§2: Introduction to Linear Circuits
§2.0
§2.1
§2.2
§2.3
§2.4
§2.5
§2.6
§2.7
§2.8
Introduction
Wire, Open-Circuits, and Switches
Power Sources
Averages
Passive Elements
Dividers
Combining and Splitting Sources
Units in Electrical Engineering
Summary
This chapter deals with the various components that can be used to make up a linear circuit. We begin by introducing the
actual circuit elements, proceed to combining them in simple circuits, and finish with basic applications of Kirchoffs and Ohms laws.
§2.1: Wire, Open Circuits, and Switches
The most fundamental portion of any circuit, and often most neglected, is wire. An ideal wire is a connection between other
elements that perfectly conducts electrical current. Since the ideal wire (depending on context, a wire may also be called a short
circuit) has no resistance, it cannot support any differences of potential, and thus always has a voltage of V = IR = I · 0 = 0
Volts. Practical wires, commonly made of copper due to cost and physical ductility, have a small yet finite resistance.
Figure 1. An ideal wire.
Thought experiment: What happens if you try to force a voltage across an ideal wire? Mathematically, the current tends
to ∞; physically, the current grows increasingly large until something breaks or burns–just consider a short circuit placed directly
across your car battery terminals or a paper clip in a wall socket...
Open circuits are the dual of wire–open circuits prevent the flow of any current between two points. The value of the resistance
is ideally infinite, and can therefore support any voltage even without current. Practical open circuits have a finite, although very
large, resistance and make the flow of current very difficult.
Figure 2. An ideal open-circuit.
Thought experiment: Ever think your car battery had gone dead, but found out the problem was “crud” on one of the
battery terminals? A chemical reaction forms an oxide (a strong insulator) between the battery and the contact resulting in no
current flowing. Therefore there is an open circuit between the battery and the rest of your car.
A switch is the controlled combination of a wire and an open circuit. Provided some external control mechanism (perhaps a
human in simple systems), an ideal switch changes between an ideal wire and ideal open circuit.
1
2
Figure 3. An ideal switch.
Though experiment: Consider the light switch in the room you are sitting in: turn the switch “on,” and it closes, allowing
current to flow as a wire to the lightbulbs; turn the switch “off,” and it becomes an open circuit, stopping any current from
reaching the bulbs.
Application: What exactly is a fuse? A fuse is a specially designed wire conductor encased in vacuum that will burn (“fuse”)
if the power dissipated in it (measured more commonly as a current) exceeds a chosen amount, converting the electrical path to
an open circuit and thereby protecting subsequent devices from being damaged by surges.
Application: How does a circuit breaker work? A circuit breaker is another protection device used to limit current, which
activates by throwing a mechanical switch to open or break the circuit. Rather than replacing a circuit breaker like you do a fuse,
you have to physically reset the breaker so that it will conduct.
§2.2: Power Sources
Voltage Sources
An ideal voltage source is an electrical device that maintains a fixed potential difference independent of how much current (or
subsequently, power) it is supplying. The IV curve of a voltage source is shown in Figure 2.4; notice that the current can be
positive, negative, or even zero and yet always produces the same voltage.
Figure 4
1
You may also show that the internal resistance, the slope of the ratio voltage/current, is (∞) −1 = ∞
= 0 Ω; we will later see
that this internal resistance is the equivalent resistance seen when the voltage source is turned off.
The different flavors of voltage sources are direct current (DC), alternating current (AC), and controlled. The circuit symbols
for each of these voltage sources are shown in Figure 2.5.
Figure 5
3
DC voltage sources maintain a fixed numerical constant for their voltage: think of either the 9V battery in your alarm clock
or the 12V battery in you car. AC voltage sources output a time-dependent voltage that is usually in the form of a sinusoid,
square, or triangle wave. Simple examples are the 110V wall outlet voltage (rms values) used in the United States and the 220V
outlet voltage used in Europe. The controlled voltage source outputs a voltage proportional to other linear variables in a circuit,
usually other voltages or currents: the two forms are voltage-controlled voltage sources (VCVSs) and current-controlled voltage
sources (CCVSs).
Thought experiment: An example of a mechanical pressure-controlled pressure source (analogous to a VCVS) is a SCUBA
regulator. The regulator will reduce the tank’s pressure (3000 p.s.i.) to the ambient pressure of the attached mouthpiece that
varies constantly with water depth.
Exercise: Describe the charging and discharging of a rechargable AA battery. Model the battery as a DC voltage source and
describe the flow of physical charge carriers during the charging and discarging processes.
Current Sources
An ideal current source produces a known current independent of what load or connection is made to it. An obvious problem
here is that current sources cannot physically exist independent of a circuit (as did voltage sources in terms of batteries). The IV
curve in Figure 2.6 is that of an ideal current source; again notice that the value of the voltage can be positive or negative, but
always with a constant value for the current.
Figure 6
The internal resistance of the ideal current source is calculate as the slope of voltage to current, or (0) −1 = 10 = ∞ Ω.
Current can also be supplied by DC, AC, and controlled sources. A DC current source outputs a fixed flow of charge to create
constant current, AC sources produce time-dependent waveforms of one form or another, and controlled current sources depend
on other circuit parameters for their magnitudes.
Figure 7
Thought exercise: When was the last time you encountered a mechanical version of a controlled current source? We have
defined voltage to be analogous to mechanical pressure, while current is on par with mass flow. Still stumped? When you took a
shower this morning, you probably turned a knob (reducing the pressure at the valve) that allowed for water to flow out of the
shower head.
4
Since voltage and current sources are supposed to be a “source” of electrical energy, we think of the energy/power generated to
be positive (or, alternatively, the energy/power dissipated to be negative). As a convention, we denote the positive orientation of
current as going from the lower potential to the higher potential; a possibility remains for the power generated to have a negative
value–just consider the rechargeable battery again.
Figure 8
Exercise: A battery recharger might be modeled as a DC current source that pumps current (charge) onto a depleted battery.
If 8 AA batteries, each drawing 10 mA of current from this current source, determine the power generated by each of the batteried
and the recharger.
§2.3: Averages
The average and root-mean-square (RMS) values of a voltage or current source are often desired to convert an analysis to and
from AC (fluctuating) and DC (constant). The average of any time-dependent quantity is calculated as
Xavg (t) =
1
t
Z
t
x(τ ) dτ
0
We will often simplify this average for periodic or repetitious waveforms by averaging over one full interval, T.
Xavg =
1
T
Z
T
x(τ ) dτ
0
Example: What is the average voltage of a regular wall socket in the U.S. power system? The
time-dependent AC voltage is 177 cos(2π · 60t).
1
Solution: The AC voltage is a periodic waveform with period T = 60
seconds. Plugging in, we
calcluate an average value of zero.
Z
1
sin(2π60t) 60
1
177 cos(2π60t) dt = (60 · 177) ·
|0 = 0 V
Vav = 60
2π60
0 60
Does this calculation surprise you? We know the average value of any sinusoid is zero, so the mathematics is sound. Physically,
the power systems desire to have zero average power in order to minimize power losses in the resistive power lines.
5
Example: What is the average power generated by an AC current source, I(t) = 10 cos(2π·100t) mA
if only connected to a 20 Ω resistor?
Solution: Since we know (via Ohm’s law) that the voltage across a resistor is the product of the
current through it and its resistance, we know that an AC voltage v(t) = 200 cos(2π100t) mV will
be across both the resistor and the parallel current source. The AC power is then p(t) = v(t) · i(t) =
2 cos2 (2π100t) mW , and the average power generated is
Pav = 100
Z
1
100
2 cos2 (2π100t) dt = 100
0
Z
1
100
(1 + cos(2π200t)) dt = 1 mW
0
After the use of a double-angle trig substitution, we have again used the fact that the average value
of a sinusoid is zero.
Exercise: What is the average voltage of a 9V DC voltage source?
Exercise: What is the average current of a 200 Ω resistor that has a voltage v(t) = 3 cos(2π200t) across it?
The RMS value of an AC waveform is calculated as the magnitude average of the waveform–we will only consider periodic
waveforms with period T.
s
Z T
1
Xrms =
X 2 (t) dt
T 0
The RMS value of a waveform is equivalent to the DC level that corresponds to the same power dissipation in a resistor. The
calculation is equal to a vector space two-norm. 1
Example: Calculate the RMS value of a sinusoidal voltage with value v(t) =
√
2 cos(ωt).
Solution:
Vrms =
s
1
T
Z
T
0
2·
cos2 (ωt) dt
We see that the RMS equivalent of any sinusoid of amplitude A is
=
s
1
T
Z
T
(1 + cos2 (2ωt)) dt =
√
1=1
0
A
√
.
2
Example: Calculate the RMS power delivered to a 10 Ω resistor that has a sinusoidal voltage of
magnitude 3V across it.
Solution: To calculate the power delivered to the resistor, we must determine the DC equivalent
or magnitude average of the voltage, and then use that for the power.
3
VDC ≈ Vrms = √ V
2
√
2
(3/ 2)2
VDC
=
= 0.45 W
PR =
R
10 Ω
Exercise: Calculate the RMS value of a vertically shifted sinusoidal voltage of v(t) = 1 + 3 cos(2πt) V .
1The easiest way to remember the formulation is to take “R-M-S” backwards: square the function, calculate the mean (average), and then the
square root.
6
§2.4: Passive Elements
Passive elements are those circuit elements that do not have the capacity to generate power. Common passive elements include
resistors, capacitors, inductors, and diodes. We will wait to consider diodes until a later chapter, and will now focus on the other
three elements, all of which have linear IV relationships. The most important thing to remember about passive elements is that
positive current flows into the side of higher potential and out the side of lower potential (element is dissipating/storing energy)
as shown in Figure 2.9.
Figure 9
Resistors
We have already discussed the material characterization of resistance: resistors are the actual physical devices that provide a
given value of resistance. Outside of the world of theory and textbooks, almost every practical device uses resistors; many like
heating coils, wire, and speakers can be modeled as single resistors in basic calculations. The circuit symbol for a resistor, R, is
shown in Figure 2.10.
Figure 10
Using Ohm’s law, we may always say that the voltage across a resistor is proportional to the current through it: V = IR, and
that the power dissipated in a resistor 2 may be calculated as P = V I = V 2 /R = I 2 R. Further, we notice that the resistor has a
constant IV relationship, indicating no ability to store energy.
Capacitors
Capacitors are passive elements that are capable of storing energy in an electrical field. The most commonly analyzed capacitor
is that of two parallel metal plates with a charge imbalance that induces an electric field, as depicted in Figure 2.11.
Figure 11
2Since a resistor is a passive element, it may only dissipate power.
7
The value of the capacitance (word comes from the capacity to store energy in electrical fields) depends on the cross-sectional
Area of the two plates, the distance between the plates and the permittivity, ², of the material between them.
²A
d
Total permittivity, ², is calculated as a product of relative permittivity, ²r , and the permittivity of free space, ²o .
C=
F
m
Various dielectrics, all with different relative permittivities, can be placed between the plates to increase the nominal capacitance.
² = ² r ²o
²o = 8.854 · 10−12
Example: Given a parallel plate capacitor with plates separated by 1 mm and having a dielectric
with relative permittivity ²r = 3 between the plates, calculate the necessary plate area to create a
capacitance of 1 Farad.
Solution:
A=
(1 F ) (0.001 m)
Cd
= 3.76 · 107 m2
=
F
²
3 · 8.854 · 10−12 m
Thus we need an area of 3.76 · 107 m2 or the area of more than 6000 football fields! Practical values
of capacitance are on the order of 10−12 − 10−6 Farads.
Exercise: Determine the capacitance of a parallel plate capacitor with vacuum between the plates, which are 1 cm on a side
and separated by 1 mm.
The capacitance of a parallel-plate capacitor may also be related to the ratio of charge to the voltage that creates the electric
field.
C=
VC
Q
=⇒
Q = CVC
Example: Calculate the voltage necessary to store a 1 mC charge on a 1 µF capacitor.
VC =
1 mC
Q
=
= 1000 V
C
1 µF
From this last equation, we derive the general IV relationship for the capacitor, which has circuit symbol as shown in Figure
2.12.
Figure 12
d
d
d
QC (t) = CVC (t) = C VC (t)
dt
dt
dt
We may calculate the instantaneous power dissipated in the capacitor as the product of its voltage and current.
IC (t) =
PC (t) = IC (t) · VC (t) = CVC (t)
d
VC (t)
dt
8
Exercise: Show that the average power dissipated in a capacitor is zero for sinusoidal voltages or currents.
We may then obtain the energy stored in the capacitor by integrating the power.
«
Z
Z „
d
1
ECap = PC (t) dt = C
VC (t) VC (t) dt = CVC2 (t)
dt
2
Example: How much energy is stored in a 1 µF capacitor if it has a charge of 1 mC built up?
Solution: The energy may be re-expressed in terms of charge
E=
(1 mC)2
1
1 Q2
CVC2 =
=
= 0.5 J
2
2 C
2 · (1 µF )
To determine whether a capacitor is charging or discharging, look at the sign of the instantaneous power: if positive, the
capacitor is charging, while the capacitor is discharging if the instantaneous power is negative.
Exercise: Calculate the instantaneous energy stored in a 10 nF capacitor if the volatge across it is 10 cos(100t) V .
Exercise: Determine when the capacitor in the previous example is charging and discharging.
Inductors
Inductors are passive elements that are capable of storing energy in a magnetic field. The physical description of an inductor
usually takes the form of a coiled wire: from Faraday’s law, the current flowing through the wire induces a magnetic field in a
direction perpendicular to flow current (use the right-hand rule to determine orientation of the magnetic field lines).
Figure 13
The value of an inductance may be calculated as the total magnetic flux (flux, φ, induced per turn times Number of turns)
per unit of electrical current.
Φ
Nφ
=
IL
IL
The voltage across an inductor (circuit symbol shown in Figure 2.14) is related to the change of total flux, corresponding to a
rate-of-change in the electrical current through the inductor;
L=
Figure 14
9
d
d
d
d
Φ(t) = N φ = LIL (t) = L IL (t)
dt
dt
dt
dt
This IV relationship for the inductor can then be used to calculate the instantaneous power dissipated by the inductor
VL (t) =
PL (t) = VL (t) · IL (t) = L IL (t)
d
IL (t)
dt
and the energy stored in the inductor’s magnetic field.
Z
Z
d
1
EL = PL (t) dt = L
IL (t) IL (t) dt = LIL2 (t)
dt
2
Question: An engineer has 0.1 meters of wire to shape into an inductor; how can he maximize the inductance?
Although the equations do not explicitly reflect the dependance on magnetic permeability, we can increase the inductance of
a wire coil by placing a ferro-magnetic material (iron or special alloys) inside the coil.
Example: What is the instantaneous energy stored in a 1 mH inductor if the voltage across it is
10 cos(10t) mV ?
Solution: The current through the inductor will be
Z
Z
VL (t)
10 cos(10t) mV
IL (t) =
dt =
dt = sin(10t) A
L
1 mH
yielding an energy storage of
EL =
1
(1 mH) sin2 (10t) (A2 ) = 0.5 sin2 (10t) mJ
2
Exercise: Determine when the energy stored in the magnetic field of the inductor from the previous example is increasing
and decreasing.
Exercise: How much energy is stored in an 1 mH inductor that has a constant current of 1 A through it?
Passive Element Equivalents
Often we will have collections of resistors, capacitors, and inductors in configurations that can be easily reduced to a mathematical equivalent, thereby making the analysis much easier. The following derivations concern physical resistors, capacitors,
and inductors, but we will later consider a more generic form where all can be combined together as an equivalent impedance.
We will use the facts that series elements have the same current flowing through them and that parallel elements have the same
voltage across them to consistently calculate the equivalents.
Equivalent Resistance
We define the equivalent resistance seen by the source to be the value of a single resistor with the same effective resistance as
any combination of resistors.
First we consider the equivalent resistance of the two series resistors shown in Figure 2.15.
Figure 15
10
The total voltage, VT , appears across the two resistors together, while the total current, I T , is identical for the two resistors.
Applying Ohm’s law to each of the resistors,
VR1 = I T R 1
VR2 = I T R 2
and KVL to the loop,
VT = VR1 + VR2 = IT R1 + IT R2 = IT (R1 + R2)
to obtain the equivalent resistance as a ratio of total voltage to total current.
Req =
VT
= R1 + R2
IT
The second simple circuit is a current source in parallel with two (parallel) resistors as shown in Figure 2.16.
Figure 16
Repeating the analysis with Ohms and Kirchoffs laws, we write the currents through each of the resistors
I R1 =
VT
R1
I R2 =
VT
R2
and use KCL at the top node for the total current.
VT
1
1
VT
+
= VT (
+
)
R1
R2
R1
R2
Solving for the equivalent resistance of the parallel combination,
I T = I R1 + I R2 =
Req =
VT
1
1 −1
=(
+
)
IT
R1
R2
The resulting equivalents show that series resistors add directly, and parallel resistors parallel add. The notation R 1 kR2 is
routinely used to denote parallel addition. If we want to generalize these results to many resistors, we find that series resistors
all add directly (use induction on combining two at a time until all are combined), and that parallel equivalents all parallel add.
N
Req series =
X
VT
= (R1 + R2 + ... + RN ) =
Rk
IT
k=1
Req parallel =
N
X 1 −1
1
1
1 −1
VT
= R1 kR2 k...kRN = (
+
+ ... +
) =(
)
IT
R1
R2
RN
Rk
k=1
Another way to view parallel addition is to introduce the concept of conductances, which are the inverse of resistances and
have units Siemens or Mhos, [0].
Gk =
1
Rk
=⇒
Geq =
1
Req
Using these inverses of resistance, we can restate the combinations as an equivalent conductance.
11
Geqseries = (
N
X
1 −1
)
Gk
k=1
Geqparallel =
N
X
Gk
k=1
In most problems, you will find calculations easier if you add resistances and parallel add conductances; then convert to the
desired final form. We can even rewrite parallel notation for two resistors in the shorthand
Ri Rj
Ri + R j
Notice that this computational shortcut of product over sum does not extend to parallel combinations of more than two
resistors (think about the resulting units!).
Do any of these equations for equivalent resistance make intuitive sense? Let us make an instinctual check on the equations.
Series resistors create consecutive barriers to the flow of current, and thus multiple resistors add to an equivalent resistance greater
than any single one in the chain. Parallel resistors provide multiple paths for current to flow; current will still choose to take the
path of least resistance (proportionally), so the extra paths do nothing but reduce the effective resistance to a value less than the
smallest resistor.
Ri kRj =
Rseries ≥ max(R1 , R2 , ..., RN )
Rparallel ≤ min(R1 , R2 , ..., RN )
Examples of Equivalent Resistances
Since equivalent resistance calculations are so important, we will consider a few detailed examples of finding the equivalent
resistance for various combinations of resistors. We found previously that equivalent resistance of resistors in series is the sum of
their values, and that the equivalent resistance of resistors in parallel is the parallel sum of the values. To apply these concepts to
combinations of resistors that are not solely in series or parallel requires intermediate equivalents. Consider the circuit in Figure
2.17.
Figure 17
We can combine the resistors R2 and R3 as a parallel equivalent, replacing with a single resistor of value R2 kR3 .
Figure 18
The result is a series combination of R1 and R2 kR3 that can be added directly to obtain Req .
12
Figure 19
The single resistance, Req = R1 + R2 kR3 , has the same properties (from node a to node b) as the collection of resistors in
Figure 2.17. Now, consider the larger collection of resistors in Figure 2.20.
Figure 20
First of all, do not be surprised by the diagonal resistor, R4 . The functional definition of “parallel” is that two elements share
the same two nodes (and therefore have the same potential difference across them), while the definition of series could be seen
as two elements that are forced to share the same current. As a result, we see that R 3 is in series with R5 , and their equivalent
(R3 + R5 ) is in parallel with both R2 and R4 .
Figure 21
Combining these last two resistors in series, we obtain the overall equivalent of the original circuit.
Figure 22
The next combination of resistors also contains both an open circuit and a short circuit.
13
Figure 23
The resistance of an ideal open circuit is infinite, so the series combination of a resistor and an open circuit is infinite (or just
the open circuit). Likewise, the parallel combination of a resistor and an open circuit is just the resistor.
Figure 24
We may also obtain these values numerically by plugging into the equations and using limits as R open ckt → ∞ for series and
parallel combinations.
R + ∞ 99K ∞
R·∞
Rk∞ =
99K R
R+∞
The resistance of an ideal short circuit is zero, so the series combination of a short circuit with a resistor is just the resistor,
and the parallel combination is just the short circuit.
Figure 25
Calculating the numerical values with limits as Rshort ckt → 0,
R+0=R
Rk0 =
R·0
=0
R+0
14
Using these simplifications, we see in Figure 2.23 that R6 and R7 are negligible since they are in series with an open circuit,
R5 is negligible due to being in parallel with the short circuit, and the rest of the resistors are unaffected. The simplified circuit
appears in Figure 2.26; we can write Req = R1 + R2 kR3 kR4 by inspection.
Figure 26
A more challenging configuration of resistances, called an infinite resistive ladder, is next in Figure 2.27.
Figure 27
The trick to solving the ladder circuit is not to take too big a gobble all at once. If you look to the right of the first shunt
resistor, you see the same resistance as you do at the input (the chain from 2 to ∞ is the same as the chain from 1 to ∞).
Replacing everything to the right of the dashed line by Req , we obtain a much simpler circuit.
Figure 28
Writing the equation for Req from the input, we obtain a quadratic polynomial in Req .
Req = R + RkReq + R
=⇒
2
Req
− 2RReq − R2 = 0
To solve this polynomial, we must realize that the equivalent resistance, Req , should vary directly with the individual resistances,
R. That is, if we double the value of every resistor in the chain, we ought to solve for an equivalent resistance that is exactly
double the original equivalent. Mathematically, this condition can be stated: Req = cR, reducing the quadratic to:
c2 − 2c − 2 = 0
=⇒
c=1±
√
3
Clearly, the resistance of the circuit in Figure 2.27 cannot be negative, so the equivalent resistance is R eq = (1 +
√
3) · R.
Example: Modify the previous example to solve for the geometrically increasing (a > 1) infinite resistive ladder in Figure
2.29.
15
Figure 29
Repeating the same solution mathod, we replace the ladder to the right of the dashed line, but this time with a · R eq .
Req = R + (aR)k(aReq )
= R + a · RkReq
2
Req
− aRReq − R2 = 0
Req = cR
√
a
a2 + 4
c2 − ac − 1 = 0
=⇒
c= ±
2
2
Again, the equivalent resistance must be positive, so we determine that
Req =
„
a
+
2
√
«
a2 + 4
R
2
We summarize the fundamental steps in determining any equivalent resistance:
1. Replace any portions containing open/short circuits by the appropriate simplification.
2. Iteratively reduce the circuit until you have only one resistance.
3. In the case of infinite chains, truncate the chain and replace by an appropriate equivalent.
4. If all else fails, use Thevinin equivalents or Nodal analysis to find the ratio of voltage to current (discussed in the next
chapter).
Whenever in doubt whether circuit elements are in series or parallel, label the nodes (not connections) and visualize where
current flows in the circuit.
Example: Determine the equivalent resistance of the resistor configuration shown in Figure 2.30.
Figure 30
The short circuit in parallel with R5 shorts out the resistor, while the open circuit has no other effect. By inspection,
Req = R1 + R2 k(R3 + R4 kR6 )
Example: Determine the equivalent resistance of the resistor configuration shown in Figure 2.31.
16
Figure 31
The series combinations (R13 + R15 ) and (R14 + R16 kR17 ) are in parallel to give:
Req = R7 k [R8 + R9 + R10 k (R11 kR12 + (R15 + R13 )k(R14 + R16 kR17 ))]
Equivalent Capacitance
Physical capacitors combine slightly differently from the forms of resistors and inductors. Let us consider Kirchoffs laws and
the differential form of the capacitors IV characteristics to determine the equivalent capacitance. The circuit in Figure 2.32 shows
two series capacitors.
Figure 32
Mimicing the procedure used before, we first write out the IV relationship for each individual capacitor.
d
d
I C 2 = C 2 VC 2
VC
dt 1
dt
We then notice that the current in each capacitor is identical since the two are in series, so we substitute into a KVL loop
equation.
IC1 = C 1
IC
IC
1
1
d
d
+
)
VT = (VC1 + VC2 ) = 1 + 2 = IT (
dt
dt
C1
C2
C1
C2
Looking at the IV relationship between total voltage and total current, we obtain the equivalent capacitance of series capacitors.
1 −1
C1 C2
1
+
) =
C1
C2
C1 + C 2
The circuit in Figure 2.33 shows two capacitors in parallel.
Ceq = (
Figure 33
Repeating the statement of the IV relationship for the individual capacitor,
17
IC1 = C 1
d
VC
dt 1
IC2 = C 2
d
VC
dt 2
we calculate a KCL equation for the total current.
d
d
d
VC + C2 VC2 = (C1 + C2 ) VT
dt 1
dt
dt
Looking at the IV relationship between total voltage and current, we obtain the equivalent capacitance of parallel capacitors.
IT = I C1 + I C2 = C 1
Ceq = C1 + C2
These results should not be too surprising after all: parallel capacitors correspond to linearly increased cross-sectional areas
for the equivalent, while series capacitors require charge to transition in between the capacitors, thus reducing the capacity to
store energy in an electric field. When we generalize to combinations of N series or parallel capacitors, we obtain the expected
relations.
1
1
1 −1
+
+ ... +
)
C1
C2
CN
= (C1 + C2 + ... + CN )
Cseries = (
Cparallel
Example: Determine the equivalent capacitance of the two circuits shown in Figure 2.34.
Figure 34
The process for determining the equivalent capacitance of a combination of capacitor closely follows that of the analysis for
equivalent resistances–the caveat is that (as explained directly above) physical capacitances will add differently than impedances.
Ceq1 = C1 k[C2 + (C3 + C4 )kC5 ]
Do not forget that the notation xky means
xy
x+y
Ceq2 = C1 k(C2 + C3 + C4 )
and not “x parallel y.”
Equivalent Inductance
The equivalent inductance of a combination follows directly from the analysis of equivalent resistance. The differential relation
for an inductor is linear (derivatives are linear operators) between V and I and in the same form as are resistors by Ohms law.
The first circuit in Figure 2.35 shows two series inductors.
Figure 35
18
Using the individual IV relationships,
V L1 = L 1
d
IL
dt 1
V L2 = L 2
d
IL
dt 2
and writing a single KVL loop equation,
V T = V L1 + V L2 = L 1
d
d
d
IL + L2 IL2 = (L1 + L2 ) IT
dt 1
dt
dt
we obtain the relation for series inductors.
Leq = L1 + L2
The circuit in Figure 2.36 contains a pair of parallel inductors.
Figure 36
Again stating the individual IV relationships,
V L1 = L 1
d
IL
dt 1
V L2 = L 2
d
IL
dt 2
and writing a single KCL node equation,
V L1
d
V L2
1
1
d
+
= VT (
+
)
IT = (IL1 + IL2 ) =
dt
dt
L1
L2
L1
L2
we obtain the relation for parallel inductors.
Leq = (
1
1 −1
L1 L2
+
) =
L1
L2
L1 + L 2
We can again generalize to combinations of N series or parallel inductors to obtain:
Lseries = (L1 + L2 + ... + LN )
Lparallel = (
1
1
1 −1
+
+ ... +
)
L1
L2
LN
Verifying these equations intuitively is easy: series inductors correspond to a single inductor having an increased number of
turns, while parallel inductors result in a reduced magnetic flux.
Example: Determine the equivalent inductance of the two circuits shown in Figure 2.37.
19
Figure 37
By inspection,
Impedance
Leq1 = L1 + L2 k(L3 + L4 kL5 )
Leq2 = L1 + L2 kL3 kL4
Impedances are a complex-valued form or resistance that allows us to easily calculate the sinusoidal stead-state response of
circuits containing mixed combinations of resistors, capacitors, and inductors. The analysis of impedances in circuits is identical
to that of resistances. We will mention impedance along with resistance many times while analyzing DC circuits containing only
resistors in order to ensure a more complete understanding and to allow for review from other chapters–for the time being, you
may take “impedance” and “resistance” to be one and the same.
§2.5: Dividers
The goal of a divider is to break down a quantity, perhaps the total voltage across a set of series impedances or the current
into a bunch of parallel impedances, to reflect a value for a specific impedance. We do not need to actually know the value of the
total quantity to determine what fraction appears elsewhere in the circuit; in fact, we would rather not calculate every voltage
and current in the circuit, only those of interest.
Voltage Dividers
A voltage divider is a shortcut to determining the voltage across a single series impedance given the total voltage across the
series combination. Consider the circuit in Figure 2.38 consisting of N series resistors, where we want to know the voltage across
only the last resistance RN (again, analysis for resistance and impedance are identical).
Figure 38
Since all resistances are in series, they must each share the same current, IT . We can write the voltage across RN and the set
of series resistances using Ohms law and this current IT .
20
VT = IT · (R1 + R2 + ... + RN )
and
V RN = I T R N
We then take the ratio of these two quantities to obtain the fraction of the total voltage that appears across R N .
VRN
IT R N
RN
=
=
VT
IT R1 + IT R2 + ... + IT RN
R1 + R2 + ... + RN
The form of the solution is simply the resistance of interest divided by the sum of all resistances in series.
Example: Determine the unknown voltage Vx across the 20 Ω resistor in Figure 2.39 using a voltage divider.
Figure 39
Vx =
20
(10 V ) = 2 V
20 + 30 + 50
Example: Determine the voltage across the 25 Ω resistor in Figure 2.40. V y is the voltage at the node between the two
resistors, relative to the bottom ground node.
Figure 40
Vy =
25
(20 V ) = 5 V
25 + 75
Example: Determine the voltage across each of the resistors in the circuit shown in Figure 2.41. Use the convention V x to
represent the voltage across the resistor Rx .
21
Figure 41
Solution: The easiest way to solve a problem like this is to obtain the equivalent resistance to the right of each point in the
circuit and then use that value in the voltage dividers. The circuit in Figure 2.42 shows the values for each of the equivalents,
followed by the appropriate dividers (you should double-check each calculation.)
Figure 42
R1 = 20 Ω
V40
V15
V12
R2 = 30 Ω
R3 = 5 Ω
R4 = 20 Ω
R5 = 10 Ω
R5
V20 =
(10 V ) = 2 V
40 + R5
R3
V6 =
V20 = 0.5 V
15 + R3
R1
V18 =
V6 = 0.3 V
12 + R1
40
(10 V ) = 8 V
=
40 + R5
15
=
V20 = 1.5 V
15 + R3
12
V6 = 0.2 V
=
12 + R1
Exercise: Use the voltages obtained in the previous example to verify that the conservation of energy hold for the circuit in
Figure 2.41.
Current Dividers
Current dividers allow us to determine the current through a single impedance out of many in parallel. The circuit in Figure
2.43 shows N resistances in parallel, and we want to determine the current through the last resistor R N .
Figure 43
22
Since all of the resistances are in parallel, they must share the same voltage, which we will call V T . Therefore, we can write
the current through RN using Ohms law and VT .
VT
VT
VT
VT
+
+ ... +
I RN =
R1
R2
RN
RN
The ratio of the two currents will then give us the value of the divider, or simply the fraction of the total current coming into
the parallel combination that flows in RN .
IT =
( R11 + R12 + ... + R1N )
I RN
GN
=
=
IT
RN
G1 + G2 + ... + GN
For the case of N = 2 (or by combining R1 through RN −1 into a single parallel equivalent), we may use another shortcut to
speed the computation of the divider.
I2
R1
=
IT
R1 + R 2
The form of the current divider is counterintuitive at first glance: the divider is the ratio of the opposite impedance (may
need to be a parallel equivalent in itself) divided by the sum of RN and that opposite. Closer inspection verifies that this makes
sense: current takes the path of least resistance (impedance), so the current through an impedance depends as much on the other
elements in parallel than on itself.
Example: Determine the current in the 10 Ω resistor shown in Figure 2.44 using a current divider.
Figure 44
I10 Ω =
6k30
5
(3 A) =
(3 A) = 1 A
6k30 + 10
5 + 10
Example: Calculate the current in the 30 Ω resistor of Figure 2.45 using current dividers.
Figure 45
I30 =
I30
60
6
I25
·
· (3 A) =
·
· (3 A) = 0.333 A
I25 + I60 I30 + I6
30 + 60 6 + 30
23
Example: Determine the current in each of the resistors from Figure 2.46 using only equivalent resistances and current
dividers. Use the notation Ix to denote the current through Rx .
Figure 46
We will repeat almost the exact same process as before: we will determine the equivalent resistance to the right of each point
in the circuit and then use those values in the dividers.
Figure 47
R1 = 18 Ω
I40
I15
I12
R2 = 7.2 Ω
R5
(9 A) = 3.629 A
=
40 + R5
R3
=
I20 = 2.514 A
15 + R3
R1
I6 = 1.714 A
=
12 + R1
R3 = 13.2 Ω
R4 = 7.02 Ω
R5 = 27.02 Ω
40
I20 =
(9 A) = 5.371 A
40 + R5
15
I6 =
I20 = 2.857 A
15 + R3
12
I18 =
I6 = 1.143 A
12 + R1
Exercise: Determine the current through each of the resistors shown in Figure 2.46 using current dividers. Verify that the
conservation of energy holds.
Properties of Dividers
We will consider two separate cases of the dividers: first, the case where all impedances are resistors, thereby reducing our
analysis to that of real (and positive) numbers, and second, a general analysis of all impedances.
With the restriction on our dividers to have all purely real and positive arguments, we can place upper and lower bounds on
the value. Since all numbers are positive, the divider can never be negative, and since the sum of all series resistances is greater
than the largest resistor (parallel sum of parallel resistors is less than any individual resistor), the fraction must be less than one.
To state these conditions more precisely, we must use limits to obtain the boundary values.
VRN
VRN
I RN
I RN
=0
lim
=1
lim
=1
lim
=0
RN →∞ VT
RN →0 IT
RN →∞ IT
VT
To consider impedances in general, we no longer have the nice conditions of positivity or bounded equivalent values (the
complex numbers are unordered, so it is impossible to say α > β with any meaning). The only conclusion we are able to draw is
lim
RN →0
24
that the power delivered to any passive element is, on average, positive and less than the total power delivered by the sources.
We will revisit complex power and dividers once we cover complex numbers.
§2.6: Combining and Splitting Sources
Another useful technique is to be able to combine series voltage sources as well as parallel current sources. If two voltage
sources are in series, the effective voltage value is simply the sum (or difference if the polarities are opposite).
Figure 48
The same can be said for parallel current sources by application of Kirchoff’s current law.
Figure 49
The most interesting case (and also most useful) is when we superimpose AC and DC sources: the result is a time-varying
source centered around a non-zero constant. We will deal with sources like these extensively when analyzing audio devices–the
output of a microphone might be an AC voice signal overlaid on a small DC battery voltage.
25
Figure 50
Question: What is the AC, or time-varying, part of a DC source’s output? Precisely zero! A DC source is defined to have a
constant output voltage/current, so there is no time-varying component.
One remaining trick for voltage and current sources is that they may be split as well as combined. Voltage sources may be
split into parallel voltage sources of the same value, while current sources may be split into series sources of identical value.
Figure 51
§2.7: Units in Electrical Engineering
We have so far described many physical quantities in terms of electrical circuits; just as important is a description of what
numerical values are reasonable. Listed in the table below are the fundamental units we will consider along with some numerical
ranges.
26
[Voltage] = Volts
Microphone output ≈ 1 mV
AA battery = 1.5 V
Static shock = 250 kV
[Current] = Amps
Low-power amplifier = 10 µW
Transistor amplifier = 1 mA
Car alternator = 20 Amps
[Resistance] = Ohms
Real wire = mΩ
Resistors = 1 Ω → 1 M Ω
[Power] = Watts
Vacuum = ∞ Ω
Cellphones = mW
Lightbulb = 100 W
Power generator = 10 MVA3
[Capacitance] = Farads
[Inductance] = Henries
[Impedance] = Ohms
[Charge] = Coulombs
Typical range: 1pF → 100 µF
Typical range: 10 µH to 100 mH
Entirely variable: −∞ → ∞ Ω
Milliken’s oil experiment: ± q
Laptop battery = mC
§2.8: Summary
Many different circuit elements can be used to build circuits. The most fundamental ones are wire connections, open circuits,
and switches, which ultimately guide the flow of electrons. Then we include voltge and current sources that input energy into a
circuit. Various types of impedances (so far resistors, capacitors, and inductors) may be used to model real devices, and so form
the basis of passive elements–those that cannot generate power.
Combining circuit elements may provide series elements, which share the same current; parallel elements which share the
same voltage; or configurations that have no easy simplification. We have also seen that open circuits and short circuits may be
simplified to radically reduce a set of impedances to a simpler version.
Beyond the configurations of individual circuit elements, we have seen ways of combining them mathematically into equivalent
circuits. Note that these equivalents do not physically change any circuit; an equivalent is just the mathematical model used to
ease analysis from outside the circuit. All that is required to determine equivalents is application of Ohm’s and Kirchoff’s laws.
The other side of the equivalents analysis is dividers, which allow you to take a large circuit and relate a particular voltage or
current in a circuit to the total input voltage/current by simple ratios of impedances. Dividers in DC circuits (those containing
only resistors as impedances) are guaranteed to be bounded by 0 and 1.
The next step is to put everything together at once!
27
Problem 2.1: What is the voltage across an ideal wire? An ideal open circuit? How much current can flow through an ideal
wire? Ideal open circuit?
Problem 2.2: Real wire is often sold by guage and length; X-guage wire is named so because X diameters of the wire equal
one inch, while length is dependent on the application. Which guage wire costs more to construct? Which guage wire would be
desired in a speaker system that assumes that the wire is ideal?
Problem 2.3: An alarm clock is capable of being used in either the United States (outlet voltages of 110 V) or in England
(outlet voltages of 220 V). If the alarm clock uses 100 mW of power, how much current is delivered by wall sockets in each
location? What if the power goes out and the alarm clock runs on the 9V backup battery?
Problem 2.4: What is the equivalent circuit of a current source in series with a voltage source? What is the equivalent circuit
of a current source in parallel with a voltage source? Consider the definitions of ideal sources.
Problem 2.5: What is the equivalent of two AA batteries in series? in parallel?
Problem 2.6: Why might a circuit designer use multiple batteries in parallel? Consider the limitation of real versus ideal
voltage sources.
Problem 2.7: What is the average voltage of a source that has value v(t) = 20 cos(1000t)? What is the average power
delivered to a 1 kΩ resistor? What if two such resistors are placed in parallel?
(
3 · t2
0≤t≤1
.
Problem 2.8: A voltage source has a periodic voltage: v(t) =
3 · (t − 2)2 1 < t < 2
(a) Plot the voltage assuming v(t) repeats every 2 seconds.
(b) What is the average value of v(t)?
(c) What is the RMS value of v(t)?
(d) How would each of these averages change if v(t) equals 0V for 2 < t < 4 seconds and repeats every 4 seconds?
Problem 2.9: A voltage source with value v(t) = A cos(ωt) supplies 10 W to a 30 Ω resistor. What is the RMS value of the
voltage?
Problem 2.10: Define passive and active circuit elements in layman’s terms.
Problem 2.11: A DC voltage source that has a value of 10 V is attached to a 20 Ω resistor? How much power is delivered
to the resistor? If two such 20 Ω resistors are placed in series and parallel, how much power is delivered to the resistors?
Problem 2.12: (a) Why must the physical voltage across a capacitor be continuous? Consider the magnitude of current at
the discontinuity.
(b) Why must the physical current through an inductor be continuous? Consider the magnitude of voltage at the discontinuity.
(c) What must be true of currents and voltages of capacitors and inductors in series? in parallel?
Problem 2.13: Calculate
(a) A 10 kΩ resistor is placed in series with a resistor of value R. What is the maximum and minimum value of the equivalent
resistance?
(b) A 10 kΩ resistor is placed in parallel with a resistor of value R. What is the maximum and minimum value of the equivalent
resistance?
(c) For both cases in (a) and (b), plot the equivalent resistance of the series/parallel combinations versus resistor value, R.
Problem 2.14: A potentiometer is a type of variable resistor, whose resistance changes with a wiper (small dial) from a
minimum value to a maximum value. Describe how a physical potentiometer (often referred to as a “pot”) that changes linearly
28
with wiper displacement may be constructed.
Problem 2.15: A linear potentiometer may be modeled as the two series resistors shown in Figure 2.52, where the wiper
bypasses a fraction of the total resistance, giving an overall equivalent resistance of αR, (0 ≤ α ≤ 1).
Figure 52
(a) Plot the equivalent resistance of a linear 10 kΩ pot versus α for 0 ≤ α ≤ 1.
(b) If this 10 kΩ pot is used in series with a 20 kΩ resistor, plot the equivalent resistance versus α.
(c) If this 10 kΩ pot is used in parallel with a 20 kΩ resistor, plot the equivalent resistance versus α.
Problem 2.16: For the resistor configuration in Figure 2.53, plot the equivalent resistance as a function of the potentiometer’s
resistance.
Figure 53
Problem 2.17: A varactor is a type of variable capacitor: by turning a small dial, the capacitance is changed.
(a) If the change is linear, describe how the varactor may be physically constructed.
(b) If the change is hyperbolic, describe how the varactor may be physically constructed.
Problem 2.18: Determine the equivalent resistance of the circuit shown in Figure 2.54.
Figure 54
Problem 2.19: Determine the equivalent resistance of the circuit shown in Figure 2.55.
Figure 55
29
Problem 2.20: Determine the equivalent resistance of the circuit shown in Figure 2.56.
Figure 56
Problem 2.21: Determine the equivalent resistance between nodes A and B of the circuit shown in Figure 2.57.
Figure 57
Problem 2.22: Determine the equivalent resistance between nodes C and D of the circuit shown in Figure 2.57.
Problem 2.23: Determine the equivalent resistance between nodes E and F of the circuit shown in Figure 2.57.
Problem 2.24: Determine the equivalent resistance between nodes A and B of the circuit shown in Figure 2.58.
Figure 58
Problem 2.25: Determine the equivalent resistance between nodes C and D of the circuit shown in Figure 2.58.
Problem 2.26: Determine the equivalent resistance between nodes E and F of the circuit shown in Figure 2.58.
Problem 2.27: Determine the equivalent resistance between nodes A and B of the circuit shown in Figure 2.59.
Figure 59
30
Problem 2.28: Determine the equivalent resistance between nodes C and D of the circuit shown in Figure 2.59.
Problem 2.29: Determine the equivalent resistance between nodes E and F of the circuit shown in Figure 2.59.
Problem 2.30: Determine the equivalent resistance between nodes A and B of the circuit shown in Figure 2.60.
Figure 60
Problem 2.31: Determine the equivalent resistance between nodes C and D of the circuit shown in Figure 2.60.
Problem 2.32: Determine the equivalent resistance between nodes E and F of the circuit shown in Figure 2.60.
Problem 2.33: Find the equivalent resistance of the circuit shown in Figure 2.61. Assume all resistances are in Ohms.
Figure 61
Problem 2.34: Find the equivalent resistance of the circuit shown in Figure 2.62. Assume all resistances are in Ohms.
Figure 62
Problem 2.35: Find the equivalent resistance of the circuit shown in Figure 2.63. Assume all resistances are in Ohms.
Figure 63
Problem 2.36: Determine the equivalent resistance between each of the nodes in the circuit shown in Figure 2.64.
31
Figure 64
Fill in a matrix with your answers as shown below; what must be true of the entries?
Resistance
A
B
C
D
E
F
G
A
0
B
C
D
E
F
G
0
0
0
0
0
0
Problem 2.37: Use voltage dividers to determine the voltage across each of the resistors in Figure 2.65.
Figure 65
Problem 2.38: Verify that the conservation of energy holds for the previous circuit.
Problem 2.39: Use voltage dividers to determine the voltage across each of the resistors in Figure 2.66.
Figure 66
Problem 2.40: Use voltage dividers to determine the voltage across each of the resistors in Figure 2.67.
Figure 67
Problem 2.41: Use current dividers to determine the current through each of the resistors in figure 2.68.
32
Figure 68
Problem 2.42: Verify that the conservation of energy holds for the previous circuit.
Problem 2.43: Use current dividers to determine the current through each of the resistors in figure 2.69.
Figure 69
Problem 2.44: Use current dividers to determine the current through each of the resistors in figure 2.70.
Figure 70
Problem 2.46: Use current dividers to determine the current through each of the resistors in figure 2.71.
Figure 71
Problem 2.47: Use current dividers to determine the current through each of the resistors in figure 2.72.
Figure 72
Problem 2.48: Use voltage dividers to determine the voltage across each of the resistors in Figure 2.73.
33
Figure 73
Problem 2.49: Verify that the conservation of energy holds for the previous circuit.
Problem 2.50: Use voltage dividers to determine the voltage across each of the resistors in Figure 2.74.
Figure 74
Problem 2.51: Use current dividers to determine the current through each of the resistors in figure 2.75.
Figure 75
Problem 2.52: Verify that the conservation of energy holds for the previous circuit.
Problem 2.53: Use current dividers to determine the current through each of the resistors in figure 2.76.
Figure 76
Problem 2.54: Use voltage dividers to determine the voltage across each of the shunt resistors in Figure 2.77. How does the
sequence of voltages V1 , V2 , ... decay?
34
Figure 77
Additional Problems:
Problem 2.55: Determine the equivalent resistance of the resistor configuration shown in Figure 2.78.
Figure 78
Problem 2.56: Determine the equivalent resistance of the resistor configuration shown in Figure 2.79.
Figure 79
Problem 2.57: Determine the equivalent resistance of the resistor configuration shown in Figure 2.80.
Figure 80
Problem 2.58: Determine the equivalent resistance of the resistor configuration shown in Figure 2.81.
35
Figure 81
Problem 2.59: Determine the equivalent resistance of the resistor configuration shown in Figure 2.82.
Figure 82