A T I^ B for 0 < r < 5 t, where T is the
9. Consider the circuit of Figure P8.9, in which
= 100 Q and /?2 = 50 Q. Let
(a) Let a
= 500 mV.
= 7. Compute the equivalent
resistance seen by the capacitor. Find
i/^r) for f > 0. Plot
for 0 < ? < 5 t
time constant o f the circuit.
(b) W ith a = 0.1, compute
(c) Repeat part (a) for a = 0.02. Determine
the time, say r,, when the inductor has
lost 99% o f its initial stored energy.
Chapter 8 • First Order RL and R C Circuits
368
-
+
t=4RC
A
i,(t)
R.
R
—
1=0
.4R
\4R
R,
Figure P 8 .1 1
Figure P8.14
12. In Figure P8.12, the current source
has
been applied for a long time before the switch
opens at r = 0. Find /^(O^) and /^(r) for f > 0 in
terms o f /^, R, and L, where is in A, R in
and L in H. Sketch /^(r) for 0 < ^ < 4x where x
is the circuit time constant for f > 0.
15. In Figure P8.15, the current excitation is
= V^ii{-t) V. Find
and
V(^t) for ^ > 0 in terms o f
R, and C, where
R is in Q and C in F. Sketch v^^t) for 0 < f < 3x,
given by
where x is the circuit time constant for f > 0.
t=RC
.4R
Figure P8.15
Figure P8.12
13. In Figure P8.13, the current excitation is
given by
A. Find /^(O^) and i^it)
for r > 0 in terms o f /^, R, and L, where R is in
and L in H. Sketch /^(r) for 0 < r < 4x, where
X
4R
is the circuit time constant for r > 0.
16. Repeat Problem 14, except find
i(^t) for /■> 0.
and
17. Repeat Problem 15, except find
and
i(^t) for r > 0.
RESPONSE OF DRIVEN
CIRCUITS
18. Consider the RC circuit o f Figure P8.18 in
which R = 10 k n and C = 0.4 niF.
Figure P8.13
(a) If
14. In Figure PS. 14 the voltage source
=
has been applied for a long time before the
switch opens at r = 0. Find
and t»^r) for
r > 0 in terms o f V^, R, and C, where
is in V,
R in Q, and C in F. Sketch V(^t) for 0 < r < 3x,
where x is the circuit time constant for / > 0.
= 0 and
V, find
U(^t). Plot your answer for 0 < r < 4x,
where x is the circuit time constant.
(b) II /^(^-(O) = 10 V and
= 0, find
Plot your answer for 0 < / < 4x, where x
is the circuit time constant.
Now making use o f linearit)' and its
associated properties, compute the indi­
cated responses without any further cir­
cuit analysis.
(c) If V(^Q) = 10 V and
find V(^t).
= 2i)ti(t) V,
Chapter 8 • First Order RL and RC Circuits
(d) If «c<0) = - 2 0 V and i/,.„(») = -I0 « (» ) V,
find V(it).
369
20. In Figure P 8.20,
= 50 Q , 7?2 = 200 Q, C
= 2.5 mF, and the voltage excitation is given by
(e) If i/^O) = 10 V and v-J^t) = 20«(r) V.
find i({t) without differentiating your
answer to part (c). Plot your answer for
0 < r < 4 t , where x is the circuit time
constant.
where
= -1 0 V
and V^2 = 20 V.
(a) Find ^(;^0'*^) and V(it) for / > 0.
(b) Sketch v^^t) for 0 < f < 5x, where x is the
circuit time constant for r > 0.
(c) Identify the zero-input response (f > 0)
and the zero-state response (r > 0) for the
yoj
answer computed in part (a).
'.w O
(d) Now compute
for f > 0 assuming
the switch opens at r = 0.2 5 sec. Plot
your result ForO < f < 0.5 sec.
Figure P8.18
19. Consider the RL circuit o f Figure P 8.19.
Suppose /? = 100 Q, Z = 0.2 H.
(a) If /^(O) = 0 and
= 20«(^) V, find
i.
I
'j» (D
Plot your answer for 0 < / < 5t ,
where x is the circuit time constant.
Figure P8.20
(b) If ij{G) = - 5 0 mA and v-J^t) = 0, find
/^(r). Plot your answer for 0 < ^ < 5x,
where x is the circuit time constant.
21. In Figure P 8.21, R^ = 50 Q , i?2 = 200 Q, £
Now making use o f linearity and its
= 2 H , and the voltage excitation is given by
Kl = -1 0 V
associated properties, compute the indi­
cated responses without any further cir­
cuit analysis.
(c) If/^(O) = - 5 0 mA and
find
= 20«(r) V,
Plot your answer for 0 < r <
5x, where x is the circuit time constant.
(d) If/^(O) = 25 mA and y.„(r) = -1 0 « (f) V,
find
Plot your answer for 0 < ^ <
5x, where x is the circuit time constant.
(e) If /^(O) = - 5 0 mA and v-^{t) = 20«(/) V,
find Vj{t) without differentiating your
and 1^2 = 20 V.
(a) Find /^(O^) and ijit) for t> 0.
(b) Sketch i^{t) for 0 < r < 4x, where x is the
circuit time constant for t> 0.
(c) Identify the zero-input response (t > 0)
and the zero-state response (^ > 0) for the
answer computed in part (a).
(d) Now compute Vj{t) for ? > 0 assuming
the switch opens at ^ = 0,0 4 sec. Plot
your result for 0 < f < 0 .2 sec.
answer to part (c). Plot your answer for
0 < ^ < 5x, where x is the circuit time
constant.
I'
'»<b O
Figure P8.21
Figure PS. 19
22. Consider the RC circuit o f Figure P8.22a in
which
= V’q «((), where Kq = 100 V.
(a) Find
'Vw>
the time constant o f the cir-
Chapter 8 • First Order RL and RC Circuits
370
cuit, Rp and /?2
circuit
response v^^t) is given by Figure P8.22b.
n
Assume C = 0.25 mE
and C so that the circuit
(b) Find
response v^t) is given by Figure P8.22b.
Assume i?2 = 10
n
+
;vc(t)
'.( s O
r\
(b)
Figure P8.23
(a)
24. In the circuit o f Figure P 8.24, V(^Qr) = 2$
V,
= 50u{t) mA, and
= 25«(/) mA
(a) Find the zero-input response, i.e., the
response due only to the initial condi­
tion.
(b) Find
for r > 0 due only to
(c) Find V(^t) for ? > 0 due only to
(d) Find the zero-state response.
(e) Find the complete response
for t >
r>
0.
Time in seconds
(f) Suppose the initial condition is doubled
(b)
and each independent source is cut in
Figure P8.22
half Find the new complete response
23. Consider the RL circuit o f Figure P 8.23a in
which
using linearity.
= Vqu(/), where Vq = 100 V.
(a) If Z = 2 H , find
the circuit time
constant, R^, and /?2 so that the circuit
1 kn
response i^(t) is given by Figure P8.23b.
(b) If /?2 = 2 k n , find ij^(0*), R-^, and L so
that the circuit response ij{t) is given by
Figure P8.23b.
V„u(t)
r>
i,(t)
d)
Figure P8.24
25. In Figure P 8.25 Ri = 2 0 0 Q, R2 = 6 0 0 Q,
(a)
T?3 = 650Q , Z = 20 H,
= -1 0 0 « (-r ) +
50u{t) V, and
= 5 0 « ( ? - 0.5) V. Compute
i^it) for ^> 0. Plot your answer using MATLAB
or its equivalent for 0 < r < 8x.
Chapter 8 • First Order RL and RC Circuits
371
directly without differentiating your
answer to part (a).
(c) W hat are the new responses if the value
o f each source is doubled?
Figure P8.25
26. Repeat Problem 25, except compute VjjJ)
for t>Q.
27. In Figure P 8.27
= 2 0 0 Q, /?2 = 6 0 0 Q,
= 8 5 0 a . C = 2.5 mF, v^^{t) = - 5 0 « M +
100«W V, and
Figure P8.30
= - 5 0 « ( f - 5 ) V. Compute
31. The switch in the circuit o f Figure P8.31
V(^t) for t > 0. Plot your answer using
has been open for a long time before it is closed
MATLAB or its equivalent for 0 < ? < 6x.
at t = 0 . Suppose
= 6R, R2 = 30i?, R^ = 20R,
and
= Vqu{T - /). In terms of Vq, R, C,
and T = 6RC,
(a) Find V(^0~) and t/(;;(0'^).
(b) Find the Thevenin equivalent seen by
the capacitance for 0 < ^ < T.
(c) Using the Thevenin equivalent found in
pan (b), find an expression for Vf^t)
Figure P8.27
valid over 0 < t < 7’.
28. Repeat Problem 27, except compute the
(d) Find the expressions for V(iT~) and
capacitor current i^t) for t > 0 .
29. For the circuit o f Figure P 8.29,
=
-2Qu{-t) + 20«(^) V. Find V(iO~) and v^^t) for
r > 0. Plot V({t) for 0 < ? < 40 msec.
(e) Find the time constant valid for t> T.
(f) Find V(^t) for t> T.
(g) Plot
for 0 < r < 4 7 using MATLAB.
20 msec
2kO
'. w ( D
8kO
=0
1.6 kn
+
Vc(t).
S jjF
o
20 V
+
.Vc(t)
vjt)
Figure P8.29
Figure P8.31
30. Consider the circuit o f Figure P 8.30 in
3 2 . For the circuit of Figure P 8.32, -V^ = - 1 0
which /?j = 300j^, /?2 = 800 Cl, R^ - 600 Q, L
= 4 H,
- -24u{-t) + 24«(/) V, and
=
V, Kj = 20 V, R^ = 6 0 0 Q, /?2 = 2 0 0 Cl, and C
24u{t) V.
= 12.5 hF(a) Find y^O"^).
(a) Compute the response ii{t) for ? > 0.
Plot for 0 < ^ < 4x where x is the circuit
time constant.
(b) Find the inductor voltage v^{t) for r > 0
(b) Using the initial condition computed in
part (a), find
< t< 160 msec.
Plot the result for 0
Chapter 8 • First Order RL and RC Circuits
372
34. The voltage waveform
of Figure
P8.34a drives the circuit o f Figure P8.34b. The
voltage-controlled switch Si closes when the
capacitor voltage goes positive and opens when
the capacitor voltage v^^t) goes negative.
Compute the voltage Vf^t) across the capacitor.
Assume that
has been at - 1 0 V for ?< 0
for a very long time. Hint: Use the elapsed time
formula as needed.
>
20V
o
v„,(t)
5
t(Msec)
2.5
(b)
--10V
Figure P8.32 (a) Pulse driving
RC circuit of part (b).
----------(a)
IMegO
33. For the circuit of Figure P 8 .3 3 ,, -V^ = - 1 0 V,
Kj = 20 V, /?! = 80 Q.7?2 = 20 Q, andZ = 4 H.
(a) Find
(b) Using the initial condition computed in
part (a), find
Plot the result for 0
< r < 160 msec.
(b)
t
Figure P8.34 (a) Pulse waveform exciting
RC circuit in part (b).
V
35. Repeat Problem 34, except find i(^t) for t>
80
t(msec)
40
36. Consider the circuit o f Figure P8.36. Suppose
--V.0
(a)
V ({0 ) = 0
and find V(^i) for
0
as follows:
(a) Find the Thevenin equivalent circuit
seen by the capacitor.
(b) Find the complete response V(^t) for t >
0. W hat is y(j(oo)?
- v,(t) +
400
(b)
6
5kO
C = 0.1 F
IV
V,
Figure P 8 .3 3
Figure P 8 .3 6
lOlv.
373
Chapter 8 • First Order RI. and RC Circuits
OP AMP CIRCUITS
41. Figure P 8 .4 la shows an op amp integrator
with positive gain, and Figure P 8 .4 lb shows a
= K
37. In the circuit o f Figure P8.37,
sin(cor)«(r) V and all capacitor voltages are zero
at f = 0. Find
and
of/e, C,/C and (0.
differentiator with positive gain for the con­
stant K>
(a) For each o f the circuits find a literal
for / > 0 in terms
expression for
in terms of
(b) For /?j = 10 k n and (7 = 0.1 mF, find
Vo,„{t) when
= 100sin(20r)//(r) mV
assuming that
= 0 in each case.
v.(t)
R.
O v/W
+
Vjt)
A .
Figure P8.37
KR
38. In the circuit o f Figure PS.38, /? = 10 klT2,
C = 10 jiF,
= 10 sin(50f)«(r) mV, and all
capacitor voltages are zero at / = 0. Find
and plot it for 0 < r < 6 sec.
Figure P8.41
M ISCELLAN EOUS
Figure P8.38
42. Although most o f the first-order circuits
39. Repeat Problem 38 for
= \QOe~~‘ii{t)
mV, But plot from 0 to 6 seconds.
considered in the text have only one capacitor
or one inductor, it is possible to have a firstorder circuit containing more than one energ}'
40. In the circuit o f Figure P8.40, /?j = 10 kH,
storage element. Consider the situation depict­
Rf = 40 kQ, C = 12.5 |.iF, and v^{t) = -lOOw(f)
ed in Figures P8.42a and b. Here
mV.
Vcii^*) are given. The networks A^, and N j are
(a) If V({0~) = 0, compute
for r > 0.
and
equivalent under two conditions:
(b) Repeat part (a) for V(jS^~) = 50 mV.
- v,(t) +
r
C, + Co
Prove this equivalence using the integral rela­
tionship o f a capacitor to show that the i-v ter­
minal conditions are the same for both
N,.
Figure P 8 .4 0
and
374
Chapter 8 * First Order RL and RC Circuits
C .d - v ,
N1
(a)
(b)
Figure P8.42
43. As mentioned in Problem 42, although
most of the first-order circuits considered in the
text have only one inductor or one capacitor, it
is possible to have a first-order circuit contain­
ing more than one energy storage element.
Consider the situation depicted in Figures
P8.43a and b. Here
and
given. The networks
and A/'j are equivalent
under two conditions:
^eq -
~
Prove this equivalence using the int^ral relation­
ship of an inductor to show that the i-v terminal
conditions are the same for both A^j and A^2 -
(a)
45. In the circuit of Figure P8.45, Cj = C2 = 2
F and the switch closes at time ^= 0. The initial
conditions on the two capacitors are t/^(0 ) =
4 V and vc2(0-) = 0 V.
(a) For = 0.5 ft, find an expression for the
current /^(f) for ^> 0.
(b) For /? = 0.5 ft, find
for ? > 0 and
for f > 0. Note that
^ 0 for
all r > 0.
(c) Compute the energy stored in each
capacitor at r = 0^. Also compute the
energy stored in each capacitor at r = c».
Finally, compute the decrease in total
energy stored in the capacitors fi-om t =
O'^ to ^= 00.
(d) Compute the energy dissipated in the
0.5 ft resistor fi-om r = 0"^ to r = 00. Verify
that the energy dissipated in the resistor
equals the decrease in total energy stored
in the capacitors fi-om r = 0* to ^= 00.
(e) Does the dissipated energy depend upon
the value of P?. What does R affect?
Verify that conservation of energy holds
for the circuit.
(b)
‘r
Figure P8.43
Figure P8.45
44. In the circuit of Figure P8.44, suppose /?j =
50 Q, /?2 = 200 a , q = 0.06 F, Cj = 0.3 F,
yQ(0“) = 15 V, and i'qCO") = 5 V. Let
=
40tt(r) V. Use the equivalence set forth in
Problem 42 to compute
for r > 0.
C. - ±
C, - t
V,
46. Repeat Problem 45 for the circuit of Figure
P8.45 when q = 1 F, Cj = 0.25 F, i/^iCO-) = 3
V, and t/f^(0~) = 8 V.
47. In the circuit of Figure P8.47,
=
1 lOtt(-f) + 220u(t) mV, Zj = 110 mH, ^2 = 11
mH, and /? = 10 ft. Compute and plot the
waveforms for inU) and /^(/). Hint: Adapt the
results of Problem 43 to ^e case of two paral­
lel inductors with initial currents.
Figure P 8 .4 4
r>
375
Chapter 8 • First Order RL and RC Circuits
and K 2 . Find
for appropriate
'JO
and
K2 in terms of K^.
iA n i'“
R
(b) Suppose the input v^{t) = -\2u{-t) +
Q
24«(/). Find y(0“ ) by inspection of the
L
circuit, and v(0+) by the principle of
conservation of charge.
Figure P8.47
(c) Use equation 8.1 7 to write down direct­
48. Repeat Problem 4 7 for the case where v-J^t)
ly the answer for v{t), f > 0. Had y(0“ )
= 220«W mV,
been used, would the answer still be cor­
= 4 4 mA, and
=
rect?
11 mA.
51. The solution to the basic RL or RC differ­
49. Consider the circuit of Figure P 8.49 in
which Cj = 1 F, Cj = 4 F, v-^ = 10 V, and R = 2
ential equation in this chapter, equation 8.3,
a.
builds on the integral solution o f equation 8.4,
(a) Compute v^Qr) and Vj^{Q*). Hint: How
which is valid for arbitrary y(/). This powerful
does the charge distribute over the two
formula will be studied in a course on differen­
capacitors at ^ = 0^?
tial equation theory. Wheny(r) = F, a constant,
(b) Compute Vjfi) for r > 0.
it is possible to develop an alternative deriva­
tion o f equation 8 .1 7 using no more than some
basic knowledge of calculus. Since the solution
to the source-free case is the exponential, it is
reasonable to expect (or to try) a solution for
the constant input case o f the form
'> ^ '
( 1)
Figure P8.49
and K2 are two constants to be deter­
where
mined. The constant K 2 arises from the con­
50. The circuit o f Figure P 8.50 contains two
stant input, suggesting that the response would
capacitors.
intuitively contain a constant term.
(a) Substitute equation 1 into equation
8 .16. You should obtain the result
=
Pi.
(b) With K2 determined, evaluate x(/) at f =
tQ* to obtain an expression for
. Your
result should be
Figure P8.50
(a) Suppose vj^i) = K^u{t) V. Show that for t
W
> 0, the voltage v{i) satisfies the firstorder differential equation
dv
It
= -K
iv + K 2
x Uq
) - F t
(2)
(c) Finally, substitute K2 = Fx and equation
2 into equation 1.
Chapter 8 • First Order RL and RC Circuits
3 "6
(ii) S is at position B when v is greater than
APPLICATIONS
52. An approximate sawtooth waveform can be
60 V and decreasing, and it moves to
produced by charging and discharging a capac­
position A when i drops to 1 mA and v
itor with widely different time constants. The
drops to 60 V.
circuit o f Figure P8.52 ilkistrates the idea. Vj„{t)
= 20 kQ,
= 10 V,
Assume that at r = 0, switch S is at A,
= 1 kQ, and C= 10 pF.
The switch S is operated as follows: S has been
and z^„,^;(0) = 60 V. Find
at position B for a long time, and S is moved to
cycle o f operation (i.e., charging and dis­
for one
position A at /■ = 0 to charge the capacitor.
charging the capacitor), and roughly
When V(~increases to 9 V, switch S is moved to
sketch the waveform. Whau is the fre­
position B to discharge the capacitor. When
quency o f the sawtooth waveform?
decreases to 1 V, switch S is moved to position
A to charge the capacitor again. The process
repeats indefinitely.
(a) Compute the waveform o f
for four
switchings.
(b) Plot
Is the name “sawtooth wave­
form” appropriate? What is the frequen­
cy in hertz o f the sawtooth waveform?
i (mA)
idealized i-v curve
ofa neon lamp
• (not to scale)
/:
A
/
slope
= 1m U
/
Figure P8.52
V
slope
53. The sawtooth waveform is used in T V sets
negative resistance
region
i
:
= 0.5 mU
60
(b)
:
N fc i
v(V)
----------- ►
90
(0
Figure P8.53
and oscilloscopes to control the horizontal
motion of the electron beam that sweeps across
m is based on a hypothetical
the screen. One method o f generating such a
energy storage system using an inductor and a
waveform is to repeatedly charge a capacitor
solar cell. Consider the circuit o f Figure P S.54.
with a large time constant and then repeatedly
During the day, the solar cell stores energy by
discharge it with a very small time constant.
increasing the current in the inductor. During
The circuit in the shaded box o f Figure P8.53a
the night, the stored energy is used to power
is a crude functional model for the neon bulb
lights and appliances. Energy from the solar cell
in Figure P8.53b (type 5AB, costing about 75
is scored in the inductor during 0 < r < 7’j. At /
cents), whose i-v characteristic is shown in
= 0, the beginning o f the storage interval, /^(O”)
Figure PS.53c. The switch S in the model oper­
= 0. At r =
ates as follows;
storing energy in the solar cell via the source
(i) S is at position A when v is less than 90
the device
is switched from
V and increasing, and it moves to B when
Vsolar <^0 powering a light denoted by Ry Note
that diere is some overlap in the switching
reaches 90 V (the breakdown volt­
movement; this is to ensure continuity o f the
V
age).
inductor current. At r = T^, the TV, represent­
ed by Rj, is also turned on.
Chapter 8 • First Order RL and RC Circuits
377
Remark: All answers to parts (a) to (f) should be
in terms o f
/?,. and R^.
photo tim er used for timing the light in pho­
(a) Draw a simplified equivalent circuit
tographic enlarger and printing boxes. Briefly,
5 5 . The circuit o f Figure P8.55 is a transistor
with three circuit elements for 0 < ? < T j,
the circuit operates as follows. When the relay
indicating all device values.
contact closes, the lamp is lit. When the contact
(b) Construct an expression for
0 < ^<
opens, the lamp is turned off. The relay has a
4 0 0 0 Q dc resistance and a negligible induc­
(c) Draw a simplified equivalent circuit
tance. The pickup current is 2 mA, and the
with two circuit elements for T-^<t<
dropout current is 0.5 mA; i.e., the contact
indicating all device values.
closes when the relay current increases from
(d) Construct an expression for /^(/), T^^<t
< Tj. You will need a value or expression
for
zero to 2 mA, and it opens when the current
drops below 0.5 mA.
After obtaining your expres­
sion, for simplicity, let
denote
To use the timer, switch S2 is closed first.
Switch Sj is normally in the B position. When
(e) Draw a simplified equivalent circuit
it is thrown momentarily to position A, the bat­
with two circuit elements for T2 < t,
tery charges the 1000 pF electrolytic capacitor
C to 1.5 V. When Sj is then thrown back to
indicating all device values.
(f) Construct an expression for
Tj < t.
position B at /^= 0, the capacitor discharges and
You will need an expression for
produces a current iy, which, after amplifica­
After obtaining your expression, for sim­
tion by the transistor, actuates the relay and
plicity, let ^Tl- denote /^(jT2 )•
Remark: For the remaining parts, all
amplified current drops below a point for the
answers are to be given in terms of
relay to open and the lamp is turned off.
K kf
tori’
^1>
^ 2> “ d
This is to prevent the substitution of
Compute
possibly incorrect answers from prior
the middle of its full range (i.e., only 5 kQ is
parts for
used in the circuit).
(g) For each o f the four devices
^ston>
^store
down an expres­
sion for the power absorbed at time t.
Call the results
and PRstore’
(h) For time t give an expression for the
energy Wj(J) stored in the inductor if
W i( r = 0 ) = 0.
^soljr
-N
-------------- ^
Figure P 8 .5 4
turns on the lamp. At some later instant
the
if the 10 kH potentiometer is set at
378
Chapter 8 • First Order RL and RC Circuits
1.5V
lO kn
E
Potentionmeter
crude transistor
circuit model
Figure P8.55
56. The circuit o f Figure P 8.56 suggests a way of generating a sustained sinusoidal oscillation. All
op amps are assumed to be ideal. Capacitors, C = 0.1 fiF are uncharged at f = 0. The first two op
amps are differentiators and the last is an inverting amplifier.
(a) With switch S at position A and v^{t) = sin(lOOOr) V, find vj^t),
and
for r > 0.
(b) If at a later instant switch S is quickly moved to position B, what would you expect
to be?
Figure P8.56
C
H
A
P
T
E
R
Second Order Linear Circuits
Warming up snacks in a microwave oven is
a common activity in student dormitories.
It works much faster than a conventional
s B s m
oven: heating a sandwich takes about 30
seconds. How does the microwave oven do
this? While a precise explanation is beyond
the scope o f this text, the basic principle
can be understood through the properties
o f a simple LC circuit.
Recall that two conducting plates separated
by a dielectric (insulating material) form a
capacitor. Suppose some food were placed
between the plates in place o f an ordinary
dielectric. The food itself would act as a dielectric. Ordinar)' food contains a great number o f
water molecules. Each water molecule has a positively charged end and a negatively charge end,
with their orientations totally random for uncharged plates, as illustrated in part (a) o f the figure
below. Applying a sufficiently high dc voltage to the plates sets up an electric field produced by
the charge deposited on the plates. This causes the water molecules to align themselves with the
field as illustrated in parr (b). If the polarity o f the dc voltage is reversed, the molecules will realign
in the opposite direction as illustrated in part (c). If the polarity o f the applied voltage is reversed
repeatedly, then the water molecules will repeatedly flip their orientations. In doing so, the water
molecules encounter considerable friction, resulting in a buildup o f heat, which cooks the food.
Microwave cooking is therefore very different from conventional cooking. Instead of heat coming
from the outside, the heat is generated inside the food itself.
380
Chapter 9 • Scconcl Order Linear Circuits
food
food
conducting plate
conducting plate
(b)
(a)
food
0
Reversal of the polarity o f the applied voltage at a low frequency can be easily achieved with the
circuit elements studied in earlier chapters: the resistor, the capacitor, and the inductor. However,
the friction-induced heat production is inefficient at low frequency, l b produce a useful amount
o f heat for cooking purposes, ver)' high frequencies must be used. The t}'pical frequency used in
a microwave oven is 2.45 gigahertz, i.e., the water molecules reverse their orientations 2 x 2.45 x
10^ times per second. At such a high frequency, capacitors and inductors are quite different in
their behavior from their conventional forms. For example, the LC circuit becomes a “resonant
cavit}” and the connecting wire becomes a “waveguide.” These microwave components will be
studied in a future field theory course. T he theory studied in this chapter will enable us to under­
stand the low-frequenc}' version o f the phenomenon, i.e., how a connected inductor and capaci­
tor can produce oscillator)' voltage and current waveforms.
CHAPTER OBJECTIVES
1.
Investigate the voltage-current interactions that occur when an ideal inductor is con­
nected to an ideal capacitor with initial stored charge.
2.
Use a second-order differential equation for modeling the series RLC and parallel RLC
circuits.
3.
Learn to solve a second-order differential equation circuit model by first finding the nat­
ural frequencies o f the circuit, then looking up the general solution form, and finally
determining the associated arbitrary constants.
Chapter 9 • Sccond Order Linear Circuits
3<S I
4.
Define and understand the concepts of underdamped, overdamped, and critically
5.
damped responses.
Investigate and understand the underlying principles o f various oscillator circuits.
SECTION HEADINGS
1.
Introduction
2.
Discharging a Capacitor through an Inductor
3.
Source-Free Second-Order Linear Networks
4.
Second-Order Linear Networks with Constant Inputs
5.
Oscillator Application
6.
Summary
7.
Terms and Concepts
8.
Problems
1. IN TRO DUCTION
The previous chapter developed techniques for computing the responses of first-order linear net­
works, either without sources or with dc (constant) sources, having first-order linear differential
equation models. Recall that the source-free response contains only real exponential terms.
This chapter focuses on second-order linear networks having second-order linear differential equa­
tion models. Usually, but not always, a second-order net\vork contains t^vo energ)^ storage ele­
ments, either {L, Q , (C Q , or {L, L). Second-order circuits have a wide variety o f response wave­
forms: exponentials
sinusoids (/l,cos(coy) + y4-,sin(coy)), exponentially damped
sinusoids, and exponentially growing sinusoids, among others. Tables 9.1 and 9.2 catalogue the
various response types. W ith no sources or with constant-value sources, some straightforward
extensions o f the solution methods o f Chapter 8 are sufficient to compute the various responses.
The behavior of second-order circuits is a microcosm o f the behavior o f higher-order circuits and
systems. Many higher-order systems can be broken down into cascades o f second-order systems or
sums o f second-order systems. This suggests that our exploration o f second-order circuits can
build a core knowledge base for understanding the behavior o f higher-order, more complex phys­
ical systems.
Many introductory texts discuss only parallel and series RLC circuits, stating separate formulas for
the responses o f each. Our approach seeks a unified treatment. To this end, we formulate a basic
second-order differential equation circuit model. The associated solution techniques become
applicable to any second-order linear nersvork and, for that matter, to second-order mechanical
systems.
An oscillator circuit (section 5) motivates our study o f second-order linear networks. The chapter
contains several other practical examples illustrative o f the wide variety o f second-order circuit
382
Chapter 9 • Second Order Linear Circuits
applications. Some advanced applications pertinent to higher-level courses include low-pass, highpass, and bandpass filtering (covered later in the text); dc motor analysis; position control; and
many others. Most important, the concepts presented in this chapter are common to a host o f
engineering problems and disciplines. Hence, the techniques and concepts described here will
prove useful time and time again.
2. DISCHARGIN G A CAPACITOR TH RO UGH AN IN DUCTO R
Chapter 8 showed that the voltage o f an initially charged capacitor in parallel with a resistor
decreases exponentially to zero: the capacitor discharges its stored energy through the resistor.
When an inductor replaces the resistor, as in Figure 9.1a, very different voltage waveforms emerge
for V(-{t) and
In order to construct these new waveforms, we first develop a differential equa­
tion model o f the LC circuit.
EX A M PLE 9 .1 . The goal o f this example is to develop a differential equation model o f the cir­
cuit in Figure 9.1b. In Figure 9.1a, with the switch S in position A, the voltage source charges the
capacitor to
volts. At / = 0, the switch moves to position B, resulting in the new circuit o f
Figure 9.1 b, valid for f > 0.
A
B
+
+
(a)
(b)
FICiURF 9.1 (a) A voltage source charges a capacitor, (b) An LC second-order linear network in
which the cnerg)’ stored in the capacitor in part (a) is passed back and forth to the inductor.
S
o l u t io n
.
Step 1. Write down the terminal i-v relationship for the capacitor and itiductor; then apply KCL and
KVL, respectively. Using the i-v relationships for L and C (see Chapter 7) in conjunction with KCL
and KVL, it follow's that
ih C
_ >c _
dt
and
±L_
c
V
J V
C definition
KCL
di
(9.1a)
(9 .1 b )
V
L definition
c
KVL
Chapter 9 • Second Order Linear Circuits
383
Step 2. Obtain a dijferential equation in the capacitor voltage, V(j^t). For this, first differentiate
equation 9.1a to obtain
1 dii^
dt
C dt
Substituting equation 9.1b into this equation yields
^
= - — VC
dt
LC ^
(9.2)
Equation 9.2 is a second-order linear differential equation circuit model o f Figure 9.1b in terms o f
the unknown capacitor voltage, v^t). Equation 9.2 stipulates that the second-order derivative
j
o f the unknown function, v^it), must equal the function itself multiplied by a negative constant,-----
LC
Step 3. Obtain a differential equation in the current, ij{t). An alternative circuit model in i^ is obtained
by first differentiating equation 9.1b and substituting equation 9.1a into the result to produce
Equation 9.3 has precisely the same form as 9.2: the second-order derivative o f the unknown
function, ij{t) , equals the function itself multiplied by a negative constant, -----. This similarity
suggests a similarity o f solutions, which we shall pursue further.
Exercise. Fill in the details o f the derivation o f equation 9.3 from 9.1.
Our next goal is to construct the waveforms V(^t) and ij{t), which are the solutions o f the differ­
ential equations 9.2 and 9.3. Although differential equations are not usually part o f the common
background o f students in a beginning course on circuits, the solutions o f equations 9.2 and 9.3
do not demand this background. Some elementary knowledge o f differential calculus is sufficient.
Specifically, recall the differential properties o f the sine and cosine functions:
d
d
dt
dt
— sin(co/+ 0 ) = (o co s(o )/-t-0 ) and — c o s ( o ) / + 0 ) = - a ) s i n ( c o / + 0 )
Differentiating a second time yields
d2
^y2
— 7 sin((or-f-0) = - 0) “ sin(co/+ 0 ) and — 7 cos(o)/+ 0 ) = - c a “ cos(co/+ 0 )
dt-
dt-
In both cases, the second derivative equals the function itself multiplied by a negative constant.
This is precisely the propert}' required by equations 9.2 and 9.3. Thus one reasonably assumes that
the solutions o f equations 9.2 and 9.3 have the general forms
V(^t)
and
=
K cos(ojr
+ 0)
(9 .4 a )
38^
Chapter 9 • Sccoiul Order Linear Circuits
/•^(r) = /^cos(tor+0)
(9.4b)
7'hese forms are general because the cosine function can be replaced by the sine function with a
proper change in the phase angle. Specifically, we note that A'sin((i)/ + (j)) = A'cos(co/ + (j) - 0.5ti)
= K cos((or + 0) with 0 = ([) - ().5ti. Computing values for (o, A', and 0 specifies the solutions to
equations 9.2 and 9.3.
E X A M PLE 9.2. Find A"and B for the capacitor voltage in equation 9.4a.
So
lu t io n
.
Step 1. Differentiate equation 9.4a to obtain
dvrit)
—
—
(It
^
= -A c o sin (O )/+ 0 )
/'o S'!
Step 2. Dijferentiate n second time. Differentiating equation 9.5 (the second derivative o f 9.4)
yields
d~Vr
^
«
">
-----^ = -A^O)“ cos(O)/+0) = -(O “ i ’<7
dt~
___
(9.6)
Step 3. Match the coefficients o f equation 9.6 with those o f 9.2 to specify O). Under this matching,
' o1r CO =
(0 ^ = -----
LC
7)
— j= =
fo
fiC
^ ^
^
Equation 9.7 specifies co, the angular frequency o f oscillation, in rad/sec, o f the capacitor voltage.
Step 4. Compute K an d f) in equation 9.4a. These two constants depend on the initial conditions
as follows: when the switch is at position A, the capacitor is charged up to I^q volts and the induc­
tor current is zero; immediately after the switch moves to position B, i.e., at r = 0+, the continu-
^
it>' properry o f the capacitor voltage ensures that y^^O*) = Vj, and^the continuity property o f the
inductor current ensures that //(O^) = 0. The initial value,
^ , ’ is now calculated from
equation 9.1a as
^/r(--(0 ) i( ^ ( 0 )
------------ = ----------- = ------------- = ()
dt
C
^
C
Evaluating equations 9.4a and 9.5 at / = 0^ , we have
= K c o m = v;,
(9.8a)
and
</v'r(0"’ )
— ^
= -A o )s m (e ) = 0
(9 g y
From equation 9.8b, 0 = 0. Consequently from 9.8a, K = V^^. Hence the capacitor voltage, i.e.,
the solution o f the second-order differential equation 9.2, is
Vc(f) = VqCOS
[J
lc
)
(9.9)
^
Chapter 9 • Second Order Linear Circuits
38S
As per equation 9.1a, one can obtain ij{t) directly by differentiating equation 9.9 and multiply­
ing by —C. However, one could aLso solve equation 9.3 by repeating the above steps to arrive at
the same answer.
Exercise. Assuming that /^(r) = K coslcor + 6), solve for (o, K, and 0 in terms o f the initial condi­
tions, and show that
Several very interesting and significant facts about this parallel ZCcircuit and the solution method
are apparent:
(1)
For the source-free LC circuit o f Figure 9.1, the voltage and current responses are sinusoidal
waveforms with an angular frequency equal to —_L _ . Since the amplitude o f sinusoidal
oscillations remains constant (i.e., does not
VZc
damp out), the circuit is said to be undamped.
(2)
rhe frequenq^ (o, depends on the values o f L and C only, while the amplitude K and the
phase angle 0 depend on L, C, and the initial values o f the capacitor voltage and inductor
current.
(3)
Although the instantaneous energ}" stored in the capacitor, \V^{t), and the instantaneous
energy stored in the inductor,
both vary with time, their sum is constant. (This is
investigated in a homework exercise.) Physically there is a continuous exchange o f the energ)' stored in the magnetic field o f the inductor and that stored in the electric field of the
capacitor, with no net energy loss. This is analogous to a frictionless hanging mass-spring
system: because o f the absence o f friction, the up-and-down motion of the mass never
stops; in such a mechanical system there is a continuous interchange between potential and
kinetic energy.
Figure 9.1 shows what is, in theory, the simplest circuit that generates sinusoidal waveforms. Such
an electronic circuit is an (idealized) oscillator circuit. Oscillator circuits play an important role
in many communication and instrumentation systems.
3. SOURCE-FREE SECON D-ORDER LINEAR N ETW ORKS
Unlike their ideal counterparts, real capacitors and inductors have resistances. A better under­
standing o f a realistic oscillator entails the analysis o f an RLC circuit. This section investigates
source-free RLC circuits having two energ)' storage elements. Our investigation begins with the
development o f the differential equation models o f the series and parallel /^//.'circuits. Both mod­
els are special cases o f an undriven general second-order linear diflerential equation. Hence we will
discuss the solution o f a general second-order linear differential equation and adapt the solution
to the series and parallel RLC circuits. We will also illustrate the theory with a second-order cir­
cuit that is not a parallel or series RLC.
386
Chapter 9 • Second Order Linear Circuits
Developmeut o f Dijferential Equation Models for Series/Parallel RLC Ciraiit
The first goal o f this section is to develop differential equation models for series and parallel RLC
circuits as detailed in the following example.
EXA M PLE 9 .3 . For the series and parallel RLC circuits shown in Figure 9.2, develop two sec­
ond-order differential equation models (one in
and one in
for each circuit.
+
+
V
''c = V''l
= V
''r
(b)
FIGURE 9.2 (a) Series RLC circuit, (b) Parallel RLC circuit. Passive sign convention is assumed as usual.
S
o l u t io n
Wc do this in “parallel” rather than in “series.”
Step 1. Apply KVL to the series RLC.
Step 1. Apply KCL to the parallel RLC.
Step 2. Choose i^^as circuit variable
and express
v^, and Vq in terms
Step 2. Choose
and express i^,
as circuit variable
and iQ in terms o f
^ f‘L(ii
1 ‘
Ril + L — + — / ^ ( t V t = 0
dt
C—
•oo
'
\’f^ I f
,
„ dv/"
— +—
vrix)dx + C — ^ = 0
R
L ^ ^
dt
Differentiate, rearrange terms,
Differentiate, rearrange terms, and
and divide by L to obtain
divide by C to obtain
d^il
Rdi^
1 .
^
— ^ + ------ ^ + ----- 1, = 0
dr
L dt
LC ^
d^Vf
dt-
d\>r
RC dt
1
H-------= 0
LC
Step 3. Choose Vq as the circuit
Step 3. Choose ij as the circuit
variable and express Vj^and
terms o f Vq
variable and express
terms o f i^.
in
Again using the KVL o f step 1,
d ir
Rin -f" L ------- 1= 0
^
dt
^
arid i^ in
Again using the KCL o f step 1,
— -H/, + C — ^ = 0
R
^
dt
Chapter 9 • Second Order Linear Circuits
38‘
= C dv^^dt,
Hence, substituting for
L dij/dt,
Hence, substituting for
rearranging, and dividing through by
rearranging, and dividing through by
LC yields
LC yields
R dvr
L dt
d t‘
llL
dr
LC
RC dt
+—
LC
Each circuit has two second-order differential equation models, one each for
= 0
and
as the
unknown quantity.
Exercise. Show that
and Vj^ satisfy second-order differential equations similar to those
derived in Example 9.3.
Solution o f the General Second-Order Dijferential Equation Model
The final differential equations o f Example 9.3 force the current
or the voltage
to satisfy cer­
tain differential constraints. All four (differential) equations have the general form
d^x
dx
— ~ + h ----- h (.'.V = 0
dtdt
(9 . 10)
for appropriate constants b and c, where .v is either ij or
Equation 9.10 stipulates that the sec­
ond derivative o f the function x{t) plus h times the first derivative o f x{t) plus c times x(t) itself
adds to zero at all times, t. Unlike the example o f section 2, where a sinusoidal solution was easi­
ly predicted, the present differential equation requires a more careful mathematical analysis. Recall
from elementary calculus that the derivative o f an exponential is an exponential. Thus the first
and second derivatives o f an exponential are proportional to the original exponential. This sug­
gests postulating a solution o f the form at(^) = Ke^‘ where we make no a priori assumptions about
s. If it is truly a solution, it must satisfy equation 9.10. Under what conditions will x{t) = Ke^‘ sat­
isfy equation 9.10?
E X A M PLE 9.4. Determine conditions under which the postulated solution x{t) = Ke^‘ satisfies
equation 9.10.
S
o l u t io n
.
Step 1. Substituting Ke^‘ for ,v(/) in equation 9.10 produces
K
rfV '
dr
+ bK —
<h
+ cK e” = Ke" (,v^ + hs + c) = 0
^
I
(9.11)
Step 2. Interpret equation 9.11. For nontrivial solutions, K\s nonzero. The function e^‘ is always
different from zero. Hence the quadratic in s on the right side o f equation 9.11 must be zero. This
necessarily constrains $ to be a root of
r + bs + c =0
(9 .1 2 )
3<S8
Chapter 9 * Sccond Order Linear Circuits
Step 3. Solve equation 9.12. From rhe quadratic formula, rhe roots o f equation 9.12 are
.V|, .v. =
-b ±
- 4c
(9.13)
C O N C LU SIO N : x(/) = Ke^' satisfies equation 9.10 provided s tatces on values given by equation
9.13. Equation 9.10 does not constrain K\ however, the initial conditions will.
Equations such as 9.12 whose solution is given by equation 9.13 are a common characteristic o f
second-order networks. Hence, equation 9.12 is called the characteristic equation o f the secondorder linear circuit. The associated roots, equation 9.13, are called the natural frequencies o f the
circuit. These are the “natural” or intrinsic frequencies o f the circuit response and are akin to the
natural frequencies o f oscillations o f a pendulum (for small swings) or o f a bouncing ball.
From elementary algebra, a quadratic equation (the above characteristic equation) can have dis­
tinct roots or equal roots. Distinct roots can be real or complex. Thus
and Sj can be two dis­
tinct real roots, two distinct conjugate complex roots, or two repeated (equal) roots, depending
on whether the discriminant, Ip- - 4r, is greater than, less than, or equal to zero. This trifold
grouping separates the solution o f equation 9.10 into three categories, listed below as cases 1, 2,
and 3:
Case 1. Real and distinct roots, i.e., b~ -A c> 0. If the roots are real and distinct, then for arbitrar}'
and K2 , both
constants
.v( o = .vi (/) =
a:,6^"''
and
xif) = X2it)=K2e-'^-'
^
satisfy the second-order linear differential equation 9.10, i.e., are solutions to the differential equation. Since equation 9.10 is a
differential equation, by superposition the sum x(r) = x,(^) +
X2 (t) is also a solution, a fact easily verified by direct substitution. Therefore, whenever
s-y, the
most general form o f the solution to equation 9.10 is
xit)=K^e^^' + K2e^^-‘
The constants
(9.14)
and K-, depend on the initial conditions o f the differential equation, which
depend on the initial capacitor voltages and inductor currents. For example, if a.*(0'^) and a''(0'*')
are known, then from equation 9.14,
.v (0^ )=
A y '*'-h
and
.v-(O^) = — ^
(It
= .s'l
1= 0 ^
+ stK')
^
Chapter 9 * Second Order Linear Circuits
389
These are simultaneous equations solvable for
If
and
and $2 are negative, the response given by equation 9 .1 4 decays to zero for large f and the cir­
cuit is said to be overdamped.
Case 2. The roots,
and $2 , o f the characteristic equation are distinct hut complex, i.e., iP- -A c <Q.
Since Jj ^
general form of the solution to equation 9.1 0 is again given by equation
9.14, i.e.,
=
with complex
and
$2
given by
+K2e"‘^^
_______
-b
. y l^ c -b ^
^2 = Y ± J ------ ------- =
,
± J^ d
(9.15)
y l^ c-b ^
where a = bH and co^ = ------ ^ S i n c e s^ and S2 are conjugates, so are d^i^and e^2^in equation 9.14.
For x(t) to be real, the constants /Cj and A'j in equation 9.1 4 must also be complex conjugates,
i.e.,
Using Euler’s formula,
giy = cos
+ j sin y
the two terms in equation 9.14 combine to yield a real time function:
= e
co s(co jf) + jKisinioi^t)] + e
ATj cos((0^/) - yATj sin(cojO
(A"! + A"! )cos(to^f) + (y^i + jK i )sin((Ojf)
Thus the solution to equation 9.1 4 with
and S2 complex is given by the (damped) sinusoidal
response
x(r) = e~^^ [A cos((o/) + B sin((0/ )]
where ^4 =
+^
= 2 Re[^j] =
+ A^2 and 5
(9.16)
- y Af] = - 2 ImATJ = /A j - yATj are real con­
stants and where Re[ ] denotes the real part and Im[ ] denotes the imaginary part. The solution
expressed in equation 9.16 is completed by specifying ^4 and B. As before, A and B depend on the
initial conditions, x(0‘^) and xXO"*^) as follows:
jc(0'*‘) =
[Acos(tOjf) + Bsin((0^r)])^_^^ = A
390
Chapter 9 • Second Order Linear Circuits
o
jr'(0'*’) =
-a e ’‘^^^y4cos (o)^0 + 5sin (co^o) + e
co^y4sin (co^O + (O^Bcos (co^o)
= - o A + (o^B
and
which are easily solved for A and B.
Making use of a standard trigonometric identity, the general solution of equation 9.16 has the
equivalent form
x{t) =
[A cos(coj) + B sin(o)^)] = Ke~^^ cos(o)^ + 0)
(9.17a)
where
K = ylA^ + B^ ,Q = tan"‘
f-B \
(9.17b)
KA)
and the quadrant of 0 is determined by the signs o f - B and A. In MATLAB, one uses the command
“atan2(-BvA)” to obtain the angle in the proper quadrant. Note that the response waveforms have
oscillations with angular frequency O)^ These oscillations are bounded by the envelope ±Ke~^*. If
Re[jj] = - a < 0, the amplitude of the oscillations decays to zero and the response is said to be underdamped. If Re[jj] = - a > 0, the amplitude of the oscillations grows to infinity.
C ased. The roots are real and eqml, i.e.,lP' - 4 c = 0 . When the two roots of the characteristic equa­
tion are equal, equation 9.1 4 does not represent the general solution form because if
two terms collapse into a single term. However, the general solution for
the
is
x(t) = {K^+K2t)e^^^
(This is investigated in a homework exercise.) Calculation of
and
in equation 9.18 is
straightforward:
x{0-) =
and
i(0+) =
Substituting the value o f
into x'(O^) yields a simple calculation for Kj.
If^l = i2 is negative, the response decays to zero and is said to be critically damped. “Critically
damped” defines the boundary between overdamped and underdamped. This means that with a
slight change in circuit parameters, the response would almost always change to either over­
damped or underdamped.
T he discussion o f the three cases is summarized in Table 9.1.
Chapter 9 • Sccond Order Linear Circuits
391
TABLE 9.1. General Solutions for Source-Free Second-Order Networks
General solution o f the homogeneous differential equation
d~x
dx
dr
dt
— ^ + h — + cx - 0
having characteristic equation
+ r = (j - 5 j)(j -
= 0, where
- h ± yjb" —4 c
^2=
-----------
Case 1. Real and distinct roots, i.e., ^ - 4 c > 0:
x {t)=
+ Kie^-
where
40^) =
and .v (0") = s^K^ + s^K^,
Case 2. The roots, i| = - a + p i^ an d ^2 =
~j^d>
tion are distinct but complex, i.e., tr - Ac < Q:
characteristic equa­
x{t) = e~^‘ [A cos(o)/) + B sin(o)/)] = Ke~^' cos(to/ + 0)
where
x(0^) = A . x'(0+) = - 0 .4 + CO/
and
+
, e = tan"^
Case 3. The roots are real and equal, i.e.,
(-B \
A
= Sj and Ir - Ac
xU) = (K^ + K2t)e^'’
where
x{0*) =
and a-'(O^) = s^K^ + Kj
Figure 9.3 displays the various response forms described above for the case where Re[xj] and Re[j-,]
are negative or zero. Because o f their similarity, it is not possible to distinguish between the over­
damped and the critically damped responses by merely looking at the waveforms. Both types o f
response may have at most one zero-crossing.
392
Chapter 9 • Second Order I.incar Circuits
> t
t
FIGURE 9.3 (-'tncric waveforms corresponding to the four cases o f damping: (a) undamped (sinu­
soidal) response, (b) undcrdamped (exponentially decaying oscillatory) response, (c) overdamped
(exponentially decaying) response, and (d) critically damped (exponentially decaying) response.
The terms “undamped,” “underdamped,” “overdamped,” and “critically damped” stem from an
intuitive notion o f “damping.” The sourcc-frce response o f an undamped second-order linear sys­
tem, whether electrical or mechanical, has an oscillatory response (waveform) o f constant ampli­
tude. Damping, due to system elements that consume cnerg\', means a monotonic decrease in the
amplitude o f oscillation. In electrical circuits, resistances produce the damping effect. In mechan­
ical systems, friction causes damping. When the amount o f damping is just enough to prevent
oscillation, the system is critically damped. Less damping corresponds to the underdamped case,
where oscillation is present but eventually dies out. A greater amount o f damping corresponds to
the overdamped case, where the waveform is non-oscillatory, and a ver\’ small perturbation o f any
circuit parameter will not cause oscillations to occur.
In summary, once the roots o f the characteristic equation are found and the expression for the
general solution selected from the above cases or Table 9.1, it remains to find the constants A', and
Chapter 9 • Sccond Order Linear Circuits
3^)3
K j (or A and 5 ) from the initial conditions on the circuit. In the above development,
and Kj
(or A and B) are given in terms of a:(0^) and x'(O^). Since jf(/) represents either a capacitor voltage
or an inductor current, its value at f = 0"^ is usually given, or can be determined from the past his­
tory o f the circuit. (See Example 9.5.) The value o f x'(O^), on the other hand, is often unknown
and must be calculated. If >:(/) =
then the capacitor v-i relationship implies that
xX0*) = v'c{0*) = ^ £ ^ .
^
If x{t) =
then the v-i relationship of an induaor implies that
The problem then reduces to finding an unknown capacitor current,
, or an unknown
inductor voltage, y^(O^).
To find
or y^(O^), we construct an auxiliary resistive circuit valid at r = 0^. Since the initial
values,
and /^(O^), are known, we replace (each) capacitor in the original circuit by an inde­
pendent voltage source o f value
and (each) inductor in the original circuit by an inde­
pendent current source o f value /£(0'*^). Here the current /(;^0^) retains its original direction and
the voltage ^^(0^) retains its original polarity. After the replacements, the (new) circuit is resistive.
Values for
and y^(O^) follow by applying any of the standard methods of resistive circuit
analysis learned earlier. This allows us to specify x(0^) and x'(O^) in terms of the initial conditions
on the circuit. Two equations in the two unknowns A'j and K2 (or A and B) result. Example 9.5
and, in particular. Figure 9.4c illustrate this procedure.
Response Calculation o f Source-Free Parallel and Series RLC Circuits
Before any additional circuit examples, let us summarize the solution procedure.
Procedurefor Solving Second-Order RLC Circuits
Step 1. Determine the differential equation model of the circuit.
Step 2 . From the differential equation model, construct the characteristic equation and
find its roots using the quadratic root formula.
Step 3. From the nature of the roots (real distinct, real equal, or complex), determine the
general form of the solution from Table 9.1; the solution form will contain two unknown
parameters.
Step 4. Find the two unknown parameters using the initial conditions on the circuit.
The following example illustrates these calculations for the three cases described in Table 9.1.
W
39-'i
Chapter 9 • Second Order Linear Circuits
E XA M PLE 9 .5 . In the circuit o f Figure 9.4a, the 1 pF capacitor is assumed to be ideal, and the
inductor is modeled by a 10 mH ideal inductor in series with a 20 Q resistor to account for the
resistance o f the coiled wire. Suppose the switch S in Figure 9.4a has been in position A for a long
time. The capacitor becomes charged to 10 V. Then the switch moves to position B at f = 0. Find
and plot V(\t) for r > 0 for the following three cases; (1)
= 405
(2) /?2 = 0, (3)
180 ti.
Each o f these cases produces a different response type.
:L=10mH
lO O
+
10V
C=1
R=200
Practical Inductor
mF
(a)
i,(0*) = 0
L=10mH
+
C=1
mF
(b)
FKJURH 9.4 (a) Discharge of a capacitor through a practical inductor in series with a resistance /?,.
(b) Kquivalent circuit for t > 0. (c) Equivalent circuit at / = 0^ for calculating
in which the
inductor has been replaced by an independent current source of value //(O^) and the capacitor by an
independent voltage source of value
.
SoL u rioN
From the problem statements,
= 10 V. When the switch moves to position B,
=
- 10 V by continuit)' o f the capacitor voltage; the circuit now becomes a series RLC, for t
> 0, with / ?= / ?!+ R-y as shown in Figure 9.4b. The first step in the calculation o f the circuit
response is to find a second-order differential equation in the unknown
From Example 9.3,
for the series RLC,
-
R dvf'
*- + -------^ +
dt~
L dt
\’r = 0
LC ^
(9.19)
Since L and C are known, the series RLC characteristic equation is
o R
1
s~ H— .y H------ = .v“ + ( 2 0 + /?-,)10^y-r-1 0 ^ = 0
L
LC
With this framework, we can separately investigate each o f the three cases.
( 9 .2 0 )
Chapter 9 • Second Order Linear Circuits
Case I: /?2 = 405 Q or /? = 425 O.
Step 1. Find the characteristic equation and the generalform o f the response using Table 9.1. If R-, =
405
the characteristic equation o f 9.20 is
+ 42,500x + 10^ = 0. Solving for the roots by the
quadratic formula yields
2 = - 2 1 ,2 5 0 ± 18,750 = -2 5 0 0 , - 4 0 ,0 0 0 sec’ ^
Real distinct roots imply an overdamped response o f the form
v c(0 =
+ K2e^~ =
(9.21)
valid for r > 0.
Step 2. Fijid
and K-,. Evaluating at f = 0"^ implies
z;^0") = 10 = /Tj + /f2
(9.22a)
Differentiating equation 9.21 implies
vL(0+) = -^^^^ = -2.5xlO^/r, - 4 0 x \ 0 ^ K 2
C
From the circuit o f Figure 9.4c, v’ ^ (0^ ) =
C
^
^
C
C
(9.22b)
^ = 0 , where //^(O'*') = i[{0~)
by the continuity o f the inductor current. Solving equations 9.22a and b after substituting the
above values yields
AT, = 10.667 and K-, = - 0 .6 6 7
Step 3. Set forth the solution for V(^t). For / > 0,
v^t) = 10.667^-2’500^-0.667e’-^0’®®°'V
This function is plotted in Figure 9.5.
Exercise. You may verify' the above answer with the Student Edition o f MATLAB (version 4.0 or
later) by typing the command: y = dsolve(‘D 2y+42500*D y + leS^y = 0,y(0) = 10, Dy(0) = O’).
Case 2: /?, = 0 or /? = 20 Q
Step 1. If /?2 = 0> then from equation 9.20, the characteristic equation is
+ 2,000^ + 10^ = 0.
Since tr - Ac = -3 9 6 ,0 0 0 ,0 0 0 < 0, the roots are complex. From the quadratic root formula,
= - 1 0 0 0 + y9950 = - a + yo)^ and Sj = - 1 0 0 0 -y '9950 = - a
From Table 9.1, the underdamped response form is
v^t) =
[A cos(w/) + B sin(co/)] = ^’"1000/
cos(9950f) + B sin(9950f)]
(9.23)
396
Chapter 9 • Sccond Order Linear Circuits
Step 2. Fiiid A and B. Ir remains to determine A and B in equation 9.23. From equation 9.23
and its derivative,
i / ^ 0 ^ ) = 1 0 = /I
(9 .2 4 a )
and
= -oA + ( .) / = -1000/1 + 9950/y
(9.24b)
^ = — — - = 0. Substituting into equations 9.24 and
As in case 1, V(- '(O"^) = — — ^
solving yields
C
C
C
A = 10 and
= — = 1.005
CO,/
Step 3. Set forth the solution for v^{t). For t > 0,
v^t) = ^-'OOOrjio cos(9950?) + 1.005 sin(9950^)] = 10.05^'-^^®^' cos(9950^ + 5.7°) V
This waveform is also plotted in Figure 9.5.
Exercise. You may verify the above answer with the Student Edition oF MATLAB (version 4.0 or
later) by typing the command: y = dsolve(‘D 2y+2000*D y + le8*y = 0,y(0) = 10, Dy(0) = O’).
Case 3: ^2 = 180
Step 1. If
=180
or R= 200 Q
then the characteristic equation from 9.20 is
+ 2 0 ,0 0 0 j + 10^ = 0, whose
roots are
= -lO '*, implying a critically damped response. From Table 9.1, the general criti­
cally damped response form is, for r > 0,
v^(/) = (/r, + K2t)e^^' = (A", + K2t)e~^^'^
(9-25)
Step 2. Find /Tj and Kj. From equation 9.25, its derivative, and the known initial conditions
from cases 1 and 2,
v^iO^) = 10 = A",
(9.26a)
and
^yO ") =
+ K, = -10^ A', + A'2 = 0
(9.26b)
Solving equations 9.26 yields
A', = 10 and Kj = --^lA", = 10^
Step 3. Set forth the sohition for v^i). For /^> 0,
v'c(n = ( l0 + 1 0 ^ ) e " '^ ‘’ '
The waveforms o f V(fJ) for the three cases (underdamped, critically damped, and overdamped)
are plotted in Figure 9.5.
39'
Chapter 9 • Second Order Linear Circuits
FIGURU 9.5 Waveforms of vj,t) in Example 9.5 for three different degrees of damping. Critical
damping represents the boundary between the overdamped condition and the oscillatory behavior of
underdamping.
Exercise. Verify the answer calculated in Example 9.5 using the Student Edition o f MATLAB and
the “desolve” command.
On a practical note, commercially available resistors come in standard values each with an associ­
ated tolerance. Tolerances vary from ±1% (precision resistor) to as much as ±20% . Further,
because o f heating action over a long period o f time, resistance values change. Given the above
example, in which the type o f response depends on the resistance, one can imagine the care need­
ed in the design o f such circuits: without consideration o f precision and long-term heating effects,
a desired critically damped response could easily become oscillator)'.
Not all second-order circuits arc RLC. Some are only RC but with two capacitors and some are
RL with two inductors. Passive RC or RL circuits cannot have an oscillatory response. The proof
o f this assertion can be found in texts on passive network synthesis. However, with controlled
sources a second-order RC or RL circuit can have an oscillatory response that is not characeristic
o f a first-order circuit, but o f a second- or higher-order circuit. The example below illustrates the
analysis o f a second-order RC circuit containing controlled sources that has an oscillator)^
response.
398
Chapter 9 • Second Order Linear Circuits
EX A M PLE 9 .6 . This example illustrates the analysis o f the second-order RC circuit shown in
Figure 9.6. The objective is to find
= 10 V and
= 0.
'Cl
1 kn
I
for / > 0 given the initial conditions V(^{0)
and
r
mF
FIGURE 9.6 Sccond-order RC circuit with controlled sources that has an oscillatory response.
S o l u t io n
Step 1. Write a dijferential equation in V(^^{t). From the properties o f a capacitor and KCL at the
left node,
di
Multiplying through by 10*^ yields
d\
■^^ — 10
— 10^ v^2 ~ 1 0 'v’f'i
dt
(9.27a)
We expect a second-order differential equation, so differentiating a second time yields
_ 1()5
, Q3
dt
dt~
To obtain a differential equation in
(9.27b)
dt
equation 9.27b must be eliminated. This
requires another relationship between t»Qand
. At the right node,
10“6 — C l _
_ -0 .
dt
or equivalently,
dt
= -1 0 ^ Vc,
(9.28)
Substituting this expression into equation 9.27b produces
dt
*2
After rearranging terms, we obtain a second-order differential equation in
'2
dt^
dt
(9.29)
Step 2. Determine the characteristic equation, its roots, and the form o f the response. The differential
equation 9.29 has characteristic equation
Chapter 9 • Second Order Linear Circuits
+ 1 0 ’° = 0
From the quadratic formula, the complex roots are
= - 5 0 0 + 799,998.75 = - a ± y W
From Table 9.1, complex roots imply an underdamped response o f the form
[A cos(co/) + B sin(co/)] = ^’" 500' [A cos(99,998.75r) + B sin(99,998.75r)] (9.30)
Step 3. Find A and B. \x. t = 0, z^qCO) = \() = A. Also, from equation 9.27a and the initial con­
ditions,
^^^'Cl(Q) ^ i0 - \ ;^ .,(0 ) - iq 3 v ^ ,(0 ) = - 10^
dt
(9.31)
Differentiating equation 9.30, evaluating at r = 0, and equating the result with equation 9.31
produces
dl
"
in which case B = 5.0001 x 10“^.
Step 4. Set forth the final form o f V(-^{t). The final form o f the response is
v^{t) = ^'-500^ [10 cos(99,998.75^) + 5.0001 x 10"2 sin(99,998.75r)] V
= 1 0 ^ 5 0 0 'c o s (9 9 ,9 9 8 .7 5 r- 0.2865°) V
Step 5. Plot the response. A plot o f the (underdamped) response is given in Figure 9.7.
Chapter 9 • Second Order Linear Circuits
Exercise. Construct a parallel RLC circuit to have the same second-order difFerential equation
model as 9.29. Note that there is no unique solution.
C H EC K : /?C= 10-^ and ZC = \0~^^
It is important to observe here that the design o f Example 9.6 achieves a second-order RLC
response without the use of an inductor, which is important for integrated circuit technolog)'.
4. SECON D-ORDER LINEAR NETW ORKS WITH CON STANT
INPUTS
The preceding section studied source-free second-order linear networks. When independent
sources are present, such as in the circuit o f Example 9.7 below, the network (differential) equa­
tions are similar to the source-free case except for an additional term that accounts for the effect
o f the input:
^
clf-
+ / ; ^ + c-.v = /(/)
dt
(9.32)
wherey{r) is a scaled sum o f the inputs and/or their first-order derivatives. Ordinarily one might
expecty(f) to be the value o f the input. A homework problem illustrates that j{t) can depend not
only on the source input, but also on the derivatives o f the input. For general circuits, those not
reducible to parallel or series RLC circuits, constructing equation 9.32 can be a challenge. Further,
the solution o f 9.32 for arbitrary inputs and initial conditions is no less challenging but is best
obtained via the Laplace transform method, which is a topic studied in a second circuits course.
However, when the input excitations are constant,/r) = F, the solution to 9.32 is a straightfor­
ward modification o f the source-free solution, as explained in the remainder o f this section.
Since the expressions o f Table 9.1 satisfy the homogeneous diflferential equation 9.10, the gen­
eral solution to equation 9.32 follows by adding a constant
to each o f the solution forms given
in Table 9.1. Specifically, the general solution o f the driven differential equation
tl~x
dx
- + h — ^ cx = F
(9 . 33 )
xit) = .V,//) + Xj:
(9.34)
dr
dt
where-v^j(f) is the solution to the homogeneous equation 9.10 (equivalently, equation 9.33 with F
= 0). Recall that the form o f x j t ) is determined by the roots of the characteristic equation r + bs
+ r = (i - ^])(j - ^2) = 0, given by the quadratic formula
h yjh~ - 4 c
i |-) = — ± -------------9
9
*0 1
Chapter 9 * Scconcl Order Linear Circuits
To verify that the structure o f equation 9.34 is a solution to 9.33 and to compute the value o^Xp sub­
stitute the structure given by equation 9.34 into 9.33. Since
satisfies the homogeneous equation
9.10, it contributes zero to the left-hand side. What remains is cXp= F. Therefore,
= — which
is independent o f the roots o f the characteristic equation. However, if Re[j|] and Re[^2] < 0, then
tends to zero for large t. Hence x{t) tends to Xp for large t. Consequently Xp is termed the
final value o f the response.
Because o f the trifold structure o f
as summarized in Tible 9.1, the solution form o f equation
9.34 once again breaks dow'n into three distinct cases. We summarize this trifold structure for the
constant-input case in Table 9.2.
TABLE 9.2 General Solutions for Constant-Source Second-Order Networks
General solution o f the driven differential equation
-z- + h — + cx = F
dr
dt
having characteristic equation p- + bs + c = (j - Jj)(^ - ^2) = 0 with roots
h yjb^ - 4 c
-7 = ---- ± --------------“
2
2
Case 1. Real and distinct roots, i.e.,lP' —4 r > 0:
x {t)=
+ Xp
F
with Xp = —. Further,
c
x(0'^) =
+ K2 + A'y^and x'(0'^) = s^K^ + s^K^
Case 2. The roots,
= - a + y'co^y and S2 = - o
o f the characteristic equation
are distinct but complex, i.e.,
- 4 c <Q. The general solution form is
x(r) = e~^^ [A cos(oj/) + B sin (to/)] + Xp = Ke~^^ cos(to/ + 0) + Xp
F
where again Xp = —, with
c
x{0^) = A + X p x(0^) = - g A +
K
V
)
Case 3. The roots are real and equal, i.e., s, = s, and Ir -A c= Q . The solution
form is
.v(/) = (/ r ,+ ^ 2 0 ^ '
where Xp = —, and
c
x{Q^) = K^+Xp
and .v W "-) = .v ,^ i-h AT2
402
Chapter 9 * Second Order Linear Circuits
The interpretation Xp = F/c. is a mathematical one. When the differential equation describes a lin­
ear circuit with constant inputs, there is a physical interpretation o f Xp and a circuit theoretic
= Xp
method for computing its value, even without writing the differential equation. Since
= a constant” satisfies the differential equation 9.33, it is also a constant solution to the circuit.
Hence Xp\s either a constant capacitor voltage or a constant inductor current. If a capacitor volt­
age is constant, its current is zero; this is interpreted as an open circuit. Similarly, if an inductor
current is constant, its voltage is zero; this is interpreted as a short circuit. Therefore, Xp is an
appropriate (capacitor) voltage or (inductor) current obtained when the capacitor (or capacitors)
are open-circuited and the inductor (or inductors) arc short-circuited. The value o f Xp can be
obtained by analyzing the resistive network resulting when all capacitors are open-circuited and
all inductors are short-circuited. Recall that if Re[^j] and Re[^2] < 0, then x{t) tends to the constant
value Xp Physically speaking, then,
equals either
or /^(o))when Re[;,] and Re[^2] < 0.
Once the proper general solution structure is ascertained from Table 9.2 and the constant Xp is
found, the parameters
(or A and B) are computed by the same methods used in the
and
source-free case. The following example illustrates the procedure for a parallel RLC circuit.
= u{t) A, excites the parallel /^ZCcircuit o f Figure 9.8,
E XA M PLE 9.7 . A step current input,
whose initial conditions satisfy /)(0) = 0 and y^^O) = 0. This simply means that the current source
turns on with a value of 1 amp at r = 0 and maintains this constant current excitation for all time.
The objective is to find the inductor current,
for / > 0, for three values o f R: (i) R = 500 Q,
(ii) R = 25 n , and (iii) R = 20 Q..
I'IGURE 9.8 Parallel RLC circuit cxcited by a step current input.
So l u t io n
Because the circuit is a parallel RLC, the characteristic equation is
. r + - L , + - L = ,2 + 1 0 l,v + 4 x l 0 ' " = 0
RC
LC
(9.35)
R
For all positive values o f R, the roots of the circuits characteristic equation have negative real parts.
Thus for large t or ideally at “t = oo,” the inductor looks like a short circuit and the capacitor like
an open circuit. Hence, for all cases o f this example, Xp = i[{^) = 1 A; note that y^co) = 0 because
the inductor looks like a short at t = <x>.
Case I. For R = 500 Q, the characteristic equation 9.35 reduces to r + 2 0 ,0 0 0 j + 4
From the quadratic formula, the roots are
2 = - 1 .0
X
lO'^ ± 7 I .9975
X
105 = - a ±yco^
x
10*^^ = 0.
'103
Chapter 9 • Sccond Order Linear Circuits
which indicates an underdamped response o f the form (Table 9.2)
[A cos(coy) + B sin(o)^)] + Xp= Ke~^‘ cos(to^ + 0) +
/^(r) =
= A + Xp = A + 1, then A = -\. Further,
Since 0 =
elicit)
0=
dt
^
\
= -0/4 + (HjB
1 = 0^
From derivative o f
expression for ij^t)
From physical circuit
This implies that
fi = — = - 5 .0 0 6 3 x 1 0 " “
(0,y
Hence for / > 0,
/•^(r) = ^.-10.000^ [ cos(1.9975 x IQ5f) + 5.0063 x 10"2 sin(1.9975 x lO^r)] + 1
= 1 .0 0 13^>-*^’‘^^®'cos( 1.9975 x 105^+ 2.866'’) + 1 A
Case 2. For R = 25 O., the characteristic equation 9.35 reduces to
From the quadratic formula, the roots are
+ 4 x 10^5 + 4 x lO'® = 0.
j, 2 = - 2 . 0 x 105
indicating a critically damped response o f the form (Table 9.2)
iL^t) = (K ^ + K 2t)e‘ '' + Xp
Since 0 = /^(O^) = A^, + Xp =
Q
+ 1, then
= - 1 . Further,
_ vc(0'*') _ vz.(Q~^) _ d ilit )
L
L
dt
= SiKi + K2 = 2xW ^ -hK2
1=0*
This implies that K y - - I y. 10^ and for r > 0,
-2xio-\
Case 3 . For R = 20 Q, the characteristic equation 9.35 is
quadratic formula, the roots are
= - 1.0
X
10^ and S2 ~ ~
specifying an overdamped response o f the form (Table 9.2)
+ 1A
+ 5 x 1O^j + 4 x 10^® = 0. From the
^
40 4
Chapter 9 • Sccond Order Linear Circuits
+ Xp. Furtiier,
Evaluating this response at f = 0+ yields
L
L
dt
t=(f
1
Equivalently, -K^
~ Solving these two equations yields K\ = ---- and K-, = —.
Therefore the actual response for r > 0 is
^
^
Figure 9.9 displays a graph o f the response for each o f the three cases.
FIGURE 9.9 Underdamped, critically damped, and overdamped
response curves for the parallel RLC circuitof Example 9.7.
Exercises. 1. Show that for t > 0 , the differential equation for the circuit o f Example 9.7 with R
= 500 Q is
2
^
^
clr
+ 2 x l 0 ‘' ^
dt
+ 4xl0'»,',.<n = 4 x l 0 JO
'
2. Use MATLAB s “dsolve” command to verify the solution obtained for case 1 in Example 9.7.
Chapter 9 • Second Order Linear Circuits
In a linear circuit or system, the response to a step input often indicates the quality o f the system
performance. The problem o f measuring a batter}' voltage using a voltmeter is illustrative o f this
indicator. Here the battery dc voltage is the input and the output is the meter pointer position.
Connecting the meter probes to the batter)'^ terminals amounts to applying a step input to the
voltmeter circuit that drives a second-order mechanical system consisting o f a spring and mass
with friction. Naturally, one would like the pointer to settle on the proper voltage reading quick­
ly. If the mechanical system is underdamped, then the pointer oscillates (undesirably) for a short
time before resting at its final position. On the other hand, if the mechanical system is over­
damped, the pointer will not oscillate but may take a long time to reach its final resting point.
This also is undesirable. A near critically damped response is the most desirable one: the pointer
will come to rest at the proper voltage as quickly as possible without being oscillatory, and small
changes in the mechanical system will not make it oscillatory.
In the next example, we reverse the process o f analysis and ask what the original circuit parame­
ters are given a plot o f the response that might have been taken in a laboratory.
E X A M PLE 9 .8 . Consider the circuit o f Figure 9.10, which shows the response,
of a
(relaxed) series RLC circuit to the voltage input i/y^(r) = 10«(r) V. In laboratory, you have meas­
ured the capacitor voltage values (approximately). If the response has the form v^it) = Ke~^^
cos(ojy + 0) + vYp find
a , 0, K, and the values o f R and C i f it is known that L = 0.5
H. Your lab instructor has told you that to^and a are integers.
m
(a)
\
___________,
406
Chapter 9 • Second Order Linear Circuits
(b)
F IG U R t 9.10 (a) Series RLC\ (b) Response to
= 10 u{t) V.
TABLE 9.3
Tim e (sec)
0.316
0 .5 2 3 6
0.839
1.5708
vcKt) (V)
10
13.509
10
10.432
First crossing
First
Second crossing
Second
o f 10 V
peak
o f 10 V
peak
S olution
Step 1. FindXp. By inspection, the curve is settling out at X p - 10 V.
Chapter 9 • Second Order Linear Circuits
Step 2. Find
40
Now obscr\'e that the first two crossings o f v^^t) = 10 occur at r = 0 .3 1 6 sec,
0.839 sec (Table 9.3). This means that a full k radians is traversed by tlie cosine over [0.316,
0.839] , which is a half c>de or half period. So the period o f the cosine is 7 = 2(0 .8 3 9 - 0.316)
2 ti
= 1.046 sec, making
CO^/ =
= 6.007 = 6 rad/sec.
Step 3. Find O . From Table 9.3, we know that two successive “peaks” occur at
= 0.523 sec
and t-y = 1.5708 sec. This means that for ^ = 1 ,2 ,
cos(COj//. + 0 ) + X/r
(9.36)
After some manipulation, equation 9.36 implies
COS(Ci),y/l + 0 )
C0.s((0^/r2 + 0 )
Thus
cos((0 ,//| +Q) V c i t j ) - ^ ^,-a(/2-/,)
Xf,-
cos(co^y/'> + 0 )
(9.37)
Equation 9.37 simplifies because two adjacent positive peaks must be 2 tu radians apart, i.e.,
(co/2
+ ^) =
which means cos(to^^^ + H) = cos((i)y, + 0). It follows that
X’c d o - ^ F
3 .509
(9.38)
Solving leads to a = 2.
Step 4. Find 0 and K. At the first crossing o f 10 V, we have
0 =
0.316
0 . 3 1 6 + B)
Thus 6 X 0.316+ 0 must equal 0.5?! or 1.5tt radians. We also know that since
=0 =K
cos(0) + 10, we must have K cos(0) = - 1 0 . Since A"> 0 (always by convention), the value o f
cos(0) must be negative. This means 0.57t < 0 < 1.5Tt. So therefore, it must be that at the first cross­
ing
o f 10 V
D
6 X 0 .3 1 6 + 0 = —
or 0 = 2.1864 rad.
Therefore K = — ^— = 10.553.
cos(0)
Step 5. Find R and C. We know that the characteristic equation o f the series RLC circuit must be
Therefore
f?
1
02
^
^
s~ H—- . V.V +H-------=
2Rss +H—— = (5' + 2)*' + 6 " = s~ + 4.V + 40
------- = .v“ + IR
L
LC
C
/? = 2 n and C = 0 .0 5 F
-H)K
Chapter 9 • Sccontl Order Linear Circuits
In the previous examples one obsen'e that the characteristic equations are independent o f the
source values. I'his is a general property of linear circuits with constant parameters. Hence when
constructing the characteristic equation we may without loss of generality set independent source
values to zero; i.e., independent voltage sources become short circuits and independent current
sources become open circuits. With this operation, some circuits that appear to be non-series/parallel, become series/parallel. This allows us to easily compute the characteristic equation and then
use Table 9.2 and physical reasoning to obtain the solution without having to construct the dif­
ferential equation explicitly. The following example illustrates this procedure for a pseudo-parallel/.series RLC. The example will also illustrate the computation o f initial conditions due to past
excitations and the computation o f the complete response w'hen the input changes its dc level.
EXA M PLE 9.9 . The circuit of Figure 9. l i b is driven by the input o f Figure 9.11 a, i.e., vj^t)
-60//(-r) + G^u{t) + 60//(/‘ - 1) V. Our goal is to find the response
for / > 0.
FIG URE 9.11 (a) Input cxcitation whose dc level changes at r = 0 and t = 1 second, (b) A pseudoparallel RLC circuit; i.e.,when the voltage source is replaced by a short, the circuit reduces to a paral­
lel RLC whose characteristic equation is p- + —
RC
s+—
= 0.
LC
So l u t io n
Step 1. Analysis at 0“. Here the circuit has been excited by a constant - 6 0 V level for a long
time. Therefore at ^ = 0", the capacitor looks like an open circuit and the inductor a short cir­
cuit. Because the inductor looks like a short, the entire - 6 0 V appears across the 6 i l resistor,
making
= - 60 V and /^(O") = - 60/6 = — 10 A.
Step 2. Analysis at 0^. By the continuity o f the capacitor voltage and the inductor current, the
equivalent circuit at 0"^ is given in Figure 9.12.
Chapter 9 • Second Order Linear Circuits
409
= -6 0 V
^ (0 *)= i,(0 )
= -10A
FICJURE 9.12 Equivalent circuit for analysis at 0^; the capacitor is replaced by a voltage source
o f value
and the inductor by a current source o f value /^(O*) = //;(0“).
From the circuit diagram o f Figure 9.12, v^{0*) = 60 - ( - 60) = 120 V,
= - 60/6 = - 1 0 A, and
iff-) = if^{0*)l5 = 40 A. It follows that
Step 3. Find the characteristic equation and the form ofthe response using Table 9.2. To find the char­
acteristic equation, we set the independent voltage source to zero. I'he resulting circuit is a paral­
lel RLC with characteristic equation
-)
1
I
T
-)
.v“ + ----- .V+ ------= .v“ + 4.V + 4 = (.V + 2 )“ = 0
RC
LC
where R = 2 LI is the parallel combination o f 6 Q and 3 i i . The characteristic roots are s^i = - 2 ,
which correspond to a critically damped response o f the form (Table 9.2)
V(4,t) = (/f,
+
Kjt)
exp(j,
t) + Xf:
Step 4. Find constants in the response form for 0 < /■< 1. The input is constant for 0 < ^ < 1, but
changes its value to 120 V at r = 1 sec. However, the circuit does not know the input is going to
change, and so its response behaves as if the input were to remain at 60 V for all time: the circuit
cannot anticipate the future, and thus its response over 0 < r < 1 behaves as if no further switch­
ing were going to occur. If no further switching were to occur and if the input remained at 60 V,
then in Figure 9.10b for large t the capacitor is an open circuit and the inductor is a short circuit;
hence Xp= 60 V. Under these same conditions we find
and K-y. To find A^j, observe that from
= - 6 0 V. Evaluating the response form o f step 3 yields
+ Xp. Equating
these rwo expressions produces - 6 0 = ^'(;(0■^) =
+ X^ which implies thar
= - 1 2 0 . To calcu­
late Kj, observe that from step 2, /(-(O^) = 40 A. Since C = 0.125, it follows that
step 2,
c!t
H en ce,
K-,
C
/={)■'
= 320 =
+ A'2 = 240 + ^2
= 8 0 . T h u s, the response o f the circu it for 0 < r < 1 is
V({t)
= (-1 2 0 +
m)e~^‘ +
60 V
Chapter 9 • Sccond Order Linear Circuits
410
Similarly, one can compute, for 0 < r < 1,
il{t) = (-2 0 +
+ 10 A
Step 5. Analysis at t = 1“. Although the circuit does not know the input will change at r = 1 sec,
we do and we must prepare for the analysis for f > 1. To do this we must evaluate the initial con­
ditions at / = 1“ and then use the continuity o f the capacitor voltage and inductor current to
obtain the initial conditions at / =
At r = 1“, using step 4 we have V(^\~) =
= 54.59 V
and/^d") = /^ (r) = 10 A.
Step 6. Analysis att= I
This step mimics step 2 for r = P . The capacitor is replaced again by an inde­
pendent voltage source and the itiductor by an independent current source as shown in Figure 9.13.
Here v^{V) = 1 2 0 - 54.59 = 65.41 Vand
=-in\ +
+ //(O'*') =
= - 9 .0 9 8 + 21.8 + 10 = 22.71 A.
’
^ 120 - 54.59 ^
I
6
3
>
Step 7. Computation o f the responsefor t> 1. Because the characteristic equation is independent o f
the input excitation, the form o f the response is almost the same as in step 3, except for the
replacement o f t by (/ - 1); this substitution follows by the time invariance (constant parameter
values) o f the circuit. Thus, for r > 1,
v^t) = \K^ + K\{t- 1)1 e x p U ,(r- 1)]
Since the source excitation for r > 1 is 120 volts, by inspection o f Figure 9.1 lb
V. To find /f], 54.59 = ^'c^l"^) = ^p+ A',. This implies
f^^oo) = 120
= -6 5 .4 1 . Finally, to find K-, consider
that
=
dt
w hich makes
1 8 1 .7 = .ViA:, + K2 = 1 3 0 .8 + K 2
/=r
= 5 0 .9 . T h u s, for / > 1,
v^t) = [ - 6 5 .4 1
+ 5 0 . 9 ( r - l)]^>-“( ' -
+ 120 V
(9 .3 9 )
Chapter 9 • Scconcl Order Linear Circuits
41
Time in seconds
FIGURE 9.14 Complete analytical response of the capacitor voltage for 0 < r < 3 sec.
Exercise. Fill in the details for the computation o f i^it) = (-2 0 + 20/)^^ + 10 A for 0 < r < I and
then compute /y(/) for 2.5 > ^ > 1. Also, compute /^(/) for t> 2.5.
Despite the idea illustrated in Example 9.9, many second-order RLC circuits are not reducible to
series or parallel RLC circuits when the independent sources are set to zero. Furthermore, when a
dependent sourcc is present, the circuit is generally not reducible to a series or parallel RLC. In
such cases one ordinarily uses a systematic methodolog)' to compute the circuit’s differential equa­
tion and, subsequently, the characteristic equation. This systematic procedure is described in more
advanced texts and in the second edition o f this text. Nevertheless, for some situations one can use
the earlier method s integro-differential equations, which must be differentiated again to eliminate
the integral. This is illustrated in the example o f the next section.
•il2
Chapter 9 • Sccond Order Linear Circuits
5. OSCILLATOR APPLICATION
An imporrant difference between first-order and second-order linear networks is the possibilit)' o f
oscillatory responses in the latter. In some applications sinusoidal oscillations are intended
responses, while in other applications oscillations arc undesirable. This section presents an exam­
ple o f a Wien bridge oscillator circuit.
The goal is to build a circuit that generates a pure sinusoidal voltage waveform at a specified fre­
quency. In theory, as per section 2 of this chapter, this is achievable by discharging a capacitor
through an inductor. In practice, both capacitor and inductor have losses. Losses cause the oscil­
lation amplitude to decay eventually to zero. For sustained sinusoidal oscillations, some “active”
element such as a controlled source or op amp must replenish the lost energy. Note that these
active elements require a dc power supply for their operation. Ultimately the dc power supply
replenishes the power losses due to various resistances in the circuit.
E XA M PLE 9 .1 0 . Figure 9.15 shows a Wien bridge oscillator constructed with an op amp as the
active clement. Find the condition on the circuit parameters R^,
and C for sustained sinusoidal
oscillation, and the frequency' o f oscillation.
r
R.
(b)
FIGURE 9.15 (a) Wien bridge oscillator, (b) Equivalent circuit.
S o l u t io n
From the principles described in Chapter 4, the non-inverting amplifier enclosed in the dashed box of
Figure 9.15a is equivalent to a voltage-controlled voltage source with a gain equal to {2Rj- + RJ}IRr= 3.
(See Chapter 4.) Replacing the dashed lx)x with this equivalent yields the simplified circuit o f Figure
9.15b. Using the simplified circuit, the first task is to derive die differential equation model of the circuit.
Step 1. Write a single-loop equation.
(9 .4 0 a )
Chapter 9 • Second Order Linear Circuits
413
To eliminate rhe integral, we differentiate again to obtain
dt
(9.40b)
7?1 dt
/?,C
Step 2. Express /^’j in terms o f V2 - By inspection o f Figure 9.15b we obser\'e that
,
V’2
(9.41a)
^c\ ~ ^ ----- '---dt
/?,
dt^
l<2
and thus, differentiating again,
(9.41b)
dt
Step 3. Substitute equations 9.41 into equation 9.40b. Substituting as indicated yields
dt~
/?2 dt
R^C ^
dt
Rj y
- ± ^ =0
/?, dt
(9.42a)
Grouping terms and dividing by C produces
d Vf
1 dv->
1 dv-y
- + ----------- ^ + ---------- ^ +
dt~
Ro_C dt
/?,C dt
R^R^C-
R^C dt
which simplifies to
dvo
d-V2
dt~
RjC
Vo
=
0
(9.42b)
R\C}
Step 4 . Compute the characteristic equation and determine the conditions for sustained oscillations.
The resulting characteristic equation is
s~ + hs + c = s~ +
1
1
\ R ,C
= 0
R ,C )
(9.43)
For sustained sinusoidal oscillations to occur, the roots must be purely imaginary. Thus the coef­
ficient o f s must be zero, i.e.,
b=
R2C
/?,c
-
^
/e|/?2C
=0
(9.44)
Thus the condition for sustained sinusoidal oscillations reduces to /?, = R-,.
Step 6. Find the frequency o f oscillation. Under the condition R^ = Rj, the roots o f the character­
istic equation are
/?,C
We conclude that the frequenq^ o f oscillation (in rad/sec) is
(Oo =
(9 .4 5 )
/?,C
A14
Chapter 9 • Second Order Linear Circuits
> R^, then ^ > 0 and the unforced response is
an exponentially decreasing sinusoid. On the other hand, if R^ < Rj, then b <0, and the unforced
An examination o f equation 9.44 shows that if
response is an exponentially growing sinusoid. For the oscillations to start, the value of/?| should
be designed to be slightly smaller than Rj- Then the value for b in equation 9.58 will be negative,
producing an exponentially growing sinusoidal response. If all circuit parameters are truly con­
stant, the amplitude o f oscillation would theoretically grow to infinit)'. In real oscillator circuits,
such growth is limited to a finite amplitude by saturation effects or nonlinearities that clamp the
response when the voltage swing grows large. The resulting waveform then only approximates a
pure sine wave. The analysis o f this nonlinear effect is beyond the scope o f this book. However,
the next example illustrates the growing oscillation when
< /?2 and also shows the effect o f sat­
uration to produce an approximate sinusoidal oscillation.
EXA M PLE 9 .1 1 . The circuit o f Figure 9.16a is a B2 Spice schematic for the Wein bridge oscilla­
tor o f Figure 9.15. The op amp is a 741 with
= 0. Observe that
= 10
= 15 V. Suppose that
(0) = 10 m V and
= 9.5 kQ. According to the analysis o f Example 9.10, the out­
put voltage labeled IVout should be a growing sinusoid. The output response o f Figure 9.16b
shows this growth and the saturation effects induced by the op amp. The waveform is not a pure
sinusoid due to these saturation effects. Also note that the frequency o f oscillation is approximately
16 Hz, which is consistent with equation 9.45, i.e..
=
2k
C2
R1
(a)
2 k ^R^R2 C
16.3 Hz
15
Chapter 9 • Second Order Linear Circuits
Example 9.11 Oscillator-Transient-4
Time(s)
(b)
FIGURE 9.16 (a) Schematic diagram of Wein bridge oscillator, (b) Voltage response showing grow­
ing oscillation clamped at ±15 V due to saturation effects of op amp.
An alternative approach to initiating oscillations and simultaneously limiting amplitude is to use
a temperature-sensitive resistor, R^, with a positive temperature coefFicient. Any incandescent
lamp is an example o f a temperature-sensitive resistor. For small voltages the temperature o f an
incandescent lamp is lower than for larger voltages because the dissipated power is lower. Hence
the lamp temperature (and thus its resistance) increases with increasing voltage. In the case o f our
oscillator, we have a desired output voltage swing. The nominal value o f /?, is designed to be
slightly less than R-y when the output voltage swing is bclow' a pre-specified voltage less than
This causes a growing oscillation. As the voltage swing increases, the temperature o f R^ and thus
its resistance increase. When the resistance o f /?, reaches Rj, the amplitude will settle (stabilize) at
the pre-specified voltage swing, at least theoretically. If /?, happens to increase beyond R^, a decay­
ing sinusoid w^ill result, decreasing the temperature and hence the resistance o f R^. Should the
amplitude o f oscillation decrease for any reason, /?j will decrease, causing a growing sinusoid.
Although the resistance o f R^ may dither about /?2>
amplitude o f oscillation will nevertheless
restore itself to the equilibrium level. In practice this equilibrium level only approximates the spec­
ified value due to imperfections in the circuit parameter values. The resulting waveform is almost
a pure sinusoid.
4 16
Chapter 9 • Sccond Order Linear Circuits
6. SUMMARY
'I'his chapter has explored the differential equation modeling and response computation of^ sec­
ond-order linear circuits having either no input or constant input excitation. Such second-order
circuits contain at least two dynamic elements, either an LC, CC, or LL combination. Secondorder circuits may also contain active elements such as op amps. In contrast to first-order circuits,
second-order linear circuits allow for the possibilit)^ of damped and undamped sinusoidal oscilla­
tions.
Analysis o f second-order linear circuits has two phases. Pha.se 1 entails the formulation o f the sec­
ond-order differential equation circuit model. For simple I C parallel RLC, or series RLC, the cir­
cuit model can be found by inspection.
Phase 2 o f the development centers on the solution o f the second-order differential equation
model o f the circuit. I'he first step here is to compute the (quadratic) characteristic equation and
then solve for the two roots. The roots o f the characteristic equation determine the t}'pe o f
response. The three t)'pes o f roots for a quadratic— real distinct, real identical, and complex—
specify the three response types of overdamped, critically damped, and underdamped, respective­
ly. These three types o f responses characterize all second-order linear differential equation models,
be they o f electrical circuits, mechanical systems, or electro-mechanical systems.
Since sinusoidal waveforms are germane to many electrical systems, this chapter presented an
oscillator circuit that generates a sinusoidal waveform. O f the many types of oscillator circuits, we
chose one containing an RC circuit built around an op amp, avoiding the use o f an inductor.
Chapter 9 • Second Order Linear Circuits
41
7. TERM S AND CO N CEPTS
Characteristic equation: for a linear circuit described by a second-order difTerentia! equation
= j{t), the algebraic equation
,v"(f) +
+ r = 0 is called its characteris­
tic equation.
Characteristic roots: roots o f the characteristic equation, also called the natural frequencies o f the
linear circuit.
Critically damped circuit: a second-order linear circuit having characteristic roots that are real
and identical. The source-free response o f such a circuit has a non-oscillatory waveform,
but is on the verge o f becoming oscillator)'.
Damped oscillation frequency: in an underdamped second-order linear circuit, the source-free
respon.se has the form [Kc^^' cos(o)y + ()}. The angular frequency
is the damped o.scil-
lation frequenc)', which is the magnitude o f the imaginary part o f the characteristic roots.
Homogeneous differential equation: a differential equation in which there are no forcing terms.
For example, x"{t) + bx\t) + cx{t) = 0.
Natural frequencies: the characteristic roots.
Oscillator circuit: an electronic circuit designed to produce sinusoidal voltage or current wave­
forms.
Overdamped circuit: a second-order linear circuit having a characteristic equation whose charac­
teristic roots are real and distinct.
Scaled sum o f waveforms: let
...
be a set of waveforms. A scaled sum of these wave­
forms is an expression o f the form /r) =
+ ... +
for real (possibly complex)
scalars ^/j, ... ,
Second-order linear circuit: a circuit whose input-output relationship may be expressed by a second-order differential equation o f the form .v"(/') +
+ cx{t) = /(r).
Source fi-ee: there are no independent sources, or all independent sources have zero values.
Step function: a function equal to zero for r < 0 and equal to 1 for r > 0.
Step response: the response o f a circuit to a step function input when all capacitor voltages and
inductor currents are initially zero.
Undamped circuit: a second-order linear circuit where the characteristic roots are purely imagi­
nary and the unforced response is purely sinusoidal.
Underdamped circuit: a second-order linear circuit whose characteristic roots are complex with
nonzero real part.
' T h e notations
and KO arc used interchangeably in the literature to den ote the first derivative ol v{t).
Chapter 9 • Sccond Order Linear Circuits
418
PROBLEMS
where all coefficients are real (but not necessar­
ily positive).
(a)
Prove that x(r) = 0 at some r = T, 0 <
TH EO RY RELATED
r < 00, only if
1. In section 2, the solution to the undriven LC
circuit is given by v^^t) = K cos(cof + 0) V,
Observe that
and /f, have oppo­
site signs.
(b)
Prove that the x(^) vs. t curve has at
most one zero crossing for r > 0.
(c)
cos(tt)0 = - W “ cos(0)/)
State the necessary and sufficient condi­
tions on the coefficients
>
and Sj for the presence o f one zero
and
d
— 7 sin((0 / ) = - 0 )
2
crossing.
sin(coO
dt^
5. The voltage or current in a second-order
imply that A cos(tijr) and B sin(tof) are both
solutions to the differential equation
eral form
x(,) = (A:, + K 2 t)e-'>^'
d^\>C
dr
source-free critically damped circuit has the gen­
where all parameters are real, but not necessar­
LC
ily positive. Prove that x(/) = 0 at some t = T <
By superposition, then, v^t) = A cos(ojr) + B
sin((or) V. Show that for a given A and B, there
exist K and 0 such that Vf\t) = A cos(tof) + B
6. (a)
underdamped
v^t) =
and
satisfies the differential equation
x"{t) + 2}^'{t) + h~x{t) = 0
and AT, are arbitrar)^ constants,
4. The voltage or current in a second-order
source-free overdamped circuit has the general
[10 cos(9950^) + 1.005
the voltage waveform before the peak
W^{t) is constant.
x{t) = (A'l + K^j)e-^‘
the
+ 5.7°)
How many cycles o f “ringing” occur in
energy stored in C and L for the circuit o f
3. By direct substitution, show that
and
sin(9950/)] = 10.05^’" ‘®‘^®' cos(9950f
2. Find the expressions o f the instantaneous
form
is
response is given by
the two solution forms are equivalent.
where
Consider case 2 o f Example 9.5. The
circuit
sin(cof) = K cos(tor + 0). One concludes that
Figure 9.1b. Show that the sum o f
and K j have opposite signs.
COif and only if
value drops from its largest value o f
10.05 to 10.05/f = 0.3 6 8 X 10.05?
(b)
Suppose the characteristic polynomial
is written as
response
+ 2as +
form
x{t)
+ to^ with
=
cos(w y + 0). Prove that for the under­
damped case, the circuit will ring for
N = iy ijiln o) cycles before the ampli­
tude decreases to Me o i its initial
value.
7. When a dc voltage o f
volts is applied to a
series LC circuit with no initial stored energy,
the voltage across the capacitor reaches a peak
value o f twice the source voltage. To investigate
x{t) =
419
Chapter 9 • Second Order Linear Circuits
this phenomenon, consider the circuit o f Figure
UNDRIVEN RLC PROBLEMS
P9.7 where switch S is closcd at r = 0. Assume
9. The switch S in the circuit o f Figure P9.9 has
the inductor current and the capacitor voltage
been closed for a long time and is opened at / =
are zero at / = 0.
0. Express vj,t) and /^(r) for r > 0 in terms o f
the literals R, L, C, and /q. Also compute the
t =0
initial stored energy in the inductor and capac­
S ^
itor.
L
Figure P9.7
Show that for f > 0
Figure P9.9
1
.V l
]
c
',
and
10. In the circuit o f Figure P9.10, suppose
Vi„{t) = 10 V, /? = 10 n , C = 0.4 mF, Z = 0.25
H, and the switch opens at /^= 0.
yc(t) = Vo 1 -
COS
f
(a)
Compute
, //(0~), and
(b)
Compute the energ)^ stored in the
(c)
Using only energy considerations,
1
■Jlc\
inductor and the capacitor at / = 0.
8. The circuit in Figure P9.8 is a dual o f the
compute the maximum value o f
previous problem.
f > 0.
(d)
Find the analytical expression for
and verify the maximum value
of
Figure P9.8
computed in part (c).
R
i,(t)
,v (t)
v„(t) I
The switch S is opened at r = 0. Both the induc­
tor current and the capacitor voltage are zero at
r = 0. Show that for / > 0
Figure P9.10
11.
vU)
Consider the circuit o f Figure P 9 .ll in
which i r jf ) = -2 0 u (-r ) V, R = 10 Q, C = 0.4
mF, and L = 0.25 H.
and
' l ( 0 = / o 1 - cos
f
1
(a)
Compute
/^(O"), and
(b)
;,( 0 -) .
Compute the energ}' stored in the
inductor and the capacitor at r= 0.
(c)
Find the analytical expression for
Plot using MATLAB for 0 < / <
200 msec.
Chapter 9 • Sccond Order Linear Circuits
420
(d)
Find the analytical expression for
Plot using MATLAB for 0 < r < 200
msec.
i,(t)
v„(t)
(a) (t>0)
Figure P 9.11
12. Reconsider the circuit o f Figure P 9 .ll
= 50«(-/) V, R = 25
Q, C = 0.8 mF, and Z, = 1 H. Repeat Problem
under the conditions
11.
13. Reconsider the circuit o f Figure P 9 .11 under
the conditions v-^{t) = -5 0 « (-r) V, R = 25 O., C
Figure P9.15
= 0.8 mF, and Z, = 2 H. Repeat Problem 11.
16. Figure P9.16 shows an overdamped source14. For the circuit o f Figure P9.14, suppose C
= 0.8 mF and determine L so that the fre­
quency o f the sinusoidal response, for r > 0, is
2500 rad/sec. Now find y(;;(0~),
free circuit in which R = 0.4 Q., L = 0.5 H, and
C = 0.5 F.
(a)
/(^0“),
If Vf^O) = - 2 V and /^(O) = 2.5 A, find
V(it) for r > 0. Use MATLAB or the
/^O"^), and Vf^t) for r > 0.
equivalent to plot the v^^t) waveform
and verify that there is no zero cross­
^
t=0/\
L
25 0
_L
100 0
©
ing.
25 mV
(b)
If v^O) = 2 V and /^(O) = 2.5 A, find
v^t) for t> 0 . Use MATLAB or equiv­
alent to plot the V(^t) waveform and
verify that there is no zero crossing.
Figure P9.14
15. Consider the circuit o f Figure P9.15, in
which
= 2 QV, R^ = 2 Cl, Rq = %Q., R = 2 Q,
Z. = 0.5 H, and C = 62.5 mF.
(a)
Figure P9.15 a shows a source-free
parallel RLC circuit whose past history
Figure P9.16
is depicted by Figure P9.15b where
the switch S has been at position A for
17. In Figure P9.17, the switch S has been at
a long time before moving to position
position A for a long time and is moved to posi­
B at r = 0. Find
y^^O'*^), /^0~),
/^O"^), and V(^t) for r > 0. Plot V(^t)
(b)
tion B at t = 0. Suppose
0.5 a
using MATLAB for 0 < f < 1.25 sec.
Find i^{t) for / > 0. Plot V(^t) using
(a)
MATLAB forO < r< 1.25 sec.
(b)
= 100 mV, R =
/. = 1 H, and C = 0.01 F.
Find
and
for f > 0.
Plot for 0 < / < 50 sec.
Find i^{t) for r > 0. Plot for 0 < / < 50
sec.
421
Chapter 9 * Second Order Linear Circuits
20. The voltage across the capacitor for the cir­
cuit of Figure P9.20a is given by Figure P9.20b.
Suppose R = 25 k£2.
v J t)
(a)
Using the plot, estimate the values of
L and C
(b)
Figure P9.17
O '
18. For the circuit of Figure P 9.18,
current.
=
Clearly show and explain all steps in your cal­
0,5u{-t) A.
(a)
culations. Hint: You might assume a general
I f /? = 20 Q , Z = 1 H, and C = 8 mF,
find and plot V(^t) and
(b)
Now estimate the value o f the initial
capacitor voltage and initial inductor
response form v^^t) = Ae~^^ cos(co^ + 0) V and
for f > 0.
then use the plot to estimate a and O)^; what is
Repeat part (a) for R = 22.5
the relation of a and Ci)^ in the characteristic
polynomial of a parallel RLQ
0
ijt)|
Figure P 9.18
19. In Figure P 9.19. ^^<0") = 25 V, /^(O-) = 50
mA, /? = 2 kQ, Z = 0.1 H, and C = 0.1 ;/E The
switch closes at f = 0.
3
(a)
Compute y(;;(0^), ^^(0"^), and /^(O^).
(b)
Compute /^(/),
(c)
Plot i^(t) and Vjit) forO < t< 1 msec. |
(d)
and V({t).
|
|
Find the energy stored in the circuit ^
over the interval [0, 0.2 msec], i.e., in
the capacitor and the inductor over
this interval. Is this energy positive or
negative? Also compute the energy
dissipated in the resistor over this same
(b)
interval. The sum o f the energy dissi­
pated in the resistors, the energy
Figure P9.20
stored in the capacitor, and that stored
in the inductor should equal zero.
21. In the circuit of Figure P 9.21, R = AO., C
Why?
= 6.2 5 mF, and
i,(t)
/YYV<
L
for f > 0.
(a)
(b)
t=0
Figure P9.19
Compute the value of L.
Find the value of the initial condi­
tions, ^^(0*) and
(c)
'n^
= (20 - \200t)e~^^* mV
Find ijit) for t> 0.
422
Chapter 9 • Sccond Order Linear Circuits
i jt )
/ Y W
Figure P9.21
Figure P9.24
22. For the circuit o f Figure P9.22, L = 0.04 H,
C = 2.5 mF, and /^, = 10 LI.
(a)
Find the value o f R (in ohms) that
25. Figure P9.25 shows a critically damped
makes the circuit o f Figure P9.22 crit­
source-free circuit.
ically damped.
(b)
0.1 H
Given this value o f R, suppose V(^Q) =
160
mV
and
/^(O)
= -6 0
mA.
Compute V(\t).
(c)
40 0
0.25 mF
Determine the first time at which the
+
sVc(t)
capacitor voltage is zero. Plot your
result using MA'FLAB or the equiva­
F'igure P9.25
lent to veri5' your calculation.
(a)
If v^O) = - 5 V and /^(O) = 1 A, find
;^(f) for r > 0. Determine the differen­
,_ r Y Y \ =
tial equation for the circuit. Let^ = ij
and use “y = dsolve(‘D2y + 400*D y +
40e3*y = 0,y(0) = l,D y(0) = -3 5 0 ’)” in
MATLAB to verify your answer.
Figure P9.22
(b)
If i;JO ) = 5 V and /^(O) = 1 A. find
V(^t) for t> 0. Plot the V(^t) waveform
23. Reconsider the circuit o f Figure P9.22 with
and verify that there is one zero cross­
/?///?, = 0.8
ing. Again use the dsolve command in
(a)
and A = 0.04 H.
MATLAB to verif}' your calculations.
Find the value o f C so that the circuit is
underdamped with to^= 100 rad/sec.
(b)
(c)
Suppose
= 160 mV and /^(O) =
- 3 0 mA. Compute
Determine the first time at which the
26. The capacitance voltage o f a source-free
parallel RLC circuit, with R = 2.4 LX has the
form
V(^t) =
capacitor voltage is zero. Plot your
result using MATLAB or its equiva­
(a)
lent to verify your calculation.
(b)
cos(8r + B)
Find the values o f L and C.
If y^^O) = 10 and /^(O) = 0, determine
A and 0.
24. In the circuit o f Figure P9.24, /? = 20 Q and
(c)
If the values o f L and C remain
unchanged, find the value o f R for the
//:.(0 = 500f?"'^' sin (l0 V 3 /)
circuit to be critically damped, and the
mA for r > 0. Find the proper values o f L and
general
C to produce this response. Now find t'^^O'^),
response
under this condition.
Then determine the source-free
response when V(^0) = 10 and /^(O) =
y'c^O"^), and v^it) for r > 0.
0.
form
of
the
source-free
423
Chapter 9 • Second Order Linear Circuits
27. Almost 75% of fliilures in circuits, i.e., situa­
tions where a circuit dramatioilly fails to perform
as designed, are due to opens and shorts o f indi­
vidual circuit elements. Heating, c\'cling a circuit
on and off, etc., cause degradation in the circuit
parameters, resistances, capacitances, inductances,
etc. that often precipitates the short or open situa­
tion. For example, the material inside a resistor
might become brittle over a period of time and
finally crumble, leaving a break in the circuit. On
the other hand, the material might congeal or
become dense, decreasing the resistance. In the
problems below you are to determine the length of
time it takes for a circuit to move from an over­
damped behavior to an imderdamped behavior
due to changes in the resistor characteristic as a
fiinction of time.
(a)
For the parallel RLC circuit in Figure
P9.27a, suppose R = Rq + exp(/ - 5) H
where f > 0 constitutes time in years.
Determine the time f’ for which the cir­
cuit changes its behavior from over­
damped to underdamped.
(b)
For the series RLC circuit of Figure
P9.27b, the resistor satisfies R =
+
exp(r - 5)] n , where again t is time in
years. Here it is presumed that the circuit
is pan o f a larger piece o f electronic appa­
ratus, such as a TV, which is used exten­
sively over a period o f years. The tiine t'
then is not connected with the response
time o f the circuit. Determine the time
f ' for which the circuit changes its behav­
ior from overdamped to underdamped.
DRIVEN SERIES AND
PARALLEL /?/.C CIRCUITS
28. (Initial condition calculation) For the cir­
//(O'*'),
cuit shown in Figure P9.28, find
;^ 0 ^ ), and
in two steps:
Sff/> I. Find V(^0~) and /^(O") by open-circuit­
ing C and short-circuiting L.
Step 2. Construct a resistive circuit valid at t =
O'*’ and from this find
29.
and
Consider the circuit o f Figure P9.29 in
which
= -10w(-/) + 20«(f) V, 7? = 20 Q,
C = 0.1 niF, and L = 0.25 H.
(a)
Compute
/^(0“), and
(b)
Compute the energ)'^ stored in the
inductor and the capacitor at r = 0.
(c)
Find the analytical expressions for the
zero-input, zero-state, and complete
responses for
Identif}' the tran­
sient and steady-state responses. Plot
V(^t) using MATLAB over [0, 40
msec].
(d)
Find the analytical expressions for the
zero-input, zero-state, and complete
R=
R„ = 0.8 Q
responses for i/{t) . Plot ij{t) using
MATLAB over [0, 40 msec].’
i,(t)
ijO
L= 1H
R„ = i s n
C = 1/36
(b)
Figure P9.27 Parallel and series RLC circuits subjccc to resistor degradation over time.
30. Reconsider the circuit o f Figure P9.29 under
the conditions
+ 25u{t) V, /? = 25
Q, C= 0.8 mF, and Z, = 2 H. Repeat Problem 29
but construct plots over [0, 400 ms].
Chapter 9 • Second Order Linear Circuits
424
31. Reconsider the circuit of Figure P9.29
under tiie conditions
= -50«(-r) +25u{t)
V, 7? = 25 ii, C = 0.8 mF, and L = 0.2 mH.
Repeat Problem 29 but construct plots over [0,
40 ms].
32. In Figure P9.32 v-J^t) =-250«(-r) +750«(f)
mV, ^ =0.5 a , I = 1 H, and C= 0.01 F.
(a) Find /^(O^),
and the zeroinput, zero-state, and complete
responses of v^it) for r > 0. Identify
the steady-state and transient parts of
the complete response. Plot in MATLAB for 0 < ^< 50 msec.
(b) Find the zero-input, zero-state, and
complete responses of i^{t) for t > 0.
Plot in M ATLAB for 0 < r < 50 msec.
ijt)
fY Y \ h ^
36. In Figure P9.36,
Figure P9.32
T he switch closes at f =0.
(a)
(b)
(c)
Computey^O*), y^(O^), and /^(O^).
Compute the zero-input, zero-state,
and complete responses of
and v^t). Identify the transient and
steady-state parts of the complete
response.
Plot i^it) and Vjit) for 0 < r < 1 msec.
Find the energy stored in the circuit
over the interval [0, 0.2 msec], i.e., in
the capacitor and the inductor over
this interval. Is this energy positive or
negative? Also compute the energy
dissipated in the resistor over this same
interval.
o
n
i,(t)
33. Repeat Problem 32 for /? =40 Q and v-JJ)
= -0.5tt(-/) + 2«(/) V. Plots in M ATLAB
should be for 0 < f < 800 msec.
34. Repeat Problem 32 for /? = 50 Q and and
v.^{t) =-0.5«(-/) - 2u{t) V. Plots in M ATLAB
should be for 0 < ? < I sec.
=-1 0 « (-^ ) + 40«(/)
mA, /? = 4 k n , Z = 0.1 H, and C = 0.1 pF .
(d)
vJt)
o
Figure P9.35
fY Y \ k
L
+
,Vc(t)
(t)
Figure P9.36
35. For the circuit of Figure P9.35,
=
-0.5«M + 2u{t) A.
37. For the circuit of Figure P9.37, R^= 5 0.^
(a) I f /? = 2 Q, Z = 1 H, and C = 8 mF, = 2 0 a , C = 2.5 mF, L = 0.25 H, and v j t ) =
find and plot the zero-input, zero- 20«(^) - 20u(t-7) V, where T = 0.25 sec.
state, and complete responses of v^{t)
(a) Find the zero-input, zero-state, and
and ij\t) for / > 0. Identify the tran­
complete responses of v^t) for f > 0.
sient and steady-state parts of the
Plot the complete response for 0 < r <
complete response.
0.25 sec.
(b) Repeat part (a) for R= 22.5 Q.
n
o
o
n
Chapter 9 • Sccond Order Linear Circuits
(b)
■li')
Find the zero-input, zero-state, and
42. Repeat Problem 40 for
complete responses o f
200 n .
for t > 0.
= 50
and Rj =
Plot the complete response for 0 < r <
4 3 . Consider the RLC circuit in Figure P 9.43
0.25 sec.
where /?^, = 60 Q, R^2 =
= 5m F
(a)
Q, I = 4 H, and C
= 100«(-f) mA and r^2(^) =
2 0u (-t) V. Find the response,
(b)
for
t> 0. Plot for 0 < f < 1 sec.
/y,(r) = 100u(-t) + 500u(t) mA and
1/^2 ^) = 20u(-t) V. Find the response,
for ^ > 0. Plot for 0 < / < 1 sec.
= 50 Q, /?j = 200
Q, C = 0.05 mF, L = 0.5 H , and v j t ) =
- 5 0 « (- r ) + 50k(?) - 50«(r - 7) V, where T =
38. Repeat Problem 37 for
0.08 sec. However, only plot the complete
I V ,(t)
responses for 0 < f < 200 msec.
39. Repeat Problem 37 for R^= 100 Q ,
=
100 Q , C = 0.25 mF, L = 2.5 H , and v-J^t) =
Figure P9.43
-5 0 tt(-f) + 50«(f) - 50«(^ - 7) V, where T =
100 msec. However, only plot the complete
44. Repeat Problem 4 3 , except find y^(r), ^> 0.
responses for 0 < ^ < 300 msec.
45. Consider the RLC circuit in Figure P9.43
40. Consider the RLC circuit o f Figure P 9.40 in
where R^i = 20 Q, R^2 = 20 Q, L = 0.4 H, C =
which R^ = 100 Q, /?! = 4 0 0 Q , C = 0.125 mF,
4 mF.
L = 0.2 H, and v.„{t) = 50u{t) - 50u U - 7) V,
where T = 0.025 sec, v^iQr) - - 2 5 V, and
(a)
/^(0“ ) = 10 mA.
(a)
(b)
(r) = -u (-t) A and
Find the response,
= 40«(-^) V.
for t> 0 . Plot
for 0 < f < 0.8 sec.
Find the zero-state, zero-input, and
complete responses of V(it) for r > 0.
(b)
(f) =
+ 2u{f) - 2uU - 0.4) A
and v^2 (^) = 4 0 « ( -/) V. Find the
Plot for 0 < r < 60 msec.
response,
Find the zero-state, zero-input, and
r < 0.8 sec.
for ^ > 0. Plot for 0 <
complete responses of v^it) for f > 0.
Plot for 0 < r < 60 msec.
46. Repeat Problem 4 5 , except find
t> 0 .
47. Consider the RLC circuit in Figure P9.43
w
where /?j| = 16
R^2 = 32 Q, I = 0.4
H, and C = 4 mF.
w
(a)
'n- ^
f > 0. Plot for 0 < r < 0.3 sec.
(b)
Vw>
i M = - 0 .5 « ( - / ) A and v,y{t) =
24«(-^) V. Find the response,
for
41. Repeat Problem 40 for R^= 140
= 360 a
and Rj
/■^,(r) = -0.5u{-t) + 0.5«(f) A and v^2 ^t)
= 24u(-t) V. Find the response,
for f > 0. Plot for 0 < f < 0.3 sec.
426
Chapter 9 * Sccond Order Linear Circuits
r> 0.
48. Repeat Problem 47, except find
49. riic current source with /^|(^) = 5//(-/‘) mA
and the voltage source
= 10 V in Figure
P9.49 drive the circuit in which /? = 1 k^2, C =
0.5 //F, and Z. = 0.184 H.
(a)
Find
(b)
.,(0 ^ ).
Compute
t'(-(0‘"),
/^(O"^),
/^(0‘^),
and
for r > 0. Plot for 0 < r
PSEUDO SERIES AND
PARALLEL /?/.C CIRCUITS
52. Consider the circuit o f Figure P9.52 in which
/?! = 4 a
= 4 a /. = 5/12 H, C = 25 mF
(a)
Find the roots o f the characteristic
equation. C H EC K : - 8 , - 1 2
(b)
/j^(0"), ^i(0*), and
< 5 msec.
(c)
Compute
Compute
for / > 0. Plot for 0 < f
ages and currents, draw the equivalent
circuit valid for t > 0. Check your
for r > 0. Plot for 0 < r
answer for
< 5 msec.
(e)
Compute
for r > 0 . Plot for
0
for t >
0. Hint: After finding the initial volt­
< 5 msec.
(d)
If v.^,{t) = -2Qu{-t) V, find 1/(^0-),
using the “dsolve”
command in MATLAB. Plot
<t
for
0 < r < 2 sec using MATLAB or the
< 5 msec.
equivalent. Be sure you properly label
your plot.
(0
If
(d)
If y,./r) = -2 0 u (-t) + 20u(r) -2 0 u (t- 1)
= - 2 0 « (- f) + 20u(r) V, find
for / > 0.
V, find and plot v^^t) for 0 < r < 2 sec.
Figure P9.49
50. Repeat Problem 49 for the new source cur­
rent /jj(r) =
0 < r <_10 ms.
-5 ti(r-
0.005) mA. Plot for
51. The switch in the circuit o f Figure P9.51 is
in position A for a long time and moves to posi­
tion B at / = 0. Find the voltage
for t > 0
when L equals (a) 0.625 H, (b) 0.4 H, and (c)
0.2 H.
53. Repeat Problem 52, but find i^{t) and v^{t)
without differentiating
54. Repeat Problem 52 for
= 4 Cl, Rj = 4 Q,
1 = 0.2 H, a n d C = 0 .2 F .
10V
55. Reconsider the circuit o f Figure P9.52 in
which /?, = 600 Q, /?2 = \ 2 0 n ,L = 2 H, C = 1
Figure P9.51
mF and v j t ) = -72u{-r) + 72u(t) -72u{t - 1).
Find the response
for r > 0 as follows:
(a)
Find
(b)
Find z^f;(0'*') and i^CO"*").
(c)
and
Draw the equivalent circuit valid at O'*"
and find ^'/(O'^) and Z(;;;^0‘'').
^1/
Chapter 9 • Sccond Order Linear Circuits
(d)
(e)
Find die characteristic equation and
60. Reconsider the circuit of Figure P9.57.
natural frequencies o f the circuit.
Suppose /?j = 80 Q, /?2 = 4 0
^ = 2 H, and
C = 0.625 mF with
= -1 5 0 tt(-/) + 150«(/)
mA.
Determine the general form o f the
response, ij{t)y valid for 0 < ? < 1.
Determine all coefficients in the gen­
(a)
Find V(iO~) and V(^0*).
eral form o f the response.
(b)
Find ii(0~) and /^(O'*’).
(g)
Determine the form o f the response,
(c)
Find t//^(0+) and /’c(0+).
for f > 1. Find the response.
(d)
(h)
Plot the response
(f)
for / > 0 using
MATLAB or the equivalent.
Find the characteristic equation and
natural frequencies o f the circuit.
(e)
Find the response,
(f)
Find the response,
t> 0.
t> 0.
56. Repeat Problem 55, except find Vf4,t).
6 1 . Consider the circuit of Figure P9.61
57. Consider the circuit o f Figure P 9.57 in
which
= 2 £2, /?2 = 2 Q, Z = 0 .4 H, and C =
0.1 E
(a)
(b)
Suppose /?j = 80 Q, R2 = 40 Q., L = 2 H, and
C = 0.6 2 5 mF with
= 300u(f) mA, Vq =
50 V, and /?3 = 20 Q.
Find V(iO~) and y^^O'*’).
Find the roots of the characteristic
(a)
equation.
(b)
Find i^(0-) and z^(0+).
(c)
Find v^{0*) and /’c (0 ‘*‘).
(d)
Find the characteristic equation and
= -2u{-t) A. find t;c<0-),
If
^>
0, Hint: After finding the initial volt­
natural frequencies o f the circuit for t
>0.
ages and currents, draw the equivalent
circuit valid for ^ > 0. Check your
(e)
Find the response, v^t), t> 0.
answer for ij{t) using the “dsolve”
(f)
Find the response, ijXt), t> 0.
command in MATLAB. Plot
for
0 < ^ < 1 sec using MATLAB or equiv­
alent. Be sure you properly label your
plot.
(c)
If
= -2u{-t) + 2«(?) A, find
for ^ > 0.
(d)
If
= -2u{-t) + lu{t) - l u { t - 1) A,
find and plot
for ^ > 0.
62. The switch in the RLC circuit of Figure
'S->
ijt)
d)
P 9.62 opens at ^ = 0 after having been closed
+
> F
L
for a long time. The purpose o f this problem is
to find the complete response o f the capacitor
voltage, ? > 0. Suppose
Figure P9.57
58. Repeat Problem 57, except find
V^
^
(b)
(c)
59. Repeat Problem 57 with
W
(t) = 1 A, v^2 (^) = 20
V, C = 4 mF, and I = 0.625 H.
(a)
Using a dc analysis, find the initial
Using a dc analysis, find the final value
o f the capacitor voltage, Vf^oo).
= 80 £2, R2
2 0 a , L = 10 H , and C = 1/240 E
conditions /^(O") and V(^0~).
Find
and
(d)
Find the characteristic equation and
■\2H
(e)
Chapter 9 • Sccond Order Linear Circuits
compute its roots. Given tiie roots,
tion with ij{t) as the unknown. Observe that
write down the general form o f the
the derivative o f
response
hand side.
is present on the right-
Solve for the unknown coefficients in
the response form o f part (d) and write
ANSW'F.R: /"(/) + i\t) + i{t) =
^
down the exact expression for v^^t)
valid for ^ > 0.
40 fi
40 0
Figure P9.62
65. Consider the circuit shown in Figure
GENERAL SECOND-ORDER
CIRCUITS
P9.65.
(a)
roots o f the characteristic equation.
63. In the operational amplifier circuit o f
Figure P9.63 is a second-order circuit. Suppose
Write a second-order differential equa­
tion with
the unknown. Find the
(b)
If
V, find V(^t) for r > 0.
= R^ = 50 k n .
(a )‘
Determine the values o f C, and Cj
that produce a characteristic equation
having natural frequencies at - 5 and
-
(b)
10 .
Adjust the value o f /?j so that for a
step function input voltage, the value
o f the output voltage for large t is for
all practical purposes is 5.
(c)
66. Consider the circuit shown in Figure
when v^{t) = 2u{t)
P9.66.
and all capacitor voltages are zero at t
(a)
Compute
=
0.
Write a second-order differential equa­
tion with
the unknown. Find the
roots o f the characteristic equation.
Then find V(^t) when
(b)
Repeat part (a), for when
unknown.
Figure P9.63 Cascade of leaky integrator circuits
having a second-ordcr response.
64. Consider the circuit shown in Figure
P9.64. Write a second-order differential equa-
= u{t) A.
is the
Chapter 9 • Sccond Order Linear Circuits
429
67. The second-order circuit shown in Figure
nearly 1 V in about 5 nsec, it behaves approxi­
P 9.67 contains two capacitors.
mately as an open circuit. The 0.1 pF capacitor
(a)
(b)
Find the second-order differential
is then charged up with a time constant of
equation with V(^ as the unknown.
about 0.1 msec. As the larger capacitor is
Give the values o f s.
charged up, the output across the smaller one
If vci (0) = 2 V, and ^ ^ (0 ) = 4 V, find
V(^{t) for t> 0.
decreases toward zero.
0.5 0
2F
0.5 n
0.5 0
2F
Figure P9.67
Figure P9.69
68. In the circuit of Figure P 9.68, the voltagecontrolled voltage source has a gain A > 0 . Find
the ranges of
damped,
(b)
for the circuit to be (a) over­
underdamped,
(c)
critically
damped, and (d) undamped.
70. Find the value o f the negative resistance
-R„ for the circuit shown in Figure P 9.70
required to generate sinusoidal oscillations with
constant amplitude.
-R .
V,F
120
Figure P9.70
69. The second-order circuit shown in Figure
's-/'
O '
P 9.69 is of the overdamped type. Find the step
response, i.e., the expression for Vg{t) for ^ > 0,
when the input is vi{t) = u{t) V, and the capac­
itors are initially uncharged. Roughly sketch
the waveform of Vpit). Verify your sketch by
doing a SPICE simulation o f the circuit.
so
Remark: The waveform v^it) consists of a very
fest rise toward 1 V, and then a relatively slow
o
exponential decrease toward 0 V. This can be
explained using the first-order RC circuit prop­
o
erties studied in Chapter 8. During the first few
microseconds, the 0.1 jiF capacitor behaves
o
almost as a short circuit, and the 1 nF capacitor
is charged with a time constant of about 10
nsec. After the smaller capacitor is charged to
'<N->
o
71. Refer to the Wien bridge oscillator of
Example 9.11. Suppose the op amp has a satu­
ration voltage 15 V. C = 1 jiF and /?2 = 500 Cl.
is a temperature-sensitive resistor whose
resistance is a fixnction of the amplitude of the
sinusoidal current passing through
. The fol­
lowing relationships are given:
?l(f) =
sin(C0f + 0)
/?, = 500 + 1 0 0 ( 7 ^ -0 .0 1 )
(a)
Find the frequency of oscillation (iHn-
(b)
Find the amplitude of voltage at the
op amp output terminal (with respect
to ground).
(c)
Suppose /?j is a fixed resistance of 4 9 0
t^ci(O) = 100 mV, and V(^{0) = 0 in
the Wein bridge oscillator circuit.
Perform a SPICE simulation. Does
the circuit behave as expected?
430
Chapter 9 • Second Order Linear Circuits
72. In the Wein bridge oscillator example o f
this chapter, let
= R-, = \ kH, Rr= 10
and
C=0.1pF.
(a)
Determine the frequency o f oscillation
in Hz.
(b)
If
(0) = 5 V and v^{0) = 0 V, find
v^{t) for r > 0.
(c)
Use any circuit
simulation
(e.g.,
SPICE) to verify the waveform o f v^{t)
in part (b).
73. In the Wein bridge oscillator o f Figure
9.15a, /?2 = 1 k n . ^/=10
and C = 0.1 pR
The lamp resistance R^ is a function o f the peak
value o f the sinusoidal current as shown Figure
P9.73.
(a)
Determine the frequency o f oscilla­
tion.
(b )
Determine the amplitude o f the sinu­
soidal waveform v^{t).
C
H
A
P
T
E
R
Sinusoidal Steady State
Analysis by Phasor Methods
A HIGH-ACCURACY PRESSURE SENSOR APPLICATION
The control o f high-performance jet engines requires highly accurate pressure measurements, with
errors less than one-tenth o f 1% o f a full-range measurement, over a wide range o f temperatures,
- 6 5 ° to 200° F. The pressure range may be as low as 20 psia or as high as 650 psia. In jet (turbine)
engine applications, knowing pressure and temperature allows one to compute the mass (volume)
air flow, a critical aspect o f an engines performance. A pressure sensor is also a critical component
in the regulation o f aircraft cabin pressure. Such a sensor is depicted here along with a functional
block diagram o f its operation. A diaphragm consisting o f t\\'o fused quartz plates separated by a
vaciumi has a capacitance that changes as a function o f pressure and temperature. This quart/
capacitive diaphragm is an element in a bridge circuit. It is this bridge circuit, in conjunction with
detailed knowledge o f the characteristics o f a pair o f quartz capacitors over the required operating
range o f pressure and temperature, that enables accurate pressure measurements.
Functional block diagram courtcsy o f AllicdSignal Aerospace Com pany.
432
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
Because o f the small capacitances, on the order o f picofarads, associated with the quartz
(diaphragm) capacitor, the bridge circuit is driven by an ac source and is called an ac bridge.
Driving the bridge by an ac source moves its analysis outside the realm of the dc and step response
techniques studied in earlier chapters. New methods o f analysis, such as phasor analysis, are nec­
essary. Phasor methods, the primary focus o f this chapter, allow us to analrze capacitive and induc­
tive circuits excited by sinusoidal (ac) inputs. In particular, phasor techniques permit us to anal)'ze
an ac bridge circuit. Although the analysis ot the pressure sensor shown here is beyond the scope
o f this text, the chapter will end with a simplified pressure sensor circuit based on the one shown.
CHAPTER OBJECTIVES
1.
Review and elaborate on the basic arithmetic and essential properties o f complex num­
bers pertinent to sinusoidal steady-state analysis o f circuits.
2.
3.
4.
Develop two complementary techniques for computing the response o f simple RL, RC,
and RLC circuits excited by sinusoidal inputs and modeled by differential equations.
Define the notion o f a (complex) phasor for representing sinusoidal currents and voltages
in a circuit.
Using the notion ol phasor, introduce the notions of impedance, admittance, and a gen­
eralized Ohm’s law lor two-terminal circuit elements having phasor currents and voltages.
5.
Utilizing the methods o f nodal and loop analysis and the nerwork theorems o f Chapters
5 and 6, analyze passive and op amp circuits by the phasor method.
6.
Introduce the notion o f frequency response for linear circuits, i.e., investigate the behav­
ior o f a circuit driven by a sinusoid as its frequency ranges over a given band.
SECTION HEADINGS
1.
2.
3.
Introduction
Brief Review o f Complex Numbers
Naive Technique for Computing the Sinusoidal Steady State
4.
5.
Complex Exponential Forcing Functions in Sinusoidal Steady-State Computation
Phasor Representations of Sinusoidal Signals
6.
7.
8.
9.
10.
11.
12.
Elementary Impedance Concepts: Phasor Relationships for Resistors, Inductors,
and Capacitors
Phasor Impedance and Admittance
Steady-State Circuit Analysis Using Phasors
Introduction to the Notion o f Frequency Response
Nodal Analysis o f a Pressure-Sensing Device
Summary
Terms and Concepts
13.
Problems
Chapter 10 • vSinusoicial Steady State Analysis by Phaser Methods
-133
1. INTRO DUCTION
Perhaps you have experienced the bouncing motion of a car with broken shock absorbers or
watched the (mechanical) oscillations o f a sw'inging pendulum. These motions reflect the sinu­
soidal and damped sinusoidal oscillations in
circuits with conjugate poles o f the characteris­
tic equation, as detailed in Chapter 9. In this chapter we allow sources with sinusoidal forcing
functions (such as such as
cos(ov + 0) or perhaps
sin(d)/ + f))), which almost always result in
sinusoidal responses regardless o f the root locations o f the characteristic equation. A sinusoidal
voltage source models the voltage from the ubiquitous wall outlet.
If one hooks up an oscilloscope to measure a voltage in a linear circuit driven by sources with sinu­
soidal values, the voltage may not look sinusoidal at first. However, if the circuit is stable, after a
sufficiently long period o f time the screen o f the scope will trace our a sinusoidal waveform. (Here
“stable” means that any zero-input response consists o f decaying exponentials or exponentially
decreasing sinusoids.) The eventual sinusoidal behavior is not immediately apparent because at
startup, stable circuits exhibit a transient response. “Transient” means that the circuit response is
transitioning— for example, from an initial voltage or current value to another constant value.
Flickering lights during a thunderstorm illustrate the phenomenon o f transient behavior: light­
ning may have struck a transmission line or pole, causing the power system to waver briefly from
its nominal behavior.
Because sinusoidal excitations and sinusoidal responses are so common, their study falls under the
heading o f sinusoidal steady-state (SSS) analysis. Here “sinusoidal” means that source excita­
tions have the form
we take
+ 0) =
cos((or + 0) or K sin((0/ + 0). For consistency with traditional approaches,
cos(tor + 0) as the general input excitation, as shown in Figure 10.1, because
sin(to/
cos(tof + 0 - 7t/2). Steady state mean that all transient behavior o f the stable circuit has
died out, i.e., decayed to zero. Observe that every sinusoidal waveform is periodic with angular
argument (to/ + 0). In terms o f angle, each cycle o f the waveform traverses 2 n radians. In terms of
time, each cycle covers a time interval o f T = 27r/o) seconds, called the period o f the waveform.
The number o f cycles contained in 1 second is called the frequency o f the sinusoidal waveform
and is denoted by / T h e unit for/is the herrz (Hz), meaning “cycles per second.” The quantity
OJ, which specifies the variation o f the angular argument (tor + 0) in 1 second, is called the angu­
lar frequency o f the w^aveform. The unit o f (O is radians per second (rad/sec). From these defini­
tio n s,/ = 1/7'= ti)/27i and o) = 2 k /
Stable circuits driven by sinusoidal excitations produce sinusoidal voltages and currents, as illus­
trated in Figure 10.1. The output excitation in Figure 10.1 has the general form
to distinguish it from the input excitation,
cos((0/ + (j))
cos((0/ + 0). Because o f linearity, the circuit can
change only the magnitude o f the input sinusoid
is changed to K^^) and the phase angle o f the
input sinusoid (0 is changed to (j)) while ensuring that the angular frequenq^ to remains the same.
For nonlinear circuits, to can and usually does change.
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
434
Input Excitation: Vjcos(cot + 0)
CO
FlGURl'l 10.1 Graphical illustration ot steady-state sinusoidal linear circuit behavior. (K^and
could just as well be
and
or any combination thereof.) Note that 0 and (}) are often different and
that (0 is the same for both input and output excitations.
In Figure 10.1 the steady-state (voltage) response is
been a current response,
+
8
cos(a)r + ({)). Alternatively this could have
cos(a)r + (j)). Such waveforms have the equivalent structure A cos(wr)
sin(oj/), deducible from trigonometric identities,
cos({or + (}))=
cos((j)) cos(tor) -
= A cos(a)f) + B sin(tor)
where A =
obtains
cos(({)) and B = —
sin(i|)) sin(cor)
( 10 . 1)
sin((})). Conversely, by summing the squares oi'A and B, one
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
V„, =
4 3 “^
n ------ ^
+ B-
(10.2a)
By taking the inverse tangent o f the ratio o f - B and A, one obtains
(10.2b)
In using equation 10.2b it is important to adjust the resulting angle for the proper quadrant o f
the complex plane. Equations 10.1 and 10.2 turn out to be useful in developing a conceptually
simple, although naive, technique for computing the steady-state response using a differential
equation model o f the circuit, as explained in section 3.
Sinusoidal steady-state (SSS) analysis o f circuits draws its importance from several areas. The
analysis o f power systems normally occurs in the steady state where voltages and currents are sinu­
soidal. Music is a rhythmic blend o f different notes. Mathematically, a musical (voltage) signal can
be decomposed into a sum o f sinusoidal voltages o f different frequencies. The analysis o f a sound
system typically builds around the steady-state behavior o f the microphone, the amplifier, and the
loudspeakers driven by sinusoidal excitations whose frequency varies from around 40 Hz to 20
kHz. Indeed, almost any form o f speech or music transmission requires an understanding o f
steady-state circuit behavior. There are many other areas o f applicability.
This chapter will introduce three techniques for computing the SSS response. The first two, some­
what naive, approaches map out a natural motivation and path to the third, ver)' powerful tech­
nique o f phasor analysis. Phasor analysis builds on the arithmetic o f complex numbers and the
basic circuit principles studied thus far. To set the stage for phasor analysis, section 2 reviews the
necessary basics o f complex number arithmetic. O f course, the student is assumed to have stud­
ied complex numbers in high school and in prerequisite calculus courses.
2. BRIEF REVIEW OF C O M P LEX NUMBERS
Let
= a + jb be an arbitrary complex number, where ^
. The real number a is the real
part o f 2 j, denoted hy a = Re[z,]. The real number b is the im aginary part o f z^, denoted by b
= Im[ 2 j]. It is simple to verify that
« = Re ^1 ^ i L ± i L
2
and
b = Im -1
where
_
2y
a - jb is the complex conjugate* o f Zy The magnitude or modulus o f 2 ,, denoted by
|2 ,|, satisfies
|z|P =
+ Ip-
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
•»36
The number Z| =
+ jb is said to be represented in rectangular coordinates. Another represen­
tation o f 2 p called polar form or polar coordinates, follows from the simple geometry illustrated
in Figure 10.2.
FIG UIIE 10.2 Diagram showing relationship between polar and
rectangular coordinates of a complcx niunber.
In Figure 10.2, the number Zj can be thought o f as a vector o f length p = >/«“ + b" = |zj|, which
makes an angle 0 = tan"*(^/<z) with the horizontal in the counterclockwise direction. (In comput­
ing tzn~^{bln) it is important to adjust the angle (principal part) to be in the proper quadrant o f
the complex plane.) Hence Zj = /? + jb is completely specified by its magnitude p and angle 0 i.e.,
z, = /z +
= p cos(0) + yp sin(0) = p[cos(0) + j sin(0)] = peJ^ = pZ.0
where p Z. 0 is a shorthand notation for
and
t’/ ’ = cos(0) + ;s in ( 0 )
(10.3)
is the famous Euler identity. The Euler identity can be demonstrated by writing the Taylor series
for
and recognizing it as the sum of the Taylor series for cos(0) added to j times the Taylor
series for sin(0). Note that the symbol Z. has two meanings, depending on the context o f its use:
(1) L z means angle o f the complex number z, and (2)pZ. 0 means the complex number whose
magnitude is p and whose angle is 0. The properties o f the exponential immediately imply that
(10.4)
Exercises. 1. Compute the polar coordinates o f
AN SW ERS:
2. Let z = 6
/T
= -1 - j and Zj = \ + j-
r ; /4S'’
where ti/6 has units o f radians; for example, n rad equals 180°. Find the real
and imaginary parts o f z.
ANSW ER; 2 = S. 1962 + /3
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
43*
3. Show by direct computation that since eJ^\ = cos(0j) + j sin(0j) and eJ^i = cos( 02) + j sin(f)2),
then
= cos(0,+ 0-,) + j sin(0,+ (),) =
With these simple definitions, the product oF t^vo complex numbers z-^ = a +jb =
and z-y
can be found using rectangular coordinates as
z^zf = {a +jb) {c + jd) = a c - bd +j{bc + ad)
or using polar coordinates as
= p ,p 2 cos(G,+ 62 ) + yp ,p 2 sin( 0 ,+ dj)
which in shorthand notation is
~ P 1P 2
^2^
EXA M PLE 10.1. Suppose z^ = 3 -jA = 5 Z .-5 3 .1 3 ° and ^2 = 8 + ;6 = 1 0 ^ 3 6 .8 7 °. Then
£,Z 2 = (24 + 24) + y(18 - 32) = 48 - ; 1 4
Equivalently,
z ,z 2 = 5 ^ - 53.130 X 1 0 ^ 3 6 .8 7 ° = 50 Z ( - 5 3 .1 3 ° + 3 6 .8 7 °)
=
+36.87«)
= 50 cos(16.26°) - ; 5 0 sin (l6.26 0 ) = 48 - ; 1 4
Exercise. Let Zj = 2 + j l and z^= - 2 + j 6 .
(a)
Compute the polar form o f 2 , and z ,.
(b)
Compute ZjZ-, in rectangular coordinates.
(c)
Compute ZjZ2 in polar coordinates.
A N SW ERS: 2.8284^'/''^", 6.3246^>-/>"''^- *-^‘’, - 1 6 + ;8 , 17.8885^->''^-^-'''-'‘’
Similarly, in rectangular coordinates the arithmetic for the division o f two complex numbers is
Cl _ a + jb _ {a + jb ){c - jd ) _ {a + jb ){c - jd )
Z2
c + jd
ic + j d ) { c - j d )
{ac + bd) + j ( b c - a d )
~
2
^2
c- + d
c -+ d ~
-i38
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
In polar coordinates the calculation is more straightfor\vard;
il- =
=-
B
i
Z2
-
) = -Hi-cos(6, - 0 2 ) + 7 — sin(6| - G j)
P2
P2
In our shorthand notation,
Exercise. Let Zj = 2 + j l and Zj =
+76 .
(a) Compute z^tzj in rectangular coordinates.
(b)
Compute z'^lzj in polar coordinates.
A N SW ERS:
(J.i
-
/ 0 . 4 , 0 . 4 4 7 2 ^ ' / ’ ’ "*^^'’
O f particular concern in this chapter are equations involving mixed representations o f complex
numbers. For example, suppose an unknown complex number z =
satisfies the equation
{a + jb) = f + j d
Then dividing through by
+ jb yields
c + jd
a + jh
Since Vis the magnitude o f the complex number on the right-hand side o f the equal sign, it fol­
lows that
+
_ V c-+ t/ ~
+
V «^+/r
Here we have used the fact that a complex number that is the ratio o f two other complex num­
bers has a magnitude equal to the ratio o f the magnitudes. To determine the angle 0, one uses the
property that 0 equals the angle o f the complex number in the numerator minus the angle o f the
complex number in the denominator,
-1
0 = Z (c + jd ) - Z {a + jb )= tan'
tan
Exercise. Let z, = 2 - 2j and
ANSW ER: 0.471
- 6 8 .4 7 "
— -ta n
= 5.5 + jlA . Find V and 0 when
-I
—
= 2 ,/z,.
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
EXA M PLE 10.2.
Suppose v(t) =
^ a n d
-5VeJ^ + j6VeJ^ Find V, (j), and v{t).
S o l u t io n
Factoring Ve-^ out to the left and dividing by (-8 + j6) yields
- 8 + j6
Hence. K= 2, (}) = - 9 8 .1 3 ” and
v{t) =
‘j’)] = 2 cos(to/- 98.13°) V
Sometimes a function v{t) is a complex number for each t, such as if{t) =
and v{t) will
satisfy some specific algebraic or differential equation. W hen this is the case, it is possible to use
the differential equation to find values for V and (j). The next two examples illustrate this strategy.
E X A M PLE 10.3. Suppose the function
v{t)
satisfies the differential equation
^
dr
+ 2 - + 2v = 10e-'<“ ” “ °'
dl
Find the values o f A and (}) if O) is known to be 2 rad/sec.
S o l u t io n
Since the function v{t) must satisfy the differential equation, the first step is to substitute into the
differential equation. Substituting
‘1’^ into the differential equation and taking appropriate
derivatives yields
The
term, which is always nonzero, cancels on both sides, leaving
AeJ^ [2 - co2 + ;2co] =
Since cu = 2, one can equate magnitudes and angles to obtain
2 - ( 0 " + y2 (0
-2 + y4
440
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
Exercise. Repeat the above example if (o = 3 and
ANSWERS: A = 2. (|) = 13‘J.4 "
is changed to
' HI.2”)
The techniques of circuit analysis in this chapter will often require complex number arithmetic.
The voltages and currents o f practical interest are always real. T he complex arithmetic is a short­
cut to computing “real” voltages and currents. The real quantities are obtained by taking the real
part o f the complex number or complex function. The various manipulations depend on some
general properties related to the real part o f complex numbers.
Property 10.1. Re[Z| + Zj\ = Re[z,] + Rel22j.
This property has a particularly nice application to summing trigonometric waveforms. Let v^{t)
= cos(oj/ + 55°) and r-y(t) = 10 sin(d)/ - 3 0 °) = 10 cosCd)/ - 120°). Note that a - 9 0 ° shift con­
verts the sine to a cosine. Hence,
?'l(r) +
= cos(o)r + 55°) + 10 co s(to f- 120°)
= Reld’-^^^'^^
+ Ret 10^-^^^'^^“
= Rek>("*'"
= Relt’-''^^^
by the Euler identity, equation 10.3
by Property 10.1
+ lOf*"/’ ’ *' )] by equation 10.4 and then factoring
to the left
= Re{f-^‘'^'1(0.5736 +/).8192) + (-5 - 78 .66 )]} after conversion to reaangular form
= Rel^>’'(-4 .4 2 6 - ; 7 . 8 4 l ) ] = Re[9<'>(‘"^ “
after simplification and con­
version back to polar form
= 9 cos(o)/ - 119.4°) after taking the real part.
This sequence o f manipulations shows that the magnitude and phase o f two cosines at the same
frequency O) can be represented by distinct complex numbers. One can then add the complex
numbers and determine the magnitude and angle o f a third cosine equal to the sum o f the origi­
nal two cosines. This presents a shortcut for adding two cosines together.
Property 10.2 {proportioiuilityproperty). R elazj] = aRefZ]] for all real scalars a .
Properties I and 2 taken together imply
R e [a , 2 , +
= a jR e [ 2 j] + a,Re[z-,]
which is a linearity property for complex numbers with multiplication by real scalars. 'I'he next
property, which underpins the techniques of this chapter, defines how differentiation can be inter­
changed with the operation Re[-].
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
Property 10.3 {differentiation property). Lex. A = ^e^^. Then
= Re - ( / t e - '" ' )
= Re
dt
Exercise. Find Re
(10 +
cit
ANSXXTR: 1118 cosdOO ; + 116.57
Our fourth property tells us the conditions for the equality o f two complex-valued time functions.
Property 10.4. For all possibly comple.x numbers A and B, Re[/lf’-^^'^T =
only
for all t if and
A = B.
Taken together, the preceding properties imply a fifth, very important propert)'. Here note that a
complex exponential is sometimes referred to as a complex sinusoid.
Property 10.5. I he sum of any number o f (1) complex exponentials, say AjeJ^'^‘, or (2) derivatives
o f any order o f complex exponentials o f the same frequency co, or (3) indefinite integrals of any
order o f a complex exponential o f the same frequency O), is a complex exponential o f the same fre­
quency (1).
This property is another foundation stone on which the phasor analysis o f this chapter builds.
Table 10.1 summari?,es the properties o f complex numbers.
TABLE 10.1 Summar}' of Properties o f Complex Numbers
cos((jL)r + 0) = A cos(cor) +
B sin(co^)
= V-4- +
, ({) = tan"'
A
Euler identit)'
eJ^ = cos(0) + j sin(0)
Real part o f sum
Re[z, + z-y] = Re[zj] + Re[z2l
Proportional it)' property
RelfXZ]] = a Re[2 i] for all real scalars a
RelcXjZ, + a ,Z 2 ] = ct]
Linearity property
Differentiation propert)'
Equality propert)'
dt
L
J
dt^
+ «2
}
= Re
Re[y4f’-^^^T = Re[5^^'^T for all t if and only i f A = B
Sum o f complex exponentials A^eJ^^ or deriva­
Single-frequency propert)'
tives, or their indefinite integrals o f any order is a
complex exponential o f the same frequency co
442
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
3. NAIVE TECH N IQ U E FOR C O M PU TIN G THE SINUSOIDAL
STEADY STATE
Property 10.5 o f the previous section suggests a technique for computing the SSS response o f a
circuit. The technique uses a differential equation model of an RL, RQ or RLC circuit as devel­
oped in Chapters 8 and 9. In contrast to the dc sources in those chapters, suppose the source exci­
tations have the form
cos(cor+0). In addition we assume that the zero-input response consists
o f (eventually) decaying exponentials or (eventually) exponentially decaying sinusoids to ensure
that there is a valid sinusoidal steady state. Thus the form o f a first-order circuit differential equa­
tion model with a sinusoidal excitation is
^/■v(/)
cit
+ ax{t) = K^cos(co/+ 0 )
(10.5a)
or in the second-order case,
df
(10.5b)
dt
where x{t) is a desired voltage or current, such as
V(\t).
Property 10.5 guarantees that the sum o f any number o f cosines or derivatives o f any order o f
cosines o f the same frequenc)' OJ is a cosine o f the same frequency O). Hence, the circuit response
x{t) in equations 10.5 has a steady-state cosine form o f frequency O). Further, the scaled sum ofx(f)
and its derivatives on the left-hand side o f each differential equation 10.5 must equal V^cos(o)/+0),
the input excitation. This also implies that the steady-state circuit response, x{t), is a cosine o f the
same frequency as the input, but not necessarily the same magnitude or phase. We conclude that
•’^■(^) =
cos(co/‘+(j)) = A cos(tor) + B sin(to/). The SSS response is then specified upon finding A
and B. The following example illustrates this calculation.
E X A M PLE 10.4. Let the source excitation to the circuit o f Figure 10.3 be
Compute the SSS response /jr(r).
______________
i^lt)
i,» ©
F I G U R E 10.3 Parallel
RL circuit
for Exam ple 10.4.
= /^cos(o)^).
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
So
443
lu tio n
Step 1. Determine the differential equation model o f the circuit. From KCL applied to the top node
o f the circuit, i jt ) = i^(t) + //_(/)• Sincc the resistor and inductor voltages coincide, the t/-i rela­
tionship o f the inductor implies that the inductor current satisfies the difTerential equation
( 10.6)
which has the form o f equation 10.5a.
Step 2. Determine the form o f the response. Since the input is a cosine wave, the SSS response will
have the sinusoidal form
ij{t) = A cos(cor) + B sin(oj/)
(10.7)
Step 3. Substitute the form o f the response {equation 10.7) into the differential equation 10.6.
Inserting equation 10.7 into the differential equation 10.6 and evaluating the derivatives yields
—
L '
cos(co/) = — f/\cos(o)/)+ Z?.sin(o)r)l + — /4cos(co/)+ fisin(( 0 /)
(It
L
= - 0) /\sin(coO + to
RA
RB
I-t
L
cos((or) + — cos(co/) + — sin(co/)
Step 4. Group like terms and solve for A and B. Grouping like terms leads to
r
R
R
]
r
R
Bco H— A ----- 1^ cos(o)r) + — B -A ii)
L
L ',
.L
]
sin(coO = 0
(10.8)
To determine the coefficients A and B, we evaluate equation 10.8 at two distinct time instants.
Since equation 10.8 must hold at every instant o f time, it must hold at / = 0; i.e., at /= 0,
=0
or, equivalendy,
(10.9a)
In addition, equation 10.8 must hold for t = 7r/(2to), in which case
-COA + - 5 = 0
L
( 1 0 .9 b )
444
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
Solving equations 10.9 simultaneously for A and B yields
/?-/,
= — ------f — ,
„
CO/?L/
B=
R~ + L^(0-'
/?- + L“co“
Step 5. Determine the steady-state response. Since A and B are known,
.
,
,
Il (0 = — ------ ^
R~ + L~0)~
cos(o3 0 + —
R~ +
In the more common alternative form o f i^{t) =
.
O^RLL
^
SI n(co /)
cos((or + ({)) as per equation 10.2a,
where
= - tan *
(10.10b)
is adjusted to reflect the proper quadrant o f the complex plane.
This example has illustrated a procedure for finding the SSS response o f a circuit. Step 1 is to sub­
stitute an assumed sinusoidal response form, such asy4cos(cor) + Bs\n{a)t), having unspecified con­
stants A and B, into the differential equation and evaluate all derivatives. Step 2 is to group like
terms, and step 3 is to compute the constants A and B. After finding A and B, one computes the
magnitude,
and phase (j) o f the cosine
cos(oj^ + (|)) via equations 10.2.
The next section offers an alternative approach. Using complex excitation signals o f the form
y^eJ^Mt +0)^
computes
and (j) by a more direct route.
4. C O M P LEX EXPON EN TIAL FORCING FUNCTIONS IN
SIN USOIDAL STEADY-STATE COM PUTATION
Complex exponential forcing functions are simply complex exponential input excitations o f the
properties o f complex numbers in section 2, we can
form
or ^;(o)/ + 0) PfQji-j
replace the input excitation
cos{wt + 0) and the assumed circuit response
cos(to^) + j5sin(co^) with their complex counterparts
and
cos(cor + cj)) = /I
respectively, with­
out any penalt)'. To recover the actual real-valued responses, we simply take the real parts o f the
complex quantities. Again this is justified by properties 10.1 through 10.5. This process o f sub­
stitution and subsequent taking o f real parts actually simplifies the calculations developed in sec­
tion 3, because of the simple differential and multiplicative properties o f the exponential function.
The following example illustrates a more efficient calculation o f the steady-state response using
complex exponentials.
■vn
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
EX A M P L E 10.5. For the series RC circuir o f Figure 10.4 let v^it) = K^cos(cof). Compute the
steady-state response
FIG URE 10.4. Series /?Ccircuit for Example 10.5.
Step 1. Construct the dijferential equation o f the circuit. Writing a loop equation and substituting
for i(^t) yields
( 10. 11)
at
Step 2. Substitute complex forms o f the input and response into the differential equation. If v^{t) were
to be equal to the complex exponential
if vj^t) =
then the response would be
= V}cos(cor) (as is the case), then V(^t) =
o f complex numbers. Hence, for the moment, let us set vj^t) =
^
However,
from the properties
and agree that
=
appropriate real parts.
Substituting the complex expressions into the circuit differential equation 10.11 yields
After canceling the
terms, factoring V„,e^^ out to the left, and dividing through by {jixiRC +
1), we obtain
1 + jOdRC
Step 3. Determine the magtiitude
( 10 . 12)
and the angle (}). Equating magnitudes on both sides o f equa­
tion 10.12 yields
Vl + ( o V c '
(10.13a)
and equating angles yields
(j) = - tan '(to/?Q
(10.13b)
Step 4. Determine the steady-state response. Using equations 10.13 the desired response is comput­
ed by taking real parts:
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
4 i6
\'cU) = Re
(10.14)
Ks
cos[co/ -tan ha^RC)]
Vl + (0“/e-C^
In deriving the relationship 10.12 from the ciifterential equation 10.11, we utilized a complex exponen­
tial ftinction as the circuit input. A complex exponentid input is not a signal that cm be generated in the
laborator)'. Nc-vertheless, it is often used in advanced circuit theor)' to simplift' the derivation of many
important results, as was done in the preceding example. If one does not mind a more lengthy derivation,
then the s;imc result (equations 10.12 through 10.14) am be obtained without the flaitious complex
exjxjnentid excitation. For example, let the voltage source in Figure 10.4 represent a reiil signal source
v^{t) = V^cosim) = R e[K / > T
Then the steady-state response has the form
v^t) =
cosiMt + (j)) =
Substituting these expressions into the differential equation 10.11 yields
/ec— ( rc v...e
+ Re
= Re
Making use of properties 10.2 and 10.3, move the position o f the operator Re[] outside the first
term to obtain
Re
+ Re
= Re
Evaluating the derivative and using property 10.2 (linearity) produces
(10.15)
By property 10.4, equation 10.15 holds if and only if
^
+ <l>) =
This is precisely the equation following equation 10.11 that leads to equations 10.12, 10.13, and
finally 10.14.
As we can see, the use o f complex exponentials does indeed lead to a more direct calculation of
the SSS response. However, this method and the method o f section 3 require a difFerential equa-
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
44 /
tion model o f the circuit. For circuits with multiple sources, dependent sources, and many inter­
connections o f circuit elements, finding the differential equation model is ohen a nontrivial task.
In the next section we eliminate the need to find a differential equation model o f the circuit by
introducing the phasor concept.
5. PHASOR REPRESENTATIONS OF SIN USOIDAL SIGNALS
Recall
is shorthand iov
= /lcos(({)) +yy4sin((})). If the frequency co is known, then the
complex number AL^ completely determines the complex exponential
known, then AL^ completely specifies A cos(ojr +({)) =
In turn, if to is
This means that the com­
plex number A/L^ can represent a sinu.soidal function A cos(ojf + (})), whenever (O is known.
Complex number representations that denote sinusoidal signals at a fixed frequency are called
phasors. A phasor vo\iz^c or current will be denoted by a boldface capital letter. A typical voltage
phasor is V =
and a ty'pical current phasor is I =
Por example, the current i{t) =
25cos((Of + 4 5 °) has the phasor representation I = 25Z-45°. The voltage v{t) = - 1 5 sin(tor + 3 0 °)
= 15 cos(w^ + 120°) has the phasor representation V = 15/-120°.
As all voltages and currents satisfy KVL and KCL, respectively, one might expect phasor voltages
and currents to do likewise. This is not patendy clear. The following simple example demonstrates
why this is true for KCL.
Consider the circuit node drawn in Figure 10.5.
FIGURE 10.5 Single node having four incident branches.
From KCL it follows that
^4(^) =
- ijit) +
= 10 cos(tor) - 5.043 cos(cor + 7 .5 2 °) + 8 cos(cof- 9 0 °)
Using trigonometric identities or property^ 10.1 to combine terms on the right-hand side leads to
i^it) =
1 0 c o s (tO f- 6 0 °)
4 <S
Chapter 10 • Sinusoidal Steady State Analysis by Pliasor Methods
For the corresponding phasors to satisR' KCL, it must follow that
10Z-60O =
= I, - I 2 + I3
The right-hand side o f this equation requires that
I, - I2 + I3 = lOZQO - 5.043Z.7.52O + 8Z.-9()0
= 10 - (5 + >0 .66 ) + ( - ; 8 ) = 5 - 78.66 = 10/1-60° = I 4
Thus the phasors (which have both a real and an imaginar)' part) satisfy KCL. KCL is satisfied
because i^{t) = /j(^) - i-,{t) + /^(r) implies
= Re[10^>n - Re[ 5. 043^>>( ' ^' ’^] + Re(8f>i<“ ^ -^0-’)]
= Re[( 10 - 5.043<?^'7-5-" +
(10.16)
for all t. By property 10.4, equation 10.16 holds if and only if
IOZ- 6OO = 10 - 5.043e’^‘7-52° +
In phasor notation this stipulates that
It is the properties o f complex numbers and the fact that an equation is true for all t that guaran­
tee that phasors satisfy KCL. Although not general, the argument is sufficient for our present ped­
agogical purpose. A similar argument implies that phasor voltages satisf}' KVL, as illustrated by
the following example.
EXA M PLE 10.6. Determine the voltage across the resistor in the circuit of Figure 10.6 using the
phasor concept.
Vj(t) = 19.68 sin(a)t
152.8°)
V3(t) = 4 .2 1 5 cos(cot + 71.61 °)
+ . --------------
•+
v,(t) = 20co s(cot + 53.13°)
F K i U R E 10.6 Resistive circuit w ith three sourccs.
^
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
S
44^)
o l u t io n
Firsr note that 19.68 sinltDr+l 52.8”) = 19.68 cos(co/+152.8“—90") = 19.68 cos(cor+62.8°). From KVL,
Vf^it) = v^{t) - Vjit) + v^{t)
Since voltage phasors must satisfy KVL,
V/^ = V, - V 2 + ¥3 = 20 z i 5 3 . 1 3 0 - 19.68 z^62.80 + 4.215 ^ 7 1 . 6 °
Changing to rectangular coordinates and adding yields
= 12 + y i6 - (9 + y i7 .5 ) + 1.33 + j4 = 4.3 3 + jl.5
Equivalently,
= 5Z-30® V, and
= Rc[5r>(‘" ' " ^0")] = 5 co${o)t + 300) V
Exercise. In Figure 10.6, suppose yj(r) = 10 cos(cor) V, v-^ (/) = 10 co s((o r- 0.5tc) V, and
V3(/)= I 0 V 2 cos(cOf - 0.25ti ) . Find the phasorV^and then
ANSWKR:
= 20 - /20.
vrU) = 2 ()V : cos(O)/ - 4 5 " ) \’
Given that phasor voltages and currents satisfy K\^L and KCL, respectively, it is possible to devel­
op phasor O hm s law-like relationships for resistors, capacitors, and inductors operating in the
SSS. This would allow us to do SSS circuit analysis with techniques similar to resistive dc analy­
sis. The next section takes up this thread by introducing the notion of (phasor) impedance.
6. ELEMENTARY IM PEDANCE CO N CEPTS: PHASOR
RELATIONSHIPS FOR RESISTORS, INDUCTORS, AND
CAPACITORS
Ohm’s law-like relationships do exist for resistors, capacitors, and inductors operating in the SSS.
The constraint, operating in the SSS, suggests that any Ohm s law-like relationship should be
dependent on the sinusoidal frequency.
The first objective o f this section is to derive three Ohm’s law-like relationships, one each for the
resistor, the capacitor, and the inductor. The relationships each take the form V = Z(/to)I, where V
is a phasor voltage, I is a pha.sor current, and Z(/to) is called the impedance o f the device: Zy^(;tij) for
a resistor,
for a capacitor, and Z/(/co) for an inductor. The fact that the phasor voltage V is a
function Z(/co) times a phasor current I indicates a clear kinship with Ohm’s law for resistors. Indeed
the unit o f impedance is the ohm because it is the ratio of phasor voltage to phasor current. The
impedance Z(/cij) explicitly shows that the relationship is potentially frequency dependent.
450
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
The derivation o f these elementary impedance concepts will build on the assumption that all volt­
ages and currents are complex sinusoids o f the same frequency represented by complex phasors.
This is permissible because real sinusoids can be recovered from complex sinusoids simply by tak­
ing the real part. To this end consider the resistive circuit o f Figure 10.7a.
V „(t) =
i,(t) =
j
0j(u)t + 6)
(b)
(a)
FIGURE 10.7 (a) Resistive circuit driven by complcx current,
(b) Equivalent phasor representation of the circuit in (a).
From Ohms law,
vjiit) =
In terms o f the phasors
= RIj^
+ 0)
ej^\ this relationship reduces to
= 1^-^^^ and
= /?
= Zpfjia) \j^
(10.17)
and Z^(yco) = R\s the impedance o f the resistor defined by equa­
where
= Rlj^ Z.0. If
tion 10.17. Ideally the resistor impedance is independent o f frequency. Thus
=
cos(u)r + 0) = Kq[
I
^
then Vj^it) = Rlj^ cos(co/ + 0) =
This phasor
relationship restates Ohm’s law for complex excitations. The distinctiveness o f phasors comes with
their application to inductors and capacitors.
Now consider the inductor circuits o f Figure 10.8. Assume the circuit o f Figure 10.8a is in the
steady state.
i jt ) =
A
Remainder
Remalnder
of circuit
of circuit
(a)
L
-I-
V = jcoL 1^
(b)
FIGURE 10.8 (a) Inductor having complex exponential voltage and current, (b)
Phasor relationship of (a).
/is:
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
The complex current and voltage associated with the inductor are, respectively, /^(f) =
and v^it) =
Substituting these expressions into the defining equation for an inductor
yields
j = jcoz.
Canceling out
on both sides yields
the relationship is
In terms o f the phasors
and
( 10. 18)
in which case the inductor Impedance is derived as Z^(/co) = p iL . The inductor impedance clear­
ly depends on the value ol the radian frequency CO. Specifically, if U) = 0, then the impedance o f
the inductor is 0, i.e., in SSS the inductor looks like a short circuit to dc excitations. If co = oo, the
impedance is infinite, i.e., in the steady state the inductor looks like an open circuit to signals o f
very high frequency.
Equation 10.18 exhibits a frequency-dependent Ohm’s law relationship for the inductor. From the
properties o f the product o f two complex numbers, the polar form o f the voltage phasor is
\ l = (yco£)I^ = (03/./,) ^ ( 0 + 9 0 °)
Hence if
/^(/) =
^
= /, cos(tor + 0) A
then
p^(t) = Re[/wZ/^f’>(‘'^' ^ 0)] = RelcoL/^e^^^"
" ‘-^0“)] = coL/^ cos(tof + 0 + 9 0 °) V
From this relationship one sees that the voltage phase leads the current phase bv 9 0 ” . Equivalently,
one can say that the current lags the voltage by 9 0°. This leading and lagging takes on a more con­
crete meaning when one views phasors as vectors in the complex plane, as per Figure 10.9, which
shows that the voltage phasor o f the inductor always leads the current phasor by 9 0 ” .
452
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
The capacitor has a similar impedance relationship, derived as follows. Assume the circuit o f
Figure 10.1 Oa is in the steady state.
C'
+
Remainder
of
►
Remainder
circuit
of
C
L
J
V
circuit
-
(a)
J
(b)
FIGURF. 10.10 (a) Capacitor having complex exponential voltage and current.
(b) Phasor relationship o f (a).
The complex current and voltage associated with the capacitor are, respectively, i^^t) =
and V(^t) =
Substituting these expressions into the defining equation for a capacitor
yields
Vce 7(o)r+<!>)■
Canceling out
on both sides yields
In terms o f the phasors
= I(-e^ and V (-=
this relationship becomes
I^=yo)C V ^
or, equivalently.
- — — i c - ^ cO ^ ^ )Ic
ycoC
(10.19)
Equation 10.19 defines the capacitor impedance as Z(^j(a) = l/(/a)Q. if co = 0, the impedance
o f the capacitor is infinite in magnitude. This means that in SSS the capacitor looks like an open
circuit to dc signals. On the other hand, if OJ = co, then the capacitor has zero impedance and looks
like a short circuit to large frequencies.
4 S3
Chapter 10 * Sinusoidal Steady State Analysis by Phaser Methods
Looking again at equation 10.19, observe that
Vr =
70) C I c
=-
t:
(oC
- 90°)
(10.20)
Equation 10.20 has a vector interpretation in the complex plane, as shown in Figure 10.11
Imaginary axis
F ' l C l J R l i 10, 1 1
Diagram of capacitor voltage and current phasors where the voltage phasor lags the
current phasor by 90°.
The diagram o f Figure 1 0 .1 1 indicates that the capacitor voltage lags the capacitor current phasor
by 9 0 ° or that the capacitor current leads the capacitor voltage by 9 0 °, which is the opposite o f
the case for the inductor.
Exercises. 1. For the circuit o f Figure 10.12a, show that
^
ycoL ^
and that
V/
i r ( t ) = — ^COS(Cl)/ + 0 - 90°)
(OL
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
2. For the circuit o f Figure 10,12b, show that
I^ = ya)C V ^
and that
i(^t) =
+ ^ + 90”)
iiCt)
ijt )
V ,(t) =
V ,(t) =
\J
\J
0 j((i)t + 0)
g j((» t + 0)
(a)
(b)
I-ICURE 10.12 (a) inductor driven by vohagc source, (b) Capacitor driven by voltage source.
Recall that resistance has a reciprocal counterpart, conductance. Likewise, impedance has a recip­
rocal counterpart, admittance. Admittance has units o f siemens, S, as does conductance, fh e
admittance, denoted by l^yto), associated with an impedance, Z(/co), is defined by the inverse rela­
tionship
K(./co) =
( 10.21)
ZOCO)
provided Z{j(M) is not equal to zero ever)^vhere. What this means is that the phasor i-v relation­
ship o f a resistor, capacitor, and inductor satisfies an equation o f the form I = K(y(o)V. Hence, the
admittances o f the resistor, inductor, and capacitor are respectively given by
>"/eO'w) = - ,
R
r^(./co) = —
./(oL
Kc-(yco) = ycoC
( 10.22)
The impedance and admittance relationships o f the resistor, capacitor, and inductor are summa­
rized in table 10.2.
TABLE 10.2 Summary of Impedance and Admittance Relationships for Resistor,
Capacitor, and Inductor
Impedance
Admittance
2/?(y(0) - R
K^(;o)) =
i
yV-(./co) = ycoC
j(oC
JY Y V
Z Lijoi) = j(i)L
j(i)L
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
In the next section the notion o f impedance is applied to an arbitrar)' two-terminal network. This
generalization will allow us to consider the impedance and admittance o f interconnections o f
capacitors, inductors, resistors, and dependent sources.
7. PHASOR IM PEDANCE AND ADM ITTAN CE
For the resistor, the inductor, and the capacitor, the impedance equals the ratio o f the respective
phasor voltage to the phasor current. Analogously, the impedance o f any t\vo-terminal circuit, as
illustrated in Figure 10.13, is the ratio o f the phasor voltage to the phasor current, i.e..
Z „ S j^ ) = ^
= R + jX
(10.23a)
1 in
O+
Two
Terminal
Circuit
Z (j(o)orY,„(j(o)
FIGURE 10.13 Two-terminal device with phasor voltage \ p h a s o r current 1^^^,
and input impcdance Zy^^(/co).
Because impedance is the ratio o f phasor voltage to phasor current, its unit is the ohm. Inverting
the relationship o f equation 10.23a defines the adm ittance o f a two-terminal device as the ratio
o f phasor current to phasor voltage, i.e..
I;
(10.23b)
Provided Z(yto) ^ 0 for all
cd,
in contrast to a short circuit, then
As an example, the impedance o f an inductor is jwL and its admittance is \/(J(.oL). Historically,
impedance and admittance were first defined as per equation 10.23. However, with the wide­
spread use and utility o f the Laplace transform (Chapter 12) in the past several decades, imped­
‘0 6
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
ance and admittance have become understood as much broader and more useful concepts than
the steady-state presumptions o f equation 10.23, as set forth in Chapter 13.
In general, admittances and impedances are rational functions with real coefficients o f the com­
plex variable Ju). At each d) the impedance and the admittance are generally complex numbers.
Since a complex number has a real part and an imaginary part, we can further classify the real and
imaginary parts o f an impedance or an admittance. For an impedance Z(yw) the expression
lm[Z(/co)] = X is called the reactance o f the two-terminal element, while Re[Z(/(o)] = R refers to
its resistance. 1‘urther, for an admittance Vijto), Im[K(/(o)j = B is called the susceptance of the
two-terminal device whereas Re[)1[yoj)] = G is referred to as the con d u aan ce. These definitions
are summari/,cd in Table 10.3.
I'ABIJ-. 10.3 Summary Definitions of Various Terms
Admittance
Impedance
V/„
Y( jo) ) = ^
Z{J(}^) = — = R + jX
I in
V/«
= G + jB
Resistance
Reactance
Conductance
Susceptance
R = Re[Z(yw)]
X = Im[Z(;to)l
6’ = R e [n / o )]
im [K(>j)]
Using equations 10.23, one can compute the equivalent impedance
series, as in Figure 10.14a. Flere
= ^] +
o f two devices in
% Ohm’s law for impedances, V , = Z|(/to)I, and
V 2 = Zol/w) I ,. But I, = I 2 = I/„- Mence,
I.e.,
^in (
)
= —^ = Z, ( ./O)) + Z 2 (./(O)
*iit
(10.24)
This simple derivation has another consequence: given Zy^j(yco) = Z,(/to) + Z 2 (/co) and the fact that
i = 1, 2, a simple substitution yields the voltage division formula,
Z/O )
'
Z,(./(D) + Z2(7(0)
V;.,
(10.25)
Kquations 10.24 and 10.25 are consistent with our early development o f series and parallel resist­
ance.
4V
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
Y,(jco)
V.
V.
V,
oYJjco)
(b)
FIGURE 10.14 (a) Two impcclanccs in series, (b) Two admittances in parallel.
Exercises. 1. Duplicate the derivation ol" equation 10.23 for three impedances in .series.
2. Derive a formula for voltage division when there are three impedances in series.
The admittance o f two devices in parallel, as sketched in Figure 10.14b, satisfies
v,„
v,„
V,
V,
since
^1 (./“ ) = “ f
V|
Y2 ijc o ) = ^
Vo
we conclude that
(10.26)
Exercises. 1. Duplicate the derivation o f equation 10.26 for three admittances in parallel, i.e.,
show that )^y„(/co) = Kj(/o)) +
2. Show that the equivalent impedance o f two devices, Z,(/co) and Z t(/co), in parallel is given by
(10.27)
3. Show that the equivalent admittance o f two devices, Kjlyco) and Y-,{ji.o), in series is given by
458
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
>U/co)>'2(yw)
■
K|aco) + K ,0 (o )
4. Show that the admittance o f two capacitors, Cj and C^, in series is y’CO
(10.28)
C 1C 2
C\ + C-)
5. Show that the impedance o f two inductors, Z.j and Z.^, in parallel is /O)
Lt + Lo
Now the derivation o f equation 10.26 leads to a current division formula as follows. Since
= Kj(/co) +
and since
for /= 1 ,2 , one immediately obtains the current divi­
sion formula,
I; =
'
n-(7(0)
K ,(7co)+ r2(y(o)
I,
(10.29)
Since devices represented by impedances or admittances must satisfy KVL and KCL in terms of their
phasor voltages and currents, and since each device so represented satisfies a generalized Ohms law, i.e.,
V = Z(;o))I
or
I =
it follows that impedances can be 7nauipiilated in the same manner as resistances, and admittances in the
same manner as conductances. The voltage division formula o f equation 10.25 and the current divi­
sion formula o f equation 10.27 illustrate this fact. Example 10.7 further clarifies these statements.
Exercises. 1. Derive a current division formula for three admittances in parallel.
2. Find the admittance and then the impedance of each parallel connection in Figure 10.15.
AN SW FRS: Admittances are
3. Compute the equivalent inductance for Figure 10.15a and the equivalent capacitance for the
circuit ol Figure 10.15b.
ANSWI-RS:
,
L|
L.
4. Find ^ in terms o f
Cj + Ct+C^
L,
for each circuit in Figure 10.15.
I
O-
I
+
V
(a)
(b)
FIGURE 10 . 15 . (a) Set of three parallel inductors, (b) Set of three parallel capacitors.
4S9
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
EXA M PLE 10.7. For the circuit o f Figure 10.16, compute the input impedance ^/„(/Co) when co
= 500 rad/sec.
FIGURE 10.16 Series-parallel interconnection of different impedances.
S
o l u t io n
As shown in Figure 10.16, Z^-^^(/500) can be seen as the sum o f three impedances,
+ ^2 + Zy
Our approach is to first calculate Zy for each /.
Step 1. Compute Z y Since this is an I C series combination,
A = j 5 0 0 x 0 .0 0 5 -
1
5 00
X
0 .0 0 0 4 )
= - j 2 .5 Q
Step 2. Compute Z-) = MYj. From the propert)' that parallel admittances add and series imped­
ances add.
Y2 = y'500 X 0.0 0 0 2 -h
1
:—
10 + ( 1 0 -H7 IO)
= JO. 1 -f 0 .0 4 - jO .02 = 0 .0 4
H en ce,Z 2= 1/^2 = 5 - ; 1 0 a
Step 3. Compute Z 3 = 1/
=
Here
= j --------------- = o.i-H 70 . 1 - 70.2 = 0 . 1 - 70.1
5 0 0 x 0 .0 1
Hence, Z 3 = 5 + J5
Step 4. Compute Z-^^. Adding the three impedances together yields
Z.„ = Z j + Z 2 + Z 3 = - ;2 .5 + 5 - 7 IO + 5 +75 = 10 - 77.5 Q
= 12.5 Z - 3 6 .8 7 ° Q
y0.08
160
Chapter 10 * Sinusoidal Steady State Analysis by Phasor Methods
Calculations performed in this example are most easily done with an advanced calculator or in
M ATLAB.
For example,
in
M ATLAB
the command
for computing
is “Z3
=
l/(sqrt(0.02)*exp(i*pi/4) - j/(50()*0.01)).”
EXA M PLE 10.8. Compute the input impedance Zy^^(/a)) o f the ideal op amp circuit o f Figure
10.17.
I,
V,
FIGURE 10.17 Op amp circuit callcd an impedance converter.
S o l u t io n
The trick to solving this problem entails full use o f the ideal op amp properties discussed in
Chapter 4.
Step 1. From the properties o f an ideal op amp, from KVL, and from Ohms law,
V 2 - V „ , - M 3 = V,„
(10.30)
This follows because the voltage across the input terminals of each ideal op amp is zero and no
current enters the + or - terminal o f each ideal op amp. This implies that
Step 2. Using the phasor voltage division formula o f equation 10.25, it follows that
R
=
T " '
R+ j(oC
or, equivalently,
v,=
1+ -
1
Jc^RC)
Here, o f course, because o f the idealized properties of the op amp, the voltage
the resistor R in the leftmost op amp.
(10.31)
appears across
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
Step 3. Writing a node equation at the inverting terminal o f the rightmost op amp yields
again by the properties o f an ideal op amp. Simplifying this equation yields
2V,„ = Vi + V ,
(10.32)
Step 4. Substituting equations 10.30 and 10.31 into equation 10.32 yields
V.
=\. + —
+
jCdRC
-/?!■
Equivalently,
Z i„ u ^ )= ^ = m ~ c
I//J
(10.33)
Equation 10.33 suggests that the op amp circuit o f Figure 10.17 can replace a grounded inductor
whose impedance is jii)L with proper choice of R and C, i.e., L = R^C. In integrated circuit tech­
nology it is not possible to build a wire-wound inductor. Instead, inductors are “simulated” by cir­
cuits such as that o f Figure 10.17.
The next section continues to develop our skill with and deepen our understanding o f the phasor
technique by computing the steady-state responses o f various circuits.
8. STEADY-STATE CIRCU IT ANALYSIS USING PHASORS
This section presents a series o f examples that illustrate various aspects of’ the phasor technique.
Our purpose is not only to demonstrate how to compute the SSS, but also to illustrate the pha­
sor counterparts o f Thevenin equivalents, nodal analysis, and mesh analysis. Our first example
reconsiders the parallel RL circuit o f Example 10.4, together with the series RC circuit o f Example
10.6. We will demonstrate the superiority o f the phasor technique over the methods presented in
sections 3 and 4.
i6:
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
E X A M PLE 10.9. Compute the steady-state voltage V(^t) for the circuit o f Figure 10.18 when i^{t)
= cos(lOOr) A.
R=10Q
21,(t)
i ( t ) 0
C=1mF
R = 100
L = 0.1H
FIG U RF 10.18 7?ACcircuit for Example 10.9.
S
o l u t io n
Step 1. Determine I^. Since the phasor
R
= 1Z.0° A, by current division,
1
1
1
Z -4 5 "
A
(10.34)
R
Step 2. Use equation 10.34 and voltage division on the RC part o f the cirniit to compute \ q Using
voltage division and equation 10.34, the capacitor voltage phasor is
ycoC
j(oC
Step 3. Determine V(^t). Converting the phasor
o f equation 10.35 to its corresponding time
function yields
v^t) = co s(1 0 0 ^ - 9 0 °) = sin(100/) V
The next example illustrates voltage division with phasors as well as the basic impedance relationships.
E X A M PLE 10 .1 0 . Consider the circuit in Figure 10.19 where
= 2 H, and vi^t) = 10 cos(2r) V. Find Vfj^t) and ij{t) in steady state.
= 5
C = 0.1 F, /?2 =
^
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
463
SO LU TIO N
Step 1. FindZj^^{jl). Y,^cU'^) = “ +
= 0.2 + 70.2 = 0.2> / 2Z 45" . Hence
R
ZrcU2) =
Step l.F in dZ j^ iijl).
= 2 .5 V 2 Z - 4 5 " = 2.5 - J2.S
=-+ —
R
= 0 . 2 5 - ;0 .2 5 = 0 .2 5 V 2 Z - 4 5 " . Hence
j(d L
Zr,.U2) = -
4
-
= 2 v /2 Z 4 5 " = 2 + j2
Step 3. F in d Y a t id Vq (t). From volrage division
V
^ -------- Z r c (J2) -------- Y ^ ------- 2 .5 -j2 .:> ------ ^
2 r c 0 '2 ) + Z r i,0 '2 )
2 . 5 - j 2 .5 + 2 + j2
= 7 g 0 9 ^ _ 3 8 .6 6 “
4.5-J0.5
It follows that V(^{t) = 7.809 cos(2^ - 38.66°) V.
Step 4. FindYj^, 1^, a n
d From step 3,
= 10 - 7.809 Z - 38.66° = 1 0 - (6.098 - ;4 .8 7 8 ) = 3 .9 0 2 4 + ;4 .8 7 8
Hence
l,= ^
=
1 ^ 2 3 ilM iZ ! =
,5 6 ,7 Z - 3 8 .6 6 "
Thus /^(f) = 1 .5 6 l7 c o s (2 f- 38.66«) A.
The next example illustrates the computation o f a Thevenin equivalent circuit with the aid o f
nodal analysis. Because impedances may be manipulated in the same manner as resistances and
admittances in the same manner as conductances, the Thevenin theorem, the source transforma­
tion theorem (Chapter 5), and node and mesh analysis (Chapter 3) carr)' over directly.
46^
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
EXA M PLE 10.11
(a) Find cheThevenin equivalent o f the circuit o f Figure 10.20 if to = 4 rad/sec.
(b) Determine the voltage
when a 1.2
load resistor is connected across terminals a and b.
FIGURE 10.20 Z.Ccircuit for Example 10.11.
S o l u t io n
Find theThevenin equivalent circuit, and then using theThevenin equivalent, find
Vj{t).
Step 1. Establish nodal equation. A nodal equation at the left node o f Figure 10.20 in terms o f phasors is given by
I.V = — r ^ L + J^CWoc = -J^ L + y ^ o c
,/(oL
Step 2. Determine the relationship between
and
The relationship between
(10.36)
and
as
determined by the dependent source is
V z - V , , = 0 .2 5 [ ;2 V J
Equivalently,
10.37)
V , = (1 Step 3. Substitute equation 10.37 into equatio7i 10.36. Substituting yields
h = m -j)y o c
Solving for
with
= 1Z.0° yields
V^oc =
: I , = (0 .4 - y0.8)I^ = 0 .8 9 4 Z - 6 3 .4 3 ° V
0.5 + j
Step 4 . Compute the Thevenin equivalent impedance
(10.38)
Consider the circuit o f Figure 10.21,
which is the phasor version o f Figure 10.20 with the output terminals short-circuited. Hence, the
short-circuit current phasor is
1=
A
4(n
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
FIGURE 10,21 Phasor version o f Figure 10.20 with short-circuited terminals.
Therefore, from equation 10.38,
Z,/,0'4) = ^ = ( 0 . 4 - y O . S ) Q
.VC
Step 5. Interpret
to generate the Thevenin equivalent circuit. To physically interpret theThevenin
equivalent impedance, consider that
= (0-4 -yO .8 ) = {R^,, + MjAQ Q.
Thus,
= 0.4 Q and C = 0.3125 F. Hence, the desired Thevenin equivalent circuit (valid at O)
= 4 rad/sec) has the form sketched in Figure 10.22.
0 .3 1 25F
-OFIGURE 10.22 Thevenin equivalent of Figure 10.20.
Step 6. Compute
by voltage division. Using voltage division on the circuit o f Figure 10.22,
=
1.2
l.2 + ( 0 .4 - y 0 . 8 )
= (0.6 + y 0 .3 )(0 .8 9 4 Z 6 3 .4 3 °)
= 0 . 6 Z - 36.87° V
Converting the load voltage phasor to its corresponding time-domain sinusoid yields
y^^(t) = 0 . 6 cos(4/^ - 3 6 . 8 7 ° ) V
166
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
EX A M PLE 10.12. Determine the phasor voltage
and the corresponding time function vj,t)
for the circuit o f Figure 10.23 if co = 100 rad/sec.
j60Q
I'lG U R E 10.23 Phasor domain circuit for Example 10.12. Ail clement values indicate phasor
impedances at 100 rad/sec.
S o l u t io n
To solve this problem, it is convenient to execute a source transformation on the independent cur­
rent source and to combine the impedances o f the parallel combination o f the capacitor and
inductor on the right-hand side o f the circuit. After executing these rwo manipulations, one
obtains the new circuit of Figure 10.24.
FIGURE 10.24 Phasor domain equivalent circuit to that of Figure 10.23. All element values indi­
cate phasor impedances at 100 rad/sec. I denotes a phasor loop current.
For the circuit o f Figure 10.24, the indicated loop equation is
250Z.-90O = (50 - ; 2 5 ) I - 0.4(501) - ; 1 5 I = (30 - ; 4 0 ) I
Solving for I yields
I = 4 - J 3 = 5 ^ - 3 6 .8 7 ° A
Consequently,
= 501 = 2 5 0 ^ - 3 6 .8 7 ° V and vU) = 250 c o s (1 0 0 r- 3 6 .8 7 °) V.
467
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
9. IN TRO DUCTION TO THE NOTION OF FREQUEN CY
RESPONSE
The frequency response o f a circuit is the graph o f the ratio o f the phasor output to the phasor
input as a function o f frequency, i.e., as the frequency varies over some specified range. Since the
phasor input and the phasor output are complex numbers, the frequency response consists o f tw^o
plots: (1) a graph o f the magnitude o f the phasor ratio and (2) a graph o f the angle o f the phasor
ratio. Such graphs indicate the magnitude change and the angle change imposed on a sinusoidal
input to produce a steady-state output sinusoid. In steady state, the magnitude o f the output sinu­
soid is the product o f the magnitude o f the input sinusoid and the magnitude o f the frequency
response at the frequency o f the input. Similarly, the phase o f the output sinusoid in steady state
is the sum o f the input phase and the frequency response phase at the input frequency. This prop­
erty takes on greater importance once one learns that arbitrary input signals can be decomposed
into infinite sums o f sinusoids o f different frequencies, i.e., each signal has a frequency content.
This notion is made precise in a signals and systems course, where one studies Fourier series and
Fourier transforms. The frequency response o f a circuit describes the circuit behavior at each fre­
quency component o f the input signal. This permits one to isolate, enhance, or reject certain fre­
quency components o f an input signal and thereby isolate, enhance, or reject certain kinds of
information.
EX A M PLE 1 0 .1 3 . Plot the frequency response o f the RC circuit o f Figure 10.25.
-o-
-O
+
0.01 F
10
-o
FIG URE 10.25 RC circuit passing high-frequency content of an input signal.
S o l u t io n
to the input phasor voltage N^
Using voltage division, the ratio o f the output phasor voltage
is given by
___
1
out __________________yO.Ola)
Vi„ " i + ----- ! _
= m p ))
“ l + iO .O lc o
/).01(0
where we have designated this ratio as //(/co).
The two universally important frequencies are
O) =
0 and co=
oo.
At these frequencies, H{jO)
0Z .90° and //(;“ ) = 1^-0°. Asymptotically then, the magnitude |//(/ca)|
1 as
CD
^
oo
=
and
468
|//(/to)|
Chapter 10 * Sinusoidal Steady State Analysis by Phaser Methods
0 as to
0. W ith regard to angle, Z.//(/co)
0 as to
oo and Z.//(/to)
9 0 ° as oj
0. Also, a close scrutiny o f //(/w) indicates that to = 100 rad/sec is also an important frequency.
Here H{j\00) = 0.707^^45°. These values give us a prett)' good idea what the magnitude and
phase plots look like. Using a computer program. Figure 10.26a and Figure 10.26b show the exact
magnitude and phase plots. These plots are consistent with our earlier asymptotic analysis.
Frequency (rads/sec)
(a)
Frequency (rads/sec)
(b)
FIG URE 10.26 (a) Magnitude plot of frequenc)' response for Example 10.13.
(b) Phase plot of frequency response.
Do these frequency responses make sense? They should. Going back to the circuit, observe that at
to = 0, the capacitor impedance is infinite. Physically, then, in steady state, the capacitor looks like
an open circuit for dc, i.e., at zero frequenc)'. The magnitude plot bears this out. For frequencies
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
close
CO
zero, the capacitor approximates an open circuit and, hence, the magnitude remains small.
O n the other hand, for large frequencies, the capacitor has a very small impedance. This means
that most o f the source voltage appears across the output resistor. The gain then approximates 1,
as indicated by the magnitude plot. The frequency response o f the circuit is such that the highfrequency content o f the input signal is passed while the low-frequency content o f the input sig­
nal is attenuated. Such circuits are commonly called high-pass circuits.
EXA M PLE 10.14. Investigate the frequency response o f the parallel /^Z,C circuit o f Figure 10.27.
R=10
L=0.04H
C=0.25F
FIGURE 10.27 A parallel RLC circuit having a hand-pass frequency response.
S o l u t io n
The input admittance o f the circuit o f Figure 10.27 is given by
LC
R
RC
. CO
jOiL
C
Inverting to obtain the input impedance yields
J
w
C
J_ _ ^ 2 ^ ._ o L
LC
RC
Clearly,
y'4co
I00-co-+y4co
Hence the ratio o f the output phasor to the input phasor is simply
Zy^^(yoj). Once again, co = 0 and co = oo are the first two frequencies to look at. Here Zy^,(0) =
0Z .90° and -2^,„(oo) = 0Z.-90^’. Also at co = 10, the impedance is real, i.e.,
0) = I . These three
points provide a rough idea o f the magnitude and phase response. Two more points are necessary
for a real .sense o f the frequency response. At what frequency or frequencies does the magnitude
drop to 0.707 o f its maximum value or when does the phase angle equal ±45°? This will occur
when 1100 - co^| = |4co|. This is a quadratic equation. Flowever, because o f the absolute values,
there are rvvo implicit quadratics, co" - 4co - 1 0 0 = 0 and co^ + 4co - 100 = 0. Solving using the
quadratic formula yields co = ±8.2, ±12.2. Since the magnitude plot is symmetric with respect to
the vertical axis (co = 0 axis), we consider only the positive values o f co. This information provides
a good idea o f the magnitude and phase plots. A computer program was used to generate the fre-
470
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
quenc)' response plots in Figure 10.28a (magnitude) and Figure 10.28b (phase). The magnitude
plot shows that frequencies satisfying 8,2 < O) < 12.2 are passed with little attenuation. Frequencies
outside this region are attenuated significantly. Such a characteristic is said to be o f the band-pass
type, and the corresponding circuit is a band-pass circuit.
Frequency (rads/sec)
(a)
Frequency (rads/sec)
(b)
F IG U R t 10.28 (a) Magnitude plot o f frequency response for band-pass circuit o f Figure 10.27. (b)
Phase plot of frequency response.
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
•n
EX A M PLE 10.1 5 . As a final example vve consider the so-called band-reject circuit o f Figure
10.29. A band-reject circuit is the opposite o f a band-pass circuit. A band-reject circuit has a band
o f frequencies that are significantly attenuated while it passes with little to no attenuation those
frequencies outside the band. In this example our goal is to compute the magnitude and phase o f
the frequenc)' response o f the band-reject circuit o f Figure 10.29.
R =10
FIGURE 10.29 Band-reject circuit for Example 10.15.
S o l u t io n
Once again using voltage division, we obtain the phasor ratio
'
LC
t)Ul
-(0 ‘
1 0 0 - CD"
_ L _ c o 2 + /o)-5
LC
■
1 0 0 - 0 ) 2 +^25co
L
H{jLo) = 1Z.0®. Hence, as}^mptotically, |//(/to)| approaches 1 as OJ approach­
es 0 and CO. Also at (o = 10“, //(/co) = 0 Z .-9 0 ” while at to = 10+, Hijo)) = 0 Z .-2 7 0 ° = 0Z.90". For
At to = 0 and
CO = o o ,
this example, to find the frequencies where |//(/‘to)| drops to l/ V I o f its maximum value o f 1, it
is necessary to equate the magnitudes o f the real and imaginary parts o f the denominator. This
produces two quadratics whose positive roots are to = 3 .5 0 7 8 and OJ = 28 .5 0 7 8 . At these frequen­
cies the angles o f //(/w) are —45® and 45°, respectively. Fhe computer-generated plots o f Figures
10.30a and 10.30b are, o f course, consistent with these quickly computed values.
472
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
Frequency (rads/sec)
(a)
Frequency (rads/sec)
(b)
F I G U R t 10.30 (a) Magnitude plot of frcquenc)^ response for band-reject circuit of Figure 10.29. (b)
Phase plot of frequency response.
As wc can see, a wealth o f different kinds of frequency response are obtainable by different inter­
connections o f resistors, inductors, and capacitors. Historically, phasor techniques were the essen­
tial tool for the analysis and design of such circuits. Nowadays, engineers ordinarily use either
M A T L A B or SPICE to obtain frequency response plots. Two examples follow where we use MATLA B , SPICE, or both to obtain the frequenc)’ response.
4 '’3
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
E XA M PLE 10.16. Compute the frequency response o f the circuit o f Figure 10.31 using MATLAB and SPICE.
R=10Q
FIG URE 10.31 /^/.Ccircuit for Example 10.16.
S oluti on
This circuit was originally analp^ed in Example 10.9. You might want to refer to that example
before proceeding.
SPICE Part. A SPIC E simulation produces the result shown in Figure 10.32.
EX1016 FreqRsp-Small Signal AC-0
+20.000
+40.000
+60.000
+80.000
Frequency (Hz)
+100.000 +120.000 +140.000
FIGURE 10.32 SPICE plot of capacitor voltage for the circuit of Figure 10.31.
MATLAB Part. Although the analysis appears in Example 10.9, we can use MATLAB to more
easily obtain the frequency response. First define Zj(/co) = ytoA and Z-,{p)) = l/y'coC Then from
current division,
= -7 “ " r 7 —
R+
Z|(yco)
1 ,0 )
and from voltage division,
2-. (/CO)
R+
Z 2 (y co )
I
R
K2( 7(0) +
•21^( 700)
474
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
Assuming a frequency range o f 0 < co < 1000 rad/sec, the following MATLAB code will result in
a suitable magnitude frequency response plot, as shown in Figure 10.33.
»L = 0.1;R = 10;C = 0.001;
»w = 0:1:1000;
»Z1 = j*w*L;
»Y2 = j“w*C;
>>IL= R./(R+Z1);
»VC = 2 ’ 1L./(R *Y 2+1);
»plot(w/(2*pi),abs(VC),’b’)
»grid
»xlabel(‘Frequency in Hz’)
»ylabel(‘Capacitor voltage (V )’)
FIGURE 10.33 Magnitude plot of frequenc)’^response o f capacitor voltage in the circuit of Figure
10.31. The response is of the low-pass t)'pe.
Now suppose the inductor in the circuit o f Figure 10.31 is replaced by a capacitor C, = 1 mF with
the controlling current changed to /q (0- The frequency response is easily computed with a sin­
gle change to the MATLAB code, namely, “Z , = 1. ./(j*w *0.001).” The resulting plot shows a
band-pass characteristic, as illustrated in Figure 10.34.
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
47S
F IG U R E 10.34 Magnitude plot of frequency response of capacitor voltage in circuit of Figure 10.31
when inductor is replaced by a 1 mF capacitor. The response is of the band-pass type.
E X A M PLE 10.17. In Chapter 9 we investigated the Wien bridge op amp oscillator circuit,
redrawn in B2 Spice in Figure 10.35.
4" 6
Chapcer 10 • Sinusoidal Steady State Analysis by Phaser Methods
Two difTerences are notable: ( !) there is a current sourcc present across the
combination,
and (2)
is now 10 kQ, as opposed to 9.5 kH in Example 9.14, forcing /?, = /?2- This means
that the characteristic equation for the circuit is
1
■V*' + / ? i ' + ( • = .y~ +
{RlC
R,C)
which indicates a purely sinusoidal oscillation at the frequenc)'
f o = - ^ = -----!— = 15.92 Hz
"
2n
2 k R,C
for any initial condition on C ,. In fact one might recall that R^ < Rj causes a growing oscillation
that is limited by the saturation effects o f the op amp.
The current source, set at 1 A, is present in Figure 10.35 so that we can obtain the frequency
response cur\'e shown in Figure 10.36. In Figure 10.36 observ'e that the magnitude response peaks
at/q, as expected from the theoretical analysis. In an actual circuit, the current source would not
be present. Nevertheless, a sustained sinusoidal oscillation will occur because o f the presence o f
noise. W ithout going into the analysis, noise contains an infinite number o f frequency compo­
nents, each o f which has a minute magnitude. In particular, noise contains frequency components
around / q that drive the circuit into oscillation. This is precisely what the peak in the frequency
response means: a very small (noise) voltage on Cj will cause a very large-magnitude sinusoid out­
put voltage at /q. However, the presence o f nonlinearities such as saturation keep the magnitude
at an acceptable level.
Exi0.17-Small Signal AC-13
+12.000
+13.000
+14.000
+15.000
+16.000
+17.000
Frequency (Hz)
+18.000
F I G U R E 10.36 Frequeno,- response plot o f W ie n bridge oscillator.
+19.000
+20.000
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
10. NODAL ANALYSIS OF A PRESSURE-SENSING DEVICE
The bridge circuit presented in Figure 10.37, or some variation of this bridge circuit, has been and
continues to be a widely used approach to accurate measurement technology. In this section we
will analyze the ac bridge circuit o f Figure 10.37 as a pressure measurement device. The capaci­
tance C t is a diaphragm capacitor consisting o f a hollow cylinder capped on either side by fused
quartz wafers. Bersveen the wafers is a vacuum. The capacitance o f the diaphragm changes with
temperature and pressure. For our analysis we will assume that the temperature is constant and
that the pressure is constant for a time period greater than five times the longest time constant o f
the circuit. This will allow the voltages and currents in the circuit to reach steady state and thus
allow us to use phasor analysis to compute their values.
R =100 0
15cos(20,000nt)V
F1GUR1-' 10.37 Bridge circuit diagram of pressure-sensing device. The capacitance
function of pressure, which causes the voltage
changes as a
to changc as a function of pressure. This is
registered on the attached voltage meter, which has a 1 MQ internal impedance.
As a rule o f thumb, the capacitance C-, » Q.llAKAId. This means that the capacitance is inverse­
ly proportional to the distance d between the plates and proportional to the area A o f the plates
and to the dielectric constant K o f the material between the plates. Increasing the pressure on the
diaphragm decreases the distance rf'between the wafers, increasing the capacitance. Conversely, a
decrease in pressure will increase the distance between the wafers, thereby decreasing the capaci­
tance. As the capacitance changes, the magnitude o f the ac voltage appearing across the voltage
meter will vary accordingly. Hence, two relationships are necessary: (1) the relationship berween
the capacitance C , and the magnitude o f the voltage
- V e and (2) the relationship berween
the pressure applied to the diaphragm and the associated capacitance. Our first task will be to
specify the relationship between the pressure applied to the diaphragm and the resulting capaci­
tance. Following this, we will use nodal phasor analysis to determine the magnitude of
and finally, the relationship between pressure and the magnitude V ^ - V e
Pressure is measured in various units. Millimeters o f mercury (mm Hg) is a common standard; 1
mm Hg = 1 torr, and 760 torr = 1 atmosphere (atm), where 1 atm is the pressure o f the earth’s
atmosphere at sea level, which supports 76 0 mm o f mercur\' in a special measuring tube. Suppose
4 ■’8
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
it has been found experimentally that the capacitance C j (in pF) varies as a function o f pressure
according to the formuhi
C 2 ( A P ) = Q ) + ^ lo g 10
= 2 6 .5 + 68 log 10
^0
(10.39)
(16i)+A P^
760
A plot o f C2 as a function o f AP is given in Figure 10.38.
QJ
(T3
Q.
U
Change in Pressure
FIG URE 10.3H Plot of capacitancc versus pressure.
Our next task is to develop the relationship between the capacitance of^ the bridge circuit and the
magnitude o f the phasor voltage
and
In our analysis, G, = (/?,)“ ', Gj =
= (^ 3)"^
= 10~^ S is the conductance o f the meter , According to Figure 10.37, Cj = 20 pF. We
will let Cj range as 0 < C 2 < 40 pF. Finally, co = 2h x 10"^ rad/sec. The following phasor analysis
will be done symbolically so as not to obscure the methodology.
Summing the phasor currents leaving node A leads to the phasor voltage relationship
(G j + G 2 + yujCj)v^ — G-jV^ —ycoCj
= Gj 15
Similarly, summing the currents leaving node B leads to the relationship
- G 2 V ^ .( G 2 .G 3 .G J V ^ - G ,,V
c
=0
479
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
Finally, summing the currents leaving node C produces
- > C ,V ^ .;( o ( C , . q
t
G JV c-
=0
Writing these three equations in matrix form yields
G] + G 2 +
-/■(oC,
-G i
G 2 + G 3 + G ,„
-G j
-yco C ,
- G ,„
■ l5 G ,'
V /
=
~^m
G „i + ^ ( C | + C 2 )
Vc.
0
(10.40)
0
The matrix on the left is said to be a nodal admittance matrix. Its entries can be real or complex,
as indicated. It is nor advisable to solve such a set o f equations by hand over the range o f possible
C2 values. However, using MATLAB one can solve this matrix equation over the range 0 pF < C,
< 40 pF to produce the plot o f Figure 10.39.
c
CO
QJ
01
nj
*->
<U
cn
-o
o
cu
■o
D
'E
cn
fO
10
15
20
25
30
35
40
C, in pF
FIGURE 10.39 Plot of the magnitude o f the phasor voltage
V ^ as a function of capacitance.
O f course, one could measure the voltage appearing across the meter, from Figure 10.39 deter­
mine the associated value o f Cj, refer to Figure 10.38 for AP, and then determine P = 760 + AP.
This is a long route. To complete our analysis, then, we need to develop the relationship between
pressure and bridge voltage. As we have the relationship between C j and APand the relationship
between C2 and |
|, it is a matter o f using equation 10.39 to derive the value o f
in
equation 10.40. This is best done with a simple MATLAB routine, which yields the plot given in
Figure 10.40.
480
Cliaptcr 10 • Sinusoidal Steady State Analysis by Phasor Methods
>
m
>
'o
01
T3
D
'c
ro
300
400
500
600
700
900
800
1000
1100
1200
Pressure in mm Hg
I'lG U RE 10.40 Relationship herween magnitude of bridge output voltage and pressure applied to
diaphragm capacitor C-,.
An actual pressure sensor would, oFcourse, be more complex. For example, there would probably
be a difterential amplifier such as the one shown in Figure 10.41 across the terminals oFthe bridge
circuit, and this would probably drive a peak (ac) detector to determine the maximum value oF
the ac signal appearing at the output oF the dlFFerential amplifier. Fiu ther, the peak value would
probably be read by a digital voltmeter. Nevertheless, our analysis illustrates the basic principles
involved in such a measurement. OF course, one could just as easily use loop analysis to solve the
problem. This is leFt as an exercise in the problems.
kR.
-o V,
F IG U R E 10.41 Difierential amplifier having output voltage
= k { i >2 - /^i).
481
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
Exercise. Prove that
= k{v-, -
for the differential amplifier o f Figure 10.41.
11. SUM M ARY
The two primar)' goals o f this chapter were (1) to de\'elop the phasor technique for the analysis of circuits
having a sinusoidal steady state and (2) to illustrate how this technique leads to die idea of a circuit frequenc)^ response, which characterizes the circuits behavior in response to the frequency content of an input
excitation. In the development, sinusoids were first represented ;is the real part of a complex sinusoid. As
a motivation for the delineation of die phasor method, we showed how the complex sinusoids could be
urilized to compute the sinusoidal steady-state response using difierential equation circuit models. We then
pointed out that a complex (voltage or current) sinusoid is specified by a complex number or pha.sor rep­
resenting its magnitude aind phase. After introducing the notions of impedance and admittance for the
capacitor, the inductor, the resistor, and a general two-terminal circuit element, we showed how the pha­
sor voltage and phasor current for each such element satisf}" a frequency-dependent Ohms law. I'his
iillowed us to adapt the ;inalysis techniques and network dieorems of Chapters 1 through 6 to the steadystate analj'sis of circuits excited by sinusoidal inputs. For example, diere are voltage division formulas, cur­
rent division formulas, source transformations, and The\'enin and Norton theorems all valid for phasor
representations. This permits us to effectively analv/e circuits diat have a src*ady-state response.
The phasor technique opens a door to seeing how circuits behave in response to sinusoids. Given that
input excitations are composed o f different frequenc)' sinusoids, such as a music signal, phasor analy­
sis shows why a circuit will behave differently toward the different frequencies present in the input sig­
nal. This fact prompts the notion o f a circuits frequency response, which is defined as the ratio o f the
phasor output to the phasor input excitation as a function o f (u in the single-input, single-output case.
The frequency response consists of two plots. The magnitude plot shows the gain magnitude o f the
circuits response to sinusoids o f different frequencies, and the phase plot shows the phase shift the cir­
cuit introduces to sinusoids o f different frequencies. The notion o f frequency response will be gener­
alized in Chapter 14 afrer we introduce the notion of the Laplace transform.
12. TERM S AND C O N CEPTS
Admittance: o f a two-terminal device, the ratio o f the phasor current into the device to the
1;in
phasor voltage across the device, y< ( /CO) =
in
Band-pass circuit: circuit in which frequencies within a specified band are passed while frequen­
cies outside the band are attenuated.
Band-reject circuit: circuit in which one band o f frequencies is significantly attenuated while
those frequencies outside the band are passedes with little to no attenuation.
Com plex exponential forcing function: function o f the form v{t) =
^ = a +yco are complex numbers. A special case (a = 0), f(t) =
out the chapter as a shortcut for sinusoidal steady-state analysis.
, where V =
and
is used through­
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
^
Conductance: real part of a possibly complex admittance.
Current division: in a parallel connection of admittances driven by a current source, the current
through a particular branch is proportional to the ratio of the admittance of the branch
to the total parallel admittance.
Euler identitjr:
= cos(0) + j sin(0).
Frequency: in a sinusoidal function A cos((j)t + d) or B sin((Of + 0), the quantity co is the angular
frequency in radians per second (rad/sec). Equivalendy, A cos(o)f + 6) = A cos(2K/t + 0),
where/is the frequency in hertz (Hz or cycles per second). Note that O) = 2n f.
Frequency response: (of a circuit) graph of the ratio of the phasor output to the phasor input as
a function of frequency. It consists of two parts: (1) a graph of the magnitude of the pha­
sor ratio and (2) a graph of the angle of the phasor ratio.
High-pass circuit: circuit with a frequency response such that the high-frequency content of the
input signal is passed while the low-frequency content of the input signal is attenuated.
Imaginary part: the imaginary part of a complex number z = a + Jh for real numbers a and b,
denoted by Im[z], is b.
Impedance: ordinarily complex frequency-dependent Ohms law-like relationship of a two-ter­
minal device, defined as Z(/a)) = V/I, where V is the phasor voltage across the device and
I is the phasor current through the device. For the resistor,
= /?; for the capacitor,
= l/(/a)Q; and for the inductor,
= ycoZ.
Magnitude (modulus): the magnitude of a complex number z = a + jb , denoted by |z|, is
Phason complex number representation denoting sinusoidal signals at a fixed frequency. Bold&ce
capital letters denote phasor voltages or currents; a typical voltage phasor is V =
and a typical current phasor is I =
Polar coordinates: representation of a complex number z as p?-^, where p > 0 is the magnimde of z and
0 is the angle z makes with respea to the positive horizontal (real) axis of the complex plane.
Reactance: imaginary part of an impedance.
Real part: real part of a complex number z = a + jb for real numbers a and b, denoted by Re[«], is a.
Rectangular coordinates: representation of a complex number z as coordinates in the complex
plane, i.e., zs a + jb for real numbers a and b.
Resistance: real part of a possibly complex impedance.
Sinusoidal steady-state response: response of a circuit to a sinusoidal excitation after all transient
behavior has died out. This definition presumes that the zero-input response of the cir­
cuit contains only terms that have an exponential decay.
Stable circuit: circuit such that any zero-input response consists of decaying exponentials or expo­
nentially decaying sinusoids.
Susceptance: imaginary part of an admittance.
Voltage division: in a series conneaion of impedances driven by a voltage source, the voltage
appearing across any one of the impedances is proportional to the ratio of the particular
impedance to the total impedance of the connection.
Zero-input response: response of the circuit when all source excitations are set to zero.
’ In the literature, both z and z* are used to denote the conjugate o f a complex number z. However, in matrix
arithmetic, Z* usually means the conjugate transpose o f the matrix Z. We will sometimes interchange the usage.
In MATLAB, * means multiplication and conj(Z) means conjugated. So there is some ambiguity in the usage.
^
^
^
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
483
Problems
SO LUTIO N OF DIFFEREN TIAL
EQUATIONS W ITH CO M PLEX
EXPO N EN TIALS
O
'
Figure P I0.3
A N SW ER S: 0 .8 sin (2500r -
3 6 .8 6 ‘>). 80
1. Construct the difFerential equation model o f
s in (2 5 0 0 r-3 6 .8 6 » )
the series RL circuit o f Figure PI 0.1 in which L
= 0.25 H and R= 100 Q. Then use the method
4. Construct the differential equation model of
o f section 4 to compute the steady-state
response when v’,„ (/) = 20V 2 cos(400/) V.
the parallel /?Z.Ccircuit o f Figure PI 0.4 for R =
100
C = 1 pF, and Z. = 40 mH. Then use
the method o f section 4 to find the steady-state
response when / (f) = 20 cos(2500f) mA.
vJt)
R < v .J t )
Figure P I0.1
AN SW TR: 20 cos(400/ - jr/4) V
Figure P I0.4
AN SW ER: 1.6 cos(2500; + 36.8~») V
2. Find the differential equation model o f the
series RC circuit o f Figure P I0.2 in terms o f
and V(it) assuming that C = 5 pF and R
= 800 Q. Write
as a function o f
and V(^t). Then use the method o f section 4 to
determine the steady-state response when
iV„(/) = 20>/2sin(250/) V.
+ v,(t) -
KCL AND KVL W ITH
PHASORS
5. Find the phasor current I and /(/) for each
circuit o f Figure Pi 0.5 when (O = IOOti rad/sec.
(2+j4)A
+
v„(t)
R
V „(t)
Figure P i0.2
ANS^XTR: 20 sin(250/ + 0.25k) V
Figure P i0.5
AN SW TRS: (a) 10 cos( 1OOTif - 0 .9 :" )
A. fb)
S.6626 cos( 1OOtt/- 1.798) A
3. Construct the differential equation model o f
the series RLC circuit o f Figure Pi 0.3 in terms
o f /^(r) and
assuming L = 1 0 mH, C = 4
pF, and R = 100 Q. Then use the method o f
section 4 to find the steady-state response /^(t)
when
= 100 sin(2500/) V. Next compute
6. T he circuit o f Figure P i 0.6 operates in the
sinusoidal steady state with the indicated pha­
sor currents when i^{r) = 10 cos( 1OOOf) A. Find
the value of the phasor currents and and the
associated /,(r) and
484
Chapter 10 * Sinusoidal Steady State Analysis by Phaser Methods
j20
(26+j12)V
0
20
(a)
Figure P i0.6
e
C H E C K : /,(/) = 25 cos(100r + 0.9273) A
j20
(26+ jl2)V
L
20
7. Suppose that in Figure P I 0.7, v^{t) = 4
cos(o)f) V and ''2^^ ^~ 4-s/2 cos(o)/ —0.25ti ) V.
Find V[{t) = A"cos(ior+ (J)).
+
v^(t)
(b)
-
e
(26 + jl2)V
Vj(t)
v,(t)
0
j20
20
(0
Figure P I0.7
ANSW ER; vjU) = 4 cos((.»/ - ‘)0") \'
Figure P I0.9
8. Use KVL to determine the phasor voltage
10. For the circuit o f Figure P i 0.10, use KCL
in the circuit o f Figure PI 0.8.
8jV
4jV
and KVL to find the phasor voltage V^. and the
phasor current I^,. If the frequency co = 2 0 0 0 ;:
rad/sec, find the associated voltage and current
time functions.
-
(4 + ]4)V
I
Figure P10.8
(2-j10)A
-
(2+j4)A
V _y (2+
j
/ Ti
(t)
AN SW ER: - 4 V
V
(4 + j6) A
£)
(4+j6)V
1
(5
9. For the circuits of Figures PI0.9a, b, and c,
Figure PIO.IO
compute the indicated phasor currents assuming
ANSW'ERS: V^.= 18.98^17156« V. I. = 2 0 .4 ^
R = 2 0 ., L = A mH, and C = 1 mF. If oj = 500
-1 0 1 .3 '* A
rad/sec, determine the associated time functions.
BASIC IM PED AN CE AND
A D M ITTAN CE CO N CEPTS
11. (a)
A capacitor has an admittance
= y’8
mS at OJ = 400 0 rad/sec. Find the
48S
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
value o f C in pF. Compute the value of
the capacitor’s impedance at (0 = 500
rad/sec.
An inductor has an impedance Zj =
(b)
j20 Q at (0 = 4000 rad/sec. Find tlie
value o f L in mH. Compute the value
Figure P I0.14
of the inductors admittance at 10,000
A N SW ERS:
rad/sec.
3 0 ‘> A. I,
= 0 .0 2 Z 6 0 ‘> A, 1/ = 0 .0 4 Z = 0 .()2 Z ]5 0 " A. i j t ) = 0.02S3
cosi 11)00/ + 1S‘>) A
12. In the circuit o f Figure P I0.12,
cos(lOf) V, C = 0.2
(f) = 10
F, Z: = 0.1 H, R= 2.5 Q.
Determine the phasors I j, I 2, and
Find
15 . Consider the circuit o f Figure P i 0.15
where R = 200 H, A = 80 mH, C = 1 pF and
i- p ) =100 sin(2500r) niA. Find the voltage
".a/')-
phasors
r
n
I''
*
' r
t
V^,
and
and then compute
"
R
JY Y V
v„ -
l
+ V. -
Figure P i0.12
13. The circuit o f Figure PI 0.13 is operating in
the sinusoidal steady state with
= 20
Figure P i0.15
C = 1 mF.
C H E C K : v ;p ) = 28.28 co s (2 5 0 0 r- 135°) V or
(a)
v .p ) = 28.28 s in (2 5 0 0 f- 45") V
Suppose /^|(r) = 10 cos(100r + 30'*)
mA and v^^U) = 200 cos(lOOr) mV.
(b)
Find the phasor I^. and then the cur­
16. In the circuit o f Figure P IO .I 6 , C = 0.03 S,
rent ip ) .
I = 0.1 H, C = 0.4 mF, /j(r) = 1.2 cos(200r) A
Now let /j,(r) = 10 cos(50/ + 3 0 ”) mA
and
and v^2 ^t) = 200 cos(lOOr) mV. Find
(a)
the current
(b)
How does this part
= 40 sin(200^) V.
Find the phasors Ij and
Find the phasor 1^ and the associated
time function ijit).
differ from part (a)?
fi.(t)
o
v,(t)
Figure P I0.13
14. Find the phasor currents
Figure P I0.16
AN SW ER: (b) 1.2Z-9()'\ 1.2 sin(200r) A
and
and then determine i j p ) for the circuit o f
Figure P I 0.14 in which /? = 1 k li,
C = 1 pF, and
= 0.5 H,
= 20 cos(1000r + 60°) V.
17. In the circuit o f Figure P 10.17, /? = 6 f i, Z.
= 80 mH, C = 0.5 mF, Vp{t) = 8 cos(200r) V,
and I, = 0.5Z 90® A. Find the source voltage,
which operates at the same frequency o f
200 rad/sec.
■m
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
V jt)
Z Jj(o )
Figure P I0.17
ANSWHR: S.831 1 cos(200r +30.96<’) V
Figure P 10.20 Parallel Z.C circuit.
ANSWl-.RS: (a) -/1.25 12, (b) 1.6 ml-
18. The circuit o f Figure PI 0.18 operates in the
sinusoidal steady state at a frequency o f cOq =
21. Consider the circuit o f Figure P I0.21.
2000 rad/sec, /?, = 7?^ = 10 Q, V-„ = 50 V, and
(a)
Find the impedance at (o = 100 rad/sec.
= 2Z. - 53.13® A. Compute the phasor volt­
(b)
What happens to the impedance as to
age across Rj and then find the impedance
Now construct a simple series circuit
gets large?
(c)
Ifv/„(/) = IOV 2 cosdOOOV, find /.(/).
that represents this impedance at cOq.
ijt )
1 mF
v jt)
I
6
lOO
0.1 H
Figure P I0.18
Figure P I0.21
ANSWHR: Z = 2.5 +ylO Q
A N SW FR: (a) S - p U
19. (a)
Find the steady-state response o f the
circuit o f Problem 3 using the phasor
22. For the circuit o f Figure PI 0.22, suppose R
method. Discuss the relative advan­
= 100 Q, ^ = 0.5 H, C = 5 ^iF.
(a)
If
= 0.1 cos(500/) A, find v^it).
(b)
Find (O ^ 0 in rad/sec so that the input
tages o f the phasor method.
(b)
Find the steady-state response o f the
admittance is real.
circuit o f Problem 14 using the phasor
method. Discuss the relative advan­
tages o f the phasor method.
+
V ,( t)
SERIES-PARALLEL
IM PED A N CE AND
A D M ITTA N CE
CA LCU LA TIO N S
20 . Consider the circuit o f Figure P I 0.20.
(a)
(b)
(c)
Figure P i0.22
If C = 0.01 F, findZ.„(/100).
If Zy,^(/100) = 25j O., find the appro­
23. Consider the circuit o f Figure P I0.23 in
priate value o f C.
which /?j = 20
/?2 = 10
Z, = 20 mH.
(a)
I f C = 0.3 mF, find K.„(;-500).
Using the impedance o f part (b), if
ij„U) = 100 cos(100r + 45®) mA, find
v^t).
(b)
Find the value o f C that makes the input
admittance real at OJ = 500 rad/sec.
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
(c)
If C = 0.3
mF and /.„(f) = 100
487
zero. Determine the minimum value
cos(500r) niA, find /^](^) and /^-(r)
o flK „(/ 0 ))l.
using current division.
/Y Y V
L
(a)
Figure P I0.23
(b)
A N SW ERS: (a) 0.1 + yO.l S, (b) 0.1 mF
Figure P I0.25
24. Consider the circuit o f Figure P I 0.24 in
which C = 1 F. At C0= 2 rad/sec,
= 4 + jl
n.
26.
For a particular two-terminal device,
= 0.002 + y0.002 at O) = 500 rad/sec.
(a)
(b)
(c)
Find the appropriate values o f R and
Construct a parallel RC circuit having this
L
admittance at (0 = 500 rad/sec. If the circuit is
For 0 < Z < 0.2 H, specify the range o f
excited by a currcnt source with ij^t) = 10
possible reactance values for
cos(500f) niA, find the voltage appearing across
the current source.
If
= 10 cos(2/) V, find Vj^i).
A N SW ER : C = 4 uF, A* = 500 Q ..
5
\'^(/) = —;= cos 500/ - 4 5 " ) V
V2
27. For a particular two-terminal device,
R
'jt ) Q
Z .„(;1000) = 2000 + ;2 0 0 0 Q.. Construct a
series RL circuit having this impedance at CO =
1000 rad/sec. If the circuit is excited by a volt­
Z,„(jw)
age source with v^{t) = 10 cos(1000/) V, find the
current through the resistor.
Figure P I0.24
AN SW ER: Z. = 2 H, /^ = 2 kQ.,
5
/(/) = - = cos 1000/ - 4 5 " ) mA
AN SW ERS: 4 Q, 0.2 H, 0 to x
25. Consider the circuits o f Figure P I 0.25 in
which R = 5 O., L = 32 mH, and C = 5 ^F.
(a)
Find
28. The circuit o f Figure P i 0.28 operates in the
Zy^^(/(0) as a function o f (0. Then com­
sinusoidal steady state at the frequency' (0 =
pute the frequency at which Zy^^(/CO) is
purely real, i.e., the reactance is zero.
5000 rad/sec with R =4 Q., L = 0.4 mH, and C
= 0.1 mF. Find Z-^^{j5000) and ^/„(y5000).
Determine the minimum value o f
Construct a simple series circuit that is equiva­
For the circuit o f Figure P 10.25b, find
construct a simple parallel circuit that is equiv­
Ky^;(/0)) as a function o f (0. Then com­
alent to this circuit at 0) = 5000 rad/sec. In both
pute the frequency at which
cases specily the element values.
Consider
Figure
P 10.25a.
lent to this circuit at 0) = 5000 rad/sec. Finallv,
(b)
is
purely real, i.e., the susceptance is
488
Chapter 10 * Sinusoidal Stoatly State Analysis by Phasor Mcthoils
(b)
R
Find the value o f to in terms o f R and
C at which the phase angle difference
bet\veen \ a n d
(c)
is 4 5 “.
At the w computed in part (b), deter-
Figure IM0.28
-H f
C H EC K : Z.„(/‘5000) =1 + ;2 ^
■6
29. For the circuit o f Figure P i 0.29, let Z.=4
mH, C = 10 |iF, and v^{t) = K^cos(cor) V.
Compute
Find the frequency co (in
Figure P I0.31
rad/sec) at which tlie steady-.state current /^(/) =
0. At this frequenc}', what is the vohage across
32. Consider the circuit o f Figure PI 0.32.
the LC parallel combination?
(a)
If vj^t) = V^^(zos{tlRQ, find the sinusoidal
steady-state response Vf{t) in terms of
V;,, k and C.
(b)
lf/?= 1 0 Hand
10 n/2 cos(10/)
V, find the value o f C so that V(^t) =
10 cos(10r + 0) V.
(c)
For the value o f C found in part (b),
compute the corresponding value ol 0.
Figure P I0.29
AN SW FR: (I) = 5000 rad/sec
R
SERIES/PARALLEL
IMPEDANCES WITH V/l
DIVISION
^,(t)
30. Consider the circuit o f Figure P I0.30 in
which
/? =
20
Q,
Z.
=
4
H,
( 0 = 10>/2 cos(5/) niA.
(a)
Find the input impedance ■^,„(/w) and
the input admittance
(b)
Figure P I0.32
and
At OJ = 5 rad/sec determine the steady-
ANSW ER: (a) r(^(/) = ^ c o s
v2
RC
-4 5 ‘
33. Consider the circuit o f Figure P I 0.33 in
which y? = 8
(a)
state current ;’^(r).
Z, = 8 mFi, and C = 0.125 mF.
Determine the values o f the phasors
I^, and V(^ when
i,(t)
= 2 A and (d =
1000 rad/sec. Specify the correspon­
ding time functions.
©
(b)
Repeat part (a) for o) = 500 rad/sec.
Figure P I0.30
ANSW'F'.KS: (a)
.b; I (I
31. (a)
y20co
5+
/(I)
0.25
. 0.05 - /■
(I)
5; - k !4) niA
For the circuit o f Figure P I0.31, find
the ratio
in terms of/^, C, and
CO.
Express the answer in polar form.
Figure P i0.33 Parallel y?/.Ccircuit.
34. In Figure P I0.33, suppose R = 500 Q, I. =
mH,
C
=
0 .1 2 5
i:,Ar) = loV2co.s(co/+ 60^) A.
0 .5
mF,
and
•hS9
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
(a)
Compute the values o f the phasors I^
Using the phasor method, find
I^, and
v j t ) = 50 cos(4000^) mV.
when
= 2 A and OJ =
4000 rad/sec. Specify the correspon­
He
ding time functions.
(b)
Repeat part (a) for (O = 8000 rad/sec.
when
100v„,(t)
v„.(t)
35. In the circuit o f Figure P i 0.35, suppose R
= 2 0 n , L = 0.5 H, and C = 0.625 mF.
(a)
If
= 16 cos(40r) V, find
V(^t) and v^it) using phasor voltage
division.
(b)
If
Figure P i 0.38 Iw o coupled /^Ccircuits.
39. Consider the circuit o f Figure P i 0.39 in
which /?j = 200 n , Z = 0.2 H, R, = 200 Q, and
= -32sin(40r) V, find
v^t), and v^it) using phasor voltage
division. Hint: avoid repeating the cal­
C = 0.05 pF. Use the phasor method to find
i^t) when
= 10 cos(lO'^r) mA.
culations o f part (a); this can be done
by inspection.
R
6
Figure P i0.39 Two coupled circuits.
40. Consider the circuit o f Figure P i 0.40 in
Figure P 10.35 Series /?/,C circuit.
36. Reconsider Figure 10.35 lor /? = 10
Z. =
Find
0.08 H, and C = 0.02 R
(a)
If
which /?! = 500 Q, I = 0.125 H, R ,= m Q,
and C = 5 pF. Suppose Vj^^) = 120 cos(400/) V.
and
= 10 cos{25t) V, find
and v^it) using phasor voltage division.
(b)
If za//) = 16 sin(50f) V, find
and
V[ {t)
V({t),
using phasor voltage division.
37. In the circuit o f Figure PI 0.37, /?, = 20 Q.,
i = 2 H ,a n d / ? j = 1 0 a . If|V„„/V,„| = 0.2 at
co= 40 rad/sec, find the necessary value(s) o f C
(in mF).
Figure P i0.40
41. Consider the circuit o f Figure P I 0.41 in
which 7^1 = 500 QX = 0.125 H, R, = 100 iX
and C = 5 pF. Suppose
''" 6
mA. Find
Figure P I0.37
AN SW TR: 0.625 mF or 0.2083 mF
38. In the circuit o f Figure PI 0.38, /?, = 50 Q,
q
= 1 uF, Rj = 300 Q, and q
= 0.625 pR
and
= 120 cos(400r)
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
490
AN SW ERS: 2 Z 9 0 ‘>, - 2 sin(400r) A
44. For the circuit o f Figure P i 0.44, find the
and the phasor voltage Y q If
phasor current
the circuit is known to operate at a frequency o f
CO = 1000 rad/sec, find /^(f) and
and the
values o f L and C
2Q
> j2 Q
-j2 0
NETWORK THEOREMS IN
CONJUNCTION WITH V/l
DIVISION.
AN SW ERS; t'fit) = 2 cos(lOOOr) V, i/(/) = cos
(You should consider applying one or more
(1000/-O.S3T) A
Figure P I0.44
network theorems to simplify the solution to
the problems in this section.)
45. In the circuit of Figure Pi 0.45, R=20Q., L = 20
mH, C= 100 pF, and v^.(0 = 2 0V 2 cos(IOOO/)V.
42. In the circuit o f Figure P 10.42, R^ = 60
Compute the value of
in steady state.
/?2 = 40 n , and C = 0.1 mF. Find the phasor Ij
/Y Y V
L
and the corresponding steady-state current /j(/)
when /^.(/) = 5>/2cos(100/) mA. This prob­
lem can be solved by direct current division or
6
by source transformation and impedance con­
cepts. Which method is easier?
Figure P i0.45
A N SW ER: 20 co s(1 0 ()0 r- 135°) V
i(t)
0
46.
Consider the linear circuit o f Figure
PI 0.46, which operates at 50 Hz and for which
V, =
+ 1,1^2(a) Find the values o f a and b.
(b)
Figure P i0.42
If v^^{t) = 10 cos(lOOTCf) V and
200 sin(lOO)/^) mA, find v^it).
43. In the circuit o f Figure P I 0.43, CO = 400
rad/sec,
= Rj = 2
L = 5 mH, and C =
625 |.iF. Find I^^and the corresponding
steady state. vp)= 12 cos(400^) V
in
j5on
■ -©
J200Q
+ V. -
Figure P i0.46
AN SW ERS (in random order);
Fiaure P I 0.43
- 0 .8 . ./40 Q. - 1 6 cos(IO().-t O V
=
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
491
47. The linear circuit o f Figure P I 0.47 is such
that in the steady state, if
can do this with some straightforward
= 10 cos(2007U/)
reasoning without writing any equa­
A with v^2 ^t) = 0, then
\t) = 20 cos(2007ir +
45®) V. On the other hand if i^^{t) = 0 with
/;2 (/) = 10 cos(2007if + 45°) V, then v^{t) = 5
tions.
cos(200)r + 900) V
(a)
Find a linear relationship between
v^2 >and
(b)
,
.
If"/^,(/) = 5 cos(2007tr - 45«) A and
= 20 cos(2007rr) V, then in the
steady state find v^{t).
(c)
Find
and
Zj
Z - ,.
Develop simple cir­
Figure P I0.48 AC Wheatstone bridge circuit.
cuit realizations o f these impedances
valid at OJ = 20071 rad/sec.
THEVENIN AND NORTON
EQUIVALENTS
49. Find theThevenin equivalent for the circuit
o f Figure PI 0.49 when R = AQ. L = 20 niH, C
= 1.25 mF,
= 2Z.45® A and (O = 200 rad/sec.
Be sure to express the open-circuit voltage as a
Figure P I0.47
ANSWHRS: (b) r,(/) = 18.46 cos(200nr -
time function.
22.5") V; (c) Z, = 0.763 + y2.6()5 H Z, = 4 Q
OA
48. The circuit o f Figure P I 0.48 is a general
Wheatstone bridge circuit (the dc version o f
which is described in Problem 35 o f Chapter
2). Here the circuit is used to measure the value
o f the unknown inductance L.
(a)
Suppose
= 0. Show that the steady-
state voltage i'„,^f(t) = 0 when
=
UC. Note: In general the condition for
a null voltage, v{t) = 0, in the steady
state is that the products o f the cross
impedances be equal.
(b)
Again suppose
that
= 0. You are given
= 2 sec and that the voltage
source v-^j^t) is a sinusoid with a fre­
quency
of
5
rad/sec.
W ith
the
unknown inductance L inserted in the
circuit as shown, you adjust R[ until
you reach a sinusoidal steady-state
voltage null,
ing value for R^ is 3
(c)
= 0 V. The result­
Find the value
ofZ,.
Now suppose R^ ^ 0. Show that the
condition o f part (a) is still valid. You
Figure P10.49 Parallel /?ZC circuit.
A N SW ERS: Z^;, = 4 i l , /-^.(/) = 8 cos(200/ +
45*’) V
50 . For the circuit o f Figure P I0.50, let 1 = 1 0
mH, /?
mA.
(a)
=20 a, C = 20
^iF, and
l-„ = 1 0 0 ^ 0 ”
Find the Thevenin equivalent at the
terminals a and b if to = 2000 rad/sec.
(b)
If the circuit is terminated with a load
consisting o f a series connection o f a
20 Q resistor and a 20 mH inductor,
find the sinusoidal steady-state voltage
across the load.
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
492
OA
Figure P i0.53 Two coupled circuits.
54. The circuit o f Figure P 10.54 operates in the
SSS at cOq.
(a)
20 krad/sec and
= 2Z.0". Find the Thevenin
Find the Thevenin equivalent imped­
ance
51. The circuit o f Figure PI 0.51 operates at to =
(b)
Find
if
10 a , ^ = 5 a
equivalent circuit (in the phasor domain) at ter­
= 1^^^cos(w^/), R =
and
20 Q.
minals A and B. Use this Thevenin equivalent to
find the magnitude o f the phasor
when the
10 mH and 1 kH series combination load is con­
nected to A and B.
i jt )
-jx
Figure P I0.54
55. In the circuit o f Figure PI 0.55, assume co =
100 rad/sec, I = 40 mH, C = 5 mF, R = S Q.,
and a = A Q.. Find the Thevenin impedance
Figure P I0.51
seen at terminals A and B. If
CH ECK : 190 < |V J < 205 V. |V^^| . 0.5| V J
52. For the circuit o f Figure PI 0.52,
Q, Cj = 0.2 pF,
= 20
cos(lOOf) V, find
= 1000
= 500 Q, and C , = 1 \i¥.
Find the Norton equivalent circuit when v- (t)
= 50 cos(4000f) V.
e
v.(t)
-OA
R,
Figure P I0.55
■OB
GENERAL SSS ANALYSIS
(NODE OR LOOP ANALYSIS)
Figure P I0.52 Two coupled R C circuits.
53. For the circuit o f Figure P I 0.53 R = 2.5
k n . Find the Thevenin equivalent when
56. (a)
Find the phasors
in the cir­
cuit o f Figure Pi 0.56 when
=
= 2 0 V 2 Z 4 5 " V, R = AQ., L = A
mH, and oj = 1000 rad/sec. Specify
the corresponding time functions.
10 cos(4000r) mA.
(b)
Determine the value o f oj for which
the magnitude o f the output voltage
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
phasor is 20% o f rhe magnitude o f rhe
(0
input voltage.
493
At to = 100 rad/sec, determine the
Thevenin equivalent circuit in which
Z^y^(/'100) is a series combination o f
two circuit elements seen at terminals
/Y Y V
L
A and B.
'“6
-OA
2R
2R
;v,(t)
Figure P i0.56
C H E C K S: 5 A, 2o’ v
-O B
Y.(j«)
57. In the circuit o f Figure P I0.57,
= y'30 Q,
Figure P I0.59
= - ; 4 0 Q, V^, = 28 V, V^2 =
^
sinusoidal sources have been operating for a
60. Consider the circuit o f Figure P i 0.60 in
long time, and Z = 50 - j 4 0 Q. Find V^.
which v(,t) = 20 cos(lOOO^) V,
= 40 Q., Rj =
20 Q.,L = 20 m H .an d C = 7 5 \i¥.
(a)
Write and solve a nodal equation at
the top node for
6 ''“
Then write the
corresponding time-domain expres­
sion for v^t).
(b)
Figure P I0.57
z^ = y io a
z^^ = - y i o a
and then write the corre­
sponding time-domain expression for
58. For the network o f Figure P i 0.58, a = 20
a
Calculate
^ = lo a
and
= 20 V. Find the phasor current I^..
Z,
the inductor current
(c)
Find the Flievenin equivalent circuit at
the source frequenc)' relative to terminals
A and B. Draw the Thevenin equivalent
circuit showing the Thevenin impedance
as a series circuit of two elements.
Figure P I0.58
AN SW FR: O.r. + 0.2/
59. Consider the circuit o f Figure P i 0.59 for
which C = 0.8 mF, r.„(;100) = 0.01 + ;0 .0 4 S,
and v-^,(r) = 80 cos(lOOf) V.
Find R.
(a)
(b)
(c)
(d)
(e)
-
Find L.
Find
Find
At CO = 100 rad/sec, determine the
Thevenin equivalent circuit phasors
Woe and Z^;^(ylOO) at terminals A and
B.
Figure P I0.60
ANSWKRS: V^. = S - /S
= -/ 0 .:5 . Z./. =
1 0 - ; 10 12
61. The circuit o f Figure P i 0.61 operates at (o
= 2 krad/sec a n d =1. 5 mS with
= 10 0 Z 0 °
V. Find the Thevenin and Norton equivalent
circuits seen at terminals A and B.
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
494
determine the asymptotic behavior for large (0
/ 'A
J \
■9
l.s f
and for at least one other frequency without a
0-251
computer or calculator. List these properties in
.
writing along with your reasoning.
Figure P I0.61
= ;1 0 0 0 Q,
C H EC K :
'■©
= 25 + ;2 5 V,
= 25 + y25 mA
62. Consider the circuit o f Figure PI 0.62. If i^{t)
Figure P I0.64
= 40 cos(lOOO^) mA, C, = 0.25 1-iF.
= 10 mS,
/?j = 1 kQ, and R-, = 5 kH, find the Thevenin
65. Compute the magnitude and phase o f the
equivalent circuit parameters
frequency response o f the circuit o f Figure
and
P I 0.65 where I = 25 mH and /? = 50 Q. Plot
your response in MATLAB (0 < 0) < 8000
rad/sec) and determine the frequency at which
'• « '©
the magnitude is I/ V 2 o f its maximum value.
R,
Before sketching the responses, determine the
asymptotic behavior for large (0 and for at least
Figure P I0.62
C H EC K :
= 128 - ;3 3 7 .8 8 Cl,
1 4 0 .2 5 Z 9 5 .2 8 “ V
=
one other frequency without a computer or cal­
culator. List these properties in writing along
with your reasoning.
63. This problem tests whether you can synthe­
size ideas from two different parts o f the text.
L
In the circuit o f Figure P i 0.63, R = 20 Q, L =
(b
1 H, v^it) = 50 cos(100r)«(r) V (notice the step
function), and /^(O'*') = 1 A. If the response for
t > 0 has the form
i^{t) = A cos(l OOf + (j)) +
then determine the constants A, (j), X, and B.
ijt)
R
v,(t)
6
Figure P i0.65
66. Inside the black box o f Figure P I0 .6 6 a
there is a two-element circuit composed o f a
resistor o f 10 ^2, capacitors, inductors, or some
combination o f these elements. A variable-fre-
Figure P i0.63
FREQUENCY RESPONSE
64. Compute the magnitude and phase func­
tions o f the frequency response o f the circuit o f
Figure P I 0.64 in which L = 4 mH and C = 0.25
mF. Plot your response in MATLAB (0 < CO <
5000 rad/sec). Before sketching the responses,
quency voltage vp) = lOcos(O)r) V is applied to
the box and the voltage v(t) =
cos(cor + 0) is
observed. A plot o f the magnitude o f v{t) with
respect to CO is given in Figure P 10.66b.
(a)
Draw the circuit contained inside the
(b)
box. (There are two solutions.)
Specify the element values.
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
v,(t) =
10cos(cot)V
+
Black
Box
v(t)
(a)
(b)
69. Reconsider the pressure-sensing example of
Figure P i0.66
section 10. Specify a set o f mesh currents and
67. Compute the frequency response o f the cir­
write a set o f mesh equations that describe the
cuit o f Figure PI 0.67, where R= 100^2, L = 10
circuit. Solve the equations for 1 pF < € 2 ^
pF using MATLAB or some other, equivalent
mH, C = 0.1 mF, and
is the output. Use
MATLAB or its equivalent to generate the
magnitude and phase (in degrees)
plots.
Consider 0 < (O < 3000 rad/sec.
software program. Plot the magnitude o f V ^ V ^as a function o f Cj- Now construct a plot o f
the magnitude o f V ^ - V ^ a s a function o f pres­
sure in mm Hg.
OP AMP CIRCUITS
70. (a)
Figure P I0.67
a single inductor. Let v^{f) be the input excita­
tion and ip ) the circuit response. The magni­
tude frequency response is given by Figure
P I0 .6 8 b . Draw^ the circuit inside the box and
assign component values if it is known that L =
40 mH.
when
= sin(200r)
mV for the circuit of Figure P10.70a.
(b)
For the circuit o f Figure PI 0.70b, find
C so that when
68. The box labeled V{joi) in Figure P I0 .6 8 a
contains a single resistor, a single capacitor, and
Compute
(c)
= cos(400r) mV,
= sin(400r) mV.
Find the phasor transfer function,
//(yco), and plot the magnitude o f the
frequency response (using iMATL^B
or the equivalent) as a function o f 03 =
Inf, where/is in Hz and to in rad/sec.
Chapter 10 • Sinusoidal Steady State Analysis by Phasor Methods
496
r>
100 kfi
20 kO
80 kO
o
n
(a)
(b)
Figure P I0.70 Op amp differentiation circuits.
(a)
n
Compute
when
=
sin(400/) V for the circuit of Figure
P10.71a.
For the circuit of Figure P I 0.7 lb, find
C such that when
=sin(500/) V,
= 5 cos(500/) V. This represents
an integration of the input with gain.
Find the phasor transfer function,
and plot the magnitude of the
frequency response (using M ATLAB
or its equivalent) as a function of O) =
In fi where/is in Hz and O) in rad/sec.
71. (a)
(b)
(c)
n
o
(b)
Figure P I0.72 Leaky integrator circuits.
n
r s
10|j F
H e-
50 kO
+
He-
200 kn
+
o—
+
+
vjt)
JL .
(a)
(b)
Figure P I0.71 Op amp integrators.
73. (a) At (0 = 2 X 10^ rad/sec, find the
phasor voltage gain ^ouP^in
°P
circuit of Figure P 10.73.
(b) Find the phasor transfer function,
//(/to), and plot the magnitude of the
frequency response as a function of O)
= In f, where / is in Hz and (O in
rad/sec using M ATLAB or equivalent
software.
1kn
72. (a) If an 800 Hz sine wave of unit ampli­
tude excites the leaky integrator circuit
of Figure PI0.72a, determine the
steady-state output voltage.
(b)
For the circuit of Figure P 10.72a, find
the phasor transfer function, //(/(o),
and plot the magnitude of the fi*equency response (using M ATLAB or
Figure P I0.73
its equivalent) as a function of (O =
271/where/is in Hz and O) in rad/sec.
74. For the circuit of Figure P i0.74, find the
(c) If the input to the circuit of Figure
expression for the phasor transfer function
P I 0.72b is
= cos(2000)r) V,
//(/ cd) =
Assume an ideal operational
determine the values of R and C so
amplifier. Plot the magnitude of the transfer
that v i t ) = 5cos(2000ti^ +135°) V.
n
r^
o
n
Chapter 10 • Sinusoidal Steady State Analysis by Phaser Methods
O '
^
function as a function o f COusing M ATLAB or
w
•^19'
the equivalent, assuming /? = 10 k fl and C =
0.01 mE
frequency increases to infinity? What
happens as the frequency decreases to
zeroi^
VO
o
o
Figure P I0.74
Figure P I0.76 Ideal op amp circuit.
75. In the circuit of Figure PlO.75 assume the
operational amplifier is ideal and that
= 25
ki2, Cj = 1 ^F, /?2 = 5 k£2, and C2 = 0.2 pF.
Compute the gain of the circuit as a function of
(0. Then use M ATLAB or the equivalent to plot
the magnitude and phase of the fi-equency
response as the logarithm of the frequency for 1
< C0< lO'^ rad/sec.
77. Consider the circuit of Figure P I0.77 in
which
= 200 £2, Cj = 0.05 |iF, /?2 = 28 IdQ,
and C2 = 0.05 pF.
Use nodal analysis to compute the
ratio
=I
^Hz. Now
use physical reasoning to obtain the
approximate the ratio at/= 1 Hz and
/= 100 kHz.
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
w
Figure P I0.75
76. For the operational amplifier circuit of
Figure P I0.76, /?j = 5 kQ, C, = 0.02 |iF, /?2 =
5 kfi, and C2 = 0.08 pF.
(a) Write two node equations and solve to
find a relationship between the output
phasor
and the input phasor
at the frequency/= 1000 Hz. Note
that the voltage from the minus termi­
nal of the op amp to ground is
which equals the voltage from the plus
terminal to ground, assuming the op
amp is ideal.
(b) Repeat the calculation at/= 100 Hz
and/= 3000 Hz. What happens as the
Figure P I0.77 Op amp circuit having a band­
pass type of response.
Note that
is an intermediary variable useful
in the nodal analysis of the circuit.
C
H
A
P
T
E
R
Sinusoidal State State
Power Calculations
The AM or FM receiver that is often part o f a home stereo system receives signals from radio sta­
tions through an attached antenna. The intensity o f these signals or radio waves depends on the
power radiated into the atmosphere by the broadcasting station, the distance between the receiv­
ing and transmitting antennas, and the design o f the receiving antenna. The intensity or magni­
tude o f the signals picked up by the receiving antenna is very small. The power available from the
antenna and deliverable to the receiver is typically in the microwatt range. Again, this is ver)^ small.
Hence, it is important to have maximum power transfer from the antenna to the receiver input so
that the music signals received can be properly amplified and enjoyed. Since the signals in the
antenna are sinusoidal at very high frequencies, the antenna is represented by a phasor Thevenin
equivalent circuit as is the input circuit o f the receiver. Hence we must understand maximum
power transfer in the context o f sinusoidal steady-state analysis to describe and analyze this prob­
lem. An example at the end o f the chapter illustrates some impedance matching techniques to
achieve maximum power transfer from an antenna to a receiver.
CHAPTER OBJECTIVES
1.
2.
Define and investigate the notion o f average power.
Define the notion o f the effective (rms) value o f a periodic voltage or current and its rela­
3.
tionship to the average power absorbed by a resistor.
Define the notion o f complex power and its components— average, reactive, and appar­
ent power— and investigate the significance o f each and their relationships.
4.
Introduce the notion o f power factor associated with a load and describe reasons and a
5.
method for improving the power factor.
Prove the maximum power transfer theorem for the sinusoidal steady-state case, and illus­
trate its significance for the input stage o f a radio receiver.
500
Chapter 11 • Sinusoidal State State Power Calculations
SECTION HEADINGS
1.
2.
3.
4.
Introduction
Instantaneous and Average Powers
Effective Value o f a Signal and Average Power
Com plex Power and Its Com ponents: Average, Reactive, and Apparent Powers
5.
Conservation o f Com plex Power in the Sinusoidal Steady State
6.
Power Factor and Power Factor C orrection
7.
M aximum Power Transfer in the Sinusoidal Steady State
8. Summar}^
9. Terms and Concepts
10. Problems
1. INTRODUCTION
Chapter 1 defined the concept o f power. The following chapters were primarily devoted to the cal­
culation o f voltages and currents. This does not mean that the consideration of power is o f sec­
ondary importance. The very opposite is true. A homeowner pays for the energ}' used, not for volt­
age and current. The integral o f power over, say, a 30-day period determines the household ener­
gy consumed in a month. Hidden in the homeowners cost is an adjustment to cover the power
losses incurred in transmitting energy from the generating station to the home. Thus power con­
siderations have a significant impact on everyday life.
A second reason for understanding ac power usage is safety. Each appliance, and its cord that plugs
into the wall outlet, has a maximum safe power-handling capacit)'. Misunderstanding such infor­
mation and/or misusing an appliance can lead to equipment breakdown, fire, or some other lifethreatening accident.
Even for electronic equipment in which power consumption is low, such as laptops and handheld
PDAs, power consumption and, thus, battery life are important design factors. Power drainage direct­
ly determines the PDAs operating time before the battery needs recharging. In fact, optimizing power
management in laptops and hybrid electric vehicles is an important research area in todays world.
In this chapter we will investigate different notions o f power in ac circuits and discuss their sig­
nificance and application. The term “ac circuits” has a narrow meaning here. It refers to linear cir­
cuits having all sinusoidal sources at the same frequency and consideration o f responses only in
steady state. The basic analysis tool is the phasor method o f Chapter 10.
2. INSTANTANEOUS AND AVERAGE POWERS
Figure 11.1 shows an arbitrary two-terminal circuit element isolated from a larger circuit. With
the voltage (in V) and current (in A) having indicated reference directions, the instantaneous
power (in watts) absorbed by the element is given by equation 11.1:
p{t) = v{t)i{t)
(11.1)
Chapter 11 • Sinusoidal State State Power Calculations
501
FIGURE 11.1 Instantaneous power delivered to an arbitrary two-terminal element.
Evaluating battery life or the length o f operation o f your cell phone involves consideration o f a
quantit)' called average power, P (or
for emphasis), defined as the average value o f the instan­
taneous power over an interval [T*!, T^. The idea is based on the average value o f a function, say
j{t) , which is defined as
Le =
T-,
1
T
h - TM r,J
Using this idea we define the average power consumed by a rwo-terminal element as shown in
Figure 11.1 over the interval [T j, Tj] as
1
Pave(TiJ2) =
T,
piOclt
r .-r ,
( 11. 2)
When the signal is periodic with period T, we speak o f the average power consumed by an ele­
ment over the period T as
T
T
(11.3)
It is not necessary that T b e the fundamental period; the evaluation o f the integral is the same for
any integer multiple o f the fundamental period.
EXA M PLE 11.1. Compute the average power absorbed by the resistor R connected to an inde­
pendent voltage source as shown in Figure 1 1.2b with the excitation shown in Figure 1 1.2a.
v.(t) f +
F IG U R E 1 1 .2 Triangular voltage waveform
driving resistor R .
Chapter 11 • Sinusoidal State State Power Calculations
S{)2
So
lu t io n
Step 1. Compute the instantaneous power for 0 < f < 7q. Here
IV J
P {t) =
0 < / < 0 .5 ^
o .5 r „ < / < 7 ;
Step 2. Compute
Using equation 11.3 and observing that the Fundamental period is Tq, we
have
0.57n
/
‘ >ave
p {t)d t = —
rj.
TnR
6R
Exercises. 1. Suppose the sawtooth in Figure 11.2a does not drop to zero at r = 0.5 T'q , but rather
continues to increase until reaching ^ = T’q when it drops to zero and repeats. Find the average
power consumed by R.
AN SW ER:
3/^
. K?
2. Show that the average power absorbed by an R Q resistor in parallel with a Vq V dc source is—
over any time interval [T'j, 7'^].
^
O f particular importance is the average power consumed by devices in the SSS assuming all exci­
tations are at the same frequency, to. Consequently, all voltages and currents are sinusoids at the
sa7nefrequency. To compute the average power absorbed by a circuit element as depicted in Figure
11.1 (assuming a linear circuit), suppose v{t) =
cos(u)t + 9^^ and i(t) =
cos(a)r + 0^) . The
associated instantaneous power is
p ( f ) = v(f ) i { t ) = V,„ cos(co/ + e^,) X /,„ cos(cor + 0 ,-)
(11.4)
=
cosO , -
e,-) +
cos( 2 (o/+ e,, + e,.)
Equation 11.4 follows from the trigonometric identit)' cos(x) cos(y) = 0.5 c o s (x - y) + 0.5 cos(a' +
y). Observe that the instantaneous power o f equation 11.4 consists o f a constant term plus anoth­
er component varying with time at tivice the input frequency. Figure 11.3 shows typical plots o f
/>(/), v{t), and i{t).
Chapter 11 • Sinusoidal State State Power (lalculations
503
0.01
0 .005
0 .0 1 5
0.02
t in secs
FIGURE 11.3 Plots o f tit) = 10 cos(377f) A, v{t) = 2 cos(377/ + 45°) V, and p{t).
Using equation 11.3 witii T = 23t/ol), and observing that the integral o f a sinusoid over any peri­
od is zero, we obtain the following formula for average power in SSS:
T
_
*Pnvp = —
co s(0 , - 0 ; )dt + ^ j c o s ( 2 c o r + 0 , + 0,-)d! =
c o s(0 , - 0 , ) (11 -5)
If the two-terminal element is a resistance R, then v{t) = Ri{t) and 0^^ - 0y = 0 . It follows from
equation 11.5 that for a resistor
p
_ Vm‘I m _ RI^m _ m
^ave,R
2
2
2R
If the two-terminal element is an inductance L, then
^aveL ~ ^
( 11.6)
= (/(.oZ,)!^ and 0^, - 0^ = 9 0 ° . Hence,
cos(±90°) = 0. Similarly, if the two-terminal element is a capacitance C, then
c~ ^
^neans that the average power consumed by
or delivered by a capacitor or au inductor is zero. Even though an ideal capacitor and an ide;il induc­
tor neither consume nor generate average power, each may absorb or deliver a large amount ot
= (/coQV^ and 0^ - 0y = - 9 0 ° . Hence,
instantaneous power during some particular time inten-al.
Before closing this section, we need to investigate the question o f superposition o f average powder.
Is there a principle o f superposition o f average power? If so, when is it valid? When is it not valid?
The following example provides the answers.
Chapter 11 • Sinusoidal State State Power Calculations
504
EXA M PLE 11.2. Consider the circuit o f Figure 11.4, which consists o f a series connection o f two
(sinusoidal) voltage sourccs in parallel with a 1 Q resistor. For this investigation v^{t) =
cos(ojjr + 0|) (having fundamental period /'] = 2:t/tO|) and v-,(f) =
cos(t02^+ B2) (having fun­
damental period T-, = 271/(0,). For simplicit)^ we assume that v^{t) and v-,{t) have a common peri­
od o f 7 'seconds, i.e., there exist Integers ni and ;/ such that T = uT^ = niT^.
v,(t) Q
1n
< v_(t)
FIGURE 1 1.4 Circuit for investigating superposition o f average power.
So
lu t io n
Compute the average power consumed by the 1 Q resistor. First observe that the power consumed
by the 1 Q resistor with source 1 acting alone, i.e., v-j{t) = 0, is
1
1
0
0
Also note that the power consumed by the 1 H resistor with source 2 acting alone, i.e., u^{t) = 0,
is
I T
, r ^
0
0
= v^{t) +
W ith both sources active, linearit)^ (or KVL) implies that
1
1
(^ ’ 1 (/)
Pave = - \ V R {t)i,i{t)d t = - \
0
. By equation 11.3,
+ v'2 (r ))“ dr
0
vf(r)f/^ + - J v ? ( r V r + - Jv| (t)v2{t)dr
0
0
0
T
^ove.l
^a\r,2 '
^ Vj (/)V2(/)f/f
0
^ave,\
= Pave.\ + f'ave.l + '
^ove.l
T
COS((Oi /+0| )COS(C02/ +02)<^/^
j [ c o s ( ( ( 0 | + C O 2 )/ + (0 i + ©2 ) ) + C 0 s ((0 3 , - ( O 2 ) / + (0 i - 0 2 ) ) ] ^ / ^
0
Chapter 11 • Sinusoidal State State Power Calculations
S()S
When the integral term in this last equation is zero, then
, indicating that
superposition o f average power holds. When this integral term is nonzero, superposition o f aver­
age power does not hold. The next question is, under what circumstances is the integral zero and
nonzero.^ There are three cases to consider. Case 1 is when cuj
co^ , which will result in a zero
value o f the integral. In this case, the integral consists o f rwo sinusoids integrated over a common
period T. The integral of a sinusoid over any period is zero. Thus, the integral is zero and super­
position o f power holds when cOj ^ O)-,.
Case 2 is when C0 | = co-, but with (Bj - O2) = ±knl2, k an odd integer. In this case, the integral is
again 0 . This follows because the first term o f the integrand is a sinusoid whose integral is zero
over the period T. The second term o f the integrand is a constant, cos(0j - (),) = cos{±kKl2) = 0 ,
also resulting in a zero integral. Hence for case 2, superposition o f power holds.
Finally, we have case 3, for which tOj = co-, but with (6 j - 62 ) ^ ±kKll, k an odd integer; here
superposition ot power does not hold. The second term of the integrand is a constant, cos(0j - 62 )
0 , resulting in a nonzero integral over the period T. So
P^^^^ j +
For case 3, it is
desirable to use the phasor method o f Chapter 10 to compute the desired voltage and then use
equation 11.5 to compute average power.
Exercises. 1. In Example 11.2, suppose t/j(r) = 3cos(107rr) V and V2 {t) = 4 cos(15n:r + 0.25ti) V.
Compute T, a common period for the two sinusoids, and then compute the average power con­
sumed by the 1 Q resistor.
C H EC K ; T = 0.4 sec w-ill work, and
2. In Example 11.2, suppose
age power consumed by the 1
= 12.5 watts
= 3cos(107Tf) V and
= 4 sin(lOJur) V. Compute the aver­
resistor.
C H EC K ; P^..=
ave 12.5 watts
3. Now suppose v^{t) = 3cos(10Tcr) V and
=4
c o s
(1 5 7 T /
+ 0.257i) V. Compute the average
power consumed by the 1 H resistor.
C H EC K ; P^^^ = 20.99 watts
Equation 11.6 resembles equation 1.18b for the dc power absorbed by a resistor connected to a
dc source. However, in equation 11.6 the factor 1/2 is present. With the introduction o f a new
concept called the efifective value o f a periodic waveform, the formulas for the average power
absorbed by a resistor can be made the same for dc, sinusoidal, or any other periodic input wave­
forms.
Chapter 11 • Sinusoidal State State Power Calculations
S06
3. EFFECTIVE VALUE OF A SIGNAL AND AVERAGE POWER
GENERAL CONSIDERATIONS
From section 2, a resistor o f R ohms excited by a periodic voltage or current absorbs an average
power,
. T he effective value o f any periodic current, i{t), denoted by
is a positive con­
stant such that a dc current o f value /^-exciting the resistor causes the same amount o f average
power to be absorbed, i.e.,
The same holds for a resistor excited by a periodic volt­
age v{t). Mathematically,
(11.7a)
or
(11.7b)
R
Equation 11.7a suggests that
1
Iq+T'
I f " )
T
Hence, the mathematical definition o f the effective value of a periodic current i{t) is
/m+7
U>ff.R
T
(11.8a)
and, similarly, the effective value o f a periodic voltage u{t) is
h\+T
ef/M
(11.8b)
In general, the effective value o f any periodic signal y(r) is
In + T
F e jf-
(11.8c)
Observe that the expressions under the radical sign in equations 11.8 constitute the average value
ot the square o f the signal. Hence, the expressions give rise to the alternative name for the effec­
tive value, the root-m ean-square (abbreviated rms) value oiJ{t), since
mean value o f the square j{t) over one period.
the square rooto'i the
Chapter 11 • Sinusoidal State State Power Calculations
SO'
Exercises. 1. Show that the average power absorbed by an R Q resistor carrying a periodic current
2. Suppose i{t) = 3cos(2n:/) + 4 cos(47i:f) A flows through a I Cl resistor. Find
ANSWlvR:
= 1 2 .5 watts, and l^,g- =
5
7?
= 2.5\/2
and
A
EXA M PLE 11.3. Compute the efFective value o f the periodic voltage waveform sketched in
Figure 11.5.
So
lu t io n
From equation 18.b,
= T
Therefore,
1 1
.2
4J
4J
3
4J
.
3
= 2.3094 V,
Exercise. Repeat the calculation o f Example 11.3 for the case where the values on the vertical axis
o f Figure 11.5 are doubled.
A N SW l'R :
V
For a sinusoidal signal/r) =
cos(tor + 0) , the effective value can be calculated using the identit}'
cos^ (.v) = 0.5 + 0.5 cos(2.v)
as follows:
f~{t)
Fm
-
= F “jC o s “( o j/+ 0 ) = —
F~m
+ —
c o s (2 o j/+ 2 0 )
S08
Chapter 11 • Sinusoidal State State Power Calculations
Since by assumption to 7^ 0, the average value o f the cosine term is zero. The average value o f the
first (constant) term is itself Hence, by equation 11.8c,
F~
I itt
(11.9)
Eml
72
Thus, for a sinusoidal waveform, the effective or rms value is always 0 .707 times the maximum
value or, equivalently, the ma.ximum value divided by >/2— a basic fact well worth remembering.
The ac voltage and current ratings o f all electrical equipment, as given on the identification plate,
are rms values unless explicitly stated otherwise. For example, the household ac voltage is 110 V,
with a maximum voltage o f 1 lOx-s/^ = 1 5 6 V. A typical appliance such as a coffee maker will
have a 110 V rating, ac, at say, 900 watts. The effective values o f a few other periodic waveforms
are listed in Figure 11.6, with their derivations assigned as exercises.
Feff
«=F
r
dc
sinusoidal
triangular
square
F.»= F .
F IG U R E 1 1 . 6 Efleccive values o f some com m on periodic waveforms.
S09
Chapter 11 • Sinusoidal State State Power Calculations
Exercises. 1. Derive the formula for riie cfFectivc value o f a triangular waveform shown in Figure 11.6.
2. Compute the effeccive value of the waveform shown in Figure 11.2a.
AN SW FR:
SINGLE-FREQUENCY ANALYSIS WITH EFFECTIVE VALUES
We recurn now to the case of single-frequency SSS analysis. The average power as per equation 11.5
absorbed by an arbitrary rwo-terminal element mav now be rewritten in terms o f effective values:
^
cos(0,, - 0y) = - ^
2
V2 v 2
COS(0^. - 0,-) = yeff^ejf COS(0,, - 0, )
For the remainder o f the chapter, all voltage and current phasors will be taken as being effective
values unless the subscript m or
appears, indicating the maximum value. The subscript ^ w ill
be added sometimes for emphasis, however. This practice is widely accepted in the power engi­
neering literature. Omitting the subscript eff\v\ equation 11.10 yields
K / c o s (0 ^ -0 .)^ V7cos(0p
where 0^ = 0,^ - 0 y , V= 0.707
( 11.11)
, and / = 0.7071^^ . The angle 0^ is the angle o f the impedance
Z (/co) o f the two-terminal element and is also interpreted as the angle by which the voltage phasor leads the current phasor.
EXA M PLE 11.4. Figure 11.7 shows two t)'pes o f household loads connected in parallel to a 110
V, 60 Hz source,
= 1 1 0 ^ 2 cos(120:tr) V. Lainp 1 and lamp 2 have effective hot resistances
o f 202
(a)
(b)
and 121
respectively. The impedance o f the fluorescent light is Zjfjwi) = 60 + j70
Find the average power consumed by each light.
Find the average power delivered by the source.
Lamp 1
Lamp 2
FIGURK 1 1.7 An example of load current calculation.
So
lu t io n
(a) For lamp 1, Z|(/to) = 202Z.0® Cl. Hence, Ij = Vy^Zj = 0.5446Z .0° A. From equation 11.11,
^\ave ^
z\^ = 110
X 0 .5 4 4 6 cos(0°) = 59.9 watts
5 10
Cliapier 11 • Sinusoidal State State Power Calculations
This means that lamp 1 is a 60 watt bulb.
Similarly, for lamp 2, Z^C/co) = 121Z.0°
Hence, 1, = ^ iJZ j = 0.909 IZ.O^’ A. From equation
11. 11,
= 110 X 0.9091 cos(0°) = 100 watts
= 56 + y'66 = 86.56Z.49.7® O.. Hence,
Finally, for the fluorescent light,
^ jJZ jj =
1.2 7 Z .-4 9 .7 ° A. From equation 11.11,
^Jlave =
^
(b) For this part we first compute
1.27cos(49.7°) = 90.4 watts
and then apply equation 11.11 to compute the average power
delivered by the source. Here by KCL,
= I, + I 2 + I 3 = 0 .5 4 4 6 Z 0 " + 0 .9 0 9 1 ^ 0 ° + 1 .2 7 ^ -4 9 .7 °
= 2.2759 -y 0 .9 6 9 0 = 2 .4 7 3 6 -^ -2 3 .0 6 ° A
By equation 11.11, the average power delivered by the source is
Pave = |V/,,l|lm|co.s(0, - 0 , ) =
cos(6 , - 0 ,- ) = 110 X 2 .4 7 3 6 cos(23.06^ ) = 250.35 watts
Observe that the sum o f the individual average powers is 250.3 watts, which equals the power
delivered by the source within the accurac)' o f our calculations, where we have rounded our
answers.
4. COMPLEX POWER AND ITS COMPONENTS: AVERAGE,
REACTIVE, AND APPARENT POWERS
Recall the notion of a phasor. When all source excitations are sinusoidal at the same frequency,
voltages and currents in the SSS can be represented by phasors. Our question here is, can the pha­
sor method aid the computation o f power consumption in a circuit? The answer is yes. However,
the formulation will bring out several other concepts o f power associated with the sinusoidal
steady state.
In dc power calculations, the average power consumed by a two-terminal device is the product o f
the voltage and current, assuming the passive sign convention. In SSS, the complex power
absorbed by a two-terminal device, as shown in Figure 11.8, is a complex number defined by the
formula
(1 1 .1 2 )
where I^^ is the com p lex conjugate o f
51 1
Chapter 11 • Sinusoidal State State Power Calculations
eff
o +
Two
Terminal
Veff
Device
o-
FIGURE 11.8 Two-terminal device with phasor voltage and current consistent
witli passive sign convention.
The first useful result o f this definition is that
suppose v(r) =
i(t) =
= Re S = Re
. To see this result,
cos((Of-I-0J.) , which is represented by the phasor
y/l cos((or + 6,-) , which is represented by the phasor
Also suppose
The average power consumed by
a rwo-terminal device excited by this voltage-current pair (Figure 11.8) is given by equation 11.10 as
Now obseiA^e that
S = \ //C /f =
j =
h r
in which case
= Re
R e[S]= R e
The curious reader may ask why a conjugate o f the current is used in the definition o f complex
power.
Suppose
Re
one
did
not
have
the
conjugate
o f the
current.
Then
= VeffI(,ffCOs{Q^,+Qj)^I\,y^, i.e., the resulting product would have no physical
meaning. Now because S is a complex number, it has an imaginary part, a magnitude, and an
angle. T he imaginary part o f S defines a quantity called the reactive power absorbed by the twoterminal device in Figure 1.18; i.e., reactivepoioer \s defined as
Q = Im [S]=
J=
sin(0 ,. - 0,) = reactive power
^
14)
The unit o f reactive power, Q, is VAR, which stands for volt-amp-reactive. It follows immediately
that
where P =
. Also, the magnitude o f S is defined as the apparent power absorbed by the two-
terminal device o f Figure 1.18, i.e.,
________
\^\ = Kff^eff = \P ^ JQ\ = 4 p ~ ^Q~ = apparent poNver
Chapter 11 • Sinusoidal State State Power Calculations
^12
The Linir o f apparenr power is VA, short For volt-arnp. The iinerrclationship o f these different pow­
ers is illustrated by the right triangle diagram in Figure 11.9, which is often helpful in solving
problems. Observe that the apparent power is always greater than or equal to the average poiuer,
with equality applying to the case o f a purely resistive load.
FIGURE 11.9 Relationships among complex, average, reactive, and apparent powers.
I'h e distinction among these various powers is best understood by computing the powers for some
basic circuit elements. For simplicit)', except when needed or for emphasis, from this point on we
will often drop the subscripts <^’and rfw'as given in equations 1 1 .1 3 -1 1 .1 6 .
EXA M PLE 11.5. This example explores the computation o f the various powers for a simple
inductor. Given that 1^0 ) = v2/sin(CO/) in the circuit of Figure 11.10, compute
V^, S^,
Ql, the instantaneous absorbed power />/(/), and the instantaneous stored energ)' Wjit) in terms
o f L, CO, /^, and Vj. After this show that
(i)
max
Ql
(ii)
Wiit) max
Ql
(0
Remainder
of Circuit
FIG U R E. 1 1 . 1 0 Isolation o f an inductor for investigating the concept o f com plex power.
Chapter 11 • Sinusoidal State State Power Calculations
So
“SI 3
lu t io n
= -jl^,
= /oZ.1^ = cdZ./^ = V^.
jVJi^ = I\ + yQ^. This implies that
= 0 and (X = V Ji- Further, the instantaneous
absorbed power is Pi ( f ) = v
=
COLy/lIi^cosUor)x J l l i sin(o)/) = V^// sin(2(0/), which
By inspection, and noting that wc again presume effective values,
= V^I
,„a\ ~
is consistent with equation 11.4. It follows immediately that
~ Ql •Further,
W' l it ) = i ) . 5 L i l u ) = L I I sin -((o ;) = 0.5 L/; [l - cos(2w/)]
9
[l-c o s (2 c o / )]
[l-c o s (2 o )/ )]
-------------- ^l ‘ l -------- z-----------=
2co
2(0
[ l - c o s ( 2 (0 /)]
----------2(0
(f)
Since the bracketed quantity varies between 0 and 2,
\Qi\
0)
, as was to be shown.
EXA M PLE 11.6. This example, like the previous one, investigates the concept o f reactive power,
but in the case o f a capacitor. The calculations will all be dual to those o f Example 11.5. Hence,
given that ^ ^ ( 0 = >/2V(-sin((0/) in the circuit o f Figure 11.11, compute
instantaneous absorbed power
S^^
and the instantaneous stored energ)^
'r»
Qq the
terms o f C,
(O ,
Vq and Ir-. After this show that
(i)
P c (0
(ii)
lV c (0
Qc
and
Qc
(0
Remainder
of Circuit
FIGURE 11.11 Isolation of a capacitor for investigating the conccpt of complex power.
So
lu t io n
^C~
By inspection, and noting that we again presume efteaive values,
implies that P r= 0 and Q_(^ = —
S^=
absorbed power is /;^(/) =
which
is
consistent
with
~
Further, the instantaneous
)/(-(/) = (oC>/2\^(7 cos((0/) X V 2 V^7 sin((0/) =
sin(2(0/),
equation
11.4.
It
follows
immediately
that
^
= | 0 c L Funher,
\V^(i) = 0.5Cv^(/) = CVc sin“ (03/) = 0.5CVf^ [l - cos(2(0/)]
^ ,,2 [l-c o s (2 (o / )]
=
(oCVr
------------------ 2w
,, ,
= Vr^c
[l-c o s (2 (o / )]
-----------------2(0
^
= Qc
[ l - c o s ( 2 (0 /)]
-----------------2(0
Chapter 11 • Sinusoidal State State Power Calculations
^\A
Qc
Since the bracketed quantity varies between 0 and 2, W(-(/)
These quantities,
’ I
p
r
(0
o
v
e
useful for identify­
ing energy storage values in inductors and capacitors in systems where energy is to be recovered
and stored, and for modifying the power factor (to be discussed shortly) in networks with motors.
Energy storage in systems and power management are important research topics in todays world.
In Examples 11.5 and 11.6, one observes that the inductor absorbs reactive power while the capac­
itor absorbs negative reactive power or, equivalently, delivers reactive power. This follows from the
definition o f complex power (equation 11.13, i.e., S =
T he structure o f equation 11.13
derives from the convention that whenever the phasor current lags the phasor voltage (as with the
inductor), the device is considered to absorb reactive power, whereas if the current phasor leads
the voltage phasor (as with the capacitor), the devicc is considered to deliver reactive power.
Indeed, the overwhelming majority o f loads (toasters, ovens, hair dryers, motors, transformers,
TV s, etc.) have lagging currents.
When a t%vo-terminal element absorbs an average power
, there is a transformation o f elec­
trical energ)' into other forms o f energ}'— for example, heat or kinetic energy. In contrast, when a
two-terminal element absorbs reactive power Q, no energy is expended. T he energy transferred
into the two-terminal element is merely stored and later returned to the surrounding network. To
distinguish it from real (expended) power, we use VAR (volt-ampere-reactive) instead o f watt as
the unit for the reactive power Q.
EXA M PLE 11.7. This example investigates the computation o f the various powers defined above
for an /?Ccircuit. Here, consider the circuit o f Figure 11.12, where v,„(/) = 100 V 2 c o s (200071:/)
V. Find the complex, average, reactive, and apparent powers absorbed by the load.
o
100 Q
lO kO
16nF i
FIGURE 11.12 Simple RC circuit for investigating aspects of complex power.
S o lu tio n
S te p 1 .
Compute
Z,„ry20007Cj = 100
j -I- 720007T X 16x10
10
-9
Chapter 11 • Sinusoidal State State Power Calculations
Step 2. Compute
Converting
SI^
to a phasor, we have
I,v, = ^
= 10 + j9 .8 5
= 100 V. By O hm s law,
mA
^in
Step 3. Cornpute the complex power absorbed by the load. By equation 11.12,
S =
= 100(10 - y9.85)10"-'‘ = 1 - jO .985 VA
Step 4. Given the complex power, the average power is
The reactive power is
Q = I m [ S ] = - 0 .9 8 5 VAR
and the apparent power is
|S| = 1.404 VA
Before doing a more complex example, we will discuss the particulars o f the principle o f conser­
vation o f power in the sinusoidal steady state.
5. CONSERVATION OF COMPLEX POWER IN THE
SINUSOIDAL STEADY STATE
Basics and Examples
The basic principle o f power conservation is that instantaneous power is conserved.
GENERAL PRINCIPLE OF CONSERVATION OF POWER
In all circuits, linear or not, instantaneous power is conserved; i.e., the sum o f the absorbed
powers o f all the elements in a circuit is zero. If one thinks o f sources as generating power and
other elements as absorbing power, then we can rephrase this statement as “the sum o f gen­
erated powers equals the sum o f absorbed powers.”
The validity o f this principle follows from KVL and KCL. This principle leads to the particular
fact that complex power is conserved in ac circuits operating in the SSS.
PRINCIPLE OF CONSERVATION OF COMPLEX POWER IN AC CIRCUITS
In ac circuits operating in the SSS, complex power is conserved; i.e., the sum o f the absorbed
complex powers o f all the elements (operating in the steady state) in a circuit is zero.
Consequently, average power is conserved and reactive power is conserved.
Note however, that the conservation principle does not hold for apparent power, i.e., for the mag­
nitude o f the complex power. The following example illustrates a basic use o f the conservation law.
Chapter 11 • Sinusoidal State State Power Calculations
S I6
EXA M PLE 1 1.8. This example illustrates the application o f the principle o f conservation o f com­
plex power in determining power delivered by a source and the input current to a circuit. We also
show that conser\'ation o f apparent power does not hold. Consider the circuit o f Figure 11.13.
Find the power delivered by the source and the phasor input current,
given that S j = 360 +
;1 6 0 VA, S 2 = 360 - ; 1 2 0 VA, S 3 = 420 + y540 VA, S 4 = 130 + >80 VA, S 5 = 40 - ; 1 0 0 VA.
/
/
lOOV
V- \
S.
\ A
\
FIGURE 11.13 Bridge circuit where S- represents the complex power absorbed by the element.
S o lu tio n
By the principle o f conservation of power in ac circuits,
= S , + S , + S 3 + S 4 + S 5 = 1310 + j5 6 0 VA
This means that the circuit absorbs 1310 w'atts o f average power; the reactive power is 560 VAR,
and the apparent power is 1425 VA. Notice that the large component o f reactive power makes the
apparent (consumed) power larger than the actual consumed power, P
.
To compute I , recall that
^so u rce
Hence,
= 'OOI/,, = 1 3 1 0 + p 6 QVA
= 13.1 - J5.6 A.
Exercise. Repeat the above example calculations for S j = 300 + y'400 VA, S 2 = 300 - y 4 0 0 VA, S 3
= 600 + j\ 000 VA, S^ = 60 + y’80 VA, S^ = 120 -> 1 6 0 VA. What are the average and reactive pow­
ers delivered by the source?
AN SW FR: S^
= 1380 + /920 VA :uid
= 13.8 + / ).2 A. 1380
w ;itts
and 020 VAR
The next example illustrates the computation o f various powers through basic definitions and
application o f the principle o f conservation o f power.
EXAM PLE 11.9. Consider the circuit of Figure 11. 14, which depicts a motor connected to a commer­
cial pow'er source. The motor absorbs 50 kW of average power and 37.5 kVAR o f reactive power, and has
a terminal voltage
= 230 V. Find IIj, the complex power delivered by the source, S^, and IV^I.
Chapter 11 • Sinusoidal State State Power Calculations
R,line = 0 .5 0
FICfURE 11.14 Motor absorbing 50 kW and 37.5 kVAR at a terminal voltage o f 230 V; the value of
^line
exaggerated for pedagogical purposes; electrical code requires that the size of the connecting
wire be large enough that the voltage drop is only a small percentage of the source voltage.
S o l u t io n
Step 1. FiJid the apparent power, |S^^J, absorbed by the motor. Since
S,„ = ^™V-Q™ = 50t/-37.5kV A
it follows that |S,^J = 62.5 kVA.
Step 2. Find |/J. Here, |S,J =
= 230| lj. Hence, [Ij = 2 7 1 .7 4 A.
Step 3. Compute the line loss.
0-5 X 271.72 = 36 .9 2 kW
Step 4. Compute the complex power delivered by the source. From conservation o f power,
S,
= S,„ . s,,.„ = S,„ *
= 5 0 ;3 7 .5 . 3 6 .92 = 8 6 .9 2 *y 3 7 .5 kVA
Step 5. Compute |Vj.
s.
V. =- *
h
In the above example we choose
—
s.
= 3 4 8.4 V
I.
large to illustrate the calculations. In practice a line loss of
36.92 kW for a 50 kW motor operation would not be permitted.
6. POWER FACTOR AND POWER FACTOR CORRECTION
In a resistor, average power is dissipated as heat. In a motor, most ol the average consumed power
is converted to mechanical power, say, to run a fan or a pump, with a much smaller portion dis­
sipated as heat due to winding resistance and friction. The ratio o f the average power to the appar­
ent power is called the power factor, denoted by pf, i.e..
pf=
Average Power
P^ye
Apparent Power
|S|
= cos(0,-0,)
(11.25)
Chapter 11 • Sinusoidal State State Power Calculations
SIS
The right-hand portion o f equation 11.25 follows directly from equation 11.13. Equation 11.25
specifies the power factor as cos(6^, — Oy) , i.e., the cosine o f the difference between the angles o f
the voltage phasor V and the current phasor I. Clearly, 0 < pf < 1 . The angle
(11.26)
- 0y) = power factor angle (pfa)
Since cos(x) = cos(-a:), the sign o f (0^^- 0^) is lost when only the pf is given. In order to carry the
relative phase angle information along, the common terminology is p f lagging ov p f leading. A l/igging power factor occurs when the current phasor lags the voltage phasor, i.e., 0 < (0^, - 0y) < 180® .
A leaditjg power factor occurs when the current phasor leads the voltage phasor, i.e., 0 < (0y - 0^j <
180® . Practically all t)^pes o f electrical apparatus have lagging power factors. Some typical power
factor values are listed in Table 11.1.
TABLK 11.1. Power Factors for Common Electrical Apparatus
T
ype o f
L oad
P o w e r F a c t o r ( L a g g in g )
Incandescent lighting
1.0
Fluorescent lighting
0 .5 -0 .9 5
Single-phase induction motor, up to 1 hp
0 .5 5 -0 .7 5 , at rated load
Large three-phase induction motor
0 .9 -0 .9 6 , at rated load
To illustrate the idea o f leading and lagging pf, consider the circuits o f Figure 11.16. Suppose the
circuits operate at a frequency o f 400 Flz or co = 2 5 13.3 rad/sec. For the circuit o f Figure 1 1.6a,
I = (1 -jO.995) 10“^ V = 1.41 10“^/.—44.85° V. Hence, the current phasor lags the voltage pha­
sor, i.e., ( 0 J ,- 0y) = 44.8 5 ° and the pf is cos(44.85°) = 0.709 lagging. On the other hand, for the
circuit o f Figure 11.16b, I = (1 + ;2 .5 ) 10-^ V = 2.7 10“3 Z 6 8 .3 ° V. Hence, the current phasor
leads the voltage phasor by 68.3", i.e., (0 — 0^^ = 6 8 .3 ° and the pf is cos(68.3°) = 0.688 leading.
o-
O-
i0 .4 H
IH F
1 kO
1 kQ
O-
(a)
(b)
FIGURE 11.16 (a) A parallel RL circuit illustrating a lagging pf.
(b) A parallel RC circuit illustrating a leading pf
A load with a required average power demand, operating at a fixed voltage with a low pf, say 0.6,
has a relatively high reactive power component. This results in a relatively high apparent power.
Chapter 11 • Sinusoidal State State Power Calculations
SI 9
Since the operating voltage is fixed, the line current needed to drive the load is higher than if the
load operated at a higher pf, say 0.95. Relatively speaking, a higher pf has a lower reactive power
component with correspondingly lower apparent power. Figure 11.9 helps to visualize the rela­
tionships. For fixed line voltage, lower apparent power (higher pO means lower line current and
hence lower power loss in the connecting transmission line. In todays world o f energy conserva­
tion, it is important to be energy efficient. The following example illustrates how improved pf on
a load can reduce line losses and thus decrease cost o f operation.
EXA M PLE 11.10. This example reconsiders Example 11.9, involving a motor connected to a
commercial power source as illustrated in Figure 11.17. The solution process will emphasize the
basic definition o f pf and the use o f voltage and current phasors. Suppose the motor absorbs 50
kW (about 67 hp) o f average power at a pf o f 0.8 lagging. The terminal voltage,
is 230 V. The
frequency o f operation is 60 Hz or co = 120 j1 . For the first part o f the example the capacitor in
Figure 11.17 is not connected to the motor. In part (c), the capacitor is connected to the motor
to improve the p f This will reduce the magnitude o f the current supplied by the source and hence
reduce the line losses.
(a)
(b)
Find the complex power delivered to the motor.
Find I., V^, and the power delivered by the source, which might represent the power
delivered by the local electric company.
(c)
Correct the power factor o f the combined motor-capacitor load to 0.95 lagging by choos­
ing a proper value for C.
(d)
Compute the new power delivered by the source to the combined motor-capacitor load.
R,line =0.5 0
FIG U RE 11.17 Motor absorbing 50 kW and 37.5 VAR at a terminal voltage of 230 V. Again, the
value of
is exaggerated for pedagogical purposes; clectrical code requires that the size of the con­
necting wire be large enough that the voltage drop is only a small percentage of the source voltage.
S o l u t io n
(a) Find the complex power delivered to the motor.
Step 1. Use the p f o f 0.8 lagging and the given average power to fin d the apparent power. From the
definition o f pf,
= 50 kW = R e[S] = |S„,|cos(e -
6
) = |S,„| x p f = |S„,| x 0.8
As such, the apparent pow-er is
50
!s„,
0. 8
= 6 2 .5 k V A = \/„,/,„
( 1 1 .2 7 )
Chapter 1 1 • Sinusoidal State State Power Calculations
S20
where V„, = |V„J = 230 V and /„, =
Step 2. Compute
. Ligging means that current phase lags behind voltage phase, i.e., 0 ^ -
= 0 - 0^ > 0 . Consider the diagram in Figure 11.18, which shows that the current phasor I^ la g s
the voltage phasor, i.e., the current phasor makes an angle o f - 3 6 .8 7 “ = cos” ' (0.8) from the volt­
age phasor. Hence,
= 3 6 .8 7 “ > 0
S m= V IDP m
FIGIJRI- 11.18 Phasor relationship ofV^^,
Step 3. Compute the complex power, S
and
. By definition,
s,,, = | S ,„ k S m =
= 6 2 .2 Z 3 6 .8 7 ° kVA
= 5 0 + ;3 7 .5 k V A = /’„„^t>Q
(b) Fi7ici
, V., and the power delivered by the source.
Step 1. FindX.. From equation 11.27 and the fact that
62.5 X 10'^
V„,
And from Figure 11.18, again since
230
for this part.
= 2 7 1 .7 4 A
, Z.1^ = - 3 6 .8 7 ”. Hence
= 271.74 Z - 3 6 .8 7 " = 271.74 Z .-0 .6 4 3 5 rad = 217.4 - y l 6 3 A
Step 2. FiudY^. From KVL and Ohms law,
= 0.5 [217.4 - j\ 6 3 ]+ 230 = 3 3 8.7 - y 8 1.5
= 3 8 4 .4 Z - 0.2362 rad = 3 8 4 .4 Z - 13.533'" V
Chapter 11 • Sinusoidal State State Power Calculations
s :i
Step 3. Compute the complex power, , delivered by the source.
S^. = V ,I* = 3 4 8 .4 Z - 13.533" x 2 7 1 .7 4 Z 3 6 .8 7 " VA
= 94.6 6 4 Z 2 3 .3 3 7 ° kVA = 9 4 .664 LQA07 rad kVA
= (86.918 + ;3 7 .5 ) kVA
Norc that it takes 86.918 kW to run a 50 kW motor. The difference is the loss in the power line.
If we have a way o f reducing the magnitude o f
this line loss will be reduced. In fact, we do,
and this strateg)^ is the goal o f the next part o f the example.
(c) Correct the power factor o f the combined motor-capacitor load to 0.95 biggifig- Since motors are
inductive, a properly chosen capacitor can improve the pf to 0.95 lagging. The new motor con­
figuration is that o f Figure 11.17, with the capacitor connected across the motor. The proper value
o f C must be found.
Step 1. What does a p f o f 0.95 lagging require in terms ofcomplex power absorbed by the motor-capac-
itor combination'^
= —
Z c o s " l( 0 .9 5 ) = 5 2 .6 3 Z 1 8 .1 9 5 ‘’ = (50 + 7 1 6.4342) kVA
0.95
(11.28)
Recall that
= (50 + > 37.5) kVA
Step 2. Find a capacitor value to reduce the reactive power. For this step consult Figure 11.19.
FIGURU 11.19 Relationships bcrwecn new and old complex powers.
In Figure 11.19, one observes that
is the same for both the new and old complex powers since
that is what the motor requires for its operation. The reactive powers are different. The new com­
plex power with the 0.95 lagging pf has a smaller reactive power component. The capacitor must
be chosen to reduce the old reactive power to this new level. Hence,
jQoU -j^ncw = ;2 1 -0 7 kV.AR = -/reactive power o f capacitor) = -yQ^-
522
Chapter 11 • Sinusoidal State State Power Calculations
Therefore,
jQ c = - j 2 1.07 kVAR = V „,Ic = V,„ [K c(JO ))V ,„]' = -./coC| V,„p
It follows that
C = ^
= ^ H lZ iii4 = ,o 5 7 x lO - ’ F
co|V,„
12071(230)“
(d) Compute the new power delivered by the source.
Step 1. Compute the complex power, denoted
absorbed by the motor-capacitor load. The com-
plex power absorbed by the load is the sum o f the complex power consumed by the motor and
the reactive power of the capacitor, as illustrated in Figure 11.9, i.e.,
S « f'= (50 + y i6 .4 3 ) kVA
Step 2. Compute the new
denoted
Since S ”^“'is the complex power o f the combined motor-
capacitor load,
♦
^
V
m
^ 5 ( W ! M 3 ,0^ = (217 + ; 7 : .43) A
230
J
!
Step 3. Compute . From KVL and Ohm’s law,
V ;^ “’ = 0 . 5 l f + V,„ = (3 3 8 .5 - y 3 5 .7 2 ) V
Step 4. Compute the new complex power delivered by the source. By definition,
S’r ' = v f ' ( i ; '" ’’ )* = (7 6 - ; i 6 .4 8 ) kVA
Hence, the new average power delivered by the source is 76 kW with pf correction as opposed to
86.9 kW without pf correction. With this pf correction, there is a reduction o f 86.9 - 76 = 10.9
kW o f power loss in the line connecting the source to the load.
Example 11.10 illustrates how adding a parallel capacitor can improve the pf o f a load. The main
motivation for improving the pf was to reduce the power loss in
However, even if
is
negligible, another strong reason exists for improving the load pf. Example 11.11 illustrates how
an improved power factor allows a single generator to run more motors. Example 11.11 will fully
utilize the principle of conservation o f complex power and the two consequences o f equation
11.25.
From equation 11.25 and the fact that S = /^ + jQ, we can express pf directly in terms o f P an d Q
as follows:
.— ^
523
Chapter 11 • Sinusoidal State State Power Calculations
with a lagging pf for Q > 0 and a leading pf for Q < 0. Solving for Q from equation 11.29, we
obtain
e=
± P .H r - i
P f‘
(11.30)
with Q > 0 if pf is lagging and Q < 0 if pf is leading.
W ith these formulas we can simplify the process o f power factor correction.
E X A M P L E 11 .1 1 . An industrial plant has a 100 kVA, 230 V generator that supplies power to one
large motor and several identical smaller motors. The resistance o f the connecting line is assumed
negligible in the approximate analysis below'. The large motor, labeled t^-^pe A, draws 50 kW at a
pf o f 0.8 lagging. Each smaller motor, o f type B, draws 5 kW at a p f o f 0.7 lagging. The configu­
ration is illustrated in Figure 11.20.
Generator
230 V ^
60 Hz
lOOkVA
Type B Motor
Type A Motor
Type B Motor
FIGURH I 1.20 A generator supplying power to one large motor and several smaller motors.
(a)
(b)
(c)
(d)
Can the generator safely supply power to one large motor and three small motors? What
are the generator current (magnitude) and the power factor o f the combined loads?
Compute the number o f small motors (besides the one large motor) that can be run
simultaneously without exceeding
If the power factor for all motors,
ing appropriate parallel capacitors
(besides the one large motor) can
the generators rating.
large or small, is corrected to 0.9 lagging by connect­
(as done in Example 11.10), how many small motors
be run simultaneously without exceeding the genera­
tors rating?
Compute the capacitances required in part (c) for the large and the small motors.
S o lu tio n
(a) Compute the reactive power for each motor type. Using equation 11.30, the reactive power for
each type o f motor is given as
- 1
1
=50
-1
= 37.5 kVA
0.8^
and
Q b = P ,i, / A - >
y
pfg
= \
/
V
-^ 2 -
0.7
i
= .‘i . l 0 1 k V A
‘’ ■2 I
Chapter 11 • Sinusoidal State State Power Calculations
By the principle ot conservation o f power, the complex power (in kVA) supplied by the generator
is
V
^ 3(7^^ + ;Q ^ ) = (50 . 15) + ;(3 7 .5 -h 15.303)
=
= 65 + ;5 2 .8 = 8 3 .7 4 ^ 3 9 .
kVA
By inspection, the apparent power is 83.74 kVA, which is below the generator capacity o f 100
kVA, meaning that the generator can safely operate the large motor and three smaller motors.
The magnitude o f the generator current is 83,740/230 = 364 A. From equation 11.29, the pf o f
the combined loads is
pf =
.
^
P- +
= 0 .7 7 6 2
\/65“ + 5 2 . 8 “
(b) Compute the number o f stnall motors {besides the one large motor) that can be run simultaneous­
ly. When one large type A motor and n smaller t)^pe B motors are connected in parallel, the com­
plex power delivered by the generator is
^gen
7 ^
= ^^0 + « X 5) + 7(37.5 + ;/ X 5.101) kVA
rhe apparent power is
^ (50 + n x 5 ) ~ + (3 7 .5 + /; x 5 .1 0 1 )" kVA
Since the generator has a capacity ot 100 kVA, then
c
‘ = ( 50+//x 5 ) “ + (37.5+/; X 5.101)- <
100“ =
10^
(11.31)
Replacing the inequalit)' sign in equation 11.31 by an equalit)^ results in the quadratic equation
51.020/r + 882.5750/7 - 6 ,0 9 3 .8 = 0
The resulting zeros are //, = 5.288 and //2 = - 22.58. The largest positive integer satisfying the
inequality 11.31 is // = 5. Thus, at most, five small motors can be run simultaneously with the
large motor without exceeding the generators capacit)'
(c) I f all power factors are corrected to 0.9 lagging, fin d the number o f small motors {besides the one
large motor) that can be run simultaneously. We essentially repeat the calculations o f part (b) with
the new given power factor o f 0.9 lagging:
and
e :r = / '.. A
V PfA
i
=5o
y - 1 = 2 4 . 2 1 6 kVA
V 0 .9 ^
Chapter 11 • Sinusoidal State State Power Calculations
The complex power (in kVA) supplied by the generator is
= Pa + j Q T + n(Pfl + j Q T ) = (50 + « X 5 ) + j(24.216 + /, x 2.416)
The apparent power (in kVA) is
= s j (50 + /2 X 5) “ + (24.2 \6 + n x 2.4216)“
s '" "
ficn
As before, n satisfies the inequaiit)'
(50 + « X 5)2 + (24.216 + « x 2.4216)2 < 10^
To find n, we compute the largest positive root o f the quadratic equation
30.8641;/2 + 6 l7 .2 8 2 9 n - 6 ,9 1 3 .6 = 0
The roots o f this quadratic are
= 8 and «2 = —28. The largest positive integer that satisfies the
above inequalit}' is n = 8. Thus, eight small motors, as opposed to five in the earlier case, can be
run simultaneously with the large motor without exceeding the generators capacity.
(d) For the large motor, the capacitor must absorb a negative reactive power equal to
Q^.
Equivalently, the capacitor must supply a reactive power equal to
Qw - Q T =
- 24.216) X 1000 = 13284 VAR
(10.32)
From Example 11.6, the reactive power supplied by a capacitor is
I Q-CA I = ^CA^CA =
6 0 Q X 2302 VAR
(10.33)
Equating equations 10.32 and 10.33, we have In x 60C^|2302 = 13284. Solving produces
=
66 6 .1 6 X 10-<^ E
Similarly, for the smaller motors,
Q-B - Q"b ' = (5.101 - 2.4216) X 1000 = 2679.4 VAR
Also, we have
Qcb\ ~Vcb^Cb\ -
= In
X60Q
X 2302 VAR
Equating these two quantities and solving for C^, w'e obtain Cg = 134.35 x 10“^ F. We note that
in the power industry, such capacitors are usually specified only by their kVAR rating, with no
mention o f their actual capacitive value in F.
In the above example the generator capacit}^ was given in terms o f VA, the unit o f apparent power.
The example points out the importance o f reducing reactive power to more fully utilize the power
526
Chapter 11 • Sinusoidal State State Power Calculations
capacity o f the generator. Use o f VA for generator, motor, and transformer capacit)' arises out o f
safety considerations. Most ac machiner)' operates at a specified voltage depending on the insula­
tion strength. The size o f the wire and other heat transfer factors determine the maximum allow­
able current o f a machine or transformer. Also, the cost and physical size o f most ac equipment
are more closely aligned to the VA rating than to other measures. Hence, the VA rating better
reflects the safe operating capacity o f ac equipment.
Another motivation for improving the power factor is economical. A power company charges a
consumer only for the actual electrical energy used. A meter measures this energy usage in units
o f kWh (kilowatt-hour). As mentioned earlier, most clectrical loads have lagging currents. As
shown in Examples 11.10 and 11.11, for a given required average power, a higher pf means lower
transmission line losses. Also, loads that operate at low pf force power companies to pursue high­
er kVA ratings o f the generator equipment. Thus utilities companies encourage consumers to
operate their equipment and appliances at high pfs. Since power companies can supply more
power with the same equipment if the pf is high, they adjust their rates so that energy costs are
less with a high pf and are greater with a low pf.
7. MAXIMUM POWER TRANSFER IN THE SINUSOIDAL
STEADY STATE
Chapter 6 outlined the basics o f maximum power transfer for linear resistive networks. Having
introduced energ)' storage elements L and C, and having studied methods for sinusoidal steady
analysis, it is time to extend the results on maximum power transfer to general linear networks in
the sinusoidal steady state.
MAXIMUM POWER TRANSFER THEOREM FOR AC CIRCUITS
Let a practical ac source be represented by an independent voltage source V^ {voltage phasor in rnis
value) in series with an irnpedance
+ jX^. An adjustable load impedance
=
jX^j
with R l > 0, is connected to the source {Figure 11.21). In steady state, for fixed Z^, Vp and O),
the average power delivered to the load is maximum when Z[^ is the complex conjugate ofZ^, i.e.,
(11.33a)
and
(11.33b)
and the maximum average power is given by
p
^ <e ff
max
(11.33c)
>2.
Chapter 11 • Sinusoidal State State Power Calculations
To derive the conditions o f the maximum power transfer theorem, observe that the current phasor, I, is
(11.34)
1=
{R, + RO + j(X , + XO
Thus the average power delivered to the load is
(11.35)
Here Pave
a hmction o f two real variables
and Xj . To find the conditions for maximum
the partial derivadves ------ and ------ to zero and solve for
set
and X^. Differendating equation 11.35
with respect to R^ yields
V f [ ( R , + R i f + {X, + X 0 - - 2R l (R, + R , , )
dRi
{R^ + Rl)~ +
+
=
-)12
0
(1 1 .36a)
and differentiating with respect to Xj produces
dP
ax,
v "-[-1R l {X,
+ Xl )
= 0
From equation 1 1.36b, the only physically meaningful solution is
(11.36b)
528
Chapter 11 • Sinusoidal State State Power Calculations
which is equation 11.33b. Substituting this result into the numerator of equation 11.36a yields
v3 “ ^
(R s + Rl T
The only physically meaningful solution here is
^ 0 and
^
R ,-R ,
which produces equation 11.33a. (Note that this is the condition for maximum power transfer in
purely resistive circuits.) Substituting these results into equation 11.35 produces equation 11.33c,
Ir
_
4/?,
which verifies the theorem.
The theorem can be established less formally as follows. W ith any existing
connected to the
source, if the total reactance
+X^) is not zero, we can always increase the magnitude of the
current, and hence the power delivered to the load, by “tuning out” the reactance, i.e., by adjust­
ing
to be -X ^. This implies condition 11.33b. Under such a condition, the circuit becomes
resistive, and the maximum power transfer theorem of Chapter 6 may be applied to obtain equations 11.33a and c. The maximum power obtainable with a passive load, given by equation
11.33c, is called the available power of the fixed source.
The conditions for maximum power transfer, as given by equation 11.33, are valid when both
and Xj^ are adjustable. If X^ is fixed and only
is adjustable, then the condition for maximum
power transfer is
=
+
(11.37)
which is obtained by solving equation 11.36a for fixed X^ and X^ .
n
r\
If the source is a general two-terminal linear network, then its Thevenin equivalent must be found
before application of the maximum power transfer theorem. If the source is represented by a
Norton equivalent circuit, we can use a source transformation to obtain the Thevenin form and
then apply equations 11.33.
'
As pointed out in Chapter 6, maximum power transfer is not the objective in electric power systems, as the sources usually have very low impedances. On the other hand, it is a very important
factor to be considered in the design of many communication circuits, as illustrated in the following example.
o
S29
Chapter 11 • Sinusoidal State State Power Calculations
EXA M PLE 11 .1 2 . The radio receiver shown in Figure 11.22a is connccted ro an antenna. The
antenna intercepts the electromagnetic waves from a broadcast station operating at 1 MHz. For
circuit analysis purposes, the antenna is represented by the Thevenin equivalent circuit shown in
Figure 11.22b.
(a)
Find the input impedance
o f the receiver if maximum power is to be transferred
(b)
from the antenna to the receiver.
Under the condition o f part (a), find the magnitude of the voltage across the receiver ter­
minals, and the average power delivered to the receiver.
receiver input
antenna equivalent circuit
equivalent circuit
(b)
(a)
FIGURE 11.22 Example of maximum power transfer.
S o l u t io n
= 1\ i l and
(a)
From the maximum power transfer theorem, the answers are
(b)
Since the reactances in the circuit have been “tuned out,” the input current to the receiv­
er is simply 14.6/(21 + 21) = 0.348 mA. The input impedance has a magnitude
7.
= 1070.
+ 1070- = 10 7 0 .2 n
Therefore the magnitude o f the voltage across the receiver terminals is 0.348 x 1070.2 = 3 7 2.4
mV (rms). The power transferred from the antenna to the receiver is 0.348^ x 21 = 2.54 uW.
In the preceding discussions of maximum power transfer, we have assumed that the load is
adjustable. In practice the load is often fixed, as for example, in the case o f a loudspeaker having
a4
voice coil. In such cases, one designs coupling networks consisting o f lossless passive com ­
ponents. These coupling networks transform the fixed load impedance into one whose conjugate
matches the fixed source impedance. This permits maximum power transfer to the load. The fol­
lowing example illustrates the principle. A design procedure for some simple coupling networks
will be discussed in the second volume o f this text.
530
Chapter 11 • Sinusoidal State State Power Calculations
EXA M PLE 11.13.
A fixed load resistance /?^ = 100
representing the input resistance o f an amplifier, is connect­
ed to the source o f Example 11.12 through a passive coupling network, i.e., a network that does
not generate average power, as shown in Figure 11.23.
(a)
Show that the maximum voltage that can be developed across
(b)
Show that the coupling network shown in Figure 11.23 achieves this maximum voltage
across
is 0 .504 V.
.
FIGURF, 11.23. Maximum power transfer through a coupling network.
S o l u t io n
(a)
From equation 11.33c, as used in Example 11.12, the available power from the source is
2.54 j.iW. If all o f the power is delivered to R^ , then the voltage,
must be
= ylPmaxf^L = ^ 2 .5 4 X 10“^ X 100000 = 0 .5 0 4 V
(b)
The input impedance
o f the coupling network with load must be the conjugate o f
the source impedance. Specifically
^in = yw ^ + ------------- r~
j(dC +
R,
Substituting the values OJ = 10^’, L = 400.9 x 10“^, C = 109.8 x 10
and /? = 100 x 10^ into the
above expression yields
= 21 + ;1 0 7 0 Q
which is indeed the conjugate o f the source impedance. Since
is conjugate-matched to the
source impedance, the maximum power o f 2.54 |aW is transferred to the coupling network. Since
the coupling network consists o f L and C, neither o f which consumes average power, the 2.54 f,iW
power must be transferred out o f the coupling network and into the load resistance. The voltage
across the load resistor, Vj, is given by
= 4 ^
= V 2 .5 4 X 10“^ X 100000 = 0 .5 0 4 V
This verifies that the coupling network o f Figure 11.23 enables the largest voltage to appear across
the load resistor.
Chapter 11 • Sinusoidal State State Power Calculations
531
8. SUMMARY
Fundamental to the material in this chapter is the definition o f the effective value (rms value) o f
a periodic voltage or current waveform. For a sine wave, the effective value is the maximum value
divided by -v/2 • For a general periodic voltage or current, the effective value is the value o f a dc
waveform that will produce the same amount o f heat as the periodic waveform when applied to
the same resistance. Using the definition o f the effective value o f a waveform, formulas for the
average power absorbed by a linear two-terminal network in ac steady state were set forth and
derived. Recall that for a two-termlnal element with sinusoidal voltage v{t) = ^/2 V'^ cos(co/- 0y) an
current i{t) =
0/), the absorbed average power \s P = ^^^^cos(Oj; - G/), assuming
the passive sign convention. Next w'e presented the definition o f complex power and its compo­
nent parts, which include its real part or average power, its imaginary part or reactive power, and
its magnitude or apparent pow’er. Various examples illustrating the calculation o f these powers
were given. Again, for a two-terminal element w'ith sinusoidal voltage v{t) and current i{t) as
above, the reactive power absorbed is defined to be Q = ^ ^ ^ s i n ( 0 y - 0/) VAR (volt-amperereactive). After introducing these different types o f power, we proved the principle o f conserva­
tion o f complex power, which implies the consen^ation o f real power and the conservation o f reac­
tive power. This was followed by the definition o f power factor, pf, the ratio o f average power to
apparent power, which takes on values between 0 and 1. The need for improving a low power fac­
tor and a method for achieving an improved power factor were illustrated with two examples.
The maximum power transfer theorem, first studied in Chapter 6 for the resistive nervvork case,
was taken up again in this chapter for the sinusoidal steady-state case. Here, maximum power
transfer to the load requires that the load impedance be the conjugate o f theThevenin impedance
seen by the load. As pointed out earlier, the theorem has no application in electrical power sys­
tems. However, for communication circuits the maximum power transfer theorem is o f extreme
importance. The power that can be extracted from the antenna o f a radio receiver is usually in the
microwatt range, a very small value. It is therefore necessary to get as much power as possible from
the antenna system. Example 11.13 illustrates this principle.
9. TERMS AND CONCEPTS
Apparent power: the apparent power absorbed by a two-terminal element is
assuming the
use o f a passive sign convention . The unit is VA (volt-ampere).
Average power: the average value o f the instantaneous power. For a two-terminal element with
sinusoidal voltage u{t) = yfz V^^cos{o)t + Qi,) and current i{t) = ^
V^jj-cos{o)t + 0/), the
absorbed average power is P = ^ ^ ^ ^ c o s (0 y - 0/), assuming a passive sign convention.
Com plex power: for a two-terminal element absorbing average power P and reactive power Q,
the complex power is defined to be S = P + jQ. The unit o f measuremcnr is VA (voltampere). The magnitude of S is the apparent power.
Conservation o f powers: for any network, the sum o f the instantaneous powers absorbed by all
elements is zero. For any linear network in sinusoidal steady state, the sum o f the aver­
age powers, reactive powers, or complex powers absorbed by all elements Is zero. This
property is a consequence of KCL and KVL.
532
Chapter 11 * Sinusoidal State State Power Calculations
Efifiecdve value (nns value): for a sine wave, the effective value is the maximum value divided by
V 2 •For a general periodic voltage or current, the effective value is the value of a dc wave­
form that will produce the same amount of heat as the periodic waveform when applied
to the same resistance.
Instantaneous power: the power associated with a circuit element as a function of time. The
instantaneous power absorbed by a two-terminal element is p{t) = y(/)/(/), assuming that
a passive sign convention is used.
Maximum power transfer theorem: if a variable load 7.^^ = -^j^i is connected to a fixed source
Vy having a source impedance
=7?^+/A^, then the largest average power is transferred
to the load when
is the complex conjugate of Z j , i.e.,
= R^ and AT^ =-X^.
Power factor: the ratio of average power to apparent power. The pf value lies between 0 and 1.
For a passive load, the power factor is said to be lagging when 90° >0j, - 6,->0, and lead­
ing when 90° > 0^- 0^, > 0.
Real powen in ac circuits, real power means average power. It is the real pan of the complex
power.
'
Reactive power: for a two-terminal element with sinusoidal voltage v{t) = -y/2 V^cosisat + 0^)
and current i{t) = ^ /^ co s(co f + 0^, the reactive power absorbed, denoted by Q, is
defined to be Q = ^ ^ ^ sin (0 y - 0/), assuming that a passive sign convention is used.
The unit of measurement is VAR (volt-ampere-reactive).
o
n
o
o
n
n
o
Chapter 11 • Sinusoidal State State Power Calculations
^33
Problems
AV(t)(V)
20
INSTANTANEOUS AND
AVERAGE POWERS
t(sec)
1. For the source current waveform o f Figure
PI 1.1a, which drives the circuit o f Figure
-
10- -
(a)
PI 1.1b, find the average power consumed by
the 2 Q resistor.
(a)
Figure Pi 1.3 (a) Rectangular waveform, (b)
Triangular waveform.
EFFECTIVE VALUE OF
NONSINUSOIDAL SIGNALS
(b)
Figure P l l .l
AN SW FR. 0.758 watrs
4. (a)
Compute the effective value o f each o f
the periodic signals in Figures Pi 1.4a
and b.
2. Compute the average power delivered to a 1
(b)
kQ resistor by a current o f the form
(a)
1 0co s(10t)m A
(b)
10 |cos(10^)| niA
(c)
10 cos“( 1Of) mA
(d)
Plot each o f the instantaneous powers
for 0 < r < 1 sec using MATLAB or its
equivalent.
3. (a)
Compute the average power absorbed
by a 10
resistor whose voltage is
given by each o f the waveforms in
(b)
Using MATLAB, plot the instanta­
Figure Pi 1.3.
neous power associated with each
waveform for 0 < f < 3 sec.
For the circuit o f Figure P i 1.4c, find
the power absorbed by Rj if the volt­
age source v{t) is given by the wave­
form o f Figure PI 1.4a.
(c)
Repeat part (b) for the waveform o f
Figure PI 1.4b.
534
Chapter 11 • Sinusoidal State State Power Calculations
i(t)0
60 O <
= 30 Q
(c)
Figure PI 1.5
6. (a)
Find the effective value o f the source
current plotted in Figure PI 1.1a.
2Q
v(t)
(b)
6
R = 80
part (a).
Figure P11.4
7. Compute the effective value o f
(a)
Compute the effective value o f each o f
(b)
the periodic signals in Figures P11.5a
(c)
and b.
(b)
resistor in Figure PI 1.1b
using the effective value computed in
(0
5. (a)
Find the average power absorbed by
the 2
v^{t) = 10 + 2 cos(20r)
= 10 cos(2r) + 5 cos(4/)
= 10cos(2r) + 5 cos(4r) +
5 c o s ( 4 f - 4 5 ° ) V.
For the circuit o f Figure PI 1.5c, find
the power absorbed by
if the cur­
rent source i{t) is given by the wave­
form o f Figure Pi 1.5a.
(c)
Repeat part (b) for the waveform o f
Figure PI 1.5b.
Ai(t)
AVERAGE POWER
CALCULATIONS IN
SINUSOIDAL STEADY STATE
8. Using equation
44
11.10, find the average
power absorbed by the resistor in the circuit
3
shown in Figure PI 1.8, where
sin(5r) V . /? = 25 Q, C = 8 mF.
12
15
R
(a)
A
t(sec)
-4 -■
(b)
Figure PI 1.8
ANSWl-.R: 25 watts
= 50 x
Chapter 11 • Sinusoidal State State Power Calculations
9. In the circuit shown in Figure PI 1.9, i-{t) =
(b)
5 cos(30/) A, /? = 5 Q, and C = 5 niF.
(a)
Find v^it).
(b)
Compute the average power delivered
by each source.
(c)
Repeat parts (a) and (b) for R = 50 Q,
Zi = p O Q, and a = 49.
Using equation 11.10, find the instan­
taneous and average power absorbed
by the load.
/Y Y V
— ►
Load
' ■6
aV
Figure 1’ 11.1 I
ANSW'FRS: (b) in random order: - 9 0 . 100
watts; (c) 4. - 3 .9 watts
Figure P l l .9
AN SW ER: (b) 40 warts
12.
10.
= 120
R = 32 W, Z^ = ;2 0 0 Q.
Z q = y’80 ri, and ^ = 4.
Consider the circuit o f Figure PI 1.12,
where
In the circuit o f Figure PI 1,10,
5 0 z i-9 0 °
/? = 6 Q,
= ;1 2 Q, and
=
(a)
=
- ;4 Q.
(a)
Find the phasor current
(b)
and deter­
Find the average power delivered by
each source.
mine its magnitude.
(b)
Find the average power (in watts)
absorbed by the resistor.
Using equation 11.10, find the aver­
/Y Y V
z. -
age power (in watts) delivered by the
source.
(c)
O nly
R absorbs
average
power.
' •6
III is
Therefore, once
known, the
average power consumed by R is
=
y?|
(d)
bl
Figure PI 1.12
Check your answer to part (b)
using this formula.
ANSWF.RS: (b) in random order: S7.6, 230.4
Repeat parts (a) and (b) when 7? = 30
Q, Z^ = y'50 Q, and Z q = -ylO Q..
watts
13. A coil is modeled by a series connection o f
/Y Y V
L and R. When connected to a 110 V 60 Flz
R
source, the coil absorbs 300 watts o f average
pow'er. II a 10 Q resistor is connected in series
with the coil and the combination is connected
to a 220 V 60 Hz source, the coil also absorbs
300 watts o f average power. Find L and R.
AN SW FR: R = 0.9901 LI and L = 16.6 mH.
Figure Pi 1.10
A N SW FR; (b) ISO wans
11. For the circuit in Figure P i 1.11,
=
100.10° V ^ , /? = 5 n , Z^ = ;5 0 a , and ^ = 9.
(a)
Compute the current phasor I^.
14.
Consider
Figure
Pi 1.14,
where
V,•„(/) = 2 2 0 >/2cos( 120jir) V and R^^-^
=3
The inductance L is adjusted so
that IV
150 V„,„ a n d
watts. Find the magnitude o f
250
the
Chapter 11 • Sinusoidal State State Power Calculations
^36
reactance o f the coil, and the values o f L
and R.
v.(t)
Figure P 11.17
C H EC K : Complex power is 7+y’6 VA
Figure P 11.14
18. In the circuit shown in Figure Pi 1.18, to =
ANSW ER: /. = i2.9 niH, R - 14.SS3 Q
= 4 £2,
15. Repeat Problem 14 for
400 watts, and
64 rad/sec,
=
nns with all other values the
= 120/ i60"
= 20 £2, /?,
= 4 12, /., = 0.375 H, and L , = (^-125 H. Find
the complex and average powers absorbed by
the load.
same.
Load
C H EC K : L = 7.958 mH, R = 17.794 £2
—TY"YA--- Q
R.
L.
+
COMPLEX POWER
CALCULATIONS
16. For the circuit o f Figure PI 1.16, R = 5
= 3 .jA a ,
= -yiO
and
v,„(r) = 100v 2 cos(1207i:/) V.
(a)
Find the complex power absorbed by
(b)
Compute the apparent power in VA, the
Figure P 11.18
C H E C K : Complex power is 36 + p 2 VA
the load.
average power (in watts), and the reactive
19. In the circuit shown in Figure Pi 1.19, Z| =
power in VAR delivered to the load.
1
Zc
{-
+7a, z, =4 +ji2 a, z,^ 2+j i a, v, =
(104 +y50)~V, and
(a)
= (106 + y48) V at 60 Hz.
Find the voltage V , in polar and rec­
tangular forms.
C H EC K : ¥ , = 1 0 0 + ?
'-«’0
(b)
Find the complex power absorbed by
each of the three impedances and then
the
Figure P 11.16
power delivered
C H EC K : (b) 300 watts, 400 VAR
sources.
17. In the circuit o f Figure PI 1.17,
v ,(r ) = IOOV2 cos(500/ + 30^) V, /?, = 100 Q,
Z 2 is 100 + ;5 5 0 VA
by the
two
C H EC K : complex power absorbed by
R, = 700 £2, and Z. = 1.2 H. Find /^(r), the
complex power, average power, and apparent
power absorbed by the load.
(c)
Using the results o f part (b), verify
conservation o f complex power.
Chapter 11 • Sinusoidal State State Power Calculations
537
(a)
r-
Z.
Z.
power, and the average power deliv­
ered by the source.
O
v .Q
the complex power, the apparent
-I
(b)
Find the source current
in rectangu­
lar and polar form.
Figure Pi 1.19
C H EC K : complex power delivered by source h
Z,
is 10 + 260; VA.
v .Q
CONSERVATION OF POWER
20. This problem should be done without any
phasor voltage or current computations. In the
circuit o f Figure P i 1.20,
= 2300
Figure PI 1.21
C H EC K S: P,
.= 1600 watts, |Ij = 8 .6957 A
at 60
Hz and the following powers (in kVA) are con­
sumed by various impedances and resistances:
S , = 20 + y8 , S , = 20 + 7 I 8 , S 3 = 5+ 76 , and
= 3 + j4.
(a)
Find
(rectangular form) and the
complex
power delivered
by
the
POWER FACTOR AND POWER
FACTOR CORRECTION
22 . (a)
load that absorbs 2 kW o f average
source.
(b)
(c)
Determine
in polar form.
Find the complex power delivered to
(b)
(d)
(e)
Find
in rectangular and polar form.
power with pf = 0.90 lagging.
Find the complex power delivered to a
load that absorbs 4 k\V o f average
the group o f impedances Z j, Z-,, and
^4Find V , in rectangular and polar form.
Find the complex power delivered to a
power with pf = 0.90 leading.
23. For the circuit o f Figure PI 1.23, Z j absorbs
1600 watts at pf = 1, while the apparent power
absorbed by Z^ is 1000 VA at pf = 0.8 lagging,
and V;„(/)= l20V 2cos(1207cr) V. Find
(a)
the phasor current
(b)
the phasor voltage V|
(c)
the phasor voltage
in
Figure PI 1.20
C H EC K S: (a) Complex power delivered by
source: 48 + y'36 kVA; (b) |lj = 26.087 A
v,(t)
21. In the circuit o f Figure PI 1.21, V^. = 230
V.
at 60 Hz. The complex powers (in VA)
absorbed by the five impedances are S| = 100 +
yioo, S , = 200 +7 IOO, S 3 = 100 + p o , S^ = 400
+J250, and S^ = 800 + J700. Find
C H EC K :
- 1 8 .8 2 3 5 ;
Figure P i 1.23
= 20 - 5; A„„^ and V , = 75.2941
538
Chapter 1 1 * Sinusoidal State State Power Calculations
24. The circuit shown in Figure P I 1,24 is in
the sinusoidal steady state. Suppose that
absorbs 3000 W of average power at pf =
0.7905 lagging, Ri-^^ = 0.1 £2, and
= 120
V rms’
(a) Find the average power absorbed in
the transmission line resistance (in
26. The circuit in Figure P I 1.26 operates at CO
= 500 rad/sec and = 100Z0° V ^ . The com­
plex power drawn by the load without the
capacitor attached is S^; = 100Z30® = (86,6 +
y‘50) VA. This constitutes a pf of 0.866 lagging.
(a) Find the values of R and L
(b) Find the value of C in fiF that pro­
duces a pf of 0.95 lagging.
W).
(b)
Find
=A cos(1207t^ + 0) V and
the complex power delivered by the
source.
Load
0 -"0 -
!'4<
Transmission
line resistance
— 'N/S/'------ OR .,„
+
I
L
o - -oFigurePll.26
CHECK; 3 pF < C< 5 pF
Figure P I 1.24
CH ECK: (a) 100 watts
25. As shown in Figure P I 1.25b, a capacitor is
put in parallel with a motor using average
power
= 40 kW operating at a power fector of 0.7 laggmg to boost it to a power factor
of 0.9 lading. The voltage across the parallel
motor-capacitor combination is 230Z0° V ^ .
The power relationships are shown in Figure
P I 1,25a. If the frequency of operation is CO =
120tc, compute the proper value of the capaci­
tance, C (in mF).
27. The circuit shown in Figure P I 1.27 is operat­
ing in the SSS with v^(0 = 120>/2cos(1207cO
V. Device 1 absorbs 360 W with pf =0.9. Device
2 absorbs 1440 W with a pf of 0.866 lading.
Find the value of the capacitor C such that the
magnitude of the source current equals 16 A ^ .
What is the pf of the two-device-plus-capacitor
combination?
Device
1
Device
2
Figure P I 1.27
CH ECK: 0.108 mF
< C<2 mF
(b)
28. A group of induction motors is drawing 7
kW from a 240 V power line at a power faaor
of 0,65 lagging. Assume CO = 12071 rad/sec.
(a) What is the equivalent capacitance of
a capacitor bank needed to raise the
power factor to 0.85 lagging?
(b) What is the kVA radng of the capacitor
bank of part (a); i.e., what is the reactive
power supplied by the capacitor bank?
Chapter 1 1 • Sinusoidal State State Power Calculations
(c)
539
Determine the annual savings from
By what amount in kVAR must
installing the capacitor bank if a
the reactive power be reduced to
demand charge (in addition to the
produce a pf o f 0.94 lagging?
charge for the kilowatt-hours used by
(iii) Compute the needed capacitor
the induction motors) is applied at
S20.00 per kVA per month.
(iv) Compute Z^j(£)) as the ratio o f
AN SW ERS: (a) 0.1771
current
niF; (b) Minimum
the capacitor phasor voltage to
kVA rating: 3.8457 kVA; (c) S608.14
the capacitor phasor current, at
the indicated frequency.
(v) Compute the proper value o f C in
29. Consider a source that drives an electric
mF.
motor that consumes an average power o f 94
kW (about 125 hp) at a pf o f 0.65 lagging, as
show'n in Figure PI 1.29, where
= 0.07 Cl.
(i)
Compute the new I"^ .
(j)
Compute V 'f":
(k)
Compute the complex power deliv­
ered by the source and the new effi­
ciency.
— oM o to r)
—
Later
Addition of
Capacitor
for p.f.
Correction
MAXIMUM POWER
TRANSFER
Figure P11.29
30. Consider the circuit shown in Figure
The phasor voltage across the motor is ^eff~
2 3 0 Z 0 V. The sinusoidal frequency is 60 Hz.
(a)
Find the apparent power delivered to
Find the complex power,
(a)
(c)
Find the value o f the load impedance
that will absorb maximum power
at
delivered
(b)
to the motor.
CO =
delivered to the motor.
Compute I
(e)
(f)
Compute V^.
Compute the complex power deliv­
(g)
ered by the source.
Determine the efficiency o f the con­
= 100
find the average power
absorbed by the load.
u—
f
R.
v .Q
o—
figuration, i.e., the ratio o f average
power delivered to the motor to the
100 rad/sec.
Given the conditions o f part (a) and
Determine the reactive power in VAR
(d)
C=
1 mF.
the motor in kVA.
(b)
= 100 Q, R-> = 25
PI 1.30 in which
Figure PI 1.30
C H EC K : 20 + ;1 0 Q
average power delivered by the source
as a percentage.
(h)
Add a capacitor across the motor to
31. In the circuit o f Figure PI 1.31,
v^(r) = 100V 2cos(1000/) V, /?, = 80 Q, R, =
improve the power factor to 0.98 lag­
20 Q, Z. = 5 mH. Find the value o f the resist­
ging. Then
(i) Compute
ance R^ (in Q) and the capacitance C (in mF)
Compare with
sou
(ii) Recall that the role of the capaci­
tor is to reduce the reactive power.
such that maximum average power is absorbed
by the load.
Chapter 11 • Sinusoidal State State Power Calculations
540
Thevenin equivalent, i.e., a volt­
Load
age source in series with either a
/Y Y V l
L
:
series RC or a series RL as appro­
R,
priate.
(b)
Compute
the
load
impedance,
Z^(/‘150), necessary for maximum
power transfer. Show this load as
Figure P i 1.31
either a series RL circuit or a scries RC
A N S W H R : R, = 16 12 and C= 0.2 mF
circuit. Should it be the opposite of
32. The circuit o f Figure P l l .3 2 operates in the
sinusoidal steady state with W = 1000 rad/sec, R
= 1 k n , C = 1 ;<F, I = 0.5 H,
= 3 and
the case in (a)(iii)? Why?
(c)
Compute the average power con­
sumed by the load at maximum power
=
transfer.
2 ^ 0 ’A ^ (a)
Find the value o f the load imped­
ance
for maximum average
power transfer.
(b)
8Q
.
-L
Find the average power absorbed
A
3.334 mF - J 1.667 mF
16Q
by the load under the conditions
o f part (a).
Figure Pi 1.33
34. Consider the circuit o f Figure P l l . 3 4 ,
w'hich operates at
CO
=10 rad/sec. Suppose R =
l O a , L = 2 H , and /^(O = I 0 V 2 cos(lOr) A.
(a)
AN SW ER:
Q to deliver maximum average power
= 1 + \.5/ k li, 4000 warts
33. The purpose of this problem is to compute
for maximum powder transfer by following a spe­
to the load. What is this maximum
average power?
(b)
(a)
Compute the Thevenin equivalent of
the circuit at terminals A and B:
(i)
Use nodal analysis to compute
Note that there is a floating
If Rj = 30 Q, determine Q for maxi­
mum power transfer to the load. What
cific procedure. Consider the circuit o f Figure
PI 1.33 in which /^.(r) = 50V 2 cos(150/) A.
Choose the proper values o f Rj^ and
is this maximum average power?
(c)
If Q = 8 mF, then determine R^ for
maximum power transfer to the load.
What is this maximum average power?
Load
dependent voltage source.
C H EC K : V^^.= 2 8 .4 7 -791.1 V
(ii) Find the Thevenin equivalent
impedance, Z^y^(/150), seen to the
left o f terminals A and B.
C H EC K : Re[Z,//yi50)] =
8.2846Q
(iii) Show' the phasor form o f the
Thevenin
equivalent
circuit.
Then show the circuit form o f the
Figure Pi 1.34
A N SW ERS: (a) 10 12, 5 ml-, 1230 wans; (b) 5
mF, 937.5 wans
Chapter 11 • Sinusoidal State State Power Calculations
35. In rhe circuit o f Figure PI 1.35,
541
= 100
^rms
adjusted to achieve different
goals. Assume R = GO Q., Z q = - ^ 8 0 Q.
(a)
Find the value o f
that maximizes
(C H EC K : 24 Q .) What is the
value o f
(b)
Find the value o f
that maximizes
Figure PI 1.37
V /.,. W hat is the value o f IV,I
L nutx
(b)
Now suppose y ? = 4 Q ,C = Im F , and
/. = 0.1 mH. Choose R^ for maximum
—
R
— — • -------------
power transfer and find
38. Consider the circuit o f Figure PI 1.38
C
where
— • -------------
= 0.1
at
CO =
10^ rad/sec. Suppose
/?, = 100 a , /?, = 10-1 k n , Z^i = -71000.
(a)
Find the impedance
so that P j is
Figure P l l .35
ANSWHR: (a) 48 ih . r .0 3 7 watts
(b)
maximized.
Find the values o f L and C , to achieve
36. The circuit shown in Figure PI 1.36 has
the impedance Zy computed in part
v>v,(/) = 10>/2 cos(60/) V, /?^ = 4
(a).
Find the impedance Z^ such that Pj
(c)
I = 1/120 H, and Q = 1/30 F. What value o f
and IV-,1 are maximized.
R should be chosen so that maximum power is
delivered to the load? Note: It is the source resis­
tor here, not the load resistance, that is being
varied. What is this maximum average power
consumed by the load?
Load
/
Y
Y
V
L
R
zrc,
Figure P 11.38
'" ‘" Q
A N SW FRS: (a) Z^ = 100 f ylOOO Q: (b) 99 piand 0.2 niH
39. The series RLC circuit o f Figure PI 1.39 has
Figure Pi 1.36
ANS\\'|-:RS: 0. 25 waits
reached steady state and v^{t) = 110sin(l 207tr)
V.
/
37. The circuit o f Figure PI 1.37 operates in the
sinusoidal steady state with
v^(O = 5 0 V 2 co s(2 0 0 0 r) V.
(a)
Y
Y
V
6Q
v .(t )Q
-jisn
Choose R and C such that the maximum
average power is absorbed in the load
resistor R^^ = 5
when Z. = 0.1 mH.
What is this maximum average power?
Figure P I 1.39
Chapter 11 • Sinusoidal State State Power Calculations
542
(a)
Find the instancaneous stored energy
and
L^ive
at the moment when the
2(0
terminal voltage o f the source is zero.
(b)
Find the instantaneous energ)^
the moment when
(c)
at
42. (a)
For the circuit o f Figure P H .4 2 a ,
show that the powers absorbed by the
= 0.
impedance Z = R + jX are
Find the instantaneous energ)' \Vf- at
the moment when
= 0.
ANSWHRS: in random order (J), 1.14, 1.65,
and Q =
(b)
= /?|lp
where I is in
For the circuit o f Figure P H .4 2 b ,
show that the powers absorbed by the
4.115, 1.27
admittance Y= G +jB are
= (j|Vp
and Q = 5|Vp where V is in V
THEORETICAL PROBLEMS
L
40.
Let the voltage across a capacitance C b e
v{t) =
sin(cor) V.
(a)
Find p{t), the instantaneous power
delivered to the capacitance and show
that p{t) has a peak value o f
0 .5 c o C (V J2 watts and an average
value o f 0.
(b)
Find
the instantaneous energy
store in C (or, rather, in the electric
field) and show that W^t) has a peak
value o f 0.5C(1/^^)2 joules and an aver­
age value o f 0
(c)
. 2
5
j oules.
Let Q^- be the reactive power absorbed
by C. Show that
WC,ave _
Qc
2(0
41. Let the current flowing through an induc­
tance L be i{t) =
sin(co^) A.
(a)
Find p{t), the instantaneous power
delivered to the inductance and show
that
p{t)
has
a
peak
value
of
0 .5 co£(/,„)2 watts and an average
value 0,
(b)
Find
the instantaneous energy
store in L (or, rather, in the magnetic
field) and show that W^{t) has a peak
value ot
joules and an aver­
age value o f 0.25I(/^,)^ joules.
(c)
Let
be the reactive power absorbed
bv L. Show that
Z = R + jX
o --------(a)
a
1
+
V
.
Y = G + jB
a
(b)
Figure Pi 1.42. Two-terminal elements modeled
via impedance (a) and admittance (b).
C
H
A
P
T
E
R
Laplace Transform Analysis I:
Basics
HISTORICAL NOTE
The Laplace transform converts a time function into a new function o f a complex variable via an
integration process. The name Laplace transform comes from the name o f a French mathemati­
cian, Pierre Simon Laplace (1 7 4 9 -1 8 2 7 ). Pierre Laplace adapted the idea from Joseph Louis
Lagrange ( 1 7 3 6 -1 8 1 3 ), who in turn had borrowed the notion from Leonhard Euler
(1 7 0 7 -1 7 8 3 ). These early mathematicians set the stage for converting complicated diff-erential
equation models o f physical processes into simpler algebraic equations. The Laplace transform
technique allows engineers to analyze circuits and to calculate responses quickly and efficiently.
In turn engineers became better able to design circuits for radio communication and the tele­
phone, not to mention other, earlier electronic conveniences. This chapter introduces the notion
o f the Laplace transform, a mathematical tool that is ubiquitous in its application to an army o f
engineering problems.
CHAPTER OBJECTIVES
1.
2.
3.
Explain and illustrate the benefits o f using the Laplace transform tool for solving circuits.
Develop a basic understanding o f the Laplace transform tool and its mathematical prop­
erties.
Develop some skill in applying the Laplace transform to differential equations and cir­
cuits modeled by differential equations.
SECTION HEADINGS
1.
2
3.
Introduction
Review and Summary o f Deficiencies o f “Second-Order” Time Domain Methods
Overview o f Laplace Transform Analysis
4.
5.
Basic Signals
The One-Sided Laplace Transform
6.
7.
The Inverse Laplace Transform
More Transform Properties and Examples
Chapter 12 • Laplacc Transform Analysis 1; Basics
8.
Solution o f Integrodififerential Equations by the Laplace Transform
9.
10.
Summary
Terms and Concepts
11.
Problems
1. INTRODUCTION
I'his chaprer introduces a powerful mathematical tool for circuit analysis and design named the
Laplace transform. Later, more advanced courses will describe the design aspects. Use o f the
Laplace transform is commonplace in engineering, especially electrical engineering. A student
might ask why such a potent tool is necessary for the analysis o f basic circuits, especially since
many texts use an alternative technique called complex frequency analysis. Complex frequency
analysis does not permit general transient analysis; rather, it restricts source excitations to sinu­
soids, exponentials, damped sinusoids, and dc signals. This class of signals is small and does not
begin to encompass the broad range o f excitations necessary for general circuit analysis and the
related area o f signal processing. The Laplace transform framework, on the other hand, permits
both steady-state and transient analysis of circuits in a single setting. Additionally, it affords gen­
eral, rigorous definitions of impedance, transfer fimctio}!, and various response classifications perti­
nent to more advanced courses on system analysis and signal processing. Introducing the Laplace
transform early allows students an entire semester to practice using the tool and learn about its
many advantages.
Section 2 describes some of the difficulties associated with the methods o f circuit analysis intro­
duced in earlier chapters when applied to circuits o f order 3 or higher. Following this, we present
an overview o f Laplace transform analysis in section 3, define important basic signals in section 4,
and introduce the formal definition o f the one-sided Laplace transform in section 5. The inverse
Liplace transform and important properties o f the transform process arc introduced in sections 6
and 7, with numerous illustrative examples. Section 8 applies the technique to circuits modeled
by differential equations. Such models were developed in Chapters 8 and 9.
2. REVIEW AND SUMMARY OF DEFICIENCIES OF "SECONDORDER" TIME DOMAIN METHODS
Recall that the output or response o f a circuit depends on the independent source excitations, on
the initial capacitor voltages, and on the initial inductor currents. Calculation o f the output often
begins with the writing o f an algebraic or a differential equation model o f the circuit for the out­
put variable in terms o f the source excitations or inputs and element values. For first- and secondorder circuits with simple .source excitations, such as dc or purely sinusoidal, the solution o f the
differential equation circuit model has a known general form containing arbitrary constants. See,
for example, Tables 9.1 and 9.2. The arbitrary constants depend on the initial conditions and the
magnitude o f the dc excitation or on the magnitude and phase o f the sinusoidal excitation.
Specifically, the steps in finding the response o f a second-order circuit to a constant input are as
follows;
Chapter 12 • Laplace Traiisforni Analysis 1: Basics
Step 1. Generate a differential equation model o f the circuit.
Step 2. Compute the characteristic equation o f the circuit/differoitial equation and then compute its
roots {say Aj and X-y) using, for example, the quadratic fonnula in the second-order case.
Step 3. From the location ofthe roots o f the characteristic equation, determine the form o f the solution:
or if A, = X-),
Step 4. Compute the constant D by shorting itiductors, open-circuiting capacitors, and analyzing the
restdting resistive circuit.
Step 5. Compute the constants A and B using the initial conditions on the circuit.
For circuits beyond second order, the approach in the above algorithm tends to break down.
Example 12.1 demonstrates how the approach breaks down with a simple third-order circuit.
As mentioned earlier, the foregoing technique, although quite useful for simple circuits, has seri­
ous drawbacks for circuits with more than two capacitors or inductors. This is because higherorder derivatives o f circuit output variables generally have little or no physical meaning. Such
derivatives are complicated linear combinations o f initial capacitor voltages and initial inductor
currents. The following example illuminates the difficulties.
E X A M P L E 12.1. Figure 12.1 shows three
circuits coupled through the use o f dependent volt­
age sources. The goal o f this example is to construct a differential equation model, determine the
solution form in terms o f arbitrary constants, and demonstrate the difficulties w'ith the simple
recipe o f the above algorithm by attempting to relate the arbitrary constants to the initial condi­
tions.
1Q
1Q
1O
-I-
FIG URE 12.1 A cascade of three RC circuits coupled by means of dependent voltage sources. The
differential equation model of the circuit is third order.
S o l u t io n
Step 1. Construct the differential equation o f the circuit. For this task, first write a dilTerential equa­
tion relating
to
Then write one relating
to
and finally, write one relating
Some straightfor^vard algebra leads to the following three differential equations:
to
Chapter 12 • Laplacc Transform Analysis 1: Basics
^
at
+ v c i ( 0 = 0.5iv„
( 1 2 .!)
0 . 5 - ^ + it2(/ ) = 0 .5 i'c i
(12.2)
0 . 2 5 ^ . v „ „ , ( O = 0 ,5 v „
^ , ^ 3 ^
^
Successively substituring equation 12.1 into equation 12.2 and equation 12,3 into the result pro­
duces the input-output differential equation model,
^’ont
^
-7 ^~^out , 1 A
dt^
dr
, o
(12-4)
d,
Step 2. Compute the characteristic equation and its roots. The characteristic equation for differen­
tial equation 12.4 is
^ + 7s^ + \As + ^ = {s - a) {s - b) {s - d) = {)
^
which has roots a = - \, b = - 2 , and d = - 4 .
Step 3. Determine the form o f the solution. If v-^j^t) =
form
Vou, it) =
+ Be^‘
then the complete solution has the
+ E = Ae~^ + Be~~' + De~^‘ + £
(12.5)
for r > 0.
Step 4 . Compute A, B, D, and E in equation 12.5. Using the rule o f thumb mentioned earlier, a
simple calculation yields E = 0.125
Calculation o f A, B, and D specifies the solution in equa­
tion 12.5. Applying the recipe described earlier, we take derivatives o f equation 12.5. set t = 0, and
relate them to the circuit initial conditions:
^out (0) = ^
'^out (0) =
+ dD
Again, one dot over a variable means a first-order time derivative, and two dots denotes a second-order
time derivative. A, B, and D are computed by solving this set o f equations. The difficulty is in specifying
and
First,
is simply the initial capacitor voltage on the third capaci­
tor. However, v^^^^it) is proportional to the current through it, which depends on all the initial
capacitor voltages. Further, what is the physical interpretation o f
And how do v^^^{0)
and v^^fiO) relate to the initial capacitor voltages? The relationship is complex and lacks any mean­
ingful physical interpretation. Finally, even for this simple example, computation and solution o f
the differential equation 12.4 proves tedious.
Chapter 12 • Liplacc Transform Analysis I: Basics
Exercise. For Example 12.1, compute expressions for
and
(0).
A N SW ERS:
= [2iv..(0) - 4
(0) = [\6r^J0) - 1 2 ^ ,.(0 ) + 2/v^(0)]
One o f the advantages o f Laplace transform analysis is that it does not destroy the physical mean­
ing o f the circuit variables in the analysis process. Chapter 13 addresses how the Laplace transform
approach explicitly accounts for initial capacitor voltages and initial inductor currents.
3. OVERVIEW OF LAPLACE TRANSFORM ANALYSIS
Laplace transform analysis is a technique that transforms the time domain analysis o f a circuit, sys­
tem, or differential equation to the so-called frequency domain. In the frequency domain, solu­
tion o f the equations is generally much easier. Hence, obtaining the output responses o f a circuit
to known inputs proceeds more smoothly.
To apply the technique, one takes the Laplace transform o f the time-dependent input signal or
signals to produce new signals dependent only on a new' frequency variable s. In an intuitive sense,
and as precisely derived later, one also takes the Laplace transform o f the circuit. Assuming zero
initial conditions, one multiplies these two transforms together to produce the Laplace transform
o f the output signal. Taking the inverse Laplace transform o f the output signal by means o f known
algebraic and table look-up formulas yields the desired response o f the circuit. The effect o f initial
conditions is easily incorporated.
Figure 12.2 is a pictorial rendition o f the method. As just mentioned, one transforms the input
signal, transforms the circuit to obtain an equivalent circuit in the Liplace transform world, and
computes the Laplace transform o f the output by “multiplying” the two transforms together.
Inverting this (output) transform with the aid o f a lookup table or MATLAB produces the desired
output signal.
■> Output Signal
Input Signal
\
T
Laplace Transfornn
of Input Signal
y
'' r
Laplace Transform
of C I R C U I T ^ ,
Laplace Transform
of Output Signal
F I G U R E 12.2 D iagram show ing flow o f Laplace transform circuit analysis.
S4iS
Chapter 12 • Liplacc Transform Analysis I; Basics
In a mathematical context, one executes the same type o f procedure on a difFerential equation
model o f a circuit and, indeed, difFerential equations in general. Figure 12.3 illustrates the idea.
Input Signal
DIFFERENTIAL'
EQUATION
■> Output Signal
J
Laplace Transform
of Input Signal”
Laplace Transform
of DIFFERENTIAL
EQUATION
Laplace Transform
of Output Signal
FIG URE 12.3 Diagram showing flow of Laplace transform analysis for solution
o f differential equations.
The benefit o f this t)'pe o f analysis lies in its numerous uses. Some o f these uses include steadystate and transient analysis o f circuits driven by complicated as well as the usual basic signals, a
straightforward lookup table approach for computing solutions, and explicit incorporation of
capacitor and inductor initial conditions in the analysis. The forthcoming sections will flesh out
these applications.
4. BASIC SIGNALS
Several basic signals are fundamental to circuit analysis, as well as to future courses in systems
analysis. Perhaps the most common signal is the unit step function.
«(r) =
1,
r>0
0,
r<0
( 12.6)
defined in Chapter 8. The bold line in Figure 12.4, resembling a step on a staircase, represents the
graph o f u{i).
u(t)
FKiURE 12.4 Graph of the unit step function. It often represents a constant voltage or current le\'el.
The unit step function has many practical uses, including the mathematical representation o f dc
voltage levels. Any t)'pe o f sustained, constant physical phenomenon, such as constant pressure,
constant heat, or the constant thrust o f a jet engine, has a step-like behavior. In the case o f jet
engine thrust, a pilot sends a command signal through the control panel to the engine requesting
a given amount of thrust. The step function models this command signal.
Chapter 12 * Ijp lacc Transform Analysis 1: Basics
The shifted step, shown in Figure 12.5, models a rime-delayed unit step signal.
u(t-T)
A
1 --
FIGURE 12.5 Graph of a unit step shifted T units to right.
This function is often used to represent a delayed startup.
Shifted steps, u (t- 7), often represent voltages that turn on after a prescribed time period T. The
flipped step fioiction, u{T - t), o f Figure 12.6 depicts yet another variation on the unit step. Here
the step takes on the value o f unity for time t ^ T. Often it provides an idealized model o f signals
that have excited the circuit for a long time and turn o ff at time T. The key to knowing the val­
ues o f these various step functions is to test whether the argument is non-negative or negative.
Whenever the argument is non-negative, the value is 1; when the argument is negative, the value
IS zero.
u(T-t)
•1 ■
FIGURE 12.6 Graph of flipped and shifted unit step. This function is often used to model signals
that have been on for a long time and turn ofTat time ’/'.
Exercise. Represent each o f the following functions as sums o f step functions;
(0 / ( 0 =
1,
0</<2
0,
otherwise
I, -3</<6
(///■) m =
0.
otherwise
1,
/ < -l
0,
-l< r < l
1,
; > 1
ANSVC^R.S: in random t)rder: //(-I - r) +
u {t
- 1), wu! - u{i - 2),
ii{ t
+ 3) - i d t - {>)
The pube function, p-^i), o f Figure 12.7 is the product o f a step and a flipped step or, equivalent­
ly, the difference o f a step and a shifted step. Specifically, a pulse o f height A and width 7" is
=
Au{i)u{T -
1) =
Au{t) - A u { t - T)
(12.7)
Chapter 12 • Laplace Transform Analysis 1: Basics
S50
Pr(t)
>
t
T
FIGURH 12.7 Pulse of width T and height A.
This function is often used to model signals o f fixed magnitude and short duration.
A signal sharing a close kinship with the unit step is the ramp function r{t) depicted in Figure
12. 8.
r(t)
FIGURE 12.8 Graph of the ramp function, r{t) =
Ramp functions conveniently model signals having a constant rate of increase.
The ramp function is the integral o f the unit step, i.e.,
/•(0=
w h e re t
J —oo
ii(T)dx =ln{t)
( 12.8)
is s i m p l y a d u m m y v a r i a b l e o f i n t e g r a t i o n .
Exercise. Plot r { - t) and r{t - 2).
ANSW ER: i\-t) is the relleciion of ;■(;) about the vertical axis while >\t - 2) is siniplv the shift of
lit) by two units to the rinht.
EXA M PLE 12.2
Express Figures 12.9a and b in terms o f steps and ramps.
f,(t)
y t)
F I G U R E 12,9 T w o signals to be represented by steps and ramps.
Chapter 12 • Liplacc Transform Analysis I: Basics
S
5S1
o l u t io n
For the signal/j (/), observe that the signal begins with a ramp with a slope o f 2. Thus we have
f\ U) = 2r(f) + ? . At ; = 7, the signal/j(^) levels off. Since the 2r{t) part o f the signal continues to
increase, the increase must be canceled by another ramp o f slope 2, but shifted to the right by T
units. Thus,/j(r) = 2r{t) - 2r(t - T).
The signal j 2(^) replicates/j(^) up to 3T . After 37', the signal drops to zero. Hence we must sub­
tract a shifted step o f height 2 7 ’ from/j(/‘), Thus fjit) =/j(f) - 2Tu{t - 3 7 ) = 2r{t) - 2r{t - T) -
2T u {t-5 T ).
Exercises. 1. Figure 12.10 depicts a sawtooth waveform denoted hy J{t). Sequences o f sawtooth
waveforms are used as timing signals in televisions and other electronic devices.
f(t)
A
FIGURE 12.10 A sawtooth waveform.
A N SW T .R : / / ) = fit) -;•(/- 1) - u{r - 1)
2. For/r) o f Figure 12.10, p lo t/ l - t) and represent the ftmction in terms o f steps and ramps.
AN SW ER: /(I - /) = ;-(l - r) - >i-t) - u{-t)
EXA M PLE 12.3. Express Figure 12.11 in terms o f steps and ramps.
f(t)
> t
FIG URE 12.1 I Triangular waveform to be represented by steps and ramps.
S
o l u t io n
Observe that the signaly(f) begins with a ramp with a slope o f 1 at r = -T . Thus /(f) = r{t ■¥ 7) +
?. The signal falls o ff with a linearly decreasing ramp for 0 < r < T. If we subtract r{t) the signal
would become flat for r > 0. Thus we subtract 2r{t) to obtain the linear decrease. Hence, y(r) =
Chapter 12 • Liplacc Transform Analysis 1: Basics
^52
r{t + 7) - 2>it) + ?. For r > T, this signal, r{t + T) - 2Kr), continues to linearly decrease. Hence
for the signal to be zero for t > T, we cancel this decrease with an additional ramp. T h u s/ r) =
r{t + 7) - lr{t) + r { t - T).
Newtonian physics provides a good motivation for defining the ramp signal. Applying a constant
force to an object causes a constant acceleration having the functional form Ku{t). 'J'he integral o f
acceleration is velocity, which has the form
a ramp function.
A very common and conceptually useful signal is the (Dirac) delta function, or unit impulse
Function, implicitly defined by its relationship to the unit step as
h{cj)dq
(12.9)
The relationship o f equation 12.9 prompts a natural inclination to define
s/ .
, V
I-
u{t)~ ii{t - h)
h
0 {{) = — //(/)= hm----------------(h
//->{)
( 12. 10)
Strictly speaking, the derivative o f u{t) does not exist at t = 0, due to the discontinuit)' at that
point. Without delving into the mathematics, one typically interprets equations 12.9 and 12,10
as follows: define a set o f continuous differentiable functions
The derivatives,
as illustrated in Figure 12.12a.
5^(/) = -y//^(/) >of these functions are depicted in Figure 12.12b.
5A(t)
-►t
(a)
(b)
F IG U R l'. 12.12 (a) Continuous differentiable approximation to the unit step.
(b) Derivative of
the integral of 6^(/) produces
Clearly, 6^(/-) has a well-def'med area o f 1, has height I/A, and is zero outside the interval 0 s t
s A. In addition, u[t) = lim //^(r), and lim 6^ = b{t) as A -♦ 0.
Hence, although the definition
o f equation 12.10 is not mathematically rigorous, one can interpret the delta function as the limit
o f a set o f well-behaved functions. In fiict, the delta function can be viewed as the limit o f a variet)' o f different sets o f functions. A problem at the end o f the chapter explores this phenomenon.
Despite the preceding mathematics, the delta function is not a function at all, but a distribution,^
and its rigorous definition (in terms o f so-called testing functions) is left to more advanced math­
ematics courses. Nevertheless, we shall still refer to it as the delta or impulse function. The stan­
dard graphical illustration o f the delta function appears in Figure 12.13, which shows a pulse of
Chapter 12 • Laplace Transform Analysis I: Basics
infinite iieight, zero width, and a well-defined area o f unit}', as identified by the “ 1” next to the
spike. Visualization of the delta function by means o f the spike in the figure will aid our under­
standing, explanations, and calculations that follow.
FIGURE 12.13 Standard graphical illustration o f a unit impulse lunction having a wcll-dcfincd area
of 1. The function typically represents an energ)' transfer, large force, or large impact over a very
short time duration, as might occur when a bat hits a baseball.
The unit area property follow's from equation 12.9, i.e.,
bU)(lr=
5(rV/r = / / ( 0 ^ ) - / / ( 0 " ) = l
where 0“ is infinitesimally to the left and 0"^ infinitesimally to the right o f zero. I f the area is dif­
ferent from unity, a number Kalongside the spike will designate the area; i.e., the spike will be a sig­
nal Kb{t).
One motivation for defining the delta function is its abilit)' to “ideally” represent phenomena in
nature involving relative immediate energ\' transfer (i.e., the elapsed time over which energy trans­
fer takes place is very small compared to the macroscopic behavior o f the physical process). An
exploding shell inside a gun chamber causing a bullet to change its given initial velocity from zero
to some nonzero value “instantaneously” is an example. Another is a barter who hits a pitched ball,
“instantaneously” transferring the energy o f the s\vung bat to the ball. Also, the delta function pro­
vides a mathematical setting for representing the sampling of a continuous signal. Suppose, for
example, that a continuous signal v{t) is to be sampled at discrete time instants t^ r,*
••• •
v{t) is to be physically measured at these time instants. The mathematical representation o f this
measuring process is given by the sifting property of the delta function.
v(//)=
(12.11
v(l)5(X
In other words, the value o f the integral is the non-impulsive part o f the (continuous) integrand,
replaced by that value o f r which makes the argument o f the impulse zero, in this case r = tj.
Verify'ing equation 12.11 depends on an application o f the definition given in equation 12.9.
Specifically, if v{t) is continuous at f = t-, then
\'(T )6 (T
-
1: )dX =
=
v (T
)5(T 5(1 -
f j )dx
rj)(lx = v(tj)
Chapter 12 • Laplace Transform Analysis I: Basics
SS4
by equation 12.9, where
are infinitesimally to the right and left o f f,.
Exercises. 1. Compute the derivatives o f the signals in Figures 12.9a, 12.9b, and 12.11.
A N SW E R S;/,(/) = 2|//(/)-/ / (/ -T ) ] . / .(z) = 2[//(/) - / / ( / - 7 ')| - 27’6 (/ - 3 7 ').
) = //(/+
T)-2iiii)+
oo
2. Compute the integral
ANSWHR; sin(l.5.T)
//(/-
T)
sin(27T/+ 0.57t)5(/ —0.5)^//.
-oo
5. THE ONE-SIDED LAPLACE TRANSFORM
Intuitively, a transform is like a prism that breaks white light apart into its colored spectral com­
ponents. T he one-sided or tmilaternl Laplace transform is an integral mapping, somewhat like a
prism, between time-dependent signals y(r) and functions F{s) that are dependent on a complex
variable s, called complex frequency.
LAPLACE TRANSFORM
Mathematically, the one-sided Laplace transform
J{t) is
( 12. 12)
where s = (5 + ./CO(y = yf-A) is a complex variable ordinarily called a complex frequency im
the signals and systems literature.
As the equation makes plain, the Laplace transform integrates out time to obtain a new func­
tion, F{s), displaying the frequenc)' content o f the original time function/r). In the vernacu­
lar, F{s) is the frequency domain counterpart of/ r). Analysis using Laplace transforms is often
called frequency domain analysis.
Exercises. 1. Find the Laplace transform o f a scaled Dirac delta function, Kb{t).
ANSWF.R; K
2. Find £[sin(2;rr + 0 .5 jr ) 6 ( r - 0.5)].
ANSWHR: sin(l.5.T)<'-'^ ‘‘^
Chapter 12 •Laplacc Transform Analysis I: Basics
A number o f questions about the Laplace transfbrm promptly arise:
Question 1: Why is it called one-sided or unilateral?
Answer: It is called unilateral because the lower limit o f integration is 0“ as opposed to -oo. If the
lower limit of integration were -oo, equation 12.12 would be called the two-sided Laplace trans­
form , which is not covered in this text.
Q uestion 2 : Why use 0“ instead ofQ * or 0 as the lower lim it o f integration?
Answer: Our future circuit analysis must account for the effect o f “instantaneous energy transfer”
and, hence, impulses at / = 0. The use o f 0"^ would exclude such direct analysis, since the Laplace
transform of the impulse function would be zero. Using ^ = 0 is simply ambiguous.
Question 3 : What aboutJunctions that are nonzero fo r t < 0.^
Answer: Because the lower limit o f integration in equation 12.12 is 0“, the Laplace transform does
not distinguish between functions that are different for f < 0 but equal for ^ > 0 (e.g., «(/) and
u{t + 1) would have the same unilateral Laplace transform). However, since ^ = 0 designates the
universal starting time o f a circuit or system, the class o f signals dealt with will usually be zero for
t < 0 and thus will have a unique (one-sided) Laplace transform. Conversely, each Laplace transform F{s) will determine a unique time fiinaion J{t) with the property that f^t) = 0 for ^ s 0.
Because o f this dual uniqueness, the one-sided Laplace transform is said to be bi-unique for signals/^) yfirh J{t) = 0 for ^ < 0.
Question 4 : Does every signalj{t) such th a tfj) = 0 fo r f < 0 have a Laplace transform?
Answer: No. For example, the function f i ) = ^ «(/) does not have a Laplace transform because
the integral o f equation 12.12 does not exist for this function. To see why, one must study the
Laplace transform integral closely, i.e..
Observe that e~j^* = cosicot) -jsm{(Ot) is a complex sinusoid. As f approaches infinity, the real and
imaginary parts o f the integrand in equation 12.13 must blow up, due to the
term. Hence,
the area underneath the curve e^~^ grows to infinity, and the integral does not exist for any value
of a.
W h enever//) is piecewise continuous, a sufficient condition for the existence o f the Laplace trans­
form is that
I
f.
\f{t)\<kx€^^
for some constants
and
(12.14)
This bound restricts the growth of a function; i.e., the fimction can­
not rise more rapidly than an exponential. Such a fiinaion is said to be exponentially bounded. The
Chapter 12 * Uplacc Transform Analysis 1: Basics
condition, however, is not necessary for existence. Specifically, the transform exists whenever the
integral exists, even if the function/f) is unbounded. W ithout belaboring the mathematical rigor
underlying the Laplace transform, we will presume throughout the book that the functions we are
dealing with are Laplace transformable.
Question 5: Why does the existence o f the Laplace transform integral depend on the value o f a , men­
tioned in the answer to question 4?
Answer: If the condition in equation 12.14 is satisfied, then there is a range o f a s (recall that s =
a + yoo) over which the Laplace transform integral is convergent. This is explained in the follow­
ing example.
EXA M PLE 12.4. Find the Laplace transform o f the unit step. By equation 12.12,
£{u(i)\ = U(s)= \Z
O'
(12.15)
1
a +
o
provided that o > 0. Notice that if a > 0, then
+ 7 (0
.V
-* 0 as t
This keeps the area under­
neath the curve finite. For a < 0, the Laplace transform integral will not exist for the unit step.
The smallest number Oq such that for all a > Oq the Laplace transform integral exists is called the
abscissa o f (absolute) convergence. In the case o f the unit step, the integral exists for all a > 0;
hence, Oq = 0 is the abscissa o f convergence. The region a > 0 is said to be the region o f conver­
gence (RO C) of the Laplace transform o f the unit step. Figure 12.14 illustrates the R O C for the
unit step.
j (o-axis
-f->a-axis
FIGURE 12.14 Region of convergence, a > 0, o f the Laplace transform of the unit step function
(i.e., the Laplace transform integral will exist for all a > 0).
Question 6: Is the unilateral Laplace transform valid only in its region o f convergence?
Answer:
the answer is no. There is a method in the theory o f complex variables called “ana­
lytic continuation” which, although beyond the scope o f this text, permits us to uniquely and anal)aically extend the transform to the entire complex plane.- Analytically means smoothly and also
that the extension is valid ever)^vhere except at the poles (to be discussed later) o f the transform.
Thus, the region o f convergence goes unmentioned in the standard mathematical tables o f one­
sided Laplace transform pairs.
Chapter 12 • Liplace Transform Analysis I: Basics
EXA M PLE 12 . 5 . Find F{s) iorJ{t) = Ke
So
u{t), where K and <{ are scalars.
lu t io n
Applying equation 12.12 yields
K
poo /
Ke~^“ e~^‘dt = K
s + ci
(12.16)
The integral exists if Re[j + ^] > 0. If^7 is real, then the R O C is a > -a. As mentioned in the answer
to question 6, by analytic continuation, F{s) = M{s + a) is valid and analytic in the entire com­
plex plane, except at the point s = -a . The point s ■=-a\s a pole o f the rational function M{s + a)
because as s approaches -a , the value o f the function becomes infinitely large.
The preceding discussion and examples set up the mathematical framework o f the Laplace trans­
form method. Our eventual focus rests on its application to circuit theor)\ which builds on two
fundamental laws: Kirchhoffs voltage law (KVT) and KirchhofT’s current law (KCL). KVL
requires that the voltage drops around any closed loop sum to zero, and KCL requires that the
sum o f all the currents entering a node be zero. For the Laplace transform technique to be useful,
it must distribute over such sums o f voltages and currents. Fortunately, it does.
Linearity property: The Laplace transform operation is linear. Suppose j{t) =
Then
L [ m = L [ a jl{t) +
= ^,£[/i(r)] + a.L\f,{t)]
= a^F^{ s ) ^a, f , St )
(12.17)
This property is easy to verify since integration is linear;
l« l/ | (0 + «
= «l
O'
/]{t)e~^'dt + fl2 [q- fi{t)e~^'dt
This is precisely what equation 12.17 states. Hence, our curiosit)^ satisfied, we may rest peaceful­
ly in the knowledge that the Laplace transform technique conforms to the basic laws o f KVL and
KCL.
E X A M PLE 12.6. Find F{s) wheny(r) = K^ti{t) +
So
for real scalars A', ,
and a.
lu t io n
The Laplace transform o f u{t) is \h by equation 12.15 and that of
is ]/{s + a) by equa­
tion 12.16. By the linearity property (equation 12.17),
F(.v) = ^
s
+- ^ ,
s +a
with region o f convergence (a > 0} H {a > - a }, where H denotes intersection.
By analytic con­
tinuation, the transform is valid in the entire complex plane except at the poles, s = 0 and s = -a.
(Henceforth we will not mention the RO C in our calculations.)
Chapter 12 • Laplacc Transform Analysis 1: Basics
Exercise. Find the Laplace transforms o f
{i) J{t) =
+ 2 r t ( r ) + 2u(t),
(/■/)
J{t) = -2u(t) +
(iil)
J{t) = 5u(t) - 4e~^‘u{t) + 2e~"^‘u{t)
?>s - a
AN SW ERS: (/) - .
,v~ - (I~
- 2e~^^u{t), and
2
.V/- A
2
...
3
.v“ - (/“
4
-
2
, + -----,
4
The transform integral o f equation 12.12 has various properties. These properties provide short­
cuts in the transform computation o f complicated as well as simple signals. For example, the
Laplace transform o f a right shift o f the s i g n a l a l w a y s has the form e~^^F{s), T > 0. Shifts are
important for two reasons:
1.
2.
Many signals can be expressed as the sum o f simple signals and shifts o f simple signals.
Excitations o f circuits are often delayed from t = 0.
Hence, provisions for shifts must be built into analysis techniques.
Tim e shift property: If £[/{i)u{t)] = F{s), then, for T > 0,
£ [ f { , - T l u i i - T ) ] = r ‘'rF{s)
(12.18)
Verification o f this property comes from a direct calculation o f the Laplace transform for the shift­
ed function, i.e.,
£ [ f ( t -T )u O - 7)1 = J " / ( r - T )u (t- D e'^ ’di = J " / (; -T )e~ ^ d t
Let cj = t -T ,W k WJ q = dt. Noting tliat the lower limit o f integration becomes 0“ with respect to q,
L\S(t -T )u U - D 1 = r
fUl)e~'^e~^'^dq =
f ” /(qye-^ 'dq =
O'
F^s)
Observe that if T < 0, the property fails to make sense, since J { t - T )ii{t- 7) would then shift left.
Since the transform ignores information to the left o f 0“ one cannot, strictly speaking, recover
J{t) from the resulting transform.
Exercises. I. Find L\J{t - T)] w hen/f) is (i) Ad(t), (ii) Au(r), and (iii) Ae~'’'u(t).
2. pyU) = /iu(t) - Aii{t - T).
A N SW ERS: In random order:
-----
.\----- ^
s+ a
s
/
s
Chapter 12 • Laplacc Transform Analysis 1: Basics
E X A M PLE 12.7. Using the tim e shift property, find F{s) for the signal
12.15.
sketched in Figure
f(t)
3
2
1
1
2
-1
-2
FIGURE 12.15 Signal for Example 12.7.
S o l u t io n
Using step functions and shifted step functions, we obtain
p )
= 5 u {t)-5 u {t-\ )^ 2 u {t-2 )
3 5e~^
2e~
Direct application o f linearity and the time shift propert)’ yields F {s) = —------------f- - —
s
s
s
Exercises. 1. Find the Laplace transform o f the pulse signal o f equation 12.7.
2. Find F{s) when/^) = A^u[t - T^) + A2 ^(t - T-^ + A^u{t - T^).
1
AN SW ERS:
~sl
1
— v7i
.4
P j(s) = A— ----- . F ( s ) = ^ -------^ +
4
.—
---------------------
One more property allows us to revisit the signals discussed in section 3 and take their Laplace
transforms. The new propert)^ is multiplication of//) by t. This always results in a Laplace trans­
form that is the negative o f the derivative o f F{s).
M ultiplication-by-f property: Let F{s) =
Then
£[tf{t)\ = - — F(5)
ds
(12.19)
Verification o f this property follows by a direct application o f the Laplace transform integral to
with the observation that te~^^ = ------ • In particular,
ds
Chapter 12 • Laplace Transform Analysis 1: Basics
S60
-St
■oo
n o
j()
d
dt =
.d s
as
ds J
Table 12.1 lists this transform pair, as well as numerous other such pairs, without mention o f the
underlying region o f convergence. As mentioned earlier, we shall dispense with any mention of
the ROC, assuming that all functions are zero for r < 0.
EXA M PLE 12.8. Find the Laplace transform o f the ramp function, r(r) = tu(t).
S o l u t io n
Using equation 12.19,
M ds \ s ,
R {s)= L\r{t)\ =
E X A M PLE 12.9. Supposey{r) = te
( 12.20)
ds
where a is real. Find F{s).
S o l u t io n
The quickest way to obtain the answer is to apply equation 12.19. Specifically, since
^ r —nt , , T
1
s+a
L
ds s + a
ds
1
{s + a ) -1
( 12.21)
An alternative, more tedious approach is to use integration by parts as follows:
F {s)= L
where v = t and dti = e
=:
te
oo
fOO
0“
ydii = uv
•oo
_
0"
O'
udv
dt. Thus,
•oo
le
+
.V+ a
■dt
The RO C is a > -a , in which case the first term on the right-hand side is zero. Thus, in this RO C ,
evaluation of the integral term implies that
F {s)•'0“
s+a
dt =
1
(s + aY
Equation 12.21 is a special case ol the more general formula
£\t''e-^^‘u(t)] =
nl
(s +a)71+ 1
(1 2 .2 2 )
Chapter 12 • Liplace Transform Analysis 1: Basics
561
Exercise. Find the Laplace transform oij{t) = p-e
AN SW ER: F{s) =
(\ + a r
EXA M PLE 12.10. Find F(s) for the signal depicted in Figure 12.16.
f(t)
FIGURE 1 2 . 1 6 A signal to be represented by steps and ramps.
S o l u t io n
From Example 12.2,/ r) = 2r(^) - 2r(t - T) - 2 T u { t - 3 7 ). Hence by linearity, the time shift prop­
erty', and equations 12.15 and 12.20,
F {s) = L \ 2rU )- 2r{t - T ) - 2Tu(t - 3 7 )
2 - 2 £ ” '^
Exercises. 1. Note that the sawtooth o f Figure 12.17 is/ f) = t[u{t) - u {t- 1)]. Suppose
= z/(r)
- u {t- 1). Compute F (5 ) = - —-G ( i')-
d.s
f(t)
2. Use equation 12.22 to compute the Laplace transforms o f/ ;) = tr{t) for the ramp function r{t)
and forjit) = p-r{t).
2
6
.s'
s
AN SW TRS: — . —
Chapter 12 • Laplace Transform Analysis I: Basics
562
EX A M PLE 12 .1 1 . The circuit o f Figure 12.18a has two source excitations, /j(/) and /2W> shown
in Figures 12.18b and c. Compute
V o Jt)
(a)
(c)
FIGURE 12.18 (a) Resistive circuit driven by two current sources.
(b) Triangular signal, /,(r), in A. (c) Pulse signal, ijit), in A.
S o l u t io n
Step 1. Find the form ofV^^^{s). By superposition and Ohms law,
= 10/, U) + 10/2(r)
From the linearity o f the Laplace transform,
= 10/, W + lOAW
Step 2. Compute /,(j) and
Some reflective thought yields /,(?) = 2ti{t) - 2r{t) + 2r{t - 1) A
and ijit) = \.5u{t) - \ 5 u {t- 1) A. From linearity, the time shift propertys and the previously com­
puted transforms.
2
h{s) = - ~
S
Step 3. Find V..Js). Since V
l + 2e
s'
and l 2 {s) =
(s) = 10/As) + lOAU), ir follows that
your(s) = - - ^
+ e~^
20
l5^
Chapter 12 * Laplace Transform Analysis I: Basics
S63
Step 4. As an introduction to the next section, by inspection we can compute the time function
o f the output voltage:
= 35«(/) - 20r{t) - \5 u (t- 1) + 2 0 r{t- 1)
Exercises. 1. Find the Laplace transform o f (i)^(^) and (iii)^ (r) =
■>r e
(ii)
=e
+e
+ te
+ te
2) + «(t - 3).
i\NSV('T,RS: in random order: — !----- 1-----!-----j------- !------- 1------- ?----- . — !------- y — 5—
.v + c/ .v + /> (.v + f/)(.v + /?)~ (.v + f/) (.v-c/)
-— --------h
.v+ ^/
2. Recall that cos(cor) = ------------------ . Show that the Laplace transform ofJ{t) = cos{cot)u{t) is
. r + 0)
3. Recall that sin(CO/) = ------------------ •Show that the Laplace transform o f/ r) = sin(ct)f)z^(f) is
(0
F{s) = - ------,
4
4
4. Find the time functions associated with Fi ( j) = — , F) (s) = ----------^
^
(s + 2)-
AN SW ERS: /;(/) = Au{t)
= Atr"‘H{() , /^(/) =
~
4 . - “’
= ---------.
-v+ 4
- 4)
We end this section with Table 12.1, which lists a number o f Laplace transform pairs. Some o f
these will be developed later in the chapter and some in the homework exercises. We will refer to
this table in the next section when computing inverse transforms.
.V
Chapter 12 • Liplace Transform Analysis I: Basics
KV4
T A B L E 12.1 Laplace Transform Pairs
Item Number
m t)
K
Ku{t) or K
KIs
m t)
KlP+1
\I {s+ a)
]/{s+ dp-
(OqUP- + OJ^)
cos{ci)Qt)u{t)
10
e
s!{P- + co^)
(Oo
s2
s in {(O Q t)u {t)
~>
(i + a) + coq
is + a)
11
{s + Cl)~ +
(Oq
2(0o^
12
(.v^ +toi5)“
t C O s{0 )Q t)u {t)
13
is +0)o)
sin((W()/ +
14
.vsin((|)) + coo cos(<)))
(j>)tiit)
s~ +(0n
.vcos((t))-a)o sin(({))
0
1
+ (Oq
15
16
te
smiO)Qt)uit)
17
te
cosi(OQt)uit)
s +a
2(0
is + a)~-(OQ
((5 + rt)^ +C0n)^
2(0o
18
[(5+C/)^ +(OoJ“
19
20
21
Cl cos(coor) +
2^I a ^ + B ~
Cj ~ C\Ci
sin(o)o/) n it)
(.? + « )“ +0)^
A + jS
cos (Oq/- tan -1
2^A~ + B- te~^' cos cogr - tan *
C|5' + C 2
' B^
A + jB
.A)j
[s + a + ;cOo)“
A —j B
^
A -jB
{s + a - ycoo)"
Chapter 12 • Laplace Transform Analysis I: Basics
6. THE INVERSE LAPLACE TRANSFORM
For the Laplace transform tool to effectively anal)'ze circuits, one must be able to uniquely recon­
struct time functions /(r) from their frequenc}'" domain partners F{s). Theoretically, this is attained
through the inverse l-aplace transform integral.
INVERSE LAPLACE TRANSFORM
Intuitively, if £[/{()] = F{s), then J{t) = X “ '[F(s)]. Rigorously speaking, the inverse Laplace
transform integral is a complex line integral defined as
/ ( ,) = r V u ) , = ^ J ^ f ( . ) e V ,
over a particular path V in the complex plane. T he path F is typically taken to be the vertical
line Oj + jio where OJ ranges from -oo to +00 and Oj is any real number greater than Oq, the
abscissa o f absolute convergence.
This integral uniquely reconstructs the time structure o f F{s) to obia.\n J{t) in whichy(r) is zero
for ^ < 0. Conceptually, the process resembles the reverse action o f a prism, to produce white
light from its spectral components. An appreciation for the power o f this integral requires a
solid background in complex variables and would not aid our purpose, the analysis o f circuits.
In fact, the evaluation o f the integral is carried out using the famous residue theorem o f com­
plex variables. Further discussion is beyond the scope of this text.
Just as the Laplace transform is linear, so, too, is the inverse Laplace transform, as its integral
structure suggests, i.e.,
Also, the unilateral transform pair
[fit), f(j)} is uftique, where by unique we mean the following: let F^{s) = L\f^{t)] and Fjis) =
Z[^(r)] coincide in any small open region o f the complex plane. Then F^{s) = Fjii) over their com ­
mon regions o f convergence, and/j(f) = f-y{t) for almost all r > 0, “Almost all” means except for a
small or thin set o f isolated points that are o f no engineering significance. Hence, there is a oneto-one correspondence between time functionsy(f) for whichy(f) = 0 for f < 0 and their one-sided
Laplace transforms. Linearity and this uniqueness make the Liplace transform technique a pro­
ductive tool for circuit analysis.
Virtually all the transforms o f interest to us have a rational function structure; i.e., F{s) is the ratio
o f two polynomials. Rational functions may be decomposed into sums o f simple rational func­
tions. These simple rational functions are called partial fractions and their sums are known as par­
tial fraction expansions. Two o f the more common “simple” terms in partial fraction expansions
have the form K b and K!{s + a). Such simple rational functions correspond to the transforms o f
steps, exponentials, and the like. Table 12.1 lists these known inverse transforms. With the table,
direct evaluation o f the line integral in equation 12.23 becomes unnecessary. Our goal is to
describe techniques to compute the simple rational functions in a partial fraction o f F{s). Once
these are found, the transform dictionar)- in Table 12.1, in conjunction with some well-known
properties o f the Laplace transform, will allow us quickly to compute the time function y(r).
Chapter 12 • Laplace Transform Analysis I: Basics
PARTIAL FRACTION EXPANSIONS: DISTINCT POLES
Our focus will center on proper^ rational functions, i.e.,
+ --- + ^^l-y + ao _
^ (^ )_
+--- + biS + bo
5" +
where m s « and p^, ... ,
+
is - Pi )(.v - 7^2
+
- Pn )
are the zeros o f the denominator polynomial,
+ ... +
+ l?Q, and are called the finite poles o f F{s). For the most part, rational functions are sufficient for
the study o f basic circuits. There are three cases o f partial fraction expansions to consider:
(1)
the case o f distinct poles, i.e., p - p - for all i
j;
(2)
the case o f repeated poles, i.e., pj = pj for at least one i
(3)
the case o f complex poles. Although case (3) is a subcategory o f case (1) or (2) or both,
j\ and
its attributes warrant special recognition.
If F{s) is a proper rational function with distinct (equivalently, sitnple) poles />j, ... ,
F(s)=K +
(S-Pi)
{S-P2 )
+■■■+
(s-p„)
then
(12.24)
where
K = lim F(s)
(12.25a)
5-400
The numbers Aj in equation 12.24 are called the residues o f the pole p- and can be computed
according to the formula
A =
[(s - Pi)f^(s)] = [(^ - Pi)Fis)\,^p,
(12.25b)
The rightmost equality o f equation 12.25b is valid only when the numerator factor {s - p ) has
been canceled with the factor [s - p ) in the denominator of F{s)\ othenvise, one will obtain zero
divided by zero which, in general, is undefined. As intimated earlier, this partial fraction expan­
sion should enable a straightforward reconstruction o f/ r). Indeed, from Table 12.1, we immedi­
ately conclude
f i t )= Kh{t) + A^e^’^'uU) +
) + •••-!- A„e^’‘‘u(t)
(12.26)
EXA M PLE 12.12. Findy(r) when
jr(^+ a )
S o l u t io n
The solution proceeds by executing a partial fraction expansion (equation 12.24) on F(s) to pro­
duce the Laplace transform o f two elementar>- signals, a step and an exponential. Specifically,
F{s) = ----------- = - - h
5(5 + a )
s
s +a
Chapter 12 • Laplace Transform Analysis I: Basics
S6 '
where Ah is the Laplace transform o f a weighted step, Au{t), and B!{s + a) is that o f a weighted
exponential,
To find A, multiply both sides by^, cancel common numerator and denom­
inator factors, and evaluate the result at j = 0, to produce A = Ma. Similarly, to find B, multiply
both sides by i
cancel common numerator and denominator factors, and evaluate the result
at j = -ay to obtain B = -Ma. Recall that, by iinearit)', X “ ’ [aF{s)] = aL~^
Hence,
/ (r) = - H(0 - - ( - “'uir) = - (I - e - “' )» (r)
a
a
a
I ^Cl
Exercises. 1. Findy(f) when F{s) = --------- ^
AN SW ER: J{t) = 2u(t) 'j
2. Find/r) when F(.s’) =
^ + 3 6 '+ 6
A N SW ER;
{.s + \)(s+ 2){s + 3)
Jit) =
3. Find a partial fraction expansion o f /r^y) =
^A +
^
s
s+a
5(i' + a)
AN SW ER: K = c . A = - , B =
a
(I
4. Find/r) for f{s) from Exercise 3.
AN SW ER: / ( ,) = c 5 ( 0 + ^ K / ) - —
a
a
E X A M PLE 12.13. Suppose V',„(5)= 1 0 “ ^ +^>y + 2
circuit o f Figure 12.19. Find
and v„U) . Assume standard units.
2Q
V ..
'>
80>v^,
FIG URE 12.19 Series resistive circuit.
S O L U T IO N
Step 1. Detemihie
By voltage division,
= 0.8t'y„(r), in which case
2 5 “ + 35 + 2
C'haptcr 12 • Liplace Transform Analysis 1: Basics
56H
Step 2. Construct a partial fraction expansion ofV^^^^{s). Since the numerator and denominator are
both o f degree 2,
16.v" + 2 4 i-+ 1 6
.v(.v + 2)
= K + — + ------.v + 2
(12.27)
The value o f K in equation 12.27 is determined by the behavior o f F(s) at infinity (equation
12.25a), i.e..
K - lim
^ 1 6 i“ + 24.v+ 16
i —^OO
\
i(A- + 2)
To I'lnd A in equation 12.27, we use equation 12.25b:
165 “ + 24.V + 16
(.v + 2)
Ks + /4 +
,v=()
Bs
.sT 2
.v=0
Similar!)’,
16.v- + 2 4 a + 16
Ii =
= -1 6
i= -2
Step 3. Find v^^^^it). Using Table 12.1,
v v ,„ ,( o = r '
r
1
8
161
' 8 '
16
16 + -------------- = r ' [ Li 6 Ji + r ‘
- r ‘
.V .v+ 2.
s
..v + 2.
= 166(r) + 8u{t) - ]6e--'u{t) V
Exercises. 1. Repeat Example 12.13 for \A^^(.y)=
= lOuU) \'
AN SW ER:
*v(.v + « )
2. Given the circuit o f Figure 12.20, find a partial fraction expansion o f
/. (.v) = ^
'
+ {a + b + c ).v + b
assuming standard units.
.v(.v+l)
A N SW l'R: /■-,(/) =
+ ■\hn{t) + -\ce~‘u(t) A
ijt)|
1Q
40
F I G U R E 12.20 Parallel resistive circuit.
Chapter 12 • Liplacc Transform Analysis I: Basics
E X A M PLE 12.14. Compute the inverse transform oFthe function
F{s) =
So
-e
lu t io n
From Example 12.12,
1
r '
This result and the shift theorem yield
.v(.v+l)
By the linearit}' o f the inverse Laplace transform,
/rt = ( l - e - O u W - ( l - r < ' - ' ) » ( ; - ! )
A sketch
appears in Figure 12.21..
FIG URL 12.21 Sketch ofy{/) = [ 1
^]«(r) - [1 -
!)•
PARTIAL FRACTION EXPANSIONS: REPEATED POLES
Proper rational functions with repeated roots have a more intricate partial fraction expansion, and
calculation o f the residues often proves cumbersome. For example, suppose
F{s) =
fijs)
is-a)^d(s)
Chapter 12 • Liplace Transform Analysis 1: Basics
S70
- a)^ specifies a repeated root o f order k, d{s) is the remaining fac­
where the denominator factor
tor in the denominator o f the rational function F{s), and n{s) is the numerator o f F{s). The struc­
ture o f a partial fraction expansion with repeated roots is
Ak
FCv) = - ^ +
where A^, ... ,
(s-af
(12.28)
are unknown constants associated with s - a, ... ,(s - a)^, respectively, and ^i^(s)
and <^(s) are whatever remains in the partial fraction expansion o f f(s). The formulas for comput­
ing the y4y o f equation 12.28 are
A k=(s-arF(s)
n(s)
-*i-a
(12.29a)
f/(5)
n{s)
[d(s)
^A-l = y ( ( . v - « / 'F ( ^ ) )
(12.29b)
and, in general,
1
/!
ds '
njs)''
(s-afF(s)
(12.29c)
O f these expressions, only the first looks like the case with distinct roots; the others require deriv­
atives o f (.f - a)^ F{s). Computation o f high-order derivatives borders on the tedious and is prone
to error. The above formulas, equation 12.29c in particular, are included for completeness.
Computer implementation circumvents these difficulties. An example that illustrates the use o f
the preceding formulas, as well as a usefiil trick, comes next.
EXA M PLE 12.15. The goal here is to illustrate the computation
F(s) = —
when
s +2
.v“ (.v + l)“
.V
.s-
.v + 1
(.v+l)'^
(12.30)
The two easiest constants to find are A2 and Bj, as their calculation requires no differentiation.
From equation 12.29a,
A, =
s-Fis)
-I
5-1-2
-5=0
=2
i=0
and
B 2 = ( s + \r F (s )
j'-i-2
.v=-l
_ s~
.9 = - !
Finding A^ and
is more difficult, since formula 12.29b requires some differentiation.
According to equation 12.29b,
as
Chapter 12 • Laplacc Transform Analysis I: Basics
571
To implement this formula multiply both sides o f equation 12.30 by
resulting expression with respect to s, and evaluate at i = 0:
d
’
ds
5+2
_ d
(5+l)-_
y4|5 +
ds
5=0
take the derivative o f the
^ B|iA-y H----------- h
(5+1)“
S+\
i= 0
Observe that, on the right-hand side, it is not necessary to differentiate the terms that contain A 2 ,
and ^ 2» since these terms disappear at / = 0, as the formula for
d
ds
'
5+2
’
(,v+l)2
.v=0
1
.
(5+1)"
^+2
'
requires. Consequently,
= -3
“ (5+l)-\
i=()
Similarly,
Bx=ds
0
( 5 + l ) ^ F ( 5 )
”
_ d
s= - 1
ds .
5
+2
5“
=
. A--1
■|
^s + 2'
.5 * '
S'
= 3
. ,v=-l
Aj, and B2 were known, a simple trick allows a more direct computation o f
Bji merely evaluate equation 12.29 at j = 1 (in fact, any value o f s, excluding the poles, will do),
Note that since
to obtain
0.75 = - 3 + 2 + 0.25 + 0 .5 5 j
As expected, solving yields Bj = 3. Hence
-3
2
3
= — + ^ + ------ +
i “ (.v + l)5 “ s + \ (5 + 1 )2
F(s) = —
.v+ 2
The above result can also be found with the MATLAB command “residue” as follows. Let F{s) =
7i{s)ld{s). It follows that n{s) = s + 1 and d{s) =
+ r . In MATLAB,
»num = [1 2];
>>den = [ 1 2 1 0 0 ] ;
»[r, p, k] = residue(num, den)
The answers from MATLAB are:
r = -3 2 3 1 (the residues associated with the poles)
p = 0 0 -1 -1
and constant
k = 0.
Exercises. \. Find the partial fraction expansion o f F{s) =
2 5 ^ + 2 r S 3 ^ + 35 4-2
.v^(.9+n^
AN SW ERS (residues in random order): 2, 2. 2, - 1 . -1
2. The partial fraction expansion o f a rational function is given by
F{s) = -
3s^ + \0s + 9
+ 45* + 55 + 2
A
5+ 2
B
+ ------ +
5+1
C
Chapter 12 • Laplacc Transform Analysis 1: Basics
Compute A, B, and C
ANSWHRS: In random order: 2. 1. 2
3. Use MATLAB to find a partial l^raction expansion o f F{s) =
(Iis)
where n{s) = G{s + 2)^(^ - 2)“
and d{s) = s{s + 1)“(^ + 4)“. Hint: Use num = 6*poly([-2 -2 -2 2 2]) and den = poly([0 -1 -1 -4 -4]).
ANSWHR: [r,p,kj = residue(num,den) \'iclds
r = - 4 4 4 8 - 1 6 - 6 12
p = _4 .4 _| _ ] 0
,
n .r
-^8
1 his results m rlie r r h : r(.s) = ■
-1 6
-6
v+l
(v + l)~
12
^
-{------------ - h----- H--------------- -----------h 6
_______________________________ •'>+ 4
(.v + 4 )
^
________________________
The derivative formulas o f equations 12.29 are often difficult to evaluate for complicated ration­
al functions, such as
s
55^ + 955-^ + 692^^ + 2369.V- + 3715.9 + 2076
F{s) = ---------------------------------------------------- r--------------
(.v+l)(:? + 2)(.s- + 3)(.v+5r
A
B
s+\
s +2
_ C ^ D\
i + 3
.y + 5
For these functions, it is very efficient to find A, B, C, and
1)2
D3
(.v + 5 ) -
(.^•+ 5 )'*'
directly. Then one evaluates F{s) at
two values o f j, e.g., j = 0 and s = 1, to obtain two equations in the unknowns D j and Dj.
Typically, solving the resulting two equations simultaneously is much easier than evaluating Z),
and D j directly by equations 12.29. Alternatively, one can use a software program such as MAT­
LAB to compute the answers. In particular, in MATLAB:
n =[5 95 692 2369 3715 2076]
d = [l 21 176 746 1665 1825 750]
»[r,p,k]=residue(n,d)
r=
-l.OOOOe+00
-l.OOOOe+00
-l.OOOOe+00
3.0000e+00
2 .0000 e+00
l.OOOOe+00
P=
-5.0000e+00
- 5 .0000 e +00
- 5 .0000 e +00
- 3 .0000 e +00
- 2 . 0000 e +00
-l.OOOOe+00
k=
Chapter 12 • Laplacc Transform Analysis 1: Basics
573
PARTIAL FRACTION EXPANSIONS: DISTINCT COMPLEX POLES
Distinct complex roots present challenges different from those for the repeated root case. Since the
roots are distinct but not real, the methods o f equations 12.25 and 12.29 apply. Unfortunately, the
resulting partial fraction expansion has complex residues, and the resulting inverse transform has
complex exponentials multiplied by complex constants. Such imaginar}^ time functions lack mean­
ing in the real world unless their imaginar}- parts cancel to yield real-time functions. When they do,
our goal is to find a direct route for computing the associated real-time signals. To do this, consider
a rational function having a pair o f distinct complex poles as in the following equation:
F(s) =
n{s)
n{s)
|(.v + a ) - + ( 0 -](/(. 5)
(s + a + ju>)(s + a - j w ) d ( s )
(12.31)
Since the poles - a - j c o and - a + jco are distinct, the partial fraction expansion o f equation 12.24
is valid. Since the poles are complex conjugates o f each other, the residues o f each pole are com­
plex conjugates. Therefore, it is possible to write the partial fraction expansion of f{s) as
r(s)-
.v + « + yto
I
5 + a -y c o
(12.32)
d{s)
and d{s). As per equation 12.25b, the first residue in equation
for appropriate polynomials
12.32 is
(12.33)
jco
With A and B known, executing a little algebra on equation 12.32 to eliminate complex numbers
results in an expression more amenable to inversion, i.e..
C\S + C->
n^is)
/?|(.v)
F{s) = ------ \2, ~ 2 + ^ 7 T = ^0(■'■) +
(I{s)
d(s)
{s + fl) + to
(12.34)
C, =2/1
(12.35a)
C-, = 2aA + 2 ojB
(12.35b)
where
and
with A and B specified in equation 12.33. W ith Cj and Cj given by equations 12.35, it is straight­
forward to show that
F'ois) =
C \S + C
2
{s + a)~ +oj^
=C
( C2 -C ^a \
1,
x2
(.9 + a) + 0)
2
to
to
{s + « )"
From Table 12.1, item 19, or a combination o f items 10 and 11,
M O = e'
C| cos(tor) +
(C2-Cici)
sin (to/) Hit)
[ to
)
+03"
(12.36)
574
Chapter 12 • Laplace Transform Analysis 1: Basics
Exercise. Suppose F { s ) =
. Compute/(r).
a- 2 + 4
AN SW ER: C, cos(2r)//(f) + 0 . 5 ^ sin(2f)«(r)
The following example illustrates the algebra for computing C, and Cj without using complex
arithmetic.
EX A M PLE 12.16. Find/^) when
3 .r + 5 + 3
D
A + jB A - j B
D
C^s + C^
F{s) = ------------^------- = -------- + ------ =;- + ------^ = ------ + - ^ ------- (5 + 1 )(5 “ + 4 )
^+1
s + j2
s-jl
i +1
s +A
, ^
(12.37)
Step 1. Compute the coefficients D, C ,, and C2 in the partial fraction expansion o f equation 12.37.
First we find D by the usual techniques:
3 .r
+ 5
+ 3
.v^ + 4
= 1
s=-\
Given that D = 1, to find C , we evaluate F{s) at j = 0, in which case 0.75 = 1 + O.2 5 C2, or Cj =
- 1 . With D = 1 and C2 = - 1 , we evaluate F{s) at j = 1 to obtain 0.7 = 0.5 + 0.2(C j - 1) or, equiv­
alently, Cj = 2. Thus,
■V+ 1
s^ + 4
+ 4
Step 2. Compute j{t). Using Table 12.1, items 8 and 9, to compute the inverse transform yields
J{t) = [e~‘ + 2 cos(2r) - 0.5 sin(2f) ]u{t)
Alternative Step 1. Compute A and B in equation 12.37 by hand or with MATLAB. In MATLAB,
»num = [3,1 3];
»den = conv([l 1],[1 0 4])
den = [1 1 4 4]
»[r, p, k] = residue(num.den)
r=
1.0000 + 0.25001
1.0000 - 0.2500i
1.0000 + O.OOOOi
P=
-0.0000 + 2.0000i
-0.0000 - 2.0000i
1.0000
-
k =0
Chapter 12 • Laplacc Transform Analysis 1: Basics
This implies that
.y+1
^+
. A -JB
s + j2
s-jl
1
, l- y '0 .2 5 , l + yO.25
5+ 1
s+ jl
s-jl
(12.39)
Alternative Step 2. One must exercise caution here and note the difference between the MATLAB output and the form o f the partial fraction expansion. From equation 12.39, w = +2, A = 1,
and B = - 0 .2 5 . Again using MATLAB to obtain the form needed in item 20 o f Table 12.1,
»K = 2*sqrt(A^2 + B^2)
K = 2.0616
»theta = atan2(B,A)* 180/pi
theta = -1 4 .0 3 6 2
Thus
Example 12.16 illustrates not only the computation o f an inverse transform having complex poles,
but also the computation o f Cj and C , without resorting to complex arithmetic, as was needed in
equation 12.32. The trick again was to evaluate F{s) at two distinct ^-values different from the
poles o f F{s). This yields two equations that can be solved for the unknowns Cj and C ,.
5 ^ 8
4
Exercises. 1. Find Kt) when F{s) = ------- z----------.
s(s-+4)
AN SW ER://) = [1 + 4 cos(2r) - 4 sin(2r)]//(/)
,
5s" - 2 ^ + 5
2. Find/(r) when F ( s ) - — ^
.
. v ( r + 2 5 + 5)
A N SW ER://) = u(t) + 4 r ‘ [cos(2/l - sin(2/)l/<(/)
7. MORE TRANSFORM PROPERTIES AND EXAMPLES
Another handy propert)' o f the L'lplace transform is the frequenc)' shift property, which permits
one to readily compute the transform o f functions multiplied by an exponential. With knowledge
o f the transforms o f u{t) , sin(o)/), and other functions, computation o f e~^‘u{t) and f’“"^sin(to/)«(^)
becomes quite easy.
Frequency shift property: Let F{s) =
Then
L[ e~^'p)] = F{s * a)
(1 2 .4 0 )
Chapter 12 • Laplacc'Iransform Analysis I: Basics
ThisS property can be verified by a direct calculation,
= F (s + c)
Xlc’- “7 ( ' ) l =
where we have viewed the sum s + a m tlie integral as a new variable p, which leads to F{p) with
p replaced hy s + a.
EXA M PLE 12.17. Let//) = sin(wr)//(r). D efine^/) = e~‘" p ) = e~‘*‘ s\n{iot)u{t). Suppose it is
known that
Compute G{s).
So
lu t io n
By the frequency shift property, G{s) = F{s + a), or
CO
)| = L\e-"\m \ = F (s + « ) =
G( j ) = £|
CO
5“ + (0 “
Exercise. Lcijit) = cos(cor)u(i) for which r ( s ) = ___ - ___ D efin e^r) = e
Compute G{s).
J{t) = e
cos{LOt)u{t).
+C0“
.V+ a
AN’SW I-R:
i.s +
a )-
+ (0 "
Another property o f particularly widespread applicability is the time differentiation formula. Its
utility resides not onl)' in obtaining shortcuts to transforms o f signals, but also in the solution o f
differential equations. Differential equations provide a ubiquitous setting for modeling a large
variety o f physical systems— mechanical, electrical, chemical, etc. In terms o f signal computation,
recall that the velocity of a particle is the derivative o f its position as a function o f time. The accel­
eration is the derivative o f the velocity. After computing the Laplace transform o f the position as
a function o f time, one finds that a differentiation formula allows direct computation o f the trans­
forms of the velocity and acceleration. Also, as discussed at the very beginning o f this chapter, cir­
cuits have differential equation models. For example, weighted sums o f derivatives o f the response
of- the circuit are equated to weighted sums o f derivatives o f the input signal. Therefore, a differ­
entiation formula is an essential ingredient in the analysis o f circuits.
First-order tim e difiFerentiation formula: Let L\j{t)] = F{s). Then
L jfU )
(It
= sF{s)-f{Q-)
(12.41)
Chapter 12 • Liplace Transform Analysis I: Basics
The difFerenriarion property is validarcd using integration by parrs as follows:
4/(0V' ' d t = f U ) e
y / (/ )
(It
JO'
The following examples explore some clever uses o f the first-order time difTerentiation formula
E XA M PLE 12.18.^^
Recall that 6f/) = — u ( t ) . Using the sifting propert)', a direct calculation yields £ [(5(/)] = 1. Is
this consistent with the differentiation propert}^? Interpreting the delta function as the derivative
o f the step function and applying the differentiation formula yields
I
d
£ 6 { l ) = £ - H i t ) = s £ u{r) - u { 0 ) = s
dt
s)
(12.42)
which demonstrates the expected consistency.
Exercises. I. The Liplace transform o f a signal/r) is F{s) =
f o r m o f e - 2 '4 / W
— . What is the Laplace trans
?
dt'
2s + 4
AN SW I-R: -— — ^-----
(.Y+ 2 r + 4
2. £[sm{wt)6it)] = ?
AN SW ER: u
EXA M PLE 12 .1 9 . Suppose^r) = sin(wr)w(/) and we know (for example, from Table 12.1) that
Compute £ [cos(wr)//(r)] using the time differentiation formula.
F { s ) = £ sin((0/);/(/)
.V" + (0
So
lu t io n
j
Since cos(coO/^0 = ------- sin{a)/)//(/)and sin(wr)«(/)]j ^ o = 0 , the differentiation property imme­
diately implies that ^ ^^
£ cos(coO//(0
(0
(0
■)
+C0‘
.S - + W
Exercises. 1. Express//) = s\n{o)t)u{t) in terms o f the derivative o f ^ r) = cos{(Ot)u{t). Note the
presence o f the delta function.
AN SW ER: - '
(1)
^
di
(ij
= sin (w ;)//(r '
2. Now suppose it is known that £ cos(co/)/KM
-. Use the result of Exercise 1 and the
+ 0)'
differentiation propert)' to compute the Liplace transform of/r) = sin(oj/)«(/) noting that^^O ) = 0 .
-t
S~
ANSW ER; £ sin((0/);/(/)
1
(0
(!)
_
1
1
(I)
(0
(0
V +(0'
s' - (O'
Chapter 12 • Laplacc Transform AnaK'sis I: Basics
578
EXA M PLE 12.20. Lety(r) and its derivative iiave the shapes shown in Figure 12.22. Th e goal o f
this example is to explore the relationship between the Laplace transforms o f/ r) and f'{t) in light
o f the differentiation property.
FIGURE 12.22 A pulse and its derivative for Example 12.20. Observe how the derivative o f the
pulse leads to a pair o f delta functions.
Using linearity and the shift theorem on j{t) yields
X|/(01 = £ | » ( 0 - - 1)] = £ [« (f)]- X | h(/ -1)1 = - (I Applying the linearity o f the Laplace transform x.of\t) yields
£ l / ’(r)] = i : [ 5 ( 0 - 5 ( f - I ) ] = l - ^ - ^
From the differentiation formula, it must follow that L\f{t)] = sL\J{t)]. Thus,
£|/'(/)l = sL U (t)] = j i d - « - * ) = I demonstrating consistency.
As might be expected, the formula for the first derivative is a special case o f the more general dif­
ferentiation rule:
L
cit”
m
= s"F(s) -
)
(12.43)
This rule proves useKil in the solution o f general «th-order difTerential equations. O f particular
use is the second-order formula:
(1 2 .4 4 )
Chapter 12 • Laplace Transform Analysis 1: Basics
579
The inverse o f differentiation is int^ration. The following property proves useful for quantities
related by integrals.
Integradon p r o p e r ^ Let F{s) = -£[/(/)]. Then for ? > 0,
■
c Jo-/(9 )d? =—
(12.45a)
and
F (s)
(12.45b)
As with many o f the justifications o f the properties, integration by parts plays a key role. By direct
computation (using equation 12.16),
To use integration by parts, let
fiq ) d q and dv = e
u=
Then
-too
rr
JO'
For the appropriate region of convergence, the first term to the right o f the equal sign reduces to
fO'
et
f(q ) d q
f( q ) d q
JO*
Since the second term to the right of the equal sign is F{s)/s, as per equation 12.45a, the proper­
ty is verified.
Chapter 12 • Laplace Transform Analysis I: Basics
S80
E XA M PLE 12 .2 1 . Find the Laplace transform o f the signal/f) sketched in Figure 12.23a using
the integration property.
f(t)
FIGUIIE 12.23 (a) A triangular signaly(/) for Example 12.21. (b) The derivative
S o l u t io n
Observe that the triangular waveform y{r) o f Figure 12.23a is the integral o f the square wave^^).
Since ^t) is easily represented in terms o f steps and shifted steps as
^t) = u{t) - 2u{t - 7) + u{t - 2 7)
its Laplace transform follows from an application o f linearity and the time shift property:
x u (o i= The integration property implies that
g{q)dq
.S -
EXA M PLE 12.22. This example explores the voltage-current {v-i) relationship o f a capacitor in
the frequency domain by way o f the integration property. Recall the integral form o f the voltagecurrent dynamics o f a capacitor:
1 r'
Taking the Laplace transform o f both sides and applying the integration property produces
£[vcit)] = £
J
C->
I
1 fO-
ic(T)dx = 7T^c(-^) + —
Cs
Cs
But this expression depends on the initial condition
), because
S8]
Chapter 12 * Laplacc Transform Analysis 1: Basics
Therefore,
Cs
(12.46)
s
Equation 12.46 says that the voltage V(^s) is the sum of rvvo terms: a term dependent on the fre­
quency domain current I^^s) and a term that looks like a step voltage source and depends on the
constant initial condition V(^0~). The quantity Z^^s) = MCs looks like a generalized resistance—
“generalized” because it depends on the frequency variable s and a “resistance” because it satisfies
an Ohm’s law-like relationship, V^^s) = Z^^s) I({s). These analogies prompt a series-circuit inter­
pretation o f equation 12.46 as depicted in Figure 12.24. An application o f this equivalent frequenc)' domain circuit to general network analysis appears in the next chapter.
\,{S)
o — >+
ic(t)
O— >■
4-
£ [ •]
c ----- > VJs)
V,(t)
Cs
O 'f
O------FIG U RE 12.24 Equivalent circuit interpretarion of a capacitor in the frequency domain. This equiv­
alent is arrived at by applying the integration propert)' of the Laplacc transform to the capacitor volt­
age, seen as the integral of the capacitor current.
A second example interpreting the v-i characteristics o f the capacitor in the frequenc)' domain
ensues from the differentiation rule. Instead o f winding up with a series circuit, one obtains a par­
allel circuit. The interpretation is thus said to be dual to the one just described.
E X A M PLE 12.23. This example has two goals: (i) Verify that equation 12.46 is consistent with
the differentiation formula interpretation o f the capacitor; (ii) Build a dual frequenc)' domain
interpretation o f the v-i characteristic o f a capacitor analogous to that o f Example 12.22.
As a first step, recall equation 12.46:
Vc(-
Cs
s
which, after some algebra, becomes
I(is) = CsVfis) - CvfiO-)
(12.47)
Notice that equation 12.47 is consistent with the application o f the derivative formula to i(^t) =
C[civ(Jdt]. This consistency offers some reassurance in the accuracy o f our development. The
interpretation o f equation 12.47, however, is quite different from that o f equation 12.46. In the
latter equation, the current /^j) equals the sum o f two currents, CsV^s) and -C t/^ 0“). This sug­
Chapter 12 • Laplace Transform Analysis 1: Basics
582
gests a nodal interpretation, resulting in an equivalent circuit having two parallel branches. One
branch contains a capacitor with voltage V^s). The other, parallel branch contains a current
source with amperage Cv(\Qr). The current through the capacitive branch is
where
“G ” now acts like a generalized conductance because it multiplies a voltage, similar to Ohm’s law.
“Q ” is generalized because it depends on s. Figure 12.25 presents the equivalent circuit o f the
capacitor in the frequency domain and is dual to the circuit o f Figure 12.24. Chapter 13 covers
in detail the role o f these equivalent circuits in analysis.
Ic(s)
FIGURE 12.25 Equivalent circuit to a capacitor in the frequency domain
using the differentiation formula.
The last elementar)' property o f the Laplace transform that we consider in this chapter is the timescaling property, also called the frequency-scaling property. Its importance is fundamental to net­
work synthesis. Here, numerical problems, such as roundoff, prevent engineers from directly
designing a circuit to meet a given set o f specifications. Instead, the design engineer will normal­
ize the specifications through a frequency-scaling technique. Once the normalized circuit is
designed, frequency-scaling techniques arc reapplied in an inverse fashion to obtain a circuit meet­
ing the original specifications.
Time-/Frequency-scaling property: Let ^ > 0 and L\J{t)] = F{s). Then
L[J\at)\ = - F
a \ci)
(12.48)
or, equivalently, F{sln) = aL[f{nt)].
Since the proof o f this property is straightforward, it is left as an exercise at the end o f the chap­
ter.
E XA M PLE 12.2 4 . Figures 12.26a and b show impulse trains that model sampling in signal-pro­
cessing applications. The impulse train o f Figure 12.26b is the time-scaled counterpart to that o f
Figure 12.26a.
^83
Chapter 12 • Laplacc Transform Analysis 1: Basics
f(2t)
f(t)
A
2 --
A
2 --
i<
1
1 --
- -
> t
> t
1
1
FIG URE 12.26 (a) Unit impulse train, (b) Time-scaled imit impulse train. Unit impulse trains such
as these model sampling in signal-processing applications.
The time-scaled impulse train in Figure 12.26b increases the frequency at which the impulses
occur (twice as often as in the original signal). This is reflected in the Liplace transforms o f the
two signals:
'Z&U-k) = I
k=0
A-=0
\-e'
(12.49)
By the time-scaling property,
£ [/ (2/ )l = 0.5
l - e -0 .5 5
(12.50)
Notice that what occurs at, say, Sq in equation 12.49 now occurs at 2s^^ in equation 12.50. Hence,
time scaling by numbers greater than 1 concentrates more o f the frequenc)^ contcnt o f the signal
in the higher frequency bands.
Exercise. Verify, by direct calculation, that L\J{2t)\ is given by the right side o f equation 12.50.
Several more properties o f the Laplace transform are germane to our purpose. However, these
properties have a systems flavor and are postponed until Chapter 13. We close this section by pre­
senting Table 12.2, which lists the Laplace transform properties and the associated transform pairs.
Chapter 12 • Laplacc Transform Analysis 1: Basics
lA Bl.E 12.2 Liplace Transform I’ropertics
Property
Transform Pair
Linearity
L \ j{t-
Time shift
£\t/lt)n(t)]=-— F'{s)
as
Multiplication by t
n cr r j s )
Multiplication by t“
ds"
= Rs + //)
Frequency shift
Tim e differentiation
£ jfO )
(it
d - f{ t )
Second-order differentiation
£
= sF(s)-J{0-)
= r F c v ) - 5 / ( ( r ) - / '\ ( D
dr
wth-order differentiation
T> 0
7)1 =
d''fU)
dt’'
(i)X
fUl)dq
m dq
Fis)
Time integration
(ii) £
Time/Frequency scaling
.-fU l)d q
Fis)
£[f{at)\ = MaFista)
S8S
Chapter 12 • Liplace Transform Analysis I: Basics
8. SOLUTION OF INTECRO-DIFFERENTIAL EQUATIONS USING
THE LAPLACE TRANSFORM
Differential equarions provide a cross-disciplinary mathematical modeling framework. Although
difTerentia! equation models may represent only the dominant behavioral facets of a circuit or
physical process, their widespread utility and importance to circuits and control systems warrant
special discussion. To begin, recall the time differentiation formulas o f equations 12.41 and 12.43
and the integration formulas o f equations 12.45a and 12.45b. Also, recall that a differential equa­
tion relates a sum o f derivatives o f an output signal to a sum o f derivatives o f an input signal. For
example, if the input and output signals are voltages, then the relation
d'\-
+ a..V.
dl
for constants
dt'
+b
and bj might model the behavior o f a linear circuit. We may use the following steps
to solve this differential equation for
using the Laplace transform procedure:
1.
Take the Laplace transform o f both sides o f the equation, using the appropriate deriva­
2.
Algebraically solve the resulting expression for
3.
Compute a partial fraction expansion o f the expression for
4.
Inverse-transform the partial fraction expansion to obtain the time function
tive formulas, equations 12.41 and 12.43.
If the equation is an integro-differential equation, i.e., a mixture o f both derivatives and integrals
of the input and output signals, then we simply apply the same algorithm, except we use the inte­
gral formula where appropriate. Some examples ser\-e to illuminate the procedure.
EXA M PLE 12.25. Consider the pulse current excitation o f Figure 12.27a) to the RC circuit o f
Figure 12.27b. The goals o f this example are (i) to use and illustrate Laplace transform techniques
to solve a difTerential equation derived from a simple RC circuit and (11) to find the response volt­
r > 0, when V(^0~) = 1 V.
age
F IG U R E 12.27 Excitation currcnt (a) fora simple /?Ccircuit (b) for Example 12.25.
S o l u t io n
Step 1. Find L[i{t)]. Since i{t) = 0.5//(r) - 0.5//(^- 1),
X [/ (0 ] = 0 .5
\ -e
Chapter 12 • Laplacc Transform Analysis I: Basics
Step 2. Find the circuit’s dijfereutial equation model that links the excitation current i{() to the response
voltage, V(it). Since ij^it) = 0.5v(\t) and
= ^.‘b d v jd t, summing the currents into the top node
of the circuit yields
d\’c {t)
dt
After multiplying through by 2, the desired differential equation circuit model is
dv({f) + v c ( f ) = 2 i(t)
dt
Step 3. Take the Laplace transform o f both sides, apply the differentiation rule to the left side, and solve
for V(i{s). Applying the Laplace transform to both sides yields
sVcis) - v^Q-) + V^s) = ll{s)
Solving for V^s) produces
2
v r(0 ” )
.v+1
i’ + l
Vcis) = ----- :/(.v) + - ^ ^
\-e ~ ^
= -----------+
^(i-i-1) .v + 1
Some straightforward calculations show that
1
1 _
s
5(.v+ l)
1
(.v+1)
Thus, with the aid o f the shift propert)' and the transform pairs o f Table 12.1, we obtain
vcU) = r'[V c(s)\ = r '
.v(i + l)
V
s+\
/
= / / ( 0 - ( l- c '" ^ '“ '^)/K ^-l).
Figure 12.28 presents the graph o f this response. Because o f the initial condition and the magni­
tude o f the pulse input, the capacitor voltage is constant for 0 < r < 1 second. At r = 1 second, the
pulse magnitude drops to zero, making the circuit equivalent to a source-free RC circuit in which
the capacitor voltage decays to zero as shown in the figure.
V,(t)
F I G U R E 12.28 T h e response voltage
v^^t)
for Exam ple 12.25.
S8‘
Chapter 12 • I^aplace Transform Analysis 1: Basics
EXA M PLE 12.26. The goal o f this example is to compute the response, denoted here by the
given the scries RLC circuit o f
input current /,„(^), to the input voltage cxcitation
Figure 12.29. Suppose the initial conditions are /^(O-) = 1 A and V(^0~) = -2 V.
— TYYY
40
1H
FIGURE 12.29 Series RLC circuit for Example 12.26. Here the current //„(^) =
So
lu t io n
Step 1. Compute the Laplace transform o f the input. From the tables or by inspection, X[6(r)] = 1.
Step 2. Compute the integro-differential equation o f the circuit o f Figure 12.29. The first task is to
sum the voltages around the loop to obtain
Substituting for each o f the element voltages using the mesh current, ij„{t), yields the desired integro-difFerential equation,
diin
(12.51)
Step 3. Take the Laplace transform o f both sides, substitutefor R, L, C,
solve for
Vf^Qr), and
and
W ith the aid o f the differentiation and integration formulas, taking the Laplace
transform o f both sides o f equation 12.51 produces
/?/,„(.V) +Uli„ (,v) - L/i(0-) +^Cs
(s) +
= V;„(s)
s
This has the form
i - + S.s + 7li
L ------ 1----- ^ h „ U )
= K „ (S ) + i , / t ( 0 - ) -
V>-(0‘ )
S
S
Plugging in the required quantities and solving for
produces
I
•') ~
5 + 4 5 + 4
5 +2
~
5+ 2
~
( 5 + 2 )-
Step 4 . F in d iinit). Taking the inverse Laplace transform yields the desired result:
i d t ) = (2 - 2,)e-^-‘u(t) A
588
Chapter 12 • Liplace Transform Analysis I: Basics
A plot o f this response appears in Figure 12.30.
Exercises. 1. An integro-differential equation for an LC circuit is given by
cli,C
dl
C —oo
= 0, and V(^0~) = - 1 0 V. Compute
with C = 1 F,
L.(s) r . ( ( ) “ )
ANS>X^R: .v/,-.(.v) + -^----- + -^-------- = 0
^
.S
= > / ..(/ )= 10sui(/)//{/) A
.V
2. If two signals x{^) and^(^) are related by the equations
dxU)
dt
+ 2v(/) = 45(/) and 2.v(/)-
y{z)dz = 2ii(t)
where x(0 ) = 2 , u{t) is the unit step function, and 8{t) is the Dirac delta function, then findATi).
AN SW ER: X{s) = •V
EXA M PLE 12.27. The final example o f this chapter looks at the leaky integrator circuit o f
Figure 12.31, which contains an ideal operational amplifier (op amp).
resistance o f the capacitor. Given C and
represents the leakage
/?, is chosen to achieve an overall gain constant, in
this case, 1. The objective is to compute the response
assuming that t'(j(0~) = 0, and com­
pare it with that o f a pure integrator having a gain constant o f —1.
SH‘)
Chapter 12 • Liplacc Transform Analysis 1: Basics
R, = IMegO
V (t) = 5 u (t )
''ou.W
FIGURE 12.31 Leaky integrator op amp circuit.
So
lu t io n
First, note that since the op amp is ideal, -V(^t) =
equation that relates
to z^^-and solve for
The goal, then, is to write a difFerential
using the Laplace transform method.
Step 1. Determine the dijferential equation. Since the op amp is ideal, it follows that ijr= -i^. From
O hm s law, i^ = vJR^^. On the other hand,
^
/?2
dt
This leads to the difFerential equation model o f the op amp circuit,
dt
where, as indicated before,
I<2C
~
R^C
= -v^^t). Note that if /?,C= 1 and R2 is infinite, then the cir­
cuit works as a simple integrator. The circuit is called a leaky integrator because /?-,C is large but
finite. Since /?, C = 1, one expects the gain constant to be 1 as well.
Step 2. Substitute values, take the Laplace transform o f both sides, a7jd solve for Vg,,f{s). Taking the
Laplace transform o f both sides, one obtains
'w ( 0 " ) +
Since
=- -
= 0, it follows that
-5 0
5( 5 + 0 . 1 )
Step 3. Invert
to obtain
.V
+
50
.9 + 0 .1
Solving for v^^^^{t) produces
^92
Chapter 12 • Laplacc Transform Analysis 1: Basics
'Problems
5. Find the Laplace transform o f each o f the
following time functions.
f\{t) = Ke-‘^‘u {t- 7), r > 0
(a)
BASIC SIGNALS, SIGNAL
REPRESENTATION, AND
LAPLACE TRANSFORMS
=
(b )
(d)
(e)
r > 0
f^{t) = Kte-‘^‘u {t- 7), T > 0
(c)
r>0
=
f^{t) =
T>0
1. Find the Laplace transform o f each o f the
following signals assuming T- > 0.
6. Compute the Laplace transform for each of
the following signals.
(a)
(h)
/ 2 W = / / - 7 „ ) 6 U - 7-,)
(c)
h (D = e-^' cos(0.5rt/ + f )5(2/ -
(d)
/^W = A-,6W + 7 - „ )
=
(c)
(f)
(a)
/\{t) = Kt^[uU)-N{t- T)l T>()
(b )
f,U)
)
sin(2yTr - 2jr)u{t - 1)
(c )
K^6(t-T„)
/sW = c o s h (2 / )a (t-2 )^
= s in ( 2 . T / ) « ( r ) -
(d )
f^{t) = r [ s i n ( 2 . T / ) w ( r ) s i n ( 2 ; r / ‘ - 2jt)u{t - 1)]
f^{t) = 2 s i n ( 4 ; r r ) / / ( r ) « ( 2
-
r)
sinh(/ - 2 r , ) 6 ( f - r , )
(g)
7. Represent each o f the following signals using
fj{t) = sm{2m - Ji)d{2t - 4)
(sums of) steps, ramps, shifts o f basic signals,
2. Find the Laplace transform o f each o f the
etc. Then find the Laplace transform.
following signals. Use Tables 12.1 and 12.2 as
f,{t)
needed.
(a)
f^{t) = lu{t) + u{t - 1) + u{t - 2)
2K -
-Au{t-A)
(b)
f^{t) = 2 r { t ) - l r { t -
l ) - r ( r - 3 )
+
At - 5)
->
(c)
(d)
t(s)
j\{t) = cosiS)5m)u{t) +
c o s (0 .5 > -r(/ -2 ))« a -2 )
(a)
(e)
f,(t)
3. Sketch the indicated waveforms and find the
2K -
Laplace transform. Use Tables 12.1 and 12.2 as
needed.
(a)
(b)
(c)
(d)
(e)
f^{t)
fJ,t)
f^{t)
f^[t)
f^{t)
= ti{t) - l i t - \ ) .
= u{t)-r{t-\)-^r{t-2)
= 2u{t) u{2 - t).
= Ar{t) u{\ - t)
= 2rit)u{\ - t) + u { t - \)r{2 - t)
■>
2
(b)
A
K
4. Find the Laplace transform o f each o f the
following time functions.
(a)
f^{t) = Ktu{t-\)
(b)
/,(/) = AT/- ])«(/)
(c)
(d)
f^{t) = Ktr\t-\)
f^{t) ^ K{t - \)r{t)
r
ir
(c)
t(s)
S93
Chapter 12 • Laplace Transform Analysis I: Basics
f,(t)
2K
K
> t
■>
t
2T
T
(d)
i k f^ft)
’
-2
f,(t)
1
K
■> t
■> t
1
-T.
2
-K
(e)
f«(t)
(d)
9. Represent each of the following signals using
3 •
(sums oO steps, ramps, shifts o f basic signals,
2 ■
etc. Then find the Laplace transform.
1
^-------1-------1------ r—
1
2
3
4
■>
f,{t)
t
5
4K -
(f)
8. Represent each of the following signals using
(sums of) steps, ramps, shifts o f basic signals,
etc. Then find the Laplace transform.
2K
-I---------- '----------1--------- 1— ► t(s)
1 2
3
4
(a)
(a)
(b)
Chapter 12 • Laplace Transform Analysis I: Basics
5 94
r>
-► t
r^
(0
11. For the circuit o f Figure P I 2 .11, suppose
/?, = 6 0 0 Q, Rj = 1000 Q, and
= 1500 Q.
Use the Laplace transform tables to compute
\ltu{t) + 3e~^^u{t) + 18^’“ ^sin(2jK)«W A.
■> t
| U t)
i,(t)
(!)
Figure P I2,11
12. For the circuit of Figure P I 2 .12b, R^ = 600
Q, /?2 = 1000 Q, and R^ = 2 4 0 0 Q. Use the
Laplace transform tables to compute Vg^t^s) =
£
for the input given in Figure P I2 .12a.
10. Represent each of the following signals
using (sums of) steps, ramps, shifts of basic sig­
nals, etc. Then find the Laplace transform.
vJt)
f,(t)
2
■>
1
(a)
2
■> t
o
Figure P I2.12
(a)
13. For the circuit o f Figure P i 2 .13, suppose
/?! = 6 0 0 Q, /?2 = 2 0 0 0 Q, and R^ = 3 0 0 0 Q.
Find the Laplace transform of the voltage
when yjj(f) = 2Ae~^^u{i) V and v^2 ^t) =
30^-^Mf) V.
r>
>
t
o
r^
Chapter 12 • Laplace Transform Analysis I: Basics
(e)
Using the formulas o f part (c) and the
frequency shift property, compute
(i) L [Ke-^^ cos(wr)]
(ii) L [K e -‘^^s\n{o)t)]
17. Find a simple expression for each o f the
waveforms shown in Figure P i 2.17 in terms of
sines, cosines, shifted sines, and shifted cosines.
Then find the associated Laplace transforms.
Half-cycle of sin (nt)
LAPLACE TRANSFORMS VIA
TABLE 12.1 AND PROPERTIES
VIA TABLE 12.2
14. Find the Laplace transform o f J{t) =
Ke-‘"u{t)u{T- t), T> 0, as follows.
(a)
Express Ku{t)u( T - t) zs a difference o f
step functions.
(b)
(c)
Find £ [Kn{t)u{T- t)].
Apply the frequency shift property on
Quarter cycle of 2cos(0.25m)
your answer to part (b) to compute
£ W )l
15. Prove the time-/frequenc)'-scaling property
by direct calculation o f the Laplace transform
integral.
16.(a)
Using the famous Euler formula,^'^‘ =
cos((Ot) + j sin(wf), find an expression
(b)
for cos(a>/) and sin(wr) in terms o f the
Figure P I2.17
complex exponentials
(b)
Determine the Laplace transform o f
(c)
Using the formulas developed in parts
18.(a)
Represent
sin(7cO
0<t< 2
0
otherwise
(a) and (b), show that
as the difference o f a sine and a shifted
(i) £[co s((o r)]=
sine. Find G{s).
(ii) Xfsin(co/)] =
(d)
(b)
■>
~>
S +(D“
Using the formulas o f part (c) and the
multiplication-by-/ propert)', compute
(i)£[/irrcos(w r)]
(ii) L[Kts\n[cot)]
Relate j^r) in Figure P 12.18 to ^ r) o f
part (a). Then find F{s) from G{s).
S‘)6
Chapter 12 • Laplacc Transform Analysis 1: Basics
Figure P I2.18
19.(a)
22. (a)
Using the formulas cosh(/7r) =
and sinh(^/r) =
(i)
+
- e~“^, find
Idt
,
using the derivative property.
f(X )dx
(iii) Com putc£[//(0] = £
multiplication-by-r property, compute
(1 )X [/ T r c o s h (/ 7 r )]
(c)
as a sum o f appropri
(ii) Compute £ \g {t ) \=£ — /(/)
Using the formulas o f part (a) and the
(2)
Express
ate step functions. Compute F{s)
{\) £ [cosh(/7r)]
(2) L [sinh(^r)]
(b)
Considery(/) in Figure P i 2.22.
using the integral property.
£ [Kt sinHat)]
Again using the formulas o f part (a)
(b)
=J{t + 4).
Repeat part (a) for
and the multiplication-by-r property,
f(t)
compute
{\) £ [K r cosh(/7r)]
(2) £ [Kr
3 -
s m h ia t)]
2 -
20. Suppose/r) = 0 for r < 0 and F(.v) =
2.V + 4
s +1
■> t
Find the Laplace transforms o f the functions
1
below, identifying each o f the properties used
2
3
4
5
Figure P I2.22
to compute the answer. Solutions obtained by
finding/f) are not permitted.
(a)
(b)
(c)
(d)
23. The Laplace transform
g^{t) = 5 J { t - T), T> 0
g,{t) = 2r^^t)
g^it) = 2e-‘^ % t - D , T > 0
g^{t) = 5 i f i t - r i , T > 0
F{s) =
2 1. Use Liplace transform properties to find G/is)
4^ + 20
as eiven below when F{s) = --------- :r .
and
(.9+1)-
State each property that you used. Assume that
J{t) = 0 for ^ < 0 . Solutions obtained by finding
J{t) are not permitted.
(a) g^{t) = 0.5^r)
(b)
=
W
W)
^^W = 2 ^ , ( 2 f - 4 )
(c)
^5W = 2 ( r - 2 ) / ; - 2 )
(f)
\-e
is given as
-(s-a)
s-a
(a)
Find the Laplace transform o f e “%t).
(b)
Find the Laplace transform of tj{2t).
AN SW LR: (a)
.s'
24. The Laplace transform
F{s) =
(a)
-V
J{t) is given as
l-e "_ ^
Find the Laplace transform of
withy(0“) = 3.
cm
cit
(b)
Now find the Laplace transform o f
(c)
Finally, find the Laplace transform o f
dt
Chapter 12 • Laplace Transform Analysis I: Basics
25. Supposey(/) = S(t) —d{t — T), T > 0.
(a)
Find L\J{2t)\ by direct calculation of
597
transform o f / r ) and then, using the relation­
ship, find the Laplace transform o f^ f).
the Laplace transform integral.
(b)
Find L\J{2t)] by computing Hs) and
then using the scaling property.
26. Use only Laplace transform propenies to
answer the following question. Suppose that for
w
Vw/
O
-► t
t < 0 , / / ) = e^u{-t), and the one-sided Laplace
5 “f" 1
transform o f/^ ) is X [ / ( f ) ] = F {s) = — ^ .
s
L et^ r) =J{t)u(t). Find the Laplace transform of
v{t) when
(a)
(b)
(c)
v(t) = 2 g " {t)-g \ t)
=
t/ W = / a )+ J_ ^ q )d q
(d)
-► t
It is not necessary to have the answer be a
rational function.
27. (a)
W
O '
O
Find the Laplace transform of the
function f j )
sketched
in
Figure
P12.27a.
(b)
Identify a relationship between
and the function
Figure P i2,28
sketched in
Figure P I2.27b . Use your answer to
29. Develop a relationship betweenj^f) and g(t)
part (a) and the appropriate property
in Figure P I2 .29. Find the Laplace transform
from Table 12.2 to compute the
ofy(/) and then
Laplace transform o f ^ /).
tionship between the two functions. Assume
by making use of the rela­
that 0 < A < B < C. Also, determine D and E in
terms o f A, B, and C
O
->
o
Figure P I2.27
28. In Figure P I 2.28, what is the relationship
between j{t) and g^t) ? First find the Laplace
598
Chapter 12 • Laplace Transform Analysis I: Basics
33. Find (i) the partial fraction expansion and
(ii) the inverse Laplace transform for each o f
the following functions by hand. Show all
work. (No details, no credit.)
(a) F,Cv) =
s{s
a){sb)
C H E C K : One residue is a.
^
- 7.v^ + 4.V + 2
( b ) F ^ s ) = ---------------------------- ^
Figure PI 2.29
30. Let j{t) and ^t) be as sketched in Figure
P I 2.30. Find G{s) in terms o f F{s).
f(t)
20
10
C H EC K : One residue is - 2 .
. X
s 2 / + 18i-^ + 4 6 5 ^ + 4 4 5 + 12
(c) F2,{s ) = --------------------r---------- z----------(5 + l ) ^ 5 + 2 r
C H E C K : Two residues are at 2 and - 2 .
Remark: Check answers using MATLAB.
Use the help command to make sure you
-
understand the terms used. For example
■
for part (b),
■>
n = [2 -7 4 2];
d = conv([l -1],[1 -4 4]):
Figure P12.30
INVERSE LAPLACE
TRANSFORMS BY PARTIAL
FRACTION EXPANSION
31. Using partial fraction expansions and your
knowledge o f the Laplace transform o f simple
signals, find j{t) when F{s) equals
(a )
2 5 ^ + 1 3 r + 305 + 32
[r,p,k] =residue(n,d)
34. Find (i) the simplified partial fraction
expansion and (ii) the inverse Laplace trans­
form for each o f the following functions by
hand. Show all work.
(a) Fi(.v) =
{a + b ) s + l a b
(5 + fl)(5 + b )
Check: One residue is a.
(b ) F2{ s ) =
( a + b + c ) s ‘' + ( b e + 2 a b + a c ) s + a b c
s { s + a ) { s + b)
s{s^ + 6^ + 8)
C H E C K : One residue is a.
(b)
- s-6
Cv + 2 ) ( . v - - l )
(c)
cs~ + { a + 2 a c ) s +
2.v'^ + 12.v“ +22.y + 8
( r +25 + 1)( j + 2)
(d ) F4(s ) =
(d)
il + c ) a ^ - a
(c) F3(.v) =
/ +185^+ 9 8 5 -+ 2085+ 144
(5 + 2 ) “ (5 + 4 ) “
/ + 1 2 A - ^ - 2 4 r - 3 2 .y + 16
(e)
C H EC K : Two residues are at 2 and - 2 .
(e ) F ^ ( s ) =
55^ + 1 4 4 5 + 2 0 4
(5+1)
32. Inadvertently left out by the authors.
(5+ 2)“ + 6 4
599
Chapter 12 • Laplacc Transform Analysis I: Basics
'w '
Vg„f{s) and then find
35. F in d /r) when F{s) equals
(a)
(b)
(c)
(d)
for the
input current
25 + 16
+16
r
“ 45 + 9
/X
lin ( s ) = 2 0 -------------- 2— 2---------
245-72
5^ + 4 5 + 40
Check: One residue is at 20.
25^ + 885
(5 + 4)(5^ +64)
2 5 ^ + 2 5 ^ -2 5 -6
v.(t)
0
36. Find (i) the partial fraction expansion and
+
20
80
24 0
(ii) the inverse Laplace transform for each of
(a)
the following functions by hand. Show all
work.
, X r ., X
25^ + 125^ + 235 + 17
(5 + l)(5 + 2)(5 + 4 )
C H EC K : Residue at j = - 2 is - 1 .5 .
(b) F2(5) =
25^ + 9 5 ^ + 1 6 5 + 1 1
(5 + i)(5 + 2 y
C H EC K : Two o f the residues are 2 and 1.
5^ + 4 5 ^ - 2 5 ^ - 9 5 - 3
(C) ^3 ( 5 ) = -------------=-------- T-----
( 5 - 1 ) 2 ( ^ + 2)2
CHECK: Two residues are at I and two are at - 1 .
4 5 ^ - 1 2 5 ^ + 325 + 16
(d) /=4(5)=r
(5 + ir+ 1
38. Suppose F {s)
it follows that
(5 + 2)^ + 9
fj) =
cos{(ji)i)u{i) + Ar2^“'“sin(cflf)«(r). Find
tf, A^j, Kj, and CO. Now express J{t) =
cos{(Ot + 9)u{i) by finding K y a, (O and d.
(5 + 2)2 + 16
37. Find the partial fraction expansion and the
inverse Laplace transform for each of the indi­
39 .
The
Laplace
cated output voltages or currents. All answers
[A^jf“'^'cos(fi)?) +
transform
of f j )
=
sin(fi)r) + K^e~^^u{t) is
must be in terms o f real functions with real
coefficients or symbols. Show a ll work,
(a)
- 5 2 5 + 228
F {s)
For the circuit o f Figure P 12.37a, find
the partial fraction expansion o f
and then find
for die
(5 + 4 ) ( 5 + 0 ^ + 100
Find a, b, K^,
K y and 0).
input voltage
40. Consider the resistive circuit in Figure
P 12.40. Use Table 12.1 and the shift property
5(5 + 4 )
(b)
For the circuit o f Figure P 12.37b, find
the partial fraction expansion
for each Vj^(s) below. Sketch
t^ouM) by hand or with the help o f MATLAB.
to find
of
600
Chapter 12 • Liplacc Transform Analysis I: Basics
(a)
Kv,(.v) =
IOf'"'' +
(b)
\-e'
Vi„{s)=\0
- 5e'
..-4 5
-2 0 -
(b)
Use the Laplace transform method to
compute the inductor current,
for t > 0.
(c)
If the input is changed to
=
1 0 « ( r - 7) V, where 7"= 10 msec, find
/^(/), for t > 0. Hint: Your answer
should be a shift o f the answer com­
8kQ
puted in part (b).
-n/ V ^
8kQ
0
(d)
for r > 0.
1 8 kQ / 9kQ,
(e)
for / > 0. Hint: Can you
use superposition?
(0
If ij{0~) = 100 mA and
for t > 0.
tion expansions o f the rational functions listed
below. Then use Table 12.1 and MATLAB to
R
obtain the associated time function.
i,(t)
(c) F^{s) =
(d) 74(‘' ) =
(e) F^{s) =
= 10//(r
- 7) V, where 7 = 1 0 msec, find i/it),
41. Use MATLAB to compute the partial frac­
(b) F2(s ) =
= 10//(r)
If //(0~) = 100 mA and
V, find
Figure P i2.40
(a) F,(5) =
= 0, find
If /^(0~) = 100 mA and
3.v'^ + 3 0 5 - + 86.9 + 6 4
'■">6
.s'* + 8 i - +20.V+ 16
-O
+
v,(t)
- 4 6 .2 5 s - 6 9 2 . 8 125
.s-'* + 14.5.V- + 169.5625s + 510.25
-2.s-^ + 23.V- - 68.V - 3265
Figure P i2.42
.V - S 3 .55 " + 134.V + 797.5
10.5.r'^ + 47.875.s- + 151.875 - 108.5938
+ 6.5
+ 36.5625.v- + 101,5625.v + 2 07.0312
- 1 .5.s-'^ - 25.75.v‘^ - 127.5.y'* - 2 9 1 .5.y~ - 330^ - 143.75
.s-^’ + 10.5.v-'^ + 50.v-^ + 141.V-'' + 250.V- + 262.5.V + 1 2 5
CIRCUIT RESPONSES VIA
LAPLACE TRANSFORM
APPLIED TO DIFFERENTIAL
EQUATIONS
4 3 . For the circuit o f Figure Pi 2.43, suppose R
= 10 £2 and C = 0 .0 1 F.
(a)
Show that the differential equation lor
the circuit is
(h'rU)
1
1
^
+ ----- VcU) --------,ll
RC
RC
42. T he input to the circuit o f Figure PI 2.42 is
= 10//(r) V, valid for r > 0, and has an ini­
tial inductor current of /^(0“ ) = 0. Suppose R =
(b)
30 Q. and L = 0.2 H.
(a)
Show that the differential equation for
for r > 0 when
(c)
R. , .
1
, .
, ({) = - Vi„U)
L
L
= 10«(r) V, valid
for t > 0.
the circuit is
di l it )
(If
Use the Laplace transform method to
compute the capacitor voltage,
Now suppose the initial capacitor volt­
age ^'^O") = —10 V and
= 0.
601
Chapter 12 • Laplace Transform Analysis 1: Basics
(a)
Find V(^t) , for r > 0.
(d)
Find
the capacitor voltage,
for r > 0, wlien
(b)
= 10/^r) V.
- 1 0 and
Construct a differential equation in
Solve the differential equation by the
Liplace transform method; i.e., show
that V(it) =
R
v jt)
/> 0 and for appropriate constants A'j,
K-,, and to, which are to be found in
terms o f /y,
L and C.
vJt)
6
sin(ojf) + KjCosUot) for
-O
i,(t)
ijt)
\r
Figure P I 2.43
;r
v^(t)
44. The circuit o f Figure Pi 2.44 has rwo source
excitations, v^^{t) = 10//(f) V and
= lu{t)
A, both applied at / = 0. Suppose /?j = 5 Q, R-,
= 20 Q, and Z = 2 H. The initial condition on
the inductor current is /^(0“ ) = —I A.
(a)
ANSWl-R; (a) —
the circuit, assuming the response is
and
(c)
Find the response due only to
(e)
+
- \V' = 0 . ( b ) (I) = 7 =
LC ^
J lC
46. The circuit o f Figure P 12.46 is a series (loss­
less) LC circuit driven by a voltage source.
Suppo.se
(a)
= 0 and /^(0“ ) = 0.
Construct the differential equation o f
assuming /^(0“ ) = 0.
the circuit in terms o f the capacitor
Find the response due only to ip{t)
voltage,
assuming /^(0“ ) = 0.
(d)
dr
Construct a differential equation for
/^(r). Leave the inputs in terms o f
(b)
Fieurc P I2.45
(b)
Solve the differential equation using
Find the response due only to the ini­
the Laplace transform method, and
tial condition ;^(0~) = -1 A, assuming
show that
both inputs are zero.
for t > 0.
10 -
10
c o s(0
.5 jia )
for t
Find the complete response,
l = ih
> 0 by superposition.
^
'„ « ) Q
=
R.
0
' . ,(t)
4C=-,F
vJt)
Figure P I2.46
47. A pair o f (coupled) differential equations
Figure P i2.44
that represent a circuit are given as
M t )
45. Consider the Z,C circuit ol^ Figure P i 2.45,
for w'hich /^(O") = 7q and
= V^. Since
and
dt
there is no resistance present in the circuit,
there is no damping; hence, one expects a pure­
ly sinusoidal response. Such circuits are called
lossless.
dt
-t-^/|.v(/) = oiyit)
602
Chapter 12 • Laplacc Transform Analysis I: Basics
with initial conditions ;c(0~) = 1 andyO “ ) = 2.
Suppose , = 1 ,
= 1. ^3 = 1 . ^3 = 1. and/r)
= 2u{t). Find j/(^) and A-(r).
51. The op amp in the circuit o f Figure Pi 2.51
is assumed to be ideal. /?, = 20
R-, = 40 ki2,
and C = 10 |iF.
(a)
= 2,
48. Reconsider Problem 47 with
Use nodal analysis to construct a first-
= 2,
order differential equation describing
= 4, b^ = 3, and/r) = 2u(t).
the input-output relationship o f the
voltages.
49. The inductor current i{t) in a second-order
If u jt ) = 2uU) V, and j/c(0) = -1 V,
(b)
RLC circuit satisfies the following integro-dif-
and
ferential equation for / > 0.
then
=
Sketch the response in
MATLAB.
(a)
If v-it) = 2e~^-">‘tt{t) V and ^C^O) = 0,
(c)
v'c(() ) + 8 j^ _ //(X )f/r
find
If /(0-) = 8 A and vM~) = - 4 V, find
kis).
(b)
Use your answer in part (a) to find
/,(/).
50. Consider the circuit o f Figure PI 2.50.
(a)
Use KVL and KCL to show that the
differential equation relating the input
Figure P I2.51
voltage to the capacitor voltage is
ci\\
d\ 'c
(It-
1
RC clt
LC
''C
52. Reconsider the RC active circuit shown
LC
Figure
12.1
o f Example
12.1, where we
encountered difficult)' using the single third(b)
Take the Laplace transform o f both
order difTerential approach. Now we will solve
sides o f this equation to show that
the problem with Laplace transforms applied to
three first-order differential
_
Vc(A) =
RC
(c)
( .v + 5 ) v c ( 0 " ) -h v c (0 " )
I
1
s~ + ----- 5 +
1
•s + -----
RC
LC
Assuming that vc(0 ) =
) = 0,
equations
LC
(a)
Three node equations in the time
domain have been given in equations
^ = 0.8 Q, Z. = 1 H, C = 0.25 F, and
12.1, 12.2, and 12.3. Take the Laplace
= 5(r), show that
transform o f each o f these three node
v’c(/)
=
equations, accounting for initial con­
~
(b)
L=1H
—
► /
Y
Y
v„(t)
6
+
;±
R=0.8Q I
= lOu(t) V and the initial
capacitor voltages are ^^(O) = 12 V,
= 6 V, and t/^(0) = 3 V, find
Y
ijt)
ditions.
If
v,(t)
(d)
Now do a partial fraction expansion of
Vgut^^^ and determine
Figure P I 2.50
for /> 0.
C
H
A
P
T
E
R
Laplace Transform Analysis II:
Circuit Applications
A FLUORESCENT LIGHT
APPLICATION
Fluorescence
is a process for converting one
type o f ligiit into another. In a fluorescent
light, an electric current heats up elcctrodes
at each end o f a tube. T he hot clcctrodes
emit free electrons, which, for a sufficiently
high voltage between the electrodes, initiate
an arc, causing mercury contained in the
tube to vaporize. The energized mercury
vapor emits invisible ultraviolet light that
strikes a phosphorus coating on the inside o f the tube. The phosphorus absorbs this invisible
short-wavelength energy and emits light in the visible spectrum.
A starter circuit must quickly generate a sufficient quantity of free electrons and crcate a suffi­
ciently high voltage to initiate the arc that vaporizes the mercury inside the tube. One t)'pe o f
starter circuit contains a special heat-sensitive switch in series with an inductor. We will model this
special switch by an ideal heat-sensitive (bimetal) switch in parallel with a capacitor. The concepts
developed in this chapter will allow us to analyze the operation o f such a starter circuit as set forth
in Example 13.1 1 .
CHAPTER OBJECTIVES
1.
In terms o f the Laplace transform variable
s,
Z (j), and the notion o f admittance, denoted
define the notion o f impedance, denoted
y\s).
Impedances and admittances will sat­
isfy a type o f O hm ’s law. These ideas are generalizations o f the phasor-based notions o f
impedance and admittance introduced in Chapter 10.
2.
Learn the arithmetic o f impedances and admittances in the Laplace transform domain,
which is analogous to the arithmetic o f resistances and conductances in the time domain.
Chapter 13 • Laplacc Transform Analysis 11: Circuit Applications
3.
Apply the new concepts o f impedance and admittance to redevelop the notions o f volt­
age/current division, source transformations, linearity, and Thevenin and Norton equiv­
alent circuits in the /-dom ain.
4.
Define /-domain-equivalent circuits o f initialized capacitors and inductors for the pur­
pose o f transient circuit analysis.
5.
Introduce the notion of a transfer function.
6.
Define rvvo special types
7.
Redevelop nodal and loop analyses in terms o f impedances and admittances.
8.
of
responses: the impulse and step responses.
Utilize the l-aplace transform technique, especially the /-domain-equivalent circuits o f
initialized capacitors and inductors, for the solution of switched /?ZC circuits.
9.
Introduce the notion o f a switched capacitor circuit, which has an important place in
real-world filtering applications.
10.
Set forth a technique for designing general summing integrator circuits.
SECTION HEADINGS
1.
Introduction
2.
Notions o f Impedance and Admittance
3.
Manipulation o f Impedance and Admittance
4.
Equivalent Circuits for Initialized Inductors and Capacitors
5.
Notion o f Transfer Function
6.
Impulse and Step Responses
7.
Nodal and Loop Analysis in the j-Dom ain
8.
Switching in RLC Circuits
9.
Switched Capacitor Circuits and Conservation of Charge
10.
The Design of General Summing Integrators
11.
Summary
12.
Terms and Concepts
13.
Problems
1. INTRODUCTION
Chapter 12 cultivated the Laplace transform as a mathematical tool particularly useful for circuits
modeled by differential equations. This chapter adapts the Laplace transform tool to the peculiar
needs and attributes o f circuit analysis. W ith the Laplace transform methods described in this
chapter, the intermediate step o f constructing a circuit’s differential equation, as was done in
Chapter 12, can be eliminated.
Available for the analysis o f resistive circuits is a wide assortment of techniques: O hm s law, volt­
age and cu rren t division, nodal and loop analysis, linearit)', etc. For the sinusoidal steady-state
analysis o f RLC circuits, phasors serve as a natural generalization o f the techniques o f resistive cir­
cuit analysis. The Laplace transform tool permits us to extend the sinusoidal steady-state phasor
analysis methods to a much wider setting where transient and steady-state analysis are both pos­
sible for a broad range o f input excitations not amenable to phasor analysis. Recall that transient
an;ilysis is not possible with phasors.
Chapter 13 * Laplacc Transform Analysis II: Circuit Applications
T he keys to this generalization are the i-domain notions o f impedance and its inverse, admit­
tance. Instead o f defining impedance in terms ofyoj, as in phasor analysis, we will define it in
terms o f the Laplace transform variable
s. This
definition allows the evolution of a frequency- or
j-dependent O hm s law, j-dependent voltage and current division formulas, and ^-dependent
nodal and loop analysis; in short, all o f the basic circuit analysis techniques have analogous
s-
dependent formulations. W hat is most important, however, is that with the ^-dependent formu­
lation, it will be possible to define .^-dependent equivalents for circuits containing initialized
capacitors, inductors, and other linear circuit elements. These equivalent circuits make transient
analysis natural in the i-domain.
In the final section o f the chapter, we introduce the notion o f a switched capacitor circuit.
Switched capacitor circuits contain switches and capacitors, and possibly some op amps, but no
resistors or inductors. Present-day integrated circuit technolog)' allows us to build switches,
capacitors, and op amps on chips easily and inexpensively. This has fostered an important trend
in circuit design toward switched capacitor circuits. A thorough investigation o f switched capac­
itor circuits is beyond the scope o f this text. Nevertheless, it is important to introduce the basic
ideas and thereby lay the foundation for more advanced courses on the topic.
2. NOTIONS OF IMPEDANCE AND ADMITTANCE
Chapter 10 introduced an intermediate definition o f (phasor) impedance as the ratio o f phasor
voltage to phasor current, and admittance as the ratio o f phasor current to phasor voltage. In the
Laplace transform context, impedances and admittances are j-dependent generalizations o f these
phasor notions. Such generalizations do not exist in the time domain. To crystallize this idea, we
Laplace-transform the standard differential
v-i relationship
o f an inductor,
at
to obtain
V^is) ^ Lsliis),
assuming /^(0“ )
(1 3 .1 )
Ls multiplies an ^-domain current, /^(^), to yield
Vjis), in a manner similar to O hm s law for resistor voltages and currents.
Ls are ohms. The quantit)' Ls depends on the frequency variable s and gen­
= 0. Here, the quantit)' Z^(s) =
an j-domain voltage,
T he units o f Z^(j) =
eralizes the concept o f a fixed resistance, and it is universally called an im pedance. This complexfrequency or ^-domain concept has no time-domain counterpart.
Although the inductor served to motivate ^-domain impedance, in general an impedance can be
defined for any two-terminal device whose input-output behavior is linear and whose parameters
do not change with time. A device whose characteristics or parameters do not change with time
is called
time invariant.
606
Chapter 13 • Laplacc'Iransforni Analysis II: Circuit Applications
IMPEDANCE
T he
impedance, denoted Z{s), o f a linear time-invariant
rwo-terminal device, as illustrated in
Figure 1 3 .1 , relates the Laplace transform o f the current, /(s), to the Laplace transform o f
the voltage, V^j), assuming that all independent sources inside the device are set to zero and
that there is no internal stored energy at ^ = 0 “. Under these conditions.
V{s) = Z{s)I{s)
(13.2a)
and, where defined,
Z(s) =
V(s)
I(s)
(13.2b)
in units o f ohms.
l(s)
Device
vis)
Z(s) orY(s)
FIG U R E 13.1 A two-terminal device having impedance
Exercise.
For an unknown linear circuit,
ANSW'HR;
Vj,j{s) = —------
and
5“ + 4
Z{s) or admittance
I- (s)
^(^).
= 2. Com pute
4
—
.y" + 4
T he inverse o f resistance is conductance, and the inverse o f impcdance is admittance. For exam­
ple, if we divide both sides o f equation 13.1 by
Ls,
we obtain
Ls
This suggests that
\/Ls acts
is defined as follows.
as
a. generalized conductance universally
(1 3 .3 )
called an adm ittance, which
607
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
ADMITTANCE
T he
admittance,
denoted y(^), o f a two-terminal linear time-invariant device, as illustrated
in Figure 1 3 .1 , relates the Laplace transform o f the voltage, V(j), across the device to the
Laplace transform o f the current, /{^), through the device, assuming that all internal inde­
pendent sources are set to zero and there is no internal stored energy at f = 0~. Under these
conditions,
i{s) = ns)v{s)
(13.4a)
and, where defined,
Ijs)
(13.4b )
V{s)
in units o f S.
From equations 13.2 and 13.4 , impedance and admittance satisfy the Inverse relationship
Y(s) =
Exercise.
1
Zis)
(1 3 .5 )
16
=
For an unknown linear circuit,
'
C om pute Ky,//).
(.S-
----
arid
+ 2)
A/l(-^)="
(>v +
2 )(i + 4 )
2
ANSWER:
^+ 4
As a first step in deepening our understanding o f these notions, we compute the impedances and
admittances o f the basic circuit elements shown in Figure 13.2.
i,(t)
ic(t)
O-
O—
+
-I-
v,(t)
V ,(t)
v«(t)
o — >■
o-
(a)
FIGURi-l
(b)
1 3 .2
From O hm ’s law, the resistor o f Figure 13.2a satisfies
sides yields the obvious, K^(j) =
(c)
(a) Resistor, (b) Capacitor, (c) Inductor.
Rlj^is).
= Rij^it).
Laplace-transforming both
From equations 13.1 and 13.2, the im pedance o f the
resistor is
Zf^is) = R
608
Chapter 13 • l-aplacc Transform Analysis II: Circuit Applications
and, from equation 13.5, rhe adm ittan ce o f the resistor is
Here the kinship of impedance/admittance with resistance/conductance is clear.
rh e capacitor o f Figure 13.2b has the usual current-voltage relationship,
(!v(^{ t )
dt
/c ( 0 = C-
Assuming no initial conditions, the Laplace transform relationship is
I^s) = CsV^s)
From equation 13.4 , the ad m ittan ce o f the cap acito r is
Y(is) = a
and from equation 13.5, the im pedance o f the cap acito r is
Z eis) =
1
Cs
Repeating this process for the inductor o f Figure 13.2c ,
)=
L — ——
the im pedance and
ad m ittan ce o f the in d u cto r are
Z,,U) = U . YLU) = - jLs
Exercises. 1. Given the integral form o f the v-i capacitor relationship, assume no initial stored
energy and take the Laplace transform o f both sides to derive the impedance o f the capacitor. This
provides an alternative, more basic means of deriving the impedance characterization.
2. Given the integral form o f the
v-i
inductor relationship, assume no initial stored energy and
take the Laplace transform o f both sides to derive rhe admittance o f the inductor.
Throughout the rest o f the text, whenever we refer to an impedance the unit o f Ohm is assumed,
and similarly, admittance is assumed to have the unit of siemens (S). The units for KW and
I{s)
are usually not shown, although strictly speaking they are volt-second and ampere-second, respec­
tively.
609
Chapter 13 • Laplace* Transform Analysis II: Circuit Applications
3. MANIPULATION OF IMPEDANCE AND ADMITTANCE
Recall that the Laplace transform is a linear operation with respect to sums of signals, possibly
multiplied by constants. KVL and KCL are conservation laws stating, respectively, that sums of
voltages around a loop must add to zero and sums o f all currents entering (or leaving) a node must
add to zero. Since the Laplace transform is linear, it distributes over these sums, so the sum of the
Liplace transforms o f the voltages around a loop must be zero and the sum o f the Laplace trans­
forms o f all the currents entering a node must be zero. In other words, complex-frequency
domain voltages satisfy' KVL and complex-frequency domain currents satisfy' KCL. Because of
this, and because impedances and admittances generalize the notions o f resistance and conduc­
tance, one intuitively expects their manipulation properties to be similar. In fact, this is the case.
M a n ip u la t io n
r u le .
Because impedances map j-domain currents,
I{s),
to ^-domain voltages,
K(j), and because all /-domain currents must satisfy' KCL and all /-dom ain voltages must satisf)’
KVL:
1.
Impedances,
Z{s), can
be manipulated just like resistances and, like resistances, have units
o f ohms.
2.
Admittances, K(/), can be manipulated just like conductances and, like conductances,
have units o f S.
This manipulation rule suggests, for example, that admittances in parallel add. T he following
example verifies this property for the case o f two admittances in parallel.
EXA M PLE 13.1. C om pute the equivalent admittance,
general admittances, K,(/),
V^{s), and Y:^{s) in
and impedance,
o f three
parallel, as shown in Figure 1.3.3. Then develop the
current division formula.
Yji-s]
K,(.v)+Ko(.v)+r,(.v)
(1.3.6)
An(-v)
Z Js)
FIGURE 13.3 Three general admittances, Vjis), in parallel,
having an equivalent admittance K- (/) or impcdance
Z- (s).
(>10
Chapter 13 * Liplace Transform Analysis II: Circuit Applications
S o lution
W c seek the relationship
vvhich implicitly defines
o f the admittance o f a two-terminal device, /^(j) =
= AW
for
^3(5) =
From the definition
= 1 ,2 , 3. From KCL,
>3(5))
This relationship implicitly defines the equivalent admittance as
= y i(s)+ y 2 (s)+ y 3 (s)
y i„(s)=
affirming that admittances in parallel add. From the inverse relationship
Z;„ (5 ) = -------------- !-------------K,Cv)+K20v) + y3(.v)
Returning to the relationship
/j^(s) =
we now note that
/;,(,C) = Y / : ( S ) V , „ ( S ) = y , ( s ) Z , „ ( s ) / , ; , ( s ) =
Y^{s )+Y 2( s )+Y2{ s )
Equation 13 .6 has the obvious generalization to any number o f parallel elements.
Exercises.
1. Show that for t%vo impedances,
Z^{s)
and
Zjis),
in parallel, Z ,„(^) =
2. Show that the equivalent impedance o f two capacitors in parallel is
Z (5 )=
and that the equivalent capacitance is
‘
‘
C]^+C25
(C j+ C 2 )5
2 i ( ‘^) +
= Cj + C 2 .
3. Derive the following formula for the impedance o f two inductors in parallel:
L^ + Lo
4. A 2 |.iF and a 0 .5 uF capacitor are in parallel. Find the equivalent capacitance.
A N SW ER : 2.5
til
5. A 2 mH inductor is connected in parallel with a 0 .5 mF capacitor. Find the equivalent imped­
ance.
ANSWER: 2.^
'
.s- + i ( r "
61 1
Chapter 13 • Laplace Transform Analysis II: Circuit Applications
6. In the circuit o f Figure 13.3, suppose Kj(j) =
MR,
K,(j) =
M{L$),
and
=
Cs,
a resistance,
an inductance, and a capacitance. Derive the relationship
1
fin (s)
L C s ^ + -s +
R
the equivalent admittance,
and find
terms o f I;„{s).
in' ■
Ai\S\V1-:R: y j s ) =
I^{s)
s
1
^ ^
7. In the circuit o f Figure 13.3, suppose Kj(j) = 0 .5 ,
25+1
in
Find
^
l and /^(.v) =
EXAMPLE 13.2. Compute the input impedance o f the parallel RLC circuit sketched in Figure 13.4.
o—
+
VJs)
Z Js) =
Y Js)
FIG U R E 1 3 .4 Parallel
RLC circuit
for Example 13.2.
S o l u t io n
For parallel circuits, it is convenient to work with admittances, since
parallel admittances add.
Thus, for the circuit o f Figure 13.4,
^
1
r +—
RC
.v+'
1
LC
Since impedance is the inverse o f admittance,
1
1
C , 2 ^ _ 1L , +
RC
w hich is the equivalent input im pedance o f a parallel
1
LC
RLC circuit.
(1 3 .7 )
612
Exercises.
Chapter 13 • Laplace Transform Analysis II: Circuit Applications
1. C om pute the equivalent impedance o f a parallel connection o f three inductors hav­
ing values 4 m H , 5 m H , and 2 0 m H .
A N SW ER : 2 x 1()--S2. Com pute the equivalent impedance o f a parallel connection o f six elements: rsvo resistors, o f 6
kQ and 3 kQ; two inductors, o f 3 mH and 6 m H ; and two capacitors, o f 0 .2 |.iF and 0 .0 5 |.iF.
A N SW E R ; 4 X 1()^V(r + 2 x 1
i)-^s + 2 x 1 0 * ’)
1 he dual o f the parallel circuit o f Figure 13.3 is a series connection o f three impedances as shown
in Figure 13.5. T he following example verifies that impedances in series add, and simultaneous­
ly develops a voltage division formula.
E X A M P L E 1 3 .3 . Com pute the equivalent impedance,
general impedances,
Z^(s), Z^is),
and
Z^{s)
and admittance,
o f three
in series, as shown in Figure 13.5. Then develop the
voltage division formula,
Zj{s)
=
Z,(.v) + Z2(.v) + Z3(i-)
+ V,(s) -
V^n(.v)
(1 3 .8 )
+ V^(s)
r i d l J R E 13.5 Series impcdance circuit illustrating voltage division.
S
o l u t io n
O hm s law tell us that
(1 3 .9 )
for / = 1 , 2 , 3,. From KVL
(1 3 .1 0 )
Using equation 1 3 .1 0 and the definition o f input impedance, it follows that
Z,„ (s) = -^^4^ = Z| (s) + Z , ( i) + Z3(,s)
(1 3 .1 1 )
Chapter 13 * Laplacc Transform Analysis II; Circuit Applications
613
T he voltage division formula o f equation 13.8 follows from a modified form o f equation 1 3 .1 0 ,
and equation 13.9, to yield
Z i (5) + Z2(5) + Z3(5)
T he voltage division formula is easily extended to the case o f w devices in series:
Z (s )
Z ,( 5 ) + Z 2 ( s ) ...+ Z„(5)
Exercises.
1. C om pute the equivalent impedance o f two capacitors, C, and C j, in series.
1
AN SW I-R;
1
c
C’|CS
V
Cj + C .
2. Show that the equivalent admittance o f rwo capacitors, Cj and C-,, in series is
3. Suppose
Zj(s)
= 10 Q,
Z-,(s)
=
2s,
and
Z^(s) = 6^ in
Figure 13.5. Find
Y{s) = -------- — s.
C\ + Cl
V^-)W, and
.V
A N SW ER S:
Z,,(s)
= 10 + 8^. \ s ( .0 =
■
4.V + 5
. /-.(/) =
'
2
Zj(s) = 10 Q, Z-,(s) = 2s,
V^Js).
4. Suppose
terms o f
and
-7
+ 10.v + 2
A N SW ER S: Z,„(.v) = ---------------------,
■V
Z t^(s )
=-
=—
^
in Figure 13.5. Find
Z^-^^(s)
and K^(j) in
1
-------------- V;„(.v)
,v‘' + 5 .v + !
5. Verify that the equivalent inductance o f two inductors in series is
= -^i +
O f course, there are series-parallel connections of circuit elements that combine the concepts illus­
trated in Examples 13.1 through 13.3, as set forth next.
EXA M PLE 13.4. Com pute the input impedance Z-J,s) o f a series connection o f t\vo pairs o f par­
allel elements, as shown in Figure 13.6, in which
Then compute
in terms o f
If
=
= 10 Ci, C = 0.1
u{t),
find
Vjit).
¥,
= 5 O.,
and Z. = 1 H.
Chapter 13 • l^place Transform Analysis II: Circuit Applications
614
FIG U R E 13.6 Series-parallel connection of
S
RC elements
for Example 13.4.
o l u t io n
Conceptually, view the circuit as shown in Figure 13.7.
V,(s)
FIG U R E 13.7 Conceptual series structure of the circuit in Figure 13.6.
Here
Z ,U ) =
1
10
O.l+O.l.v
.y + 1
and
Z2{s) =
Is
7+2
in which case
10(.y + 2 )+ 2 .s;(.v + 1) _ 2^ “ + 1Is + 20
(5 + 1 )(5 +
(s + \)(s + 2)
2)
It fo llo w s th at
1//■ X
,
(.v + l )(i + 2 )
2.V
.v(.v+l)
2.v“ + 12.v + 20
-V+ 2
.v“ + 6 .y + I 0
Chapter 13 * Laplace Transform Analysis 11; Circuit Applications
Finally, if
61 S
= -,
s
(^■+1)
(-^ + 0
^2(^') = -3 ---------------= ---------- ^
.9“ + 6 . 9 + 10
( 5 + 3)" +
From Table 12.1, item 19,
Exercise.
Repeat Example 13.4 with the following changes: C = 0.01 F and /?, = 10 Q.
A N SW ER S:
Z Js )
= 10
VS(.v) =
\+10
Another basic and useful circuit analysis technique is the source transform ation property,
exhibited now in terms o f impedances and admittances. The first case we will examine is the voltage-to-current source transformation, illustrated in Figure 13.8,
(b)
(a)
FIG U R E
1 3 .8
Z^[s), as shown in
Z^(s), as shown in part (b).
Illustration of the transformation of a voltage source in series with
part (a), to an equivalent current with a current source in parallel with
Often, voltage-to-current source transformations provide an altered circuit topology that is more
convenient for hand or calculator analysis. Mathematically, the goal is to change the structure o f
a voltage source in series with an impedance to a current source in parallel with an admittance
while keeping both
Vjis)
and / 2 W fixed. To justify this, one starts with Figure 13. 8a, in which
voltage division implies
V,(.v) = ^
^ V^^is)=Z.is)l2is)
Z ,(5 )+ Z 2 (5 )
H en ce,
ZAs) ^ 0 ,
V ;(.v) =
Z2(.V)Z, is )
1
(V iu (s)\
Z,(.v)+Z2(.v) I 2 ,(5 )J ■r,(5)+K2(.v) U i„(.v )j
(1 3 .1 2 )
Chapter 13 • U p lace Transform Analysis 11: Circuit Applications
616
where
Yjis)
=
[Zj{s)]
^ This equation identifies the parallel structure of Figure 1 3 ./b ; i.e., Figure
13. 8b is a circuit equivalent o f equation 13.8.
Reversing these arguments leads to the current-to-voltage source transformation, illustrated in
Figure 13.9.
(a)
(b)
FIG U R E 13.9 Illustration of (a) current source to (b) equivalent voltage source transformation.
Clearly, the manipulation of impedances and admittances parallels that o f resistances and con ­
ductances, as suggested earlier. Indeed, for a rigorous statement o f the soiuce transformation tech­
nique developed above, refer to the source transformation theorem in Chapter 5 and replace
by
Z{s),
by
and
by
Ijp).
R
Indeed, all such values in Chapters 5 and 6 have i-domain
counterparts.
This section ends with a demonstration of finding a Thevenin equivalent in the /-dom ain.
E X A M P L E 1 3 .5 . C om pute the Thevenin equivalent circuit o f Figure 13.10.
VJ s )
V Js )
v„(s)
(b)
FIG U R E 1 3.10
SOLUTIO N
From the material in Chapter 6, our new concepts o f admittance and impedance, and Figure
13.10b,
(13.13a)
o r equivalently,
o s-B b )
Chapter 13 * L^placcTransform Analysis II: Circuit Applications
61
Now from Figure 13.1 Oa,
(d + 1)/(^(.V)+ GV'j-^C.v)
/(;’ (.v)+ CV^„(5) =
= {ii + i).vc[v;„(.9) - \/^(^)] + gv;„(.9)
= I {a + 1).vC + C ]
(13.14)
( 5 ) - (« + 1).vCV/^(.v)
Rewriting equation 13.14 in the form o f equation 13.13a, we have
1
(a + \)sC + G
(13.15)
'
(W +D.9C + G
Com paring equations 13.15 and 13.13a, we identify
(a + 1).vC + G
Exercises. 1.
In Example 13.5, what is the Norton short circuit current,
A N SW ER : /.^.(.v) = U +
2. Find
(fl + 1).?C + G
and
\)s(:Vjj,s)
VgJ<s) for
the circuit in Figure 13.11.
2-2t;\
..
2/ / , (. s)
2 + 2.V +
FIG U RE 13.11
3. For the circuit o f Figure 13.12, use source transformations to find I^p) and Y^jj^s) for the indi­
cated terminals.
.
A N SW ER S: I. is) = 0 . 2 .a ' (>■) and
= 0.2.^ + - + 0.4
0.2 F
(V Js )
2.5 Q
1 H
F IG U R E 1 3 .1 2
618
Chapter 13 * Laplace Transform Analysis 11: Circuit Applications
4. EQUIVALENT CIRCUITS FOR INITIALIZED INDUCTORS AND
CAPACITORS
T he notions o f impedance, admittance, and transfer function do not account for the presence of
initial capacitor voltages and initial inductor currents. Hoiv can one incorporate initial conditions
into various analysis schemes? For an answer vve look at the transform o f an initialized capacitor
and inductor and interpret the resulting equation as an equivalent circuit in the complex-frequency domain. For the capacitor and the inductor, rvvo equivalent circuits result for each; a series
circuit containing a relaxed (no initial condition) capacitor/inductor in series with a source, and
a parallel circuit with a relaxed capacitor/inductor in parallel with a source. Example 1 2 .2 3 pre­
viewed this notion.
T he capacitor has the standard voltage-current relationship
c
^
(it
= /c (0
Taking the Laplace transform and allowing for a nonzero initial condition
Cs
yields
- Cv(J,Q~) = I (is)
(1 3 .1 6 )
T he left side of equation 1 3 .1 6 is the difference o f two currents, one given by the product o f the
capacitor admittance and the capacitor voltage
{CsVf^s))
and the other by Cy^^O"). Thus the cir­
cuit interpretation o f equation 1 3 .1 6 consists o f a relaxed capacitor in parallel with a current
source, as illustrated in Figure 13.13. In the time domain the current source o f Figure 1 3 .1 3 cor­
responds to an impulse that would immediately set up the required initial condition.
1^(5)
^
.....................................................
V,(s)
Cs
FIGllRK 13.13 Parallel form of an equivalent circuit for an initialized capacitor. Here, the capacitor
within the dotted box is relaxed while the current source
Cv(4S)~) accounts
for the initial condition.
Rearranging equation 1 3 .1 6 yields
Cs
s
(1 3 .1 7 )
Example 12.22 previewed this equation by taking the transform o f the integral relationship o f the
capacitor. We observe that the right-hand side o f equation 1 3 .1 7 is the sum o f two voltages, one
o f which is the product o f the capacitor impedance and the capacitor, current and the other
z/^(0“)A'. Thus, the interpretation is a series circuit, as sketched in Figure 13.14.
Chapter 13 • Laplace Transform Analysis II: Circuit Applications
619
V Js)
FIG URE 1 3.14 The series form of an equivalent circuit for an initialized capacitor. Here the capacitor
in the dotted box is relaxed, and the voltage source accounts for the effect of the initial condition.
Initialized inductors have similar j-domain equivalent circuits analogous to those o f the capaci­
tor. W ith the voltage and current directions satisfying the passive sign convention, the differen­
tial inductor current-voltage relationship is
Transforming both sides yields
( 13 . 18 )
Again, this equation consists o f a sum o f voltages,
Lsl^is)
and - Z /^ ( 0 ). Thus equation 1 3.18 can
be interpreted as a series circuit, as depicted in Figure 1 3 .1 5 .
FIG U R E 13.15 Series form of equivalent circuit for an initialized inductor. Here the inductor with­
in the dotted box is relaxed; notice the polarity orientation of the voltage source.
To construct a parallel equivalent circuit, divide equation 1 3 .1 8
Ls
s
by Ls and
rearrange to obtain
(1 3 .1 9 )
T he right side o f Equation 13.19 is a sum o f currents that determines a parallel equivalent circuit,
as sketched in Figure 13 .1 6 .
Chapter 13 * Laplace Transform Analysis II: Circuit Applications
620
F'lGURli 13.16 Parallel form of equivalent circuit for an initialized inductor.
Again, the inductor inside the dotted box is relaxed.
Two examples illustrate the use o f these four equivalent circuits for initialized capacitors and
inductors.
E X A M P L E 1 3 .6 . This example illustrates an ^-domain application o f superposition. In the /?Z,Ccir­
= Au{t) V.
cuit o f Figure 13.17, suppose ^(--(0“ ) = 1 V, /^(0“ ) = 2 A, and
Find Vj{t) for f > 0.
+ vjt) --------- (-
F K iU R E 13.17 Circuit for Example 13.6.
S
o l u t io n
In this example, it is convenient to replace the capacitor by its (series) 5-domain voltage source
equivalent circuit, because the capacitor is in series with the input voltage source. On the other
hand, it is convenient to replace the inductor by its (parallel) /-dom ain current source equivalent
circuit, because the desired output is the inductor voltage. This results in a three-source or multi­
input circuit. O nce the equivalent circuits are in place, one can apply superposition to obtain the
answer, although there arc many other ways to solve the problem.
Using the voltage source ynodelfor the capacitor and the current source modelfor the induc­
tor, draw the equivalent s-domain circuit. Using the equivalent circuits o f Figures 1 3 .1 4 and 1 3 .1 6 ,
Step 1.
we obtain the circuit of Figure 13.18. Here we note that
V;„(.v) = - .
s
---- = - ,and
s
s
-----= - .
s
s
Chapter 13 * Laplacc Transform Analysis II: Circuit Applications
+
621
>
—O
+
1 .5 0
V,(s)
0.5s
FIG U R E 1 3 .1 8 j-domain equivalent accounting for initial conditions of the circuit of Figure 13.17.
Step 2 . Fmd the contribution to
from
From voltage division,
S- + 3S + 2
I.5 + - + 0.55
s
Step 3. Find the contribution to
from
^ = - . Again, from voltage division,
s
s
V l(s) = —
1
-.y
^
s^ + 3s + 2
X —=
.5 + - + 0.55
s
Step 4 . Find the contribution to Vj{s) from Z,;'^(0 ) = 1. Using O hm ’s law in the 5-domain,
0.5^ 1.5 + -
s/
V t{s) = -
1.5 + - + 0.55
5
2
-3 s - 2
^
^“ + 3^ + 2
--------------
X — =
Step 5. Su?n the three contributions and take the inverse transform.
V i(s) = v l u ) + v l { s ) + vl(.s) = -
-2
- + 35 + 2
2
2
-V+ 2
.V+ 1
in which case
Vj{t) =
Exercise. Find Ij{s) and
2e
^hi{t)
- le
^u{t) V
ij{t) for the circuit o f Figure 1 3 .1 7 using the equivalent circuits o f Figures
1 3 .1 4 and 1 3 .1 5 . Hint: Write one loop equation.
ANSW FR:
/ ,(,)=
./•/(/) = M ’- ' i a i ) - 2( " “ '/M/)
( .V
I )(.V + 2 )
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
622
E X A M P L E 1 3 .7 . This example illustrates a single-node application o f nodal analysis. In the
circuit o f Figure 13.1 9 , suppose
V(^Qr)
= 1 V, /^(0“ ) = 2 A, and
=
n{t) V.
Find
V(^t)
RLC
t>
for
0.
0.5 H
/Y Y \
-o
1.5 Q
+
v,(t)
v Jt)
1F
MGURH 1 3.19 Circuit for Example 13.7.
S
o l u t io n
In this example, it is convenient to replace the inductor by its (series) coniplex-frequency domain
voltage source equivalent circuit, because the inductor is in series with the input voltage source.
On the other hand, it is convenient to replace the capacitor by its (parallel) complex-frequency
domain current source equivalent circuit, because the desired output is the capacitor voltage. This
results in a three-source, or multi-input, circuit. O nce the equivalent circuits are in place, one can
combine the voltage sources and write a single node equation to find
V(As).
Using the voltage source model for the inductor and the current source ynodelfor the capaci­
tor, draw the equivalent complex-frequency domain circuit. Using the voltage source equivalent for
Step 1.
the initialized inductor and the current source equivalent for the capacitor produces the circuit o f
Figure 13.20a. Combining the voltage sources and the series impedance into single terms results
in the circuit shown in Figure 13.20b.
0.5 i^(O-) = 1
/ Y Y V
1.50
_ Q
0.5 s
CvJO-) = l
(a)
FIG U R E 1 3.20 (a) Complex-frequenc)’ domain equivalent accounting for initial conditions o f the
circuit of Figure 13.19. (b) Circuit equivalent to part (a) with voltage sources combined.
Chapter 13 * Laplacc Transform Analysis II: Circuit Applications
Step 2.
Write a single node equation for V(^s).
623
Summing the currents leaving the top node o f
yields
1
V c ( s ) ---------------- l + 5V ’c ( 5 ) = 0
s
1.5 + 0.5.y
Grouping terms produces
1
V c is )= -:
----- r+ 1
5 (0 .55'+ 1.5 )
+ 5
U . 5 + 0.5^
V(\s)
Solving for
leads to
s~ + 5s + 2
Vcis) =
5
( 5 + 0 ( 5 + 2)
Exemte a partialfraction expansion on V(^s), and take the inverse transform to obtain V(^t).
Step 3 .
Using the result o f step 2,
5^ + 55 + 2
1
2
-2
^(7(5) = -------------------- = — I----------- f-
s{s +
1) ( 5 + 2 )
S
.V +
1
5
+ 2
Inverting this transform yields the desired time response,
v^t)
Exercises.
le-'-2e-^~^u{i)V
1. In Example 13.7, change the resistance from 1.5 H to 2 .2 5
A N SW ER :
2. Find
= [1 +
v^p)
I^{s)
and
Find
V(\t)
for r > 0.
= [1 » 0.57l4i>-'^-^'- 0 .5 7 l4 f -^ q « (/) V
i^it)
for the circuit o f Example 13.7, using the equivalent circuits o f Figures
13.14 and 13.15. Hint: W rite one loop equation.
TV
A N SW ER :
1,
EXA M PLE
1 3
( 5 ) = -------- ----------- .
( 5 + I ) ( 5 + 2)
i,{t)
^
=-
le -‘ii{t)
+
Ae--'u{t)
. 8 . T he chapter opened with a discussion o f the operation o f a fluorescent light
with classical starter, com m on in residential usage. For a fluorescent light to begin operating, there
must be a sufficient supply o f free electrons in the tube and a sufficiently high voltage between the
electrodes to allow arcing to occur. During arcing, mercury particles in the tube vaporize and give
off ultraviolet light. The ultraviolet light excites a coating o f phosphorus on the inside o f the tube
that emits light in the visible range.
For a simplified analysis, assume that all resistances are negligible and refer to Figure 1 3 .2 1 . The
source
VjJ^t)
is 120 V, 6 0 Hz, i.e., ordinary house voltage, which is too low to cause arcing inside
the fluorescent tube. Prior to arcing the gas inside the fluorescent tube acts like a very large resist­
ance betvN'een the rwo electrodes. W hen the switch is turned on, the starter, a neon bulb with a
bimetallic switch inside, lights up and heats the bimetallic strip. This causes the metal to curl and
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
624
close the contacc. The bulb then looks like a short circuit, and a large current, limited by the
inductive ballast, flows through the heating electrodes o f the fluorescent tube, making them bet­
ter able to emit electrons. During this time the neon bulb is shorted out and the bimetallic strip
cools and opens the circuit after a few seconds. At this point in time, which we will call ^ = 0 , the
inductor has an initial current
Because o f the Z.Ccombination, a very high voltage will then
appear across the electrodes of the lamp, resulting in ignition or arcing. After the lamp ignites, the
voltage between the electrodes becomes “small” and is insufficient to relight the neon starter lamp.
Hence, the ac current flows between the two electrodes inside the fluorescent tube. The ballast
again serves to limit the current.
Heating
Direction
of curl when
FIGIIRK 13.21 Wiring diagram of simple fluorescent light circuit, including an inductive ballast, a
capacitor, and a starter within which is a neon bulb containing a bimctallic switch.
Suppose
L = 0 .8
H , C = 1 nF, and
= 0.1 A. For r > 0, we find the com ponent o f
due
to the initial inductor current, i.e., the zero-input response. The other com ponent, the zero-state
response, is not as important for ignition purposes. O ur strategy will be to use the ^-domain equiv­
alent circuit for
L,
as illustrated in Figure 13 .2 2 .
0 .8 s
Voltage
due to
I,(S)
Li,(0)
= 0.08
intial
inductor
current
'....................................................
•
u- u
•*
High resistance
prior to
arcing
FIGUKI! 13.22 Equivalent complex-frequenc)' domain circuit immediately
prior to arcing and normal lamp operation in fluorescent lighting.
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
625
Since we are assuming that all resistances are negligible and that the internal resistance (between
electrodes) o f the fluorescent lamp prior to arcing approximates infinity, voltage division in terms
o f impedances yields
— + Ls
Cs
^
r + —
J l .25x10^
r
+ 1 .2 5 x 1 0
LC
J l . 2 5 x lo '
2 ,8 2 8 ^
o
. r + 1.25 X 10^
Hence, immediately prior to arcing, the capacitor voltage approximates
= - 2 ,8 2 8 sin (3 5 ,3 5 5 /) V
which is sufficiently high to induce arcing and cause the fluorescent lamp to operate.
See the homework exercises for an extension o f this analysis to the case where the ballast model
includes a resistance o f 100
Q.
5. NOTION OF TRANSFER FUNCTION
Besides impedances and admittances, other quantities such as voltage gains and current gains are
critically im portant in amplifiers and other circuits. T he term transfer fu nction is a catchall
phrase for the different ratios that might be o f interest in circuit analysis. Impedances and admit­
tances are special cases o f the transfer function concept.
TRANSFER FUNCTION
Suppose a circuit has only one active independent source and only one designated response
signal. Suppose fiirther that there is no internal stored energy at f = 0~. T he transfer func­
tion o f such a circuit or system is
H{ s) =Thus if the input
X fdesignated response signal
--------- r------ . ■ V
(13.20)
£ [designated input signal
and the response is^(f), then y(^) =
which is a handy for­
mula for computing responses. Notice that if the input is the delta function, then
and
Y{s)
=
H{s).
F{s)
= 1
This means that the transfer function is the Laplace transform o f the so-
called im pulse response o f the circuit, i.e., the response due to an impulse applied at the
circuit input source when there are no initial conditions present. The idea is easily extend­
ed to multiple inputs and multiple outputs to form a transfer function matrix. This exten­
sion, however, is beyond the scope o f this text.
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
626
Exercise.
A transfer function o f a particular circuit is
H(5 ) =
response. Hint: Review Table 12.1.
—. Find the impulse
(•5' + ^^)
A N SW ER : ^'-‘"[cos(Z^/) + S sin(Z^r)]/Hf)
A transfer function, as defined by equation 1 3 .2 0 , has broad applicability to electrical and electro-mechanical systems. For example, the designated output may be a torque while the input
netiuork
driving point impedance, where the
the current source; (ii) driving point
might be voltage. However, in the context o f circuits, a transfer function is often called a
function.
T he literature distinguishes four special cases: (i)
input is a current source and the output is the voltage across
admittance,
where the input is a voltage source and the output is the current leaving the voltage
source; (iii)
transfer impedance,
where the input is a current source and the voltage is across a des­
ignated pair o f terminals; and (iv)
transfer admittance,
where the input is a voltage source and the
output is the current through another branch in the circuit. In cases (i) and (iii), the voltage polar­
ity must be consistent with the conventional labeling o f sources as set forth in Chapter 2. In gen­
eral, however, we will adopt the ordinary language o f transfer function.
EXA M PLE
1 3
. 9 . T he circuit o f Figure 1 3 .2 3 has elements with zero initial conditions at f =
0 “ Find
^out
V :Js)
S
o l u t io n
There are many ways to solve this problem. O ur approach is to execute a source transformation
on the
R-L
impedance in series with the voltage source. After the source transformation, we use
current division to obtain the necessary transfer function.
Step 1.
Execute a source transformation to obtain three parallel branches as per Figure 13.24.
627
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
FIG U R E 1 3 .2 4 Circuit equivalent to Figure 13.23 after a source transformation.
This circuit has the parallel structure o f Figure 13 .2 5 .
,(S)
V Js ) r A
©
Y,
sT T
FIG U R E 1 3 .2 5 Parallel admittance form of Figure 13.24.
Step 2 . Use current divisiofi. Since the output current,
^ current through one o f three
parallel branches, the current division formula (equation 13.9) applies, producing
>3(^)
M l
y,(A-)+r2(.s') + >3(-^'V 5 + 1
Hence,
H(s) =
Y^is)
(1 3 .2 1 )
U W + i 2 ( > ^ ) + W / s+ 1
Vinis)
Step 3 . Compute K, (^), Y2 {s), and Y^^is). Because impedances in series add, and admittance is the
inverse o f impedance (equation 13 .7 ), some straightforward algebra yields
K,(.v) = ----- K2 (.v) = ------------. = ----- ^3(5) =
S+\
5+1
5
2.5s
0 .4 5 +
, 2^1
0.4
Step 4 . Substitute into equation 13.21 to obtain H{s):
2.5 s
H(s) =
1
5+1
2.5s
1
5^ +1
2.55
+ - —I— X---- \ 5 + 1/
5
5+1
5
“ + 1 + 5 ( 5 “ + 1) +
s +l
2.5s
2.5s
(5 + 1 ) ( 5 - +
2.5s +
1)
(s +
1)(^ + 0 . 5 ) ( 5 + 2 )
2
.5 5 ( 5 + 1)
(i2«
Exercise.
Chapter 13 • Laplace Transform Analysis II: Circuit Applications
For
A N S W FR :
H{s)
as computed in Example 13.9, find the so-called impulse response
//{/) =
-
--C
-0 .5 /
-2 /
h{t)
=
u(t
3
E X A M P L E 1 3 .1 0 . C onstruct the transfer function o f the ideal operational amplifier circuit o f
Figure 13 .2 6 , where
Zp)
and
Ip)
denote a feedback impedance and feedback current, respec­
tively.
V,„(s)
FIG U R E 1 3 .2 6 Simple ideal operational amplifier circuit for Example 13.10.
S
o l u t io n
Since no current enters the inputs o f an ideal op amp, I - p ) = - I p ) . Further, the voltage at the
negative op amp terminal is driven to virtual ground; hence,
V-p)
=
Z -p )I-p ),
and
Vg^,f{s)
=
Z p ) I p ) . Combining these relationships with I - p ) = - I p ) yields
Vi„{s)
Zj„{s)
Yj{s)
(1 3 .2 2 )
Equation 1 3 .2 2 is a verv' handy formula for computing the transfer functions and responses o f
many op amp circuits.
Exercises.
Find R so
1. In the circuit o f Figure 1 3 .2 6 , suppose
that the transfer function is
A N SW E R :
R=
H{s)
=
-Ms,
Z p)
is the impedance o f a 0.1 mF capacitor.
i.e., an inverting integrator.
1{) kQ
2. In the circuit o f Figure 1 3 .2 6 , now suppose
a 0 .2 m F capacitor, and
the transfer function,
Zis)
consists o f a 10 kQ resistor in parallel with
consists o f a 4 0 k ti resistor in parallel with a 0 .4 m F capacitor. Find
the dc gain, and the gain as x
A N SW E R : -{s + 0.5)/(2y + 0 .1 2S).
-O.S
oo.
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
629
3. Find the value o f C for which the transfer function o f the op amp circuit in Figure 1 3 .2 7 is
H {s) = ----------- ---------- .
(s + 2)is + 4)
A N SW ER : C = 0 .5 F
0.2 5 Q
FICJURE 1 3 .2 7 Op amp circuit.
E X A M P L E 1 3 .1 1 . T he ideal op amp circuit o f Figure 1 3 .2 8 is called a leaky ifjtegrator. If the
input to the leaky in tegrator circu it is v-^^) = e~^u{t), find the values o f /?,,
an output response
Rj,
and Cleading to
= -lte~ ‘u{t), assuming that I'fjCO") = 0.
R,
FIG U R E 1 3 .2 8 Ideal operational amplifier circuit known as the leak)' integrator.
S
o l u t io n
Step 1.
From the given data, compute the actual transferfunction o f the circuit.
By definition o f the
transfer function,
2
H{s) =
L[resp(mse\
Voui{s)
(s + l)“
2
£\input\
Vi„(s)
1
.y+1
5+1
( 1 3 .2 3 )
(i3 0
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
Step 2 . Using Figure 13.28, fin d the transferfiinction o f the circuit in terms o f R^, Rj, and C. Here,
obsen^e that Figure 1 3 .2 8 has the same topolog)^ as Figure 13 .2 6 , where
Ky.(,9)=— !— =
c j+
Z /(.v )
-^
w
'
R2
'
«i
From equation 13.22 o f Example 13 .1 0 ,
Cs + Step 3 . Match coejfcients in equatiotis 13.23 and 13.24a to obtain the desired values ofR^, Rj, and
C. Equating the coefficients yields
_1_
Cv+—
Rl
O ne possible solution is
R^ =
0 ,5 Q, /?, = 1 Q, and C = 1 E If we rewrite equation 1 3 .24a as
H (s)= -
(1 3 .2 4 b )
CRj
other solutions are also possible. For example, for any
=
> 0,
^2new ~
^new
represents a valid (theoretical) solution. In Chapter 14 we encounter a concept called
magnitude scaling.
is called a magnitude scale factor, which leaves this transfer function
unchanged but produces more realistic values for the circuit elements.
Exercises. 1. In equation 13.24b, it is required that C = 10 uE Find appropriate values o f /?j and R-^.
AN SW ER:
R^ =
50
kLl R,
= 100
kLl
2. Given equation 13.24b , compute h{t) = L [//(^)].
A N SW ER :
_JL
t
H,C
tl(l)
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
631
6. IMPULSE AND STEP RESPONSES
H{s), with ^-domain input
Y{s) = H W F{s). Assuming that
Suppose a circuit or system lias a transfer function representation
denoted by
F{s)
and j-domain output given by y(y) in which case
all initial conditions arc zero, if /(f) =
^{t),
then the resulting^(r) is the system im pulse response.
Some simple calculations verify that the transform o f the impulse response is the transfer function,
1.e.,
X[y(r)] =
H {s)m t)]
=
His)
Hence, the impulse response o f the circuit/system, denoted /;(/“), is the inverse transform o f the
transfer function
Ht) ‘ £-'[M(s)]
(13.25a)
H{s) = £[h{t)]
(1 3 .2 5 b )
and conversely
These equivalences represent another use o f the transfer function concept.
Exercises.
l.T h e transfer function o f a certain linear network is
H{s) = (s + 5)l[{s + 1)(j + 2)].
Find
the impulse response o f the network.
A N SW E R :
[lc^‘ -
2. If the impulse response o f a circuit is a pulse ^(f) =
u{t) - u {t- T), T > 0,
compute the transfer
function.
A N SW ER : (I
3. Suppose
- e^'^)/s
t/{t) = 2b{r-
ing an impulse response
ANSXXHER: y(r) =
2hU -
1) - 3 6 (/‘- 3) is the input to a relaxed (zero initial conditions) circuit hav­
h{t)
=
2u(t) - 2u{t—5).
Find the output ^(f).
1) - 3A(r - 3)
Why is the impulse response important? hs
we will see, it is because every linear circuit having con­
stant parameter values for its elements can be represented in the time domain by its impulse
response. This is shown in Chapter 15, where we define a mathematical operation called
tion
convolu­
and show that the convolution o f the input function with the impulse response function
yields the zero-state circuit response. In addition to this significant theoretical result, the impulse
response is im portant for identification o f linear circuits or systems having unknown constant
parameters. Sometimes a transfer function is unavailable or a circuit diagram is lost. In such a
predicament, measuring the impulse response on an oscilloscope as the derivative o f the step
response is quite practical.
What is the step response o f a circuit? T he step response is merely the zero-state response o f the cir­
f^t) to the circuit is « (/), then F{s) = 1/^ and K(j)
= H[s) (1/^). By the integration propert}’ o f the Laplace transform, it follows that the step response
cuit to a step function. Observe that if the input
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
032
is the integral o f the impulse response. Conversely, the derivative o f the step response is the
impulse response. In lab, many scopes can display the derivative of a trace and hence can display
the derivative o f the step response, which is the impulse response. Alternatively, a homework prob­
lem will suggest a means o f directly generating an approximate impulse response.
Exercises.
1. If the transfer function o f a circuit is
H{s)
= 1/j, what arc the impulse and step
responses?
2. If the Laplace transform o f the step response o f a circuit is given by
Y{s)
= I/fi'U + I)], what is
the impulse response?
3. If the step response o f a circuit
\sy{t)
= [1 - 0.5^’“ “^-
cos(2r)]//(r), what is the impulse
response?
u(t),
A N SW ER S: in random order: H/).
cus(2/ + 2 6 .5 7 ‘')//(r).
E X A M P L E 1 3 .1 2 . Figure 13.29a shows the impulse response o f a hypothetical circuit. If an input
= b{r)
+
b {r -
1), com pute the response,^(f).
y(t)
A
h(t)
3
A
2
-
2
1
--
1 H-----
1
2
1
(a)
2
3
(b)
FIG URE 13.29 (a) Impulse response of hypothetical circuit, (b) Response to 6(r) +
b {t-
1).
S o lu t io n
Since X [6(r)
+ 6(/- 1)]
= I +
e~\ the
response,
is simply the sum
o f /}{[) and h (t- \)u {t-
1).
Doing the addition graphically yields the waveform o f Figure 13.25b.
E X A M P L E 1 3 .1 3 . The response o f a relaxed circuit to a scaled ra m p ,/(/) =
= ( - 6 + 4r + 8 e~' S
Com pute the impulse response,
is given by;/(^)
h{t).
o l u t io n
T he relationship between/( /) and
b{t)
identifies the strateg)' o f the solution. If the step function
is the integral o f the delta function and the ramp the integral o f the step, then the delta function
equals the second derivative o f the ramp. Hence,
the impulse response
h{t)
y\t)
b{t)
= 0 .1 2 5 /" (r ). By the linearity o f the circuit,
= 0.1 2 5 7 "(^ ), and some straightforward calculations produce
= [4 -
+ [ - 6 + 4 / + 8 ^ * -'-
(t)
Chapter 13 • Laplace Transform Analysis 11: Circuit Applications
y"{t)
But the right-hand term is zero. (Why?) Hence,
633
= [Sf*
and
hit) = [e-^-
To see the utility o f this approach, try the alternative method o f computing
V(s)/f(s).
F(s), V(s),
and
H(s) =
The algebra is straightforward, but tedious and prone to error.
As a final example, we compute a circuits step response and verify that its derivative is the impulse
response.
E X A M P L E 1 3 .1 4 . C om pute the step response o f the
RLC circuit
o f Figure 13 .3 0 .
/m
R = 40
v
. »
+
L=1H
Q
C = 0 .2 F
-o
FIG U R E 13.30
S
circuit for Example 13.14.
o l u t io n
From voltage division.
1
5
s~ -\ --s+ —
L
LC
f.v + 2 ) “ + l
1
Cs
/^ + L .V + —
Cs
(1 3 .2 6 )
From equation 1 3 .2 6 , the Laplace transform o f the step response is
His)
1
5
= - +—
.y (.y + 2 ) “ + l l
-.V - 4
(.v + 2)^ + l
Rearranging terms yields
.V
(5 + 2 ) - + 1
{s + 2)-+\
Taking the inverse transform produces the desired step response:
= [1 -
cos(f) -
2e-~‘ sin(f)];K r)
(1 3 .2 7 )
fi3^
Chapter 13 • Laplacc IVanstorm Analysis II: Circuit Applicarions
As a check, obser\'e that the derivative o f equation 1 3 .2 7 is
—
, , ( / ) = 2 e ""^ [co s(/) + 2 s i n ( 0 k ( 0 -
(It
= 5e
+ 2 c o s(/)]« /(/) + (l - 1)(5(/)
sin{t)u{t)
Thus
5
£
Cv + 2 ) - + 1
in which case
h{{) =
5^’" “^sin(r)«(f)
as expected.
6. NODAL AND LOOP ANALYSIS IN THE S-DOMAIN
This section develops ^-domain formulations of node and loop analysis. Nodal analysis o f circuits
builds around KCL, whereas mesh/loop analysis utilizes KVL. In Chapter 3 and, indeed, in most
beginning courses on circuits, loop and nodal analysis are taught first in the context o f resistanc­
es and conductances and then (in Chapter 10 here) in the phasor context. Recall that KCL
requires that the sum o f the currents leaving any circuit node be zero. Further, KVL requires that
the voltages around any loop of a circuit sum to zero. By linearity, the Laplace transform o f a sum
is the sum o f the individual Laplace transforms. Hence, a KVL equation and a KCL equation have
an j-domain formulation where elements are characterized by impedances and/or admittances.
For loop analysis, one writes a KVL equation for each loop in terms o f the transformed loop cur­
rents and element impedances. The set o f all such equations, then, characterizes the circuit’s loop
currents, which determine the ciu rents through each o f the elements. Knowledge o f the loop cur­
rents and the element impedances permits the computation o f any o f the element voltages.
In nodal analysis, one writes a KCL equation at each node in terms o f the Laplace transform o f
the node voltages with respect to a reference, the transform of the independent excitations, and
the element admittances. The set o f all such equations characterizes the node voltages o f the cir­
cuit in the ^-domain. Solving the set o f circuit node equations yields the set o f transformed node
voltages. Knowledge o f these permits the computation o f any o f the element voltages. W ith
knowledge of the element admittances, one may com pute all o f the element currents. Since nodal
analysis has a more extensive application than loop analysis, our focus will be on nodal analysis.
E X A M P L E 1 3 .1 5 . Figure 13.31 shows an ideal operational amplifier circuit called the
Key normalized low-pass Butterworth filter.
Sallen and
(See Chapter 19 for a full discussion o f filters.) A nor­
malized low-pass filter passes frequencies below 1 rad/sec and attenuates higher frequencies. As we
will see later in the text, the 1-rad/sec frequenc)' “cu toff” can be changed to any desired value by
frequency-scaling the parameter values o f the circuit. (See Chapter 14 for a discussion o f frequen-
Chapter 13 • Laplace Transform Analysis II: Circuit Applications
63 S
cy scaling.) The goal here is to utilize the techniques o f nodal analysis to compute the (normal­
ized) transfer function o f this circuit.
FIG U R E 13.31 Sallen and Key normalized Butterworth low-pass
filter circuit containing an ideal operational amplifier.
S
o l u t io n
T he solution proceeds in several steps that utilize nodal analysis techniques in conjunction with
the properties o f an ideal op amp. Recall that for an ideal op amp, the voltage across the input ter­
minals is zero and the current into any o f the input terminals is also zero. Finally, note that one
does not write a node equation at the output, which appears across a dependent voltage source
whose value depends on other voltages in the circuit.
Step 1. Find Vy. Because the voltage across the input terminals o f an ideal operational amplifier
is zero.
Step 2. 'Write a node equation at the node identified by the node voltage V^. Summing the currents
leaving the node yields
(Va -Vi„) + (V a-V t) + V2.v(l/,
Substituting
for
- V„,„) = 0
and grouping like terms produces
(■J2.S + 2 ) V „ - ( J 2 S + \)V,„„=V,„
(13.28)
Step 3. Write a node equation at the node identified by the node voltage Vy. By inspection, the desired
node equation is
5+1
(1 3 .2 9 )
Step 4 . Write the foregoittg tivo node equations in matrix form. In matrix form, equations 1 3 .2 8 and
13.29 combine to give
- (V 2 . + I)'
■
-1
-|=5+ 1
IV 2
J
K;
■
'^in
0
(1 3 .3 0 )
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
6.U>
Step 5. Solve equation 13.30 for
in terms ofV-^^ using Cramer's rule. From Cram ers rule,
del
'{ ■ J lS + l )
V;„
-1
0
(>/ 2 :i- +
2
)
-(V
2
.V + 1 )
del
-1
\42
The resulting transfer function is
K..
[.j2 s + 2 )lj^ s+ \ \ -[y l2 s+ \ )
IV2
s- + yf2s+\
N otice that for small values o f ^ = yto (i.e., low frequencies), the magnitude o f H{s) approximates
1, and for large values o f s = JiO (i.e., high frequencies, where |/b)| »
small. Since
1), the magnitude o f H(s) is
= Myo)) ^^^(yco), such a circuit blocks high-frequency input excitations and
passes low-frequency input excitations. As mentioned at the beginning o f the example, the circuit
passes low frequencies and attenuates high frequencies.
T he preceding example used matrix notation, com m on to much o f advanced circuit analysis. In
one sense, matrix notation is a shorthand way of writing n simultaneous equations: the n variables
are written only once. More generally, matrix notation and the associated matrix arithmetic allow
engineers to handle and solve large numbers o f equations in numerically efficient ways. Further,
the theory o f matrices allows one to develop insights into large circuits that would otherwise
remain hidden. Hence, many o f the examples that follow will utilize the elementary properties o f
matrix arithmetic.
T he next example uses nodal analysis to compute the response to an initialized circuit. The exam­
ple combines the equivalent circuits for initialized capacitors and inductors with the technique o f
nodal analysis.
E X A M P L E 1 3 .1 6 . In the circuit o f Figure 13.3 2, suppose /y^^(r) = 6(r), /^(O") = 1 A, and
= 1 V. Find the voltages
V(^t)
FIG U R E
and
v^Qi~)
v^it).
1 3 . 3 2
Two-node /?Z.Ccircuit for Example 13.16.
Given the indicated current direction of i[{t), what is the implied voltage polarit)' for
v^{t)}
Chapter 13 • Laplacc Transform Analysis II: Circuit Applications
63'
Step 1. Draw the s-domain ecjuivaletit circuit with an eye toward nodal analysis. Inserting the equiv­
alent current source models for the initialized capacitor and inductor in Figure 1 3 .3 0 , one obtains
the 5-domain equivalent circuit shown in Figure 13 .3 3 .
V Js)
VJs)
1o
ii(O-)
,Cv,(0 )
O
1o
_
= 1
1F
1H
FIG U R E 1 3.33 5-Domain equivalent of the circuit of Figure 13.31.
Step 2 . Write two node equations and put in matrix fonn. At the node labeled
V(is)
K C L implies
that
(1 +
s)V(is)
+
[V(is)
-
Vj{s)] = 2
Simplifying produces the first node equation:
{s^ 2 )V ^ s)-V i{s) = l
- V(is)] + {\ls)Vj{s) = - (1/i), or, equivalently,
At the node labeled
.9+1
-V c(^ ) + —
1
= —
S
5
T he matrix form o f these uvo node equations is
.v + 2
,
2 1
-1
_i
.9+1
s
Step 3 . Solve the matiix equation o f step 2 for the desired voltages. Using C ram ers rule, computing
the inverse, or simultaneously solving the equations gives
2{s+\)-\
2
Vcis)-
5“+ 2 . 9 + 2
[5 +1
1 1 ■2 ■
s
1
5+2
(s+Vr + \
(1 3 .3 1 )
_i
s
C v + D -3
(.V+ 1r +1
Step 4 , Take the inverse Laplace tratisfonn to obtain time domain voltages. Breaking up equation
13.31 into its components yields
Chapter 13 • L iplacc Transform Analysis II: Circuit Applications
638
2(^ + 1)________ 1
Vcis) =
(5 + 1)- + !
(5+ 1)^ + 1
in which case
V(^t)
=
e ^[2 cos(t)
- sin(/)]/^(r)
Also,
(.^ + 1)
3
(A-+1)^ + 1
(.V+1)^ + 1
V^(s) =
leading to
=e
Figure 13.34 presents plots o f
V(\t)
^[cos(^) - 3sin(f)]«(r)
and
H G U R E 13.34 Plots of the capacitor and inductor voltages for Example 13.16.
Dual to nodal analysis is loop analysis. In loop analysis, one defines loop currents and writes KVL
equations in terms o f these loop currents. The following example illustrates the method o f loop
analysis for computing the input impedance o f a bridged-T network.
Chapter 13 • Laplace Transform Analysis 11: Circuit Applications
639
E X A M P L E 1