EG1103 Electrical Engineering
Ben M. Chen
Associate Professor
Office: E4-6-7
Phone: 874-2289
Email: bmchen@nus.edu.sg
http://vlab.ee.nus.edu.sg/~bmchen
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Lecture Format & Course Outlines
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text
book
or
lecture
notes
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Electrical Engineering
Well known electrical engineering companies:
• Singapore Telecom (largest in Singapore)
• Creative Technology (largest manufacturer of PC sound boards)
• Disk Drive Companies (largest producer of PC hard disk Drives)
“Yan can Cook”:
• Ingredients + Recipe + (Funny Talk) = Good Food
Electrical Systems:
• Components + Method + (Funny Talk) = Good Electrical System
EG1103 Module:
• To introduce basic electrical components & analysis methods.
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Reference Textbooks
• D.
E. Johnson, J. R. Johnson and J. L. Hilburn, Electric
Circuits Analysis, 2nd Ed., Prentice Hall, 1992.
• S. A. Boctor, Electric Circuits Analysis, 2nd Ed., Prentice
Hall, 1992.
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Lectures
Lectures will follow closely (but not 100%) the materials in
the textbook.
However, certain parts of the textbook will not be covered
and examined and this will be made known during the
classes.
Attendance is essential.
ASK any question at any time during the lecture.
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Tutorials
The tutorials will start on Week 4 of the semester. (Week 1 corresponds
to the Orientation Week.)
Although you should make an effort to attempt each question before the
tutorial, it is NOT necessary to finish all the questions.
Some of the questions are straightforward, but quite a few are difficult
and meant to serve as a platform for the introduction of new concepts.
ASK your tutor any question related to the tutorials and the course.
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Examination
The examination paper is 2-hour in duration.
You will be provided with a list of important results. This list is given
under “Summary of Important Results” in the Appendix of the textbook.
To prepare for the examination, you may wish to attempt some of the
questions in examinations held in previous years. These papers are
actually the Additional Problems in the Appendix of the textbook (no
solutions to these problems will be given out to the class).
However, note that the topics covered may be slightly different and
some of the questions may not be relevant. Use your own judgement to
determine the questions you should attempt.
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Mid-term Test
There will be an one-hour (actually 50 minutes) test. It will be given
some time around mid-term (most likely after the recess week). The
test will consists 15% of your final grade, i.e., your final grade in this
course will be computed as follows:
Your Final Grade = 15% of Your Mid-term Test Marks (max. = 100)
+ 85% of Your Examination Marks (max = 100)
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Outline of the Course
1. DC Circuit Analysis
SI Units. Voltage, current, power and energy. Voltage and current
sources. Resistive circuits. Kirchhoff's voltage and current laws. Nodal
and mesh analysis. Ideal and practical sources. Maximum power
transfer. Thevenin’s and Norton’s equivalent circuits. Superposition.
Dependent sources. Introduction to non-linear circuit analysis.
2. AC Circuits
Root mean square value. Frequency and phase. Phasor. Capacitor and
Inductor. Impedance. Power. Power factor. Power factor improvement.
Frequency response. Tune circuit. Resonance, bandwidth and Q factor.
Periodic signals. Fourier series.
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Outline of the Course (Cont.)
3. Transient
First order RL and RC circuits. Steady state and transient responses.
Time constant. Voltage and current continuity. Second order circuit.
4. Magnetic Circuit
Magnetic flux and mmf. Ampere’s law. Force between surfaces.
Transformers.
5. Electrical Measurement
Current and voltage measurement. Common instruments. Oscilloscope.
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Web-based Virtual Labs
are now on-line
Developed by: CC Ko and Ben M. Chen
• Have you ever missed your experiments?
• Do you have problems with your lab schedule?
• Do you have problems in getting results for your report?
Visit newly developed web-based virtual labs available from
5:00pm to 8:00am at http://vlab.ee.nus.edu.sg/vlab/
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Chapter 1. SI Units
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1.1 Important Quantities and Base SI Units
L e n g th
m e tre
m
M a ss, m
kilo gra m
kg
T im e , t
se con d
s
am pere
A
T he rm o d yn a m ic te m p e ra tu re
ke lvin
K
P lane angle
radia n
rad
E lectric cu rren t,
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Chapter 2. DC Circuit Analysis
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2.1 Voltage Source
Two common dc (direct current) voltage sources are:
Dry battery (AA, D, C, etc.)
Lead acid battery in car
Regardless of the load connected and the current drawn, the above sources
have the characteristic that the supply voltage will not change very much.
The definition for an ideal voltage source is thus one whose output voltage
does not depend on what has been connected to it. The circuit symbol is
v
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Basically, the arrow and the value signifies that the top terminal has a
potential of v with respect to the bottom terminal regardless of what has
been connected and the current being drawn.
Note that the current being drawn is not defined but depends on the load
connected. For example, a battery will give no current if nothing is
connected to it, but may be supplying a lot of current if a powerful motor
is connected across its terminals. However, in both cases, the terminal
voltages will be roughly the same.
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Using the above and other common circuit symbol, the following are
identical:
+
1.5 V
− 1.5 V
+
1.5 V
1.5 V
−
−
Note that on its own, the arrow does not correspond to the positive
terminal. Instead, the positive terminal depends on both the arrow and
the sign of the voltage which may be negative.
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2.2 Current Source
In the same way that the output voltage of an ideal voltage source does
not depend on the load and the current drawn, the current delivers by an
ideal current source does not depend on what has been connected and
the voltage across its terminals. Its circuit symbol is
i
Note that ideal voltage and current sources are idealisations and do not
exist in practice. Many practical electrical sources, however, behave like
ideal voltage and current sources.
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2.3 Power and Energy
Consider the following device,
i
v
Device
Power Consumed
by Device
p = vi
In 1 second, there are i charges passing through the device. Their electric
potential will decrease by v and their electric potential energy will decrease
by iv. This energy will have been absorbed or consumed by the device.
The power or the rate of energy consumed by the device is thus p = i v.
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Note that p = v i gives the power consumed by the device if the voltage
and current arrows are opposite to one another. The following
examples illustrate this point:
2A
Power consumed/
= 3W
absorbed by source
Energy absorbed
= 300 W hr
in 100 hr
= 0.3 kW hr
= 0.3 unit in PUB bill
1.5 V
−2A
2A
1.5 V
Power supplied
= 3W
by source
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1.5 V
Power absorbed
= −3W
by source
Copyrighted by Ben M. Chen
2.4 Resistor
The symbol for an ideal resistor is
i
R
v
Provided that the voltage and current arrows are in opposite directions,
the voltage-current relationship follows Ohm's law:
v = iR
The power consumed is
p = vi = i
2
v 2
R =
R
Common practical resistors are made of carbon film, wires, etc.
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2.5 Relative Power
Powers, voltages and currents are often measured in relative terms
with respect to certain convenient reference values. Thus, taking
p ref = 1 mW
as the reference (note that reference could be any value), the power
p =
2 W
will have a relative value of
2W
2W
p
=
=
= 2000
3
p ref
1mW
10 W
The log of this relative power or power ratio is usually taken and given
a dimensionless unit of bel. The power p = 2 W is equivalent to
⎛ p
log⎜
⎜p
⎝ ref
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⎞
⎟ = log(2000) = log(1000) + log(2) = 3.3 bel
⎟
⎠
Copyrighted by Ben M. Chen
As bel is a large unit, the finer sub-unit, decibel or dB (one-tenth of a
Bel), is more commonly used. In dB, p = 2W is the same as
⎛ p ⎞
⎟ =10 log (2000 )=33dB
10 log ⎜
⎜p ⎟
⎝ ref ⎠
As an example:
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Reference
Actual power
Relative
Power
p ref
p
p pref
1mW
1mW
1
0 dB
1mW
2 mW
2
3dB
1mW
10 mW
10
10dB
1mW
20 mW
20 = 10 × 2
13 dB = 10 dB + 3 dB
1mW
100 mW
100
20dB
1mW
200 mW
200 = 100 × 2
23 dB = 20 dB + 3 dB
10 log( p pref
)
Copyrighted by Ben M. Chen
Although dB measures relative power, it can also be used to measure
relative voltage or current which are indirectly related to power.
For instance, taking
v ref = 0 .1V
as the reference voltage (again reference voltage could be any value),
the power consumed by applying vref to a resistor R will be
p ref =
v ref
2
R
Similarly, the voltage
v =1V
will lead to a power consumption of
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v2
p =
R
Copyrighted by Ben M. Chen
The voltage v relative to vref will then give rise to a relative power of
v2
⎛ v
p
R
=
=⎜
2
⎜v
pref vref
⎝ ref
or in dB:
2
⎞ ⎛ 1 ⎞2
⎟ = ⎜ ⎟ = 100
⎟ ⎝ 0.1⎠
⎠
R
2
⎛ p ⎞
⎛ v ⎞
⎛ v ⎞
1⎞
⎛
⎟
⎜
⎟
⎜
⎟
⎜
10log⎜
dB = 10log⎜
dB = 20log⎜
dB = 20log⎜ ⎟dB = 20dB
⎟
⎟
⎟
⎝ 0.1⎠
⎝ pref ⎠
⎝ vref ⎠
⎝ vref ⎠
This is often used as a measure of the relative voltage
v vref
.
Key point: When you convert relative power to dB, you multiply its log
value by 10. You should multiply its log value by 20 if you are converting
relative voltage or current.
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As an example:
Reference
Actual voltage
Relative
v ref
v
v v ref
0.1 V
0.1 V
1
0.1 V
0.1 2 V
0.1 V
0. 2 V
0.1 V
0.1 10 V
10
10dB
0.1 V
0.1 20 V
20 = 10 × 2
13dB=10dB+3dB
0.1 V
1V
2
2= 2 × 2
10 = 10 × 10
Voltage
20 log (v v ref
)
0dB
3dB
6dB=3dB+3dB
20dB=10dB+10dB
The measure of relative current is the same as that of relative voltage and
can be done in dB as well.
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The advantage of measuring relative power, voltage and current in
dB can be seen from considering the following voltage amplifier:
Amplifier
v
Voltage
gain
2 = 6 dB
2v
The voltage gain of the amplifier is given in terms of the output voltage
relative to the input voltage or, more conveniently, in dB:
2v
g=
= 2 = 20 log (2 ) dB = 6 dB
v
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If we cascade 3 such amplifiers with different voltage gains together:
Amplifier
v
Voltage
gain
2 = 6 dB
2v
Amplifier
Amplifier
Voltage 2.8v
gain
1.4 = 3 dB
Voltage
gain
10 = 20 dB
28v
the overall voltage gain will be
gtotal = 2 ×1.4 ×10 = 28
However, in dB, it is simply:
g total = 6dB + 3dB + 20dB = 29dB
Under dB which is log based, multiplication's become additions.
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Frequently Asked Questions
Q: Does the arrow associated with a voltage source always point at the
+ (high potential) terminal?
A: No. The arrow itself is meaningless. As re-iterated in the class, any
voltage or current is actually characterized by two things: its direction
and its value. The arrow of the voltage symbol for a voltage source could
point at the - terminal (in this case, the value of the voltage will be
negative) or at the + terminal (in this case, its value will be positive).
Q: What is the current of a voltage source?
A: The current of a voltage source is depended on the other part of circuit
connected to it.
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Frequently Asked Questions
Q: Does a volt source always supply power to other components in a circuit?
A: NO. A voltage source might be consuming power if it is connected to a
circuit which has other more powerful sources. Thus, it is a bad idea to predetermine whether a source is consuming power or supplying power. The
best way to determine it is to follow the definition in our text and computer
the power. If the value turns out to be positive, then the source will be
consuming power. Otherwise, it is supplying power to the other part of the
circuit.
Q: Is the current of a voltage source always flowing from + to - terminals?
A: NO. The current of a voltage source is not necessarily flowing from the
positive terminal to the negative terminal.
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Frequently Asked Questions
Q: What is the voltage cross over a current source?
A: It depends on the circuit connected to it.
Q: Is the reference power (or voltage, or current) in the definition of the
relative power (or voltage, or current) unique?
A: No. The reference power (voltage or current) can be any value.
Remember that whenever you deal with the relative power (voltage or
current), you should keep in your mind that there are a reference power
(voltage or current) and an actual power (voltage or current) associated
with it.
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2.6 Kirchhoff's Current Law (KCL)
As demonstrated by the following examples, this states that the
algebraic sum of the currents entering/leaving a node/closed surface
is 0 or equivalently to say that the total currents flowing into a node is
equal to the total currents flowing out from the node.
i1
i2
i1
i5
i4
i3
i1 + i 2 + i 3 + i 4 + i 5 = 0
i2
i5
i4
i3
for both cases.
Since current is equal to the rate of flow of charges, KCL actually
corresponds to the conservation of charges.
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2.7 Kirchhoff's Voltage Law (KVL)
As illustrated below, this states that the algebraic sum of the voltage
drops around any close loop in a circuit is 0.
v2
v3
v1
v5
v4
v1 +v2 +v3 +v4 +v5 =0
(note that all voltages are in the
same direction)
Since a charge q will have its electric potential changed by qv1, qv2,
qv3, qv4 , qv5 as it passes through each of the components, the total
energy change in one full loop is q ( v1 + v2 + v3 + v4 + v5 ). Thus, from
the conservation of energy: v 1 + v 2 + v 3 + v 4 + v 5 = 0
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2.8 Series Circuit
Consider 2 resistors connected in series:
v
v1
v2
By KVL: - v + v1 + v2 = 0
i
R1
R2
v1 = i R1
v2 =i R2
the voltage-current relationship is
v = v1 + v 2
v = i (R1 + R2 )
Now consider
v
i
the voltage-current relationship is
v = i (R1 + R2 )
R1 + R2
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Since the voltage/current relationships are the same for both circuits,
they are equivalent from an electrical point of view. In general, for n
resistors R1, ..., Rn connected in series, the equivalent resistance R is
R = R1 +L+ R n
Clearly, the resistance's of resistors connected in series add (Prove
it).
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2.9 Parallel Circuit
Consider 2 resistors connected in parallel:
v
i1
i
R1
i2
R2
v
i1 =
R1
v
i2 =
R2
⎛1
1 ⎞
i = i1+ i2 = v ⎜⎜ +
⎟⎟
⎝ R1 R2 ⎠
Clearly, the parallel circuit is equivalent to a resistor R with voltage/current
relationship
v
i=
R
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with
1
1
1
=
+
R
R1
R2
Copyrighted by Ben M. Chen
In general, for n resistors R1, …, Rn, connected in parallel, the equivalent
resistance R is given by
1
1
1
=
+L +
R
R1
Rn
Note that 1/R is often called the conductance of the resistor R. Thus,
the conductances of resistors connected in parallel add.
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2.10 Voltage Division
Consider 2 resistors connected in series:
i
R1
v1
R2
v2
v
i=
R1 + R2
⎛ R1
v1 =iR1 = ⎜⎜
⎝ R1 + R 2
⎞
⎟⎟ v
⎠
⎛ R2
v 2 =iR 2 = ⎜⎜
⎝ R1 + R 2
⎞
⎟⎟ v
⎠
v
The total resistance of the circuit is R1 + R2.
Thus,
v1 R1
=
v2 R2
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2.11 Current Division
Consider 2 resistors connected in parallel:
i
v
1
⎛
⎞
⎜
⎟
v
R
1
⎟ i = ⎛⎜ R2 ⎞⎟ i
i1=
= ⎜
⎜ R +R ⎟
1 ⎟
⎜ 1
R1
⎝ 1 2⎠
+
i2
⎜R
⎟
R2 ⎠
⎝ 1
i1
R1
R2
i2 =
v
R2
1
⎛
⎞
⎜
⎟
R
2
⎟ i = ⎛⎜ R1 ⎞⎟ i
= ⎜
⎜ R +R ⎟
1 ⎟
⎜ 1
⎝ 1 2⎠
⎜R + R ⎟
2 ⎠
⎝ 1
Thus,
1
1
1
=
+
The total conductance of the circuit is
R R1 R2
while the equivalent resistance is
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v= i R =
i
i1
i2
=
R2
R1
1
1
+
R1
R2
Copyrighted by Ben M. Chen
2.12 Ladder Circuit
4+(3 || 2)
5
Consider the following ladder circuit:
2
4
3
5
5 || [4+(3||| 2)]
The equivalent resistance can be
determined as follows:
The network is equivalent to a
4
resistor with resistance
5
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3 || 2
R = 5 || [4+(3 || 2)] =
1
1
1
+
5 4+(3 || 2)
=
1
1
1
+
5 4+ 1
1 1
+
3 2
Copyrighted by Ben M. Chen
2.13 Branch Current Analysis
Consider the problem of determining the equivalent resistance of the following
bridge circuit:
4
2
3
2
4
Since the components are not connected in straightforward series or parallel
manner, it is not possible to use the series or parallel connection rules to
simplify the circuit. However, the voltage-current relationship can be
determined and this will enables the equivalent resistance to be calculated.
One method to determine the voltage-current relationship is to use the branch
current method.
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Branch Current Analysis: Example One
i
i1
1. Assign branch currents (with
i − i1
4(i − i1 ) 4
i2
v
i − i1 − i2
2(i −i1 −i2) 2
3
3 i2
2 i1 2
any directions you prefer so
that currents in other branches
i1 + i2
4 ( i1 + i2 )
4
can be found)
2. Find all other branch currents
(with any directions you prefer)
(Use KCL to find them)
v
= Equivalent Resistance
i
KVL:
v = 2i1 + 4(i1 + i2 ) = 6i1 + 4i2
KVL:
KVL:
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4(i − i1 ) + 3i2 = 2i1
4(i1 + i2 ) + 3i2 = 2(i − i1 − i2 )
3. Write down branch voltages
4. Identify independent loops
Eliminate i1 and i2,
7i
i1 =
12
i
i2 = −
6
v=
17
i
6
Copyrighted by Ben M. Chen
Branch Current Analysis: Example Two
2i2
4i1
4
2
2 − i1
i2
(so that currents in other
i1 − i2
2 − i1
2A
2
i1
1. Assign branch currents
1 3(i − i )
1
2
branches can be found).
1V
3
2. Find all other branch
currents (KCL)
3. Write down voltages across components
KVL: 2 − i1 = 4i1 + 3(i1 − i2 )
This implies:
3i1 − 5i2 = 1
⇒
8i1 − 3i2 = 2
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4. Identify independent loops
(ex. 2A branch)
KVL: 1 + 2i2 = 3(i1 − i2 )
⎡3 − 5⎤ ⎡ i1 ⎤ ⎡1⎤
⎢8 − 3⎥ ⎢i ⎥ = ⎢2⎥
⎣
⎦ ⎣ 2⎦ ⎣ ⎦
⎡ i1 ⎤ ⎡3 − 5⎤
⇒ ⎢ ⎥= ⎢
⎥
i
8
3
−
⎦
⎣ 2⎦ ⎣
−1
⎡1 ⎤
1 ⎡ 7 ⎤
⎢2⎥ = 31 ⎢ − 2⎥
⎣
⎦
⎣ ⎦
Copyrighted by Ben M. Chen
2.14 Mesh (Loop Current) Analysis
i
ia − i
4
ia
4(ia − i )
i
v
2
ib − ia
ib −i
ib
2(ib − i )
1. Assign fictitious loop currents.
2ia
4
2. Find branch currents (KCL)
3. Write down branch voltages
ib
3(ib − ia )
2
KVL:
3
ia
4ib
4. Identify independent loops
5. Simplify the equations
v = 2 i a + 4 ib
obtained, we get
KVL: 4 ( i a − i ) − 3 ( i b − i a ) + 2 i a = 0
KVL:
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2(ib − i ) + 4ib + 3(ib − ia ) = 0
17i
17
⇒ Requivalence =
v=
6
6
Copyrighted by Ben M. Chen
2.15 Nodal Analysis
Va − Vb
2
2A
4
A
Va
1 − Vb
Va V a − Vb
4
1
2
B
Vb
2. Find branch voltages (KVL)
Vb
3
1 − Vb
2
3. Determine
1V
3
branch currents
4. Apply KCL to Nodes A & B
1. Assign nodal voltage w.r.t. the reference node
5Va − Vb = 8
⎡5 − 1 ⎤ ⎡Va ⎤ ⎡ 8 ⎤
⎢3 −13⎥ ⎢V ⎥ = ⎢− 6⎥ ⇐ 3V − 13V = −6 ⇐
a
b
⎣
⎦ ⎣ b⎦ ⎣ ⎦
⇒
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55
27
Va =
Vb =
31
31
Node A:
2 = Va +
Va − Vb
4
Node B:
Va − Vb 1 − Vb Vb
=
+
4
2
3
Copyrighted by Ben M. Chen
2.16 Practical Voltage Source
i
An ideal voltage source is one whose
terminal voltage does not change
v
R load
with the current drawn. However, the
terminal voltages of practical sources
usually decrease slightly as the
Practical voltage source
currents drawn are increased.
A commonly used model for a
i
practical voltage source is:
R in
To represent it as a series of an ideal
voc
v
R load
voltage source & an internal
resistance.
voc = v + iRin
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⇐
Model for voltage source
Copyrighted by Ben M. Chen
When Rload = 0 or when the source is short circuited so that v = 0 :
i=
voc
R in
R in
voc
v=0
Short circuit
R load = 0
When Rload = ∞ or when the source is open circuited so that i = 0 :
i=0
R in
voc
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v = voc
Open circuit
R load = ∞
Copyrighted by Ben M. Chen
Graphically:
v
Slope = R in
voc
Rin
voc
i
0
Good practical voltage source should therefore have small internal
resistance, so that its output voltage will not deviate very much from the
open circuit voltage, under any operating condition.
The internal resistance of an ideal voltage source is therefore zero so that
does not change with
49
v
i.
Copyrighted by Ben M. Chen
To determine the two parameters voc and Rin that characterize, say, a
battery, we can measure the output voltage when the battery is opencircuited (nothing connected except the voltmeter). This will give voc .
Next, we can connect a load resistor and vary the load resistor such
v
that the voltage across it is oc . The load resistor is then equal to Rin :
2
voc
2
i
R in
voc
50
voc
v=
2
R load = R in
Copyrighted by Ben M. Chen
2.17 Maximum Power Transfer
The power absorbed by the
load resistor is
Consider the following circuit:
2
v
R
pload =i 2 Rload = oc load 2
(Rin +Rload )
i
This is always positive. However, if
R in
voc
v
R load
Rload = 0 or Rload = ∞, Pload = 0.
pload
Model for voltage source
The current in the load resistor is
voc
i=
Rin + Rload
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0
R load
Copyrighted by Ben M. Chen
Differentiating:
⎡
1
2 Rload ⎤ 2 ⎡ Rin − Rload ⎤
dpload
= voc2 ⎢
−
= voc ⎢
2
3⎥
3⎥
dRload
(
)
(
)
(
)
+
+
+
R
R
R
R
R
R
in
load
⎣ in load
⎦
⎣ in load ⎦
The load resistor will be absorbing the maximum power or the source will be
transferring the maximum power if the load and source internal resistances
are matched, i.e.,
Rin =Rload . The maximum power transferred is given by
2
v
R
pload =i 2 Rload = oc load 2
(Rin +Rload )
⇒
voc2
pmax load =
4 Rin
pload
2
voc
4Rin
When the load absorbs the maximum
power from the source, the overall
power efficiency of 50%, which is too
0
52
Rin
Rload
low for a usual electric system.
Copyrighted by Ben M. Chen
Why is the electric power transferred from power stations to local stations
in high voltages?
resistance in wire
Power
Station
P = 300 KW
Rw
v
load
i = 300 KW
v
Power loss in the transmission line:
Ploss = i 2 Rw
2
(
300 KW ) Rw
=
v2
The higher voltage v is transmitted, the less power is lost in the wire.
53
Copyrighted by Ben M. Chen
2.18 Practical Current Source
An ideal current source is one which delivers a constant current regardless
of its terminal voltage. However, the current delivered by a practical
current source usually changes slightly depending on the load and the
terminal voltage.
A commonly used model for a current source is:
i
i sc
R in
v
R load
⇒
i sc =
v
+ i
R in
Model for current source
54
Copyrighted by Ben M. Chen
When Rload = 0 or when the source is short-circuited so that v = 0 :
i = i sc
0
i sc
R in
Graphically:
v=0
Short circuit
R load = 0
Good practical current source should therefore
v
have large internal resistance so that the current
Slope = R in
delivered does not deviate very much from the
short circuit current under any operating
0
i
i sc
55
condition.
The internal resistance of an ideal current source is
therefore infinity so that i does not change with v.
Copyrighted by Ben M. Chen
2.19 Thevenin's Equivalent Circuit
i
Key points:
Complicated
circuit
Complicated circuit
with
withlinear
linear elements
elements
v
such
suchas
asresistors,
resistors,
voc = v + i R in
1. The black box
(i.e., the part of the
voltage/current
sources
voltage/current sources
circuit) to be
simplified must be
linear.
i
2. The black box
R in
voc
must have two
v
voc = v + i R in
terminals
connected to the
rest of the circuit.
Thevenin's equivalent circuit
56
Copyrighted by Ben M. Chen
Thevenin’s Equivalent Circuit (An Example)
6 − 3i
3
1
Applying KVL:
4i
2−i
i
4
v
2
1 + (6 − 3 i) = 4 i + v
⇒
The circuit is equivalent to:
7 = v + 7i
7i
i
7
voc = 7 Rin = 7
57
⇐
7
v
Copyrighted by Ben M. Chen
Alternatively, note that from the Thevenin's
equivalent circuit:
6
0
2
3
R in
voc
Open circuit voltage= voc
Short Circuit
R in
58
1
0
4
voc=77
2
⇒
Resistance seen with
source replace by= R in
internal resistance
3
4
Rin= 7
7
Copyrighted by Ben M. Chen
2.20 Norton's Equivalent Circuit
It is simple to see
i
that if we let
R in
voc
v
voc = v + i R in
voc = isc Rin
then the
relationships of
Thevenin's equivalent circuit
if voc = i sc R in
if voc = i sc R in
both Thevenin’s
i
i sc
R in
voltage/current for
i sc = v + i
R in
v
i sc R in = v + i R in
and Norton’s
equivalent circuits
are exactly the
same.
59
Norton's equivalent circuit
Copyrighted by Ben M. Chen
From the Norton's equivalent circuit, the two parameters isc and Rin
can be obtained from:
i = i sc
i sc
Short circuit current = i sc
R in
v=0
R in
Resistance seen with
source replace by = R in
internal resistance
Open Circuit
60
Copyrighted by Ben M. Chen
Example: Reconsider the circuit
The Norton's equivalent
circuit is therefore:
3
4
1
2
7
1
6−3isc
2 − isc
3
1
4 isc
isc
4
0
2
And the Thevenin's equivalent
circuit is:
1 + 6− 3isc = 4 isc or isc = 1
7
3
7
4
R in = 7
61
Copyrighted by Ben M. Chen
Summary on how to find an equivalent circuit:
Step 1. Identify the circuit or a portion of a complicated circuit that is to be
simplified. Be clear in your mind on which two terminals are to be
connected to the other network.
Step 2. Short-circuit all the independent voltage sources and open-circuit
all independent current sources in the circuit that you are going to simplify.
Then, find the equivalent resistance w.r.t. the two terminals identified in
Step 1.
Step 3. Find the open circuit voltage at the output terminals (for Thevenin’s
equivalent circuit) or the short circuit current at the output terminals (for
Norton’s equivalent circuit).
Step 4. Draw the equivalent circuit (either the Thevenin’s or Norton’s one).
62
Copyrighted by Ben M. Chen
More Example For Equivalent Circuits:
R1
ii11
R11
R2
i
v
R
RR
vv
RR
R
R
R
2
v
2
ii11
R
iR
RR11
RR
2
R
v v2
i1
R
iR
63
R1
R
2
v
2
Copyrighted by Ben M. Chen
2.21 Superposition
1/7
3
Consider finding isc in the circuit:
4
1
isc
3
Open circuit
4
24/7
1
24/7
2(4/7)
2
3
2(3/7)
4
2
Short circuit
By using the principle of superposition, this can be done by finding
the components of isc due to the 2
isc = 2(3/7) + (1/7) = 1
independent sources on their own
(with the other sources replaced by
3
1
4
2
their internal resistances):
64
Copyrighted by Ben M. Chen
Linear Systems and Superposition
in1
Linear System
Out = 6 in1 + 7 in2
in2
16
Linear System
6 (16) = 96
0
(the definition of the
linear system)
Note that:
Linear system: linear
relationship between
0
Linear System
7 (27) = 189
inputs and outputs
27
Superposition:
Applicable only to
16
Linear System
96 + 189 = 285
linear systems
27
65
Copyrighted by Ben M. Chen
2.22 Dependent Source
Consider the following system:
CD
Player
Amplifier
This may be represented by
300
vd
1
CD Player
66
2
3k
Amplifier
2 vd
2
v
Loudspeaker
Copyrighted by Ben M. Chen
Note that the source in the Amplifier block is a dependent source. Its
value depends on vd , the voltage across the inputs of the amplifier. Using
KCL and KVL, the voltage v can be easily found:
5
11
1
3300
300
1
2
vd = 10
11
3k
2 vd = 20
11
2
v = 10
11
However, if we use the principle of superposition treating the dependent
source as an independent source (which is wrong !), the value of v will be 0:
67
Copyrighted by Ben M. Chen
1
3300
0
300
2
vd = 10
11
1
3k
0
0
0
2
0
0
300
0
2
2
vd = 0
3k
2 vd = 0
Dependent sources, which depend on other voltages/currents in the circuit
and are therefore not independent excitations, cannot be removed when
the principle of superposition is used. They should be treated like other
passive components such as resistors in circuit analysis.
68
Copyrighted by Ben M. Chen
Topics Skipped
• Nonlinear Circuit
• Delta Circuit
• Star Circuit
• All these topics are not examinable in test and examination.
69
Copyrighted by Ben M. Chen
Reading Assignment
• Appendix C.1. Matrix Algebra
• Appendix C.2. Complex Number
• Appendix C.3. Linear Differential Equation
70
Copyrighted by Ben M. Chen
Chapter 3. AC Circuit Analysis
71
Copyrighted by Ben M. Chen
Appendix Materials: Operations of Complex Numbers
Coordinates: Cartesian Coordinate and Polar Coordinate
12 + j5 = 13 e j 0.39 =
real part
imaginary part
Euler’s Formula:
magnitude
12 2 + 52 e
⎛ 5 ⎞
j tan −1 ⎜ ⎟
⎝ 12 ⎠
argument
e jθ = cos(θ ) + j sin(θ )
Additions: It is easy to do additions (subtractions) in Cartesian coordinate.
( a + jb) + ( v + jw) = ( a + v ) + j (b + w)
Multiplication's: It is easy to do multiplication's (divisions) in Polar coordinate.
re jθ
72
•
ue jω = ( ru )e j (θ + ω )
re jθ
r j (θ − ω )
=
e
jω
ue
u
Copyrighted by Ben M. Chen
3.1 AC Sources
Voltages and currents in DC circuit are constants and do not change with
time. In AC (alternating current) circuits, voltages and currents change with
time in a sinusoidal manner. The most common ac voltage source is the
mains:
1
50
θ = phase = 0.4rad
230 2
f = frequency = 50Hz
t
− 0.4
2π ( 50 )
ω =2πf =angular frequency =100π =314rad s
T=
2πt ⎞
+θ ⎟
v (t ) = 2r cos(2πft+θ ) = 2r cos(ωt+θ ) = 2r cos⎜⎛
⎝ T
⎠
= 230 2 cos(100πt+0.4 )
73
1
1
= period = = 0.02s
f
50
2r=peak value=230 2=324V
r=rms (root mean square) value=230V
Copyrighted by Ben M. Chen
How to find the phase for a sinusoidal function?
r 2
−a
v (t ) = r 2 cos(ω [t + a ])
= r 2 cos(ω t + ω a )
Phase θ = ω a
− π ≤ θ ≤ π
74
t
v (t ) = r 2 cos(ω t )
⎛ 0 .4 ⎞
⎟⎟ = 0.4
θ = ω a = 314⎜⎜
2
(
50
)
π
⎠
⎝
for previous example.
Copyrighted by Ben M. Chen
Euler’s Formula:
e jω = cos(ω ) + j sin(ω )
3.2 Phasor
A sinusoidal voltage/current is represented using complex number format:
[
]
[( )(
v (t ) = 2r cos(ωt+θ ) = 2r Re e j (ω t+θ ) = Re re jθ
2e jω t
)]
The advantage of this can be seen if, say, we have to add 2 sinusoidal
voltages given by:
π
v1 (t ) = 3 2 cos⎜⎛ ωt + ⎞⎟
6⎠
⎝
π
v2 (t ) = 5 2 cos⎜⎛ ωt − ⎞⎟
4⎠
⎝
π
⎡
⎛
⎞
j
⎛
6
⎜
v1 (t )=3 2 cos⎜ ωt+ ⎟=Re ⎢ 3e ⎟ 2e jωt
⎜
⎟
6⎠
⎝
⎢⎣⎝
⎠
(
π⎞
π
⎡⎛ j π
−j ⎞
v1 (t ) + v2 (t ) = Re ⎢⎜ 3e 6 +5e 4 ⎟ 2eωt
⎜
⎟
⎢⎣⎝
⎠
(
75
⎡⎛ − j π ⎞
π⎞
⎛
v2 (t )=5 2 cos⎜ ω − ⎟=Re ⎢⎜ 5e 4 ⎟ 2e jωt
⎜
⎟
⎝ 4⎠
⎢⎣⎝
⎠
⎤
⎥
⎥⎦
(
)
⎤
− j 0.32
⎥ = Re 6.47e
⎥⎦
)
[(
)(
⎤
⎥
⎥⎦
)
)]
2e jωt = 6.47 2 cos(ωt−0.32 )
Copyrighted by Ben M. Chen
Note that the complex time factor
2e jω t appears in all the expressions.
If we represent v1 (t ) and v2 (t ) by the complex numbers or phasors:
V1 = 3 e
j
π
6
π
representing v1 (t ) = 3 2 cos⎜⎛ ωt + ⎞⎟
6⎠
⎝
V2 = 5e
−j
π
4
π
representing v2 (t ) = 5 2 cos⎛⎜ ωt − ⎞⎟
4⎠
⎝
then the phasor representation for v1 (t ) + v2 (t ) will be
V1 + V2 = 3e
3e
j
π
6
+ 5e
−j
π
4
j
π
6
+ 5e
−j
π
4
= 6.47e − j 0.32
representing v1 (t ) + v2 (t ) = 6.47 2 cos(ωt − 0.32 )
⎛ ⎛π ⎞
⎛ π ⎞⎞ ⎛ ⎛ π ⎞
⎛ π ⎞⎞
= 3⎜ cos⎜ ⎟ + j sin⎜ ⎟ ⎟ + 5⎜ cos⎜ − ⎟ + j sin⎜ − ⎟ ⎟ = 6.14 − j 2.03 = 6.47e − j 0.32
⎝ 6 ⎠⎠ ⎝ ⎝ 4 ⎠
⎝ 4 ⎠⎠
⎝ ⎝6⎠
Euler’s Formula:
76
e jω = cos(ω ) + j sin(ω )
Copyrighted by Ben M. Chen
By using phasors, a time-varying ac voltage
[( )(
v (t ) = 2r cos(ωt+θ ) = Re re jθ
2e jωt
)]
becomes a simple complex time-invariant number/voltage V = re jθ = r
θ
r = V = magnitude/modulus of V = r.m.s. value of v (t )
θ = Arg[V ] = phase of V
Graphically, on a phasor diagram:
Imag
Using phasors, all time-varying ac quantities
V
become complex dc quantities and all dc circuit
r
analysis techniques can be employed for ac
θ
0
circuit with virtually no modification.
Real
Complex Plane
77
Copyrighted by Ben M. Chen
i ( t)
Example:
4
5 2cos (ω t − 0.2)
3 2cos (ω t + 0.1)
6
I
4
5 e-
j 0.2
6
3e
j 0.1
I
6
30 e-
78
j 0.2
Thevenin's equivalent circuit
for current soure and 6 Ω resistor
4
3e
j 0.1
Copyrighted by Ben M. Chen
I =
30 e-
− 3e
10
j 0.2
j 0.1
= 2.64 − j 0.63 = 2.71 e - j 0.23
6
4
3 e j 0.1
30 e- j 0.2
30e − j 0.2 − 3e j 0.1
I =
10
= 3 [cos( −0.2) + j sin( −0.2)] − 0.3 [cos 0.1 + j sin 0.1]
= ( 2.940 − j 0.596) − (0.299 + j 0.030) = 2.641 − j 0.626
=
2.6412 + 0.6262 e
⎛ − 0.626 ⎞
j tan − 1 ⎜
⎟
⎝ 2.641 ⎠
= 2.71e − j 0.23
⇒ i (t ) = 2.71 2 cos(ω t − 0.23)
79
Copyrighted by Ben M. Chen
3.3 Root Mean Square (rms) Value
For the ac voltage
T
v( t )
r 2
2πt
v (t ) = 2r cos(2πft + θ ) = 2r cos⎛⎜
+ θ ⎞⎟
⎝ T
⎠
−θ T
2π
t
cos(2 x )=2 cos 2 ( x )−1
v 2( t)
4πt
⎡
⎤
+ θ ⎞⎟ = r 2 ⎢1 + cos⎛⎜
+ 2θ ⎞⎟⎥
v (t ) = 2 r cos ⎜
⎝ T
⎠
⎝ T
⎠⎦
⎣
2
2
2 ⎛ 2πt
2r
2
t
The average or mean of the square value is
1 T 2⎡
⎛ 4πt + 2θ ⎞⎤ dt = 1 T r 2 dt = r 2
1
1 T 2
2
+
r
1
cos
v (t ) dt = ∫0 v (t )dt = ∫0 ⎢
⎜
⎟⎥
∫
∫
T
T 0
⎝ T
⎠⎦
T
1 period 1 period
⎣
The square root of this or the rms value of v (t) is rms value of 2r cos(ωt + θ ) = r
Side Note: rms value can be defined for any periodical signal.
80
Copyrighted by Ben M. Chen
3.4 Power
Consider the ac device:
i ( t ) = ri 2cos (ω t + θi )
v ( t ) = rv 2cos (ω t + θv )
Device
Using 2 cos( x1 ) cos( x2 )=cos( x1− x2 )+cos( x1+ x2 ,) the instantaneous power
consumed is
p (t ) = i (t )v (t ) = 2 ri rv cos(ωt + θ i ) cos(ωt + θ v ) = ri rv [cos(θ i − θ v ) + cos(2ωt + θ i + θ v )]
The average power consumed is
pav =
81
1
ri rv T ⎡
⎛ 4πt + θ + θ ⎞⎤ dt
(
)
p
t
dt
(
)
=
−
+
θ
θ
cos
cos
= ri rv cos(θ i − θ v )
∫
⎜
∫0 ⎢
i
v
i
v ⎟⎥
1 period 1 period
T
⎝ T
⎠⎦
⎣
Copyrighted by Ben M. Chen
T
In phasor notation:
rv 2
v( t )
− θv T
2π
V =rv e
t
ri 2
i ( t)
− θi T
2π
I =ri e
jθ v
jθ i
, V =rv e
*
, I =ri e
*
− jθ v
− jθ i
t
V I = rv ri e
j (θ i −θ v )
VI = rv ri e
j (θ v −θ i )
*
v( t ) i ( t )
r i rv cos ( θ i − θ v )
t
[
*
]
[ ]
[ ]
pav = rv ri cos(θ v − θ i ) = rv ri cos(θ i − θ v ) = Re rv ri e j (θ v −θi ) = Re V * I = Re VI *
82
Copyrighted by Ben M. Chen
Note that the formula pav =Re[I ∗V ] is based on rms voltages and currents.
Also, this is valid for dc circuits, which is a special case of ac circuits with
f = 0 and V and I having real values.
2.7 e - j 0.23
Example: Consider the ac circuit,
6
30e − j 0.2 − 3e j 0.1
I=
= 2.7148e − j 0.2327
4+6
(
)(
∗
3e j 0.1source : Re ⎡ 2.7e − j 0.23 3e j 0.1
⎢⎣
(
4
30 e- j 0.2
3e
)⎤⎥⎦ = Re[8.1e ]= 8.1cos(0.33) = 7.66
)(
j 0.33
)
∗
30e − j 0.2 source : Re ⎡ − 2.7e − j 0.23 30e − j 0.2 ⎤ = − 81cos(0.03) = − 80.96
⎢⎣
⎥⎦
(
)(
)⎤⎥⎦ = 6(2.7)
(
)(
)⎤⎥⎦ = 4(2.7)
6 Ω resistor : Re ⎡ 2.7e − j 0.23 6 × 2.7e − j 0.23
⎢⎣
*
4 Ω resistor : Re ⎡ 2.7e − j 0.23 4 × 2.7e − j 0.23
⎢⎣
83
*
j 0.1
2
2
=0
= 43.74
= 29.16
Copyrighted by Ben M. Chen
Ignoring the phase difference between
3.5 Power Factor
V and I, the voltage-current rating or
Consider the ac device:
I = ri e j θ i
V = rv e j θ v
apparent power consumed is
Apparent power=voltage - current rating=V I =rv ri VA
Device
However, the actual power consumed is
[ ]
Actual power=Re V ∗ I =rv ri cos(θ i −θ v )W
The ratio of the these powers is the power factor of the device:
Power factor =
Actual power
= cos(θ i −θ v )
Apparent power
This has a maximum value of 1 when Unity power factor ⇔ I and V in phase ⇔θ i =θ v
The power factor is said to be leading or lagging if
Leading power factor ⇔ I leads V in phase⇔θ i >θ v
Lagging power factor ⇔ I lags V in phase⇔θ i <θ v
84
Copyrighted by Ben M. Chen
rv = 230 r r = 2300 ⇒ r = 10
i v
i
Consider the following ac system:
AC
Generator
0.1 Ω
Electrical
Cables
0.1
e jθ i
85
AC
Generator
Electrical
Cables
230 V, 2300 VA
Electrical Machine
10 e j θ i
Unknowns
230 e j θ v
Electrical
Machine
Copyrighted by Ben M. Chen
The power consumed by the machine and power loss at different power factors are:
Voltagecurrent
rating
2300 VA
2300 VA
2300 VA
Voltage
across
machine
230 V
230 V
230 V
Current
10 A
10 A
10 A
Power
factor
0.11 leading
1
0.11 lagging
θi − θ v
cos− 1 (0.11) = 1.4 rad
0
− cos − 1 (0.11) = −1.46rad
Power
consumed
by machine
Power loss
in cables
86
(2300)(0.11) = 232 W (2300)(1) = 2300W
(0.1)(10)2 = 10W
(0.1)(10)2 = 10W
(2300)(0.11) = 232W
(0.1)(10)2 = 10W
Copyrighted by Ben M. Chen
3.6 Capacitor
A capacitor consists of parallel metal plates for storing electric charges.
Conducting plate
with area A
+ + +
+
+ + +
+
+
+ +
+ ++ + +
+
+ +
+
d
−− − − −−−
−− − −− −−
−
− − − −− −
Insulator
with a dielectric
constant ε
(permittivity)
The capacitance of the capacitor is given by C = ε A F or Farad
d
Area of metal plates required
to produce a 1F capacitor in
the free space if d = 0.1 mm is
87
A =
Cd
ε
=
1F × 0.0001m
2
=
11
.
3
(km)
8.85 × 10− 12 F/m
Copyrighted by Ben M. Chen
The circuit symbol for an ideal
For dc circuits:
capacitor is:
v (t ) = constant ⇒
i (t)
v (t)
C
dv (t )
= 0 ⇒ i (t ) = 0
dt
and the capacitor is equivalent to
an open circuit:
i (t) = 0
Provided that the voltage and current
v (t) = constant
i (t) = 0
C
arrows are in opposite directions, the
voltage-current relationship is:
i (t ) = C
88
dv (t )
dt
This is why we don’t consider the
capacitor in DC circuits.
Copyrighted by Ben M. Chen
Consider the change in voltage,
In general, the total energy stored in
current and power supplied to the
the electric field established by the
capacitor as indicated below:
charges on the capacitor plates at
v( t)
time is
Cv 2 (t )
e(t ) =
2
vf
0
1
Proof.
t
Cvf
89
−∞
dv ( x )
dx
dx
−∞
t
t
C 2
= C ∫ v ( x )dv ( x ) = v ( x )
−∞
2
−∞
C
= v 2 (t ) − v 2 ( −∞ )
2
Cv 2 (t )
=
,
if v ( −∞ ) = 0.
2
= ∫ v ( x )C
1
t
Cvf
Area = Energy stored =
2
2
0
−∞
t
p ( t) = v ( t) i ( t) = Instantaneous power supplied
consumed
Cvf
t
e(t ) = ∫ p ( x )dx = ∫ v ( x )i ( x )dx
i ( t)
0
t
1
t
2
[
]
Copyrighted by Ben M. Chen
Now consider the operation of a capacitor in an ac circuit:
i (t ) = C
v (t ) = rv 2 cos(ωt + θ v )
dv (t )
= −ωCrv 2 sin(ωt + θ v )
dt
π
= ωCrv 2 cos(ωt + θ v + )
2
In phasor format:
j
I = ω C rv e θ v e
V = rv e
jθ v
π
j2
I
j
= j ωC rv e θ v = j ω C V
1
V
=
I
jω C
C
1
j ωC
V
With phasor representation, the capacitor behaves as if it is a resistor
with a "complex resistance" or an impedance of
ZC =
1
jω C
[ ]
∗
[
∗
pav = Re I V = Re I IZ C
]
⎡ I2 ⎤
= Re ⎢
⎥=0
⎢⎣ jω C ⎥⎦
An ideal capacitor is a non-dissipative but energy-storing device.
90
Copyrighted by Ben M. Chen
Since the phase of I relative to V that of is
⎡ 1 ⎤
I
Arg[I ]−Arg[V ]=Arg ⎡⎢ ⎤⎥= Arg ⎢ ⎥=Arg[ jω C ]=900
⎣V ⎦
⎣ ZC ⎦
the ac current i(t) of the capacitor leads the voltage v(t) by 90°.
− sin(ω t )
sin(ω t )
−a
−π
2
π⎞
π⎞
⎛
⎛
− sin ⎜ ωt − ⎟ = cos(ωt ) = sin ⎜ ωt + ⎟
2⎠
2⎠
⎝
⎝
91
cos(ω t )
Copyrighted by Ben M. Chen
Example: Consider the following ac circuit:
230 V
50 Hz
319 μF
1
j
-6 = − 10
(319)
j 2π(50)
10
30 Ω
230 e j 0 = 230
30
In phasor notation (taking the source
to have a reference phase of 0):
− j 10
230
92
j 0.32
230
= 7.3 e
30 − j10
30
Copyrighted by Ben M. Chen
Total circuit impedance
Z = (30 − j10) Ω
Total circuit reactance
X = Im[Z ] = Im[30 − j10] = − 10 Ω
Total circuit resistance
R = Re[Z ] = Re[30 − j10] = 30 Ω
Current (rms)
Current (peak)
2 I = 7.3 2 = 10A
Source V-I phase relationship
I leads by 0.32rad
Power factor of entire circuit
cos(0.32) = 0.95 leading
Power supplied by source
Re (230)∗ 7.3e j 0.32 = (230)(7.3) cos(0.32) = 1.6 kW
Power consumed by resistor
93
I = 7.3A
[
)]
(
(
)(
∗
Re⎡ 7.3e j 0.32 30 × 7.3e j 0.32
⎢⎣
)⎤⎥⎦ = (7.3) 30 = 1.6 kW
2
Copyrighted by Ben M. Chen
Impedance, Resistance,
Reactance,
Relations?
Admittance, Conductance, and Susceptance
Impedance:
Z = R + jX
1
1
R − jX
=
=
Admittance: Y =
Z R + jX (R + jX )(R − jX )
R − jX
R
−X
= 2
= 2
+j 2
2
2
R +X
R +X
R + X2
= G + jB
Conductance
94
Susceptance
Copyrighted by Ben M. Chen
3.7 Inductor
For dc circuits:
An inductor consists of a coil of wires
for establishing a magnetic field. The
i (t ) = constant ⇒
di (t )
= 0 ⇒ v (t ) = 0
dt
circuit symbol for an ideal inductor is:
and the inductor is equivalent to a
i (t)
v (t)
short circuit:
L
i (t) = constant
v (t) = 0
L
v (t) = 0
Provided that the voltage and current
arrows are in opposite directions, the
voltage-current relationship is:
di (t )
v (t ) = L
dt
95
That is why there is nothing
interesting about the inductor in DC
circuits.
Copyrighted by Ben M. Chen
Consider the change in voltage,
current and power supplied to the
inductor as indicated below:
i ( t)
In general, the total energy stored
in the magnetic field established
by the current i(t) in the inductor
at time t is given by
if
0
1
t
1
t
p ( t) = v ( t) i ( t) =Instantaneous power consumed
L if 2
Area = Energy stored =
2
L if 2
96
−∞
−∞
t
L if
0
t
e(t ) = ∫ p ( x )dx = ∫ v ( x )i ( x )dx
v( t)
0
t
Li 2 (t )
e(t ) =
2
1
t
di ( x )
= ∫ i( x) L
dx
dx
−∞
t
t
L 2
= L ∫ i ( x )di ( x ) = i ( x )
−∞
2
−∞
L
= i 2 (t ) − i 2 ( −∞ )
2
Li 2 (t )
=
,
if i ( −∞ ) = 0.
2
[
]
Copyrighted by Ben M. Chen
Now consider the operation of an inductor in an ac circuit:
i (t)
v (t)
i (t ) = ri 2 cos(ωt + θ i )
L
v (t ) = L
di (t )
= −ωLri 2 sin(ωt + θ i )
dt
π
= ωLri 2 cos(ωt + θ i + )
2
In phasor:
i (t)
v (t)
I = ri e jθ
I
i
jω L
L
ZL =
V
V = ωLri e jθ e jπ / 2 = jωLri e jθ = ( jωL) I
i
V
= jωL
I
i
ZL is the impedance of the inductor. The ave. power absorbed by the inductor:
[ ] [
]
[
]
[
pav = Re I ∗V =Re I ∗ Z L I = Re jω LI ∗ I = Re jω L I
97
2
]= 0
Copyrighted by Ben M. Chen
Since the phase of I relative to that of V is
⎡1 ⎤
I
⎡ 1 ⎤
Arg[I ] − Arg[V ] = Arg ⎡⎢ ⎤⎥ = Arg ⎢ ⎥ = Arg ⎢
= − 900
⎥
⎣V ⎦
⎣ jωL ⎦
⎣ZL ⎦
the ac current i(t) lags the voltage v(t) by 90º.
As an example, consider the following series ac circuit:
230 V
50 Hz
319 μ F
3Ω
31.9 mH
We can use the phasor representation to convert this ac circuit to a ‘DC’
circuit with complex voltage and resistance.
98
Copyrighted by Ben M. Chen
− j10
3
j 2π(50) (31.9)10
230
230
= 77
3 − j10 + j10
− j 10
-3
= j 10
Summary of the circuit:
Total circuit
impedance
Z = 3 − j10 + j10 = 3Ω
Total circuit reactance
X = Im[Z ] = Im[3] = 0 Ω
Total circuit resistance
R = Re[Z ] = Re[3] = 3 Ω
3
j 10
230
Current (rms)
Current (peak)
− j 770
− j 10
230
230
3
77
j 10
j 770
2 I = 77 2 =108 A
Source voltage-current
phase relationship
0 (in phase)
Power factor of entire
circuit
cos(0) =1
Power supplied by
source
Power consumed by
resistor
99
I = 77A
[
]
Re [ (77) (3 × 77)]= 18 kW
Re (77)∗ (230) = 18 kW
∗
Copyrighted by Ben M. Chen
Note that the rms voltages across the inductor and capacitor are larger
than the source voltage. This is possible in ac circuits because the
reactances of capacitors and inductors, and so the voltages developed
across them, may cancel out one another:
Source
voltage
230
=
Voltage
across
capacitor
− j 770
+
Voltage
across
resistor
230
+
Voltage
across
inductor
j 770
In dc circuits, it is not possible for a passive resistor (with positive
resistance) to cancel out the effect of another passive resistor (with
positive resistance).
100
Copyrighted by Ben M. Chen
3.8 Power Factor Improvement
Consider the following system:
factor, the machine
I0
Mains
V = 230
Due to the small power
50 Hz
230 V
2.3 kW
0.4 lagging power factor
Electrical Machine
cannot be connected to
standard 13A outlets even
though it consumes only
2.3 kW of power.
The current I0 can be found as follows:
2300W
2300
= 0 .4 ⇒ I 0 =
= 25
(230V )( I 0 A )
(230)(0.4 )
Can we improve it?
cos{Arg[I 0 ] − Arg[V ]}= 0.4 ⎫
−1
⎬ ⇒ Arg[I 0 ] = − cos (0.4 ) = − 1.16
Arg[I 0 ] − Arg[V ]< 0
⎭
101
I 0 = I 0 e jArg [ I ]=25e − j1.16
0
Copyrighted by Ben M. Chen
EG1103 Mid-term Test
• When?
The time of your tutorial class in the
week right after the recess week.
• Where? In your tutorial classroom.
• Why?
To collect some marks for your final
grade for EG1103.
• What? Two questions cover materials up to
DC circuit analysis.
102
Copyrighted by Ben M. Chen
To overcome this problem, a parallel capacitor can be used to improve
the power factor:
25 e − j1.16
I
230
25 e − j1.16
= 9.2e j1.16
Original
Machine
Z old =
Mains V = 230
1
ZC = j 2 π(50)C
C
New Electrical Machine
I=
V
+ 25e − j1.16 = j 23000πC+10 − j 23 = 10 + j (23000πC−23)
ZC
Thus, if we choose 23000πC = 23 ⇒ C = 0.32mF then I = 10A and
Power factor of new machine = cos[Arg ( I ) − Arg (V )] = 1
By changing the power factor, the improved machine can now be connected to
standard 13A outlets. The price to pay is the use of an additional capacitor.
103
Copyrighted by Ben M. Chen
To reduce cost, we may wish to use a capacitor which is as small as possible.
To find the smallest capacitor that will satisfy the 13A requirement:
I = 102 + (72200C −23) = 132
2
132 =102 +(72200C−23)2
2
0 = 102 − 132 + (72200C − 23) = (72200C − 23) − 8.32
2
2
0 = (72200C − 23 − 8.3) (72200C − 23 + 8.3)
C = 0.2 mF or
0.44 mF
There are 2 possible values for C, one giving a lagging overall power factor,
the other giving a leading overall power factor. To save cost, C should be
C = 0.2 mF
104
Copyrighted by Ben M. Chen
Chapter 4. Frequency Response
105
Copyrighted by Ben M. Chen
4.1 RC Circuit
Consider the series RC circuit:
v(t) =
a 2cos ( 2π f t + θ )
2Ω
vC (t) =
b 2cos ( 2π f t + φ )
160 mF
Input: V = I ( 2 + Z C )
Output: VC = IZ C
I
2
VC = b e jφ
V = a e jθ
ZC =
1
jf
106
1
VC be jφ b j (φ −θ )
ZC
=
=
=
H ( f ) = = jθ = e
1
1+ j2 f
V ae
a
2 + ZC 2 +
jf
1
1
-3 =
j
j 2π f (160)10
f
Frequency
Response
Copyrighted by Ben M. Chen
The magnitude of H ( f ) is
V
V
b
H( f ) = C = C =
V
V
a
1
1
=
=
1+ 4 f 2
1 + (2 f )2
and is called the magnitude response.
The phase of H ( f ) is
V
Arg[H ( f )] = Arg ⎡⎢ C ⎤⎥ = Arg[VC ] − Arg[V ]
⎣V ⎦
⎡ 1 ⎤
= φ − θ = Arg ⎢
⎥
⎣1 + j 2 f ⎦
= − Arg[1 + j 2 f ] = − tan −1 (2 f )
and is called the phase response.
The physical significance of these responses is that H ( f )
gives the ratio of output to input phasors, |H ( f )| gives the
ratio of output to input magnitudes, and Arg[H ( f )] gives the
output to input phase difference at a frequency f.
107
Copyrighted by Ben M. Chen
Input
Frequency
V = 3e j 7
π
v(t ) = r sin[2π (4)t ] = r cos⎡⎢2π (4)t− ⎤⎥
2⎦
⎣
r − jπ 2
V=
e
2
f =5
f =4
v(t ) = 3 2 cos[2π (5)t+7]
1
1+ j10
H (4) =
1
1+ j8
1
101
H (4) =
1
65
Frequency
response
H (5) =
Magnitude
response
H (5) =
Phase
response
Arg[H (5)] = − tan−1 (10)
[
Output
]
32
vC(t) =
cos 2π (5)t +7− tan−1(10)
101
3
j [7−tan−1 (10)]
VC =
e
101
108
Arg[H (4)] = − tan−1 (8)
[
]
r
sin 2π (4)t−tan−1 (8)
65
r
π
=
cos⎡⎢2π (4)t−tan−1 (8)− ⎤⎥
65 ⎣
2⎦
−1
r
VC =
e j [−tan (8)−π 2]
130
vC (t ) =
Copyrighted by Ben M. Chen
Due to the presence of components such as capacitors and inductors
with frequency-dependent impedances, H ( f ) is usually frequencydependent and the characteristics of the circuit is often studied by
finding how H ( f ) changes as f is varied. Numerically, for the series
RC circuit:
f
H( f ) =
0
1 = 20 log(1) = 0 dB
0.5
→∞
109
1
1
1 + 4(0.5)2
=
(1 + 4 f )
2
1
1
= 20 log⎛⎜ ⎞⎟ = − 3dB
2
⎝ 2⎠
→ 0 = − ∞dB
Arg[H ( f )] = − tan−1 (2 f )
0 rad = 00
π
− tan−1 (2 × 0.5) = − rad = − 450
4
π
− tan−1 (∞) = − rad = − 900
2
Copyrighted by Ben M. Chen
H(f)
1 = 0 dB
0.7 =
− 3 dB
0.5
0
f
High
Frequency
Arg [ H ( f ) ]
0
0.5
f
− 45
o
− 90
o
Low
f
High
f
Input
Output
t
Low Frequency
Input
Output
t
110
Copyrighted by Ben M. Chen
At small f, the output approximates the input. However, at high f, the output
will become attenuated. Thus, the circuit has a low pass characteristic (low
frequency input will be passed, high frequency input will be rejected).
The frequency at which |H ( f )| falls to –3 dB of its maximum value is called
the cutoff frequency. For the above example, the cutoff frequency is 0.5 Hz.
To see why the circuit has a low pass characteristic, note that at low f, C has
large impedance (approximates an open circuit) when compare with R (2 in
the above example). Thus, VC will be approximately equal to V :
R
V
Low f
111
ZC ∝ 1 = ∞ (open circuit )
f
VC ≈ V
Copyrighted by Ben M. Chen
However, at high f, C has small impedance (approximates a short circuit)
when compare with R. Thus, VC will be small:
R
V
High f
ZC ∝ 1 = 0 (short circuit)
f
VC ≈ 0 ( small )
Key Notes: The capacitor is acting like a short circuit at
high frequencies and an open circuit at low frequencies. It is
totally open for a dc circuit.
112
Copyrighted by Ben M. Chen
An Electric Joke
Q: Why does a capacitor block DC but allow AC to pass
through?
A: You see, a capacitor is like this −−−| |−−− , OK. DC
Comes straight, like this −−−−−, and the capacitor stops
it. But AC, goes up, down, up and down and jumps
right over the capacitor!
113
Copyrighted by Ben M. Chen
4.2 RL Circuit
Consider the series RL circuit:
v(t)
f Hz
sinusoid
5Ω
160 mH
vL (t)
f Hz
sinusoid
Output: VL=I ZL
Input: V=I ( 5 + ZL )
I
5
V
f Hz
VL
-3
Z L = j 2π f (160)10 = j f
ZL
VL
jf
= H( f ) =
=
V
5 + Z L 5 + jf
114
Frequency Response
Copyrighted by Ben M. Chen
The magnitude response is
H( f ) =
2
The phase response is
2
f
f
=
52 + f 2
25+ f 2
⎡ jf ⎤
Arg[H ( f )] = Arg ⎢
⎥
⎣ 5+ jf ⎦
= Arg[ jf ] − Arg[5+ jf ]
=
Numerically:
f
f2
H( f ) =
25+ f 2
0
0 = 20 log(0) = − ∞ dB
5
→∞
115
52
1
⎛ 1 ⎞ = − 3dB
=
=
20
log
⎜ ⎟
2
25+52
⎝ 2⎠
→ 1 = 0 dB
π
f
− tan −1 ⎜⎛ ⎞⎟
2
⎝5⎠
Arg[H ( f )] =
π
2
π
f
− tan −1 ⎛⎜ ⎞⎟
2
⎝5⎠
rad = 900
π
5 π
− tan−1 ⎛⎜ ⎞⎟ = rad = 450
2
⎝5⎠ 4
π
2
− tan−1 (∞ ) = 0 rad = 00
Copyrighted by Ben M. Chen
116
Copyrighted by Ben M. Chen
117
Copyrighted by Ben M. Chen
Physically, at small f, L has small impedance (approximates a short circuit)
when compare with R (5 in the above example). Thus, VL will be small:
R
V
Low f
Z L ∝ f ≈ 0 (short circuit)
VL ≈ 0 ( small )
However, at high f, L has large impedance (approximates an open
circuit) when compare with R. Thus, VL will approximates V :
R
V
High f
Z L ∝ f ≈ ∞ (open circuit)
VL ≈ V
Due to these characteristics, the circuit is highpass in nature.
118
Copyrighted by Ben M. Chen
4.3 Series Tune Circuit
L = 0.64 H
v(t)
f Hz
sinusoid
R = 0.067 Ω
vC (t)
f Hz
sinusoid
C = 1.4 F
The total impedance is
Z = R+Z L +Z C
2
1
= + j4 f +
j9 f
30
=
2 ⎛
1 ⎞
+ j⎜ 4 f −
⎟
30 ⎝
9f ⎠
=
2
30
Input
Z L = j 2π f (0.64) = j 4 f
V
f Hz
Q factor
119
Q=
2
30
1
ZC = j 2π f (1.4) = j 91f
1
1
= ⇔ f0 =
f0 =
(4 )(9 ) 6
Resonance Frequency
Output
I
VC
1
1
=
(2πL )(2πC ) 2π LC
2πf 0 L
Reactance of inductor at f 0
23
=
= 10 ⇔ Q =
Resistance
R
2 30
Copyrighted by Ben M. Chen
The frequency response is
1
j9 f
VC Z C
1
=
=
=
2
2
1
V
Z
1
−
36
f
+ j 0.6 f
+ j4 f +
j9 f
30
The magnitude response is
H( f ) =
H( f ) =
=
1
2
(36 f ) −(72−0.6 ) f
2 2
2
2
+1
b
b
= x − 2⎛⎜ ⎞⎟ x + ⎛⎜ ⎞⎟
⎝2⎠ ⎝2⎠
2
⎛ 0.62 ⎞
0.62 ⎞ ⎛ 0.62 ⎞
2 ⎛
⎟⎟ + ⎜⎜1−
⎟⎟ +1− ⎜⎜1−
⎟⎟
−2 36 f ⎜⎜1 −
72
72
72
⎝
⎠ ⎝
⎠
⎝
⎠
(36 f ) (
)
1
2
⎡
0.6 2 ⎞ ⎤ ⎛ 0.6 2 ⎞ ⎛ 0.6 2 ⎞
2 ⎛
⎟⎟⎥ + ⎜⎜
⎟⎟⎜⎜ 2−
⎟⎟
⎢36 f − ⎜⎜1−
72 ⎠⎦ ⎝ 72 ⎠⎝
72 ⎠
⎝
⎣
2
2
1
2 2
=
=
(1−36 f ) +(0.6 f )
2 2
x 2 − bx + c
1
2
2
b
−⎛⎜ ⎞⎟ + c
⎝2⎠
2
b
b
= ⎛⎜ x − ⎞⎟ + c − ⎛⎜ ⎞⎟
2⎠
⎝
⎝2⎠
2
Since f only appears in the [•]2 term in the denominator and [•]2 >= 0, |H ( f )|
will increase if [•]2 becomes smaller, and vice versa.
120
Copyrighted by Ben M. Chen
The maximum value for |H ( f )| corresponds to the situation of [•]2 or at a
frequency f = fpeak given by:
36
2
f peak
0.62
=1−
≈1 ⇔
72
f peak ≈
1
6
⇔
f peak ≈ f 0
At f = fpeak , [•]2 and the maximum value for |H ( f )| is
H ( f peak ) =
1
⎛ 0. 6
⎜⎜
⎝ 72
2
⎞⎛
0. 6
⎟⎟⎜⎜ 2 −
72
⎠⎝
2
⎞
⎟⎟
⎠
1
≈
⎛ 0.6
⎜⎜
⎝ 72
2
H( f)
⎞
⎟⎟(2 )
⎠
= 10 ⇔
H ( f peak ) ≈ Q
The series tuned circuit has a
H( fpeak ) ≈ Q = 10
bandpass characteristic. Low- and
high-frequency inputs will get
attenuated, while inputs close to the
resonant frequency will get amplified
0
121
fpeak ≈ f0 = 1
6
f
by a factor of approximately Q.
Copyrighted by Ben M. Chen
The cutoff frequencies, at which |H ( f )| decrease by a factor of 0.7071
or by 3 dB from its peak value |H ( fpeak)| , can be shown to be given by
1 ⎞
⎛
f lower ≈ f 0 ⎜1 −
⎟
2
Q
⎝
⎠
1 ⎞
⎛
f upper ≈ f 0 ⎜1 +
⎟
2
Q
⎝
⎠
Very roughly, the
circuit will pass inputs
with frequency
H( f)
between flower and fupper.
H(fpeak ) ≈ Q = 10
Q
H( fpeak )
≈
= 7.07
2
2
The bandwidth of the
fbandwidth ≈ f0 = 0.1
Q
6
0
f bandwidth = f upper − f lower ≈
f
flower ≈ f0 −
f0 0.95
2Q = 6
fpeak ≈ f0 = 1
6
fupper ≈ f0 +
122
circuit is
f0 1.05
2Q = 6
f0
Q
and the fractional
bandwidth is
f bandwidth 1
≈
f0
Q
Copyrighted by Ben M. Chen
The larger the Q factor, the sharper the magnitude response, the bigger
the amplification, and the narrower the fractional bandwidth:
H( f)
Large Q
Small Q
0
f
In practice, a series tune circuit usually consists of a practical inductor or
coil connected in series with a practical capacitor. Since a practical
capacitor usually behaves quite closely to an ideal one but a coil will
have winding resistance, such a circuit can be represented by:
123
Copyrighted by Ben M. Chen
Equivalent circuit for coil or practical inductor
L
R
V
f Hz
C
VC
The main features are:
Circuit impedance
Z = R + j 2πfL +
Resonance frequency
f0 =
Q
Frequency response
124
1
2π LC
Q=
factor
H( f ) =
1
j 2πfC
2πf 0 L
R
1
=
2 2
1 − 4π f LC + j 2πfCR
1
2
⎛ f ⎞
j⎛ f ⎞
1 − ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟
⎝ f0 ⎠ Q ⎝ f0 ⎠
Copyrighted by Ben M. Chen
For the usual situation when Q is large:
Magnitude response
Bandpass with H ( f ) decreasing as f → 0 and f → ∞
Response peak
H ( f ) peaks at f = f peak ≈ f 0 with H ( f peak ) ≈ Q
Cutoff frequencies
H( f ) =
H ( f peak )
2
≈
Q
at
2
1 ⎞
1 ⎞
⎛
⎛
f = f lower , f upper ≈ f 0 ⎜1 −
⎟, f 0 ⎜1 +
⎟
Q
Q
2
2
⎝
⎠
⎝
⎠
125
Bandwidth
f bandwidth = f upper − f lower ≈
Fractional bandwidth
f bandwidth 1
≈
f0
Q
f0
Q
Copyrighted by Ben M. Chen
The Q factor is an important parameter of the circuit.
2πf 0 L Inductor reactance at f 0
Q=
=
R
Circuit resistance
However, since R is usually the winding resistance of the practical coil
making up the tune circuit:
Reactance of practical coil at f 0
Q=
Resistance of practical coil
As a good practical coil should have low winding resistance and high
inductance, the Q factor is often taken to be a characteristic of the practical
inductor or coil. The higher the Q factor, the higher the quality of the coil.
126
Copyrighted by Ben M. Chen
Due to its bandpass characteristic, tune circuits are used in radio and
TV tuners for selecting the frequency channel of interest:
Channel 5
f5
Channel 8
f8
L
Practical inductor or coil
To tune in to channel 5, C has to
VC
C
Amplifier
and
Other
Circuits
H( f)
be adjusted to a value of C5 so
that the circuit resonates at a
frequency given by
f5 =
127
1
2π LC5
0
f5
f8
f
Copyrighted by Ben M. Chen
To tune in to channel 8, C has to be adjusted to a value of C8 so that
the circuit resonates at a frequency given by
1
f8 =
2π LC8
and has a magnitude response of:
H( f)
0
128
f5
f8
f
Copyrighted by Ben M. Chen
Additional Notes on Frequency Response
Frequency response is defined as the ratio of the phasor of the output
to the phasor of the input. Note that both the input and output could be
voltage and/or current. Thus, frequency response could have
V (output )
,
I (input )
129
V (output )
,
V (input )
I (output )
,
I (input )
I (output )
.
V (input )
Copyrighted by Ben M. Chen
Chapter 5. Periodic Signals
130
Copyrighted by Ben M. Chen
5.1 Superposition
In analyzing ac circuits, we have assumed that the voltages and currents
are sinusoids and have the same frequency f. When this is not the case
but the circuit is linear (consisting of resistors, inductors and capacitors),
the principle of superposition may be used. Consider the following system:
i (t)
0.0025 F
5 2cos (4t − 0.2)
6
3 2cos (40 t + 0.1)
The current i(t) can be found by summing the contributions due to the two
sources on their own (with the other sources replaced by their internal
resistances).
131
Copyrighted by Ben M. Chen
i1( t )
0.0025 F
5
2 cos ( 4 t − 0.2 )
6
I1
5 e − j 0.2
132
1
− 100 j
j 4 (0.0025) =
6
Copyrighted by Ben M. Chen
I1
5 e − j 0.2 (6) = 0.3 e
=
6 − 100 j
j 1.3
− 100 j
5 e − j 0.2
i 1 ( t ) = 0.3
6
2 cos ( 4 t + 1.3 )
0.0025 F
5
133
2 cos ( 4 t − 0.2 )
6
Copyrighted by Ben M. Chen
i 2( t )
0.0025 F
6
I2
6
134
3
2 cos ( 40 t + 0.1 )
1
= − 10 j
j 40 (0.0025)
3e
j 0.1
Copyrighted by Ben M. Chen
I2
− 3 e j 0.1
=
= − 0.26 e
6 − 10 j
j 1.1
− 10 j
6
i 2 ( t ) = − 0.26
3e
j 0.1
2 cos ( 40 t + 1.1 )
0.0025 F
6
135
3
2 cos ( 40 t + 0.1 )
Copyrighted by Ben M. Chen
Lastly, the actual current when both sources are present:
i ( t ) = i1( t ) + i2 (t )
= 0.3 2cos (4 t +1.3)
− 0.26 2cos (40 t +1.1)
0.0025 F
5 2cos (4t − 0.2)
136
6
3 2cos (40 t + 0.1)
Copyrighted by Ben M. Chen
5.2 Circuit Analysis using Fourier Series
Using superposition, the voltages and currents in circuits with sinusoidal
signals at different frequencies can be found.
Circuits with non-sinusoidal but periodic signals can also be analyzed by
first representing these signals as sums of sinusoids or Fourier series.
The following example shows how a periodic square signal can be
represented as a sum of sinusoidal components of different frequencies:
137
Copyrighted by Ben M. Chen
Original periodic square waveform: v (t)
Any periodic signal can be
π
4
0
1
t
represented as an infinite
sum of sine signals.
Fundamental component: sin( 2 π t )
v (t ) = sin (2πt ) +
1
0
1
t
Same freq.
as the orginal.
1
3
sin (6πt ) sin (10πt )
+
+L
3
5
Fundamental + third harmonic:
sin( 2 π t ) + sin( 6 π t )
3
Third harmonic: sin( 6 π t )
3
t
t
Freq. is triple.
138
Copyrighted by Ben M. Chen
sin( 2 πt )
2Ω
v (t )
v (t)
sin( 6 πt)
3
s (t)
f Hz
Sinusoid
2Ω
vC (t)
160 mF
2Ω
160 mF
sC (t)
sin( 2 πt)
Fourier
representation
for
v (t )
sin( 6 πt)
3
160 mF
vC (t)
sin(10 πt)
5
2
S
f Hz
139
SC
1
1
=
j2 πf (0.16) j f
Copyrighted by Ben M. Chen
The frequency response:
H( f ) =
1
jf
SC
1
=
=
S 2 + 1 1+ j 2 f
jf
From superposition, if the input is the periodic square signal
sin (6πt ) sin (10πt )
v (t ) = sin (2πt ) +
+
+L
3
5
then the output will be
[
−1
]
sin 2nπt−tan −1 (2n )
n =1, 3,5,L
n 1+(2n )2
∑
infinite series will
not be examined.
Representation of
periodic signal is
(2 × 1)] sin[2(3)πt−tan (2 × 3)]
+
+ L skipped and
2
2
(1) 1+(2 × 1)
(3) 1+(2 × 3)
sin 2(1)πt−tan
−1
∞
=
140
[
Superposition for
Topic on Fourier
vC (t ) = 0.44 sin (2πt−1.1) + 0.05 sin (6πt−1.4 ) + L
=
Side Notes:
hence ...
Copyrighted by Ben M. Chen
Chapter 6. Transient Circuit Analysis
141
Copyrighted by Ben M. Chen
C.3 Linear Differential Equation
General solution:
differential equation
d n x (t )
d n − 1 x (t )
+a n − 1
+L+a0 x (t )=u (t )
dt n
dt n − 1
General solution
x (t )= xss (t )+ xtr (t )
n th order linear
Steady state response
with no arbitrary constant
Transient response with
n arbitrary constants
142
x ss (t )=particular integral obtained from assuming
solution to have the same form as u (t )
xtr (t )=general solution of homogeneous equation
d n xtr (t )
d n − 1 xtr (t )
+a n − 1
+L+a0 xtr (t )=0
dt n
dt n − 1
Copyrighted by Ben M. Chen
General solution of homogeneous equation:
n th order linear
homogeneous equation
Roots of polynomial
from homogeneous
equation
General solution
(distinct roots)
General solution
(non-distinct roots)
143
d n xtr (t )
d n − 1 xtr (t )
+a n − 1
+L+a0 xtr (t )=0
n
n −1
dt
dt
Roots : z1 ,L ,zn
given by (z − z1 ) L (z − zn )= z n +a n − 1 z n − 1 +L+a0
xtr (t )=k1e z1t +L+kn e z n t
xtr (t )=(k1 +k 2 t +k 3t 2 )e13t +(k 4 +k5t )e 22 t +k 6 e 31t +k 7 e 41t
if roots are 13, 13, 13, 22 , 22 , 31, 41
Copyrighted by Ben M. Chen
Particular integral:
x ss (t )
Any specific solution (with no arbitrary constant)
of
d n x (t )
d n − 1 x (t )
+a n − 1
+L+a0 x (t )=u (t )
dt n
dt n − 1
Method to determine
x ss (t )
Example to find x ss (t ) for
dx (t )
+2 x (t )=e 3t
dt
Trial and error approach: assume x ss (t ) to have
the same form as u (t ) and substitute into
differential equation
Try a solution of he 3t
dx (t )
+ 2 x (t ) = e 3t ⇒ 3he 3t +2he 3t =e 3t ⇒ h=0.2
dt
x ss (t ) = 0 . 2 e 3 t
Standard trial solutions
u (t )
eαt
t
heαt
ht
teαt
(h1 +h2 t )eαt
h1 cos(ω t )+h2
a cos(ω t )+b sin (ω t )
144
trial solution for x ss (t )
sin (ω t )
Copyrighted by Ben M. Chen
6.1 Steady State and Transient Analyses
So far, we have discussed the DC and AC circuit analyses. DC analysis can be
regarded as a special case of AC analysis when the signals have frequency f = 0.
Using Fourier series, the situation of having periodic signals can be handled using
AC analysis and superposition. These analyses are often called steady state
analyses, as the signals are assumed to exist at all time.
In order for the results obtained from these analyses to be valid, it is necessary for
the circuit to have been working for a considerable period of time. This will ensure
that all the transients caused by, say, the switching on of the sources have died
out, the circuit is working in the steady state, and all the voltages and currents are
as if they exist from all time.
However, when the circuit is first switched on, the circuit will not be in the steady
state and it will be necessary to go back to first principle to determine the
behavior of the system.
145
Copyrighted by Ben M. Chen
6.2 RL Circuit and Governing Differential Equation
Consider determining i(t) in the following series RL circuit:
t=0
3V
i (t)
5Ω
7H
v(t)
where the switch is open for t < 0 and is closed for t ≥ 0.
Since i(t) and v(t) will not be equal to constants or sinusoids for all time,
these cannot be represented as constants or phasors. Instead, the
basic general voltage-current relationships for the resistor and inductor
have to be used:
146
Copyrighted by Ben M. Chen
5 i (t)
t =0
For t < 0
i (t)
5
3
7
v(t)t =< 70
d i(t)
dt
5 i (t)
i (t) = 0
5
3
7
3
voltage cross
over the switch
147
v(t) = 7
d i(t)
dt
5 i (t) = 0
i (t) = 0
5
3
KVL
7
v(t) = 7
d i(t)
=0
dt
Copyrighted by Ben M. Chen
t ≥0
0
5 i (t)
i (t)
5
3
Applying KVL:
di (t )
7
+ 5 i (t ) = 3, t ≥ 0
dt
and i(t) can be found from determining the
general solution to this first order linear
differential equation (d.e.) which governs
the behavior of the circuit for t ≥ 0.
148
7
v(t) = 7
d i (t)
dt
Mathematically, the above d.e. is often
written as
di (t )
7
+ 5 i (t ) = u (t ), t ≥ 0
dt
where the r.h.s. is
u (t ) = 3, t ≥ 0
and corresponds to the dc source or
excitation in this example.
Copyrighted by Ben M. Chen
6.3 Steady State Response
3
iss (t ) = , t ≥ 0
5
Since the r.h.s. of the governing d.e.
7
di (t )
+ 5i (t ) = u (t ) = 3, t ≥ 0
dt
Let us try a steady state solution of
iss (t ) = k , t ≥ 0
which has the same form as u(t), as a
possible solution.
diss (t )
+ 5iss (t ) = 3
7
dt
⇒ 7(0 ) + 5(k ) = 3
3
⇒k=
5
149
7
diss (t )
d 3
3
+5iss (t ) = 7 ⎛⎜ ⎟⎞ + 5⎛⎜ ⎟⎞ = 3, t≥0
dt
dt ⎝ 5 ⎠ ⎝ 5 ⎠
and is a solution of the governing d.e.
In mathematics, the above solution is
called the particular integral or solution
and is found from letting the answer to
have the same form as u(t). The word
"particular" is used as the solution is only
one possible function that satisfy the d.e.
Copyrighted by Ben M. Chen
In circuit analysis, the derivation of iss(t) by letting the answer to have
the same form as u(t) can be shown to give the steady state response
of the circuit as t → ∞.
Using KVL, the steady state
t →∞
response is
i (t) = k
3 = 0 + 5k + 0 = 5k
5
3
7
d i(t)
v(t) = 7
dt
3
⇒k=
5
3
⇒ i (t ) = , t → ∞
5
5 i (t) = 5 k
i (t) = k
5
3
150
7
v(t) = 7
This is the same as iss(t).
d i(t)
=0
dt
Copyrighted by Ben M. Chen
6.4 Transient Response
To determine i(t) for all t, it is necessary to find the complete solution of
the governing d.e.
7
di (t )
+ 5i (t ) = u (t ) = 3, t ≥ 0
dt
From mathematics, the complete solution can be obtained from summing
a particular solution, say, iss(t), with itr(t):
i (t ) = iss (t ) + itr (t ), t ≥ 0
where itr(t) is the general solution of the homogeneous equation
7
di (t )
+ 5i (t ) = 0, t ≥ 0
dt
itr (t ) = k1e z1t =
ditr (t )
7
+ 5itr (t )
ditr (t )
dt
replaced by z
dt
z1 = −
151
5
7
t ≥0
where k1 is a constant (unknown now).
itr (t ) =
= 7 z1 + 5 z 0 = 7 z + 5
5
− t
k1e 7 ,
5
− t
k1e 7 → 0,
t→∞
Thus, it is called transient response.
Copyrighted by Ben M. Chen
6.5 Complete Response
To see that summing iss(t) and itr(t) gives the general solution of the governing d.e.
7
note that
3
iss (t ) = , t ≥ 0
5
itr (t ) =
5
− t
k1e 7 ,
t ≥0
di (t )
+ 5i (t ) = 3, t ≥ 0
dt
satisfies
7
d ⎛ 3⎞ ⎛ 3⎞
⎜ ⎟ + 5⎜ ⎟ = 3, t ≥ 0
dt ⎝ 5 ⎠ ⎝ 5 ⎠
5
5
⎛
⎞
⎛
−
t
−
t⎞
d
7
7
satisfies 7 ⎜ k1e ⎟ + 5⎜ k1e ⎟ = 0, t ≥ 0
⎟
⎜
⎟
dt ⎜⎝
⎠
⎝
⎠
5
5
5
− t
⎛
⎛
− t⎞
− t⎞
3
d
3
3
iss (t ) + itr (t ) = + k1e 7 , t ≥ 0 satisfies 7 ⎜ + k1e 7 ⎟ + 5⎜ + k1e 7 ⎟ = 3
⎟
⎟ ⎜5
5
dt ⎜⎝ 5
⎠
⎠ ⎝
5
− t
3
i (t ) = iss (t ) + itr (t ) = + k1e 7 , t ≥ 0
5
152
is the general solution of the d.e.
Copyrighted by Ben M. Chen
3
5
i ss ( t )
Steady State
Response
0
t ≥0
t <0
Switch
close
k1
−5t
7 ,
i tr ( t ) = k 1e
t≥0
k1
e
Transient Response
0
t =0
t = 7 (Time constant)
5
k1 is to be
determined later
k1 + 3
5
i ss ( t ) + i tr ( t )
Complete Response
3
5
0
153
Complete response
Copyrighted by Ben M. Chen
Note that the time it takes for the transient or zero-input
response itr(t) to decay to 1/e of its initial value is
7
Time taken for itr(t) to decay to 1/e of initial value =
5
and is called the time constant of the response or system.
We can take the transient response to have died out after a
few time constants.
154
Copyrighted by Ben M. Chen
i L( t )
6.6 Current Continuity for
2
Inductor
1
0
To determine the constant k1 in
the transient response of the RL
vL( t ) = 7
2
4
t
d i L( t )
dt
To ∞
circuit, the concept of current
continuity for an inductor has
1
7
to be used.
Consider the following example:
t
−7
i L( t ) vL( t ) =Instantaneous power supplied
i (t )
v (t )
To ∞
L
=7
7
t
155
− 14
Copyrighted by Ben M. Chen
Due to the step change or discontinuity in iL(t) at t = 2,
and the power supplied to the inductor at t = 2 will go
to infinity. Since it is impossible for any system to
deliver an infinite amount of power at any time, it is
impossible for iL(t) to change in the manner shown.
In general, the current through an inductor must be
a continuous function of time and cannot change in a
step manner.
156
Copyrighted by Ben M. Chen
Using current continuity for an
Now back to our RL Circuit:
inductor at t = 0:
i( t)
t=0
i (t = 0) =
5Ω
3V
v(t)
7H
3
3
+ k1 = 0 ⇒ k1 = −
5
5
t<0
⎧0,
⎪
i (t ) = ⎨ 3 3 − 5 t
7 , t ≥0
−
e
⎪⎩ 5 5
k1 + 3
5
3
5
3
5
0
0
i( t) = 0, t < 0
157
Switch
close
i ( t ) = i ss ( t ) + i tr ( t )
= 3 + k1
5
−5t
e 7,
t≥0
i( t) = 0, t < 0
Switch
close
i ( t ) = i ss ( t ) + i tr ( t )
3 −5t
= 3 − 5 e 7 , t≥0
5
Copyrighted by Ben M. Chen
t<0
6.7 RC Circuit
500
5
Consider finding v(t) in the following
RC circuit:
i (t) = 7
d v(t)
dt
2
3
v(t)
7
t=0
5Ω
500 Ω
i (t)
2V
3V
7F
v(t)
Taking the switch to be in this
position starting from t = −∞, the
voltages and currents will have
settled down to constant values
where the switch is in the
for practically all t < 0.
position shown for t < 0 and is
in the other position for t ≥ 0.
158
i (t ) = 7
dv (t )
d (constant )
=7
= 0, t < 0
dt
dt
Copyrighted by Ben M. Chen
t ≥0
t <0
500
5
5i (t) = 35
d v(t)
dt
500
d v (t)
i (t) = 7
dt
=0
3
v(t)
5
2
i (t) = 7
7
3
v(t)
500 i (t) = 0
500
i (t) = 0
2
v(t) = − 2
2
7
Applying KVL:
5
3
d v (t)
dt
dv (t )
35
+ v (t ) = u (t ) = 3, t ≥ 0
dt
7
which has a solution
v (t ) = vss (t ) + vtr (t ), t ≥ 0
159
Copyrighted by Ben M. Chen
(1) Steady State Response
(2) Transient Response
dvtr (t )
35
+v
dt
u (t ) = 3, t ≥ 0
vss (t ) = k , t ≥ 0
dv (t )
35 ss + vss (t ) = 3
dt
⇒ 0+k =3 ⇒ k = 3
35
dvtr (t )
+ vtr (t )
dt
(t ) = 0, t ≥ 0
dvtr (t )
replaced by z
dt
= 35 z1 + z 0 = 35 z + 1
v ss (t ) = 3, t ≥ 0
t<0
t<0 ⎧⎪− 2,
⎧− 2,
t
v (t ) = ⎨
=⎨
−
⎩vss (t ) + vtr (t ), t≥0 ⎪⎩3 + k1e 35 , t ≥ 0
160
tr
1
z1 = −
35
vtr (t ) = k1e z1t = k1e
−
t
35 ,
t ≥0
Complete Response
Copyrighted by Ben M. Chen
6.8 Voltage Continuity for Capacitor
To determine k1 in the transient response of the RC circuit, the concept
of voltage continuity for a capacitor has to be used.
Similar to current continuity for an inductor, the voltage v(t) across a
capacitor C must be continuous and cannot change in a step manner.
Thus, for the RC circuit we consider, the complete solution was derived as:
t<0
t < 0 ⎧⎪− 2,
⎧− 2,
t
v (t ) = ⎨
=⎨
−
⎩vss (t ) + vtr (t ), t ≥ 0 ⎪⎩3 + k1e 35 , t ≥ 0
At t = 0,
v (0 ) = 3 + k1 = − 2 ⇒ k1 = − 5
161
t<0
⎧⎪− 2,
t
v (t ) = ⎨
−
⎪⎩3 − 5e 35 , t ≥ 0
Copyrighted by Ben M. Chen
6.9 Transient with Sinusoidal Source
Consider the RL circuit with the dc source changed to a sinusoidal one:
i (t)
t=0
5
3 2cos (ω t + 0.1)
7
For t < 0 when the switch is open:
t<0
i (t) = 0
5
3 2cos (ω t + 0.1)
162
7
7
d i (t)
=0
dt
Copyrighted by Ben M. Chen
For t ≥ 0 when the switch is closed:
t ≥0
5 i (t)
i (t)
5
3 2cos (ω t + 0.1)
7
7
d i (t)
dt
The governing d.e. is
d i (t )
7
+ 5 i (t ) = u (t ), t ≥ 0
dt
Looking for general solution
i (t ) = iss (t ) + itr (t ), t ≥ 0
with
[
]
[(
u (t ) = 3 2 cos(ω t+0.1) = Re 3 2e j (ω t+0.1) = Re 3e j 0.1
163
)(
)]
2e jω t , t ≥ 0
Copyrighted by Ben M. Chen
Since u(t) is sinusoidal in nature, a trial solution for the steady state
response or particular integral iss(t) may be
[( )(
)]
iss (t ) = r 2 cos(ω t + θ ) = Re re jθ
[( )(
2e jω t , t ≥ 0
)]
d Re re jθ 2e jω t
diss (t )
7
+ 5iss (t ) = 7
+ 5 Re re jθ
dt
dt
[( )
)]
(
[( )(
2e jω t
)]
[( )(
2e jω t
)]
= 7 Re re jθ ( jω ) 2e jω t + 5 Re re jθ
[(
)
(
= Re re jθ ( jω 7 + 5) 2e jω t
[(
= Re 3e j 0.1
)(
This is Method One:
164
)]
)]
2e jω t = u (t )
( jω 7+5) re jθ =3e j 0.1
Copyrighted by Ben M. Chen
t ≥0
Method Two:
5 i (t)
i (t)
5
3 2cos (ω t + 0.1)
v(t) = 7 d
7
i (t)
dt
5I
I = r e jθ
5
3 e j 0.1
j ω7
j ω7I
( j ω 7 + 5) I = ( j ω 7 + 5) r e j θ = 3 e j 0.1
165
Copyrighted by Ben M. Chen
(1) Steady State Response
(2) Transient Response
ditr (t )
7
+ 5itr (t ) = 0, t ≥ 0
dt
( jω 7+5)re jθ = 3e j 0.1
⇒
re jθ
r=
3e j 0.1
=
5+ jω 7
3e
source case:
j 0.1
5 + jω 7
itr(t) will have the same form as the dc
=
3
52 +7 2 ω 2
itr (t ) =
5
− t
k1e 7 ,
t≥0
[ ]
θ = Arg e j 0.1 − Arg[5+ jω 7]
−1 ⎛ 7ω ⎞
= 0.1−tan ⎜
⎟
⎝ 5 ⎠
Complete Response
⎡
−1 ⎛ 7ω ⎞⎤
iss (t ) = r 2 cos(ω t+θ ) =
cos ⎢ω t+0.1 − tan ⎜
⎟⎥ , t ≥ 0
2
⎝ 5 ⎠⎦
⎣
25 + 49ω
3 2
166
Copyrighted by Ben M. Chen
Complete Response
i (t ) = iss (t ) + itr (t ), t ≥ 0
− t
⎡
−1 ⎛ 7ω ⎞ ⎤
7 , t ≥0
cos
0
.
1
tan
ω
t
+
k
e
=
+
−
⎜
⎟
⎥⎦ 1
⎢⎣
2
5
⎝
⎠
25 + 49ω
5
3 2
To determine k1, the continuity of i(t), the current through the inductor, can be used.
i (t ) = 0, t < 0
i (0) = iss (0) + itr (0) =
k1 = −
⎧0,
⎪
3 2
i (t ) = ⎨
⎪ 25 + 49ω 2
⎩
167
⎡
−1 ⎛ 7ω ⎞ ⎤
−
cos
0
.
1
tan
⎜
⎟⎥ + k1
⎢⎣
2
⎝ 5 ⎠⎦
25 + 49ω
3 2
⎡
−1 ⎛ 7ω ⎞⎤
−
cos
0
.
1
tan
⎜
⎟⎥
⎢⎣
2
5
⎝
⎠⎦
25+49ω
3 2
t<0
5
⎧⎪ ⎡
− t⎫
⎡
−1 ⎛ 7ω ⎞ ⎤
−1 ⎛ 7ω ⎞ ⎤
7 ⎪, t ≥ 0
ω
t
e
cos
0
.
1
tan
cos
0
.
1
tan
−
−
+
−
⎜
⎟
⎜
⎟
⎬
⎨ ⎢
⎢⎣
⎝ 5 ⎠⎥⎦
⎝ 5 ⎠⎥⎦
⎪⎭
⎪⎩ ⎣
Copyrighted by Ben M. Chen
An FAQ: Can we apply KVL to the rms values of voltages in AC circuits?
Answer: No. In an AC circuit, KVL is valid for the phasors of the voltages
in a closed-loop, i.e., the sum of the phasors of voltages in a closed-loop is
equal to 0 provided that they are all assigned to the same direction. KVL
cannot be applied to the magnitudes or rms values of the voltages alone.
For example, a closed-loop circuit containing a series of an AC source, a
resistor and a capacitor could have the following situation: The source has
a voltage with a rms value of 20V, while the resistor and the capacitor have
their voltages with the rms values of 9V and 15V, respectively. All in all, if
you want to apply KVL in AC circuits, apply it to the phasors of its voltages.
By the way, KVL is valid as well when the voltages are specified as functions
of time. This is true for any type of circuits.
168
Copyrighted by Ben M. Chen
6.10 Second Order RLC Circuit
Consider determining v(t) in the following series RLC circuit:
t =0
i (t)
t=0
5H
3Ω
500 Ω
2V
11 V
7F
v(t)
Both switches are in the position shown for t < 0 & are in the other positions for t ≥ 0.
i (t) = 0
For t < 0
3
5
500
7
d v (t)
dt
2
11
7
169
v(t)
Copyrighted by Ben M. Chen
Taking the switches to be in the positions shown starting from t = − ∞, the
voltages and currents will have settled down to constant values for
practically all t < 0 and the important voltages and currents are given by:
0
3
5
500
7
11
d v (t)
=0
dt
7
2
v(t) = 2
Mathematically:
v (t ) = 2, t < 0
170
&
i (t ) = 0, t < 0
Copyrighted by Ben M. Chen
For t ≥ 0
dv (t)
21
dt
d 2 v (t)
35
dt2
3
5
500
7
11
d ⎛ dv (t ) ⎞
d 2 v (t )
di
v L (t ) = L = L ⎜ C
⎟ = LC
dt ⎠
dt
dt ⎝
dt 2
d v (t)
dt
7
Applying KVL:
2
v(t)
d 2 v (t )
dv (t )
35
+ 21
+ v (t ) = u (t ) = 11, t ≥ 0
2
dt
dt
Due to the presence of 2 energy storage elements, the governing d.e. is a
second order one and the general solution is
v (t ) = vss (t ) + vtr (t ), t ≥ 0
171
Copyrighted by Ben M. Chen
(1) Steady State Response
u (t ) = 11, t ≥ 0
vss (t ) = k , t ≥ 0
d 2 vss (t )
dvss (t )
35
+
21
+ vss (t ) = 0 + 0 + k = 11
2
dt
dt
vss (t ) = 11, t ≥ 0
(2) Transient Response
d 2 vtr (t )
dvtr (t )
35
+
21
+ vtr (t ) = 0, t ≥ 0
2
dt
dt
d 2 vtr (t )
dvtr (t )
35
+
21
+ vtr (t )
2
dt
dt
dvtr (t )
replaced by z
dt
= 35 z 2 +21z1+ z 0 = 35 z 2 +21z+1
− 21 ± 212 − 4(35)(1) − 21 ± 17
z1 , z2 =
=
= − 0.54, − 0.06
2(35)
2(35)
vtr (t ) = k1e z1t + k 2 e z2t = k1e −0.54 t + k 2 e −0.06t , t ≥ 0
172
Copyrighted by Ben M. Chen
Complete Solution (Response)
To be determined
t<0
t<0
⎧2,
⎧2,
=⎨
v (t ) = ⎨
−0.54 t
−0.06 t
(
)
(
)
+
≥
v
t
v
t
,
t
0
11
k
e
k
e
, t≥0
+
+
⎩
⎩ ss
tr
1
2
t<0
dv (t ) ⎧0,
i (t ) = 7
=⎨
−0.54 t
−0.06 t
−
−
7
(
0
.
54
k
e
0
.
06
k
e
), t ≥ 0
dt
⎩
1
2
To determine k1 and k2, voltage continuity for the capacitor and current
continuity for the inductor have to be used.
The voltage across the capacitor at t = 0:
v (0) = 11 + k1 + k2 = 2 ⇒ k1 + k 2 = −9
The current passing through the inductor at t = 0:
i (0) = −0.54k1 − 0.06k2 = 0 ⇒ 0.54k1 + 0.06k2 = 0
173
k1 = 9
8
k 2 = − 81
8
Copyrighted by Ben M. Chen
t ≥0
General RLC Circuit:
d v ( t)
21 tr
RC
dt
d 2 v tr ( t )
35
LC
dt2
R3
5
L
for t ≥ 0
dv (t )
i (t ) = C500 tr
dt
0
C7
By KVL:
d 2 vtr (t )
dvtr (t )
+
+ vtr (t ) = 0, t ≥ 0
LC
RC
2
dt
dt
d 2 vtr (t )
dvtr (t )
LC
+
RC
+ vtr (t )
2
dt
dt
174
v tr ( t )
dvtr (t )
replaced by z
dt
− RC ± ( RC ) 2 − 4 LC
z1 , z2 =
2 LC
= LCz 2 +RCz1+1 = 0
Copyrighted by Ben M. Chen
Recall that for RLC circuit, the Q factor is defined as
2πf 0 L 2πL
1
L
LC
Q=
=
=
=
R
R 2π LC R LC
RC
Thus,
− RC ± ( RC ) 2 − 4 LC
=
z1 , z2 =
2 LC
− RC ± RC 1 − 4
2 LC
LC
( RC ) 2
− R ± R 1 − 4Q 2
=
2L
two real roots if 1− 4Q2 > 0 or Q2 < 1/4 or Q < 1/2
=
two complex conjugate roots if 1− 4Q2 < 0 or Q > 1/2
two identical roots if 1− 4Q2 = 0 or Q = 1/2
175
Copyrighted by Ben M. Chen
6.11 Overdamped Response
Reconsider the previous RLC example, i.e.,
t ≥0
d v ( t)
21 tr
dt
d 2 v tr ( t )
35
dt2
3
5
500
2
0
7
d 2 vtr (t )
dvtr (t )
35
+
21
+ vtr (t ) = 0, t ≥ 0
2
dt
dt
Q=
v tr ( t )
LC
35
=
= 0.2817 < 1
2
RC
21
− 21 ± 212 − 4(35)(1) − 21 ± 17
z1 , z2 =
=
= − 0.54, − 0.06
2(35)
2(35)
176
Copyrighted by Ben M. Chen
vtr (t ) = k1e z1t + k2 e z2t = k1e −0.54 t + k2 e −0.06t , t ≥ 0
k1
− 0.54 t
, t≥0
k1e
t
0
Due to its exponentially decaying
nature, the response itr(t) and the RLC
k2
− 0.06 t
k2 e
,t≥0
circuit are said to be overdamped.
Typically, when an external input is
t
0
suddenly applied to an overdamped
system, the system will take a long time
k1 + k2
− 0.06 t
v tr ( t ) = k 1e
− 0.54 t
+ k2e
,t≥0
to move in an exponentially decaying
manner to the steady state position.
The response is slow and sluggish, and
the Q factor is small.
177
0
t
Copyrighted by Ben M. Chen
6.12 Underdamped Response
vtr (0) = 2
t ≥0
d v tr ( t )
d 2 v tr ( t )
0.21
35
dt
dt2
0.03
5
500
2
0
7
5
Q=
= 28 > 1
2
0.03 7
v tr ( t )
d 2 vtr (t )
dvtr (t )
35
+ 0.21
+ vtr (t ) = 0, t ≥ 0
2
dt
dt
− 0.21 ± 0.212 − 4(35)(1) − 0.21 ± − 139.96
z1 , z2 =
=
= − 0.003 ± j 0.17
2(35)
2(35)
178
Copyrighted by Ben M. Chen
vtr (t ) = k1e z1 t + k 2 e z2 t = k1e ( −0.003 +
j 0.17 ) t
+ k 2 e ( −0.003 −
j 0.17 ) t
vtr (0) = k1 + k 2 = 2
itr (t ) = 7k1 ( −0.003 + j 0.17)e ( −0.003 + j 0.17 ) t + 7k 2 ( −0.003 − j 0.17)e ( −0.003 − j 0.17 ) t
= −0.042 + j1.19( k1 − k 2 ) = 0
= −0.021( k1 + k 2 ) + j1.19( k1 − k 2 )
itr (0) = 7k1 ( −0.003 + j 0.17) + 7k 2 ( −0.003 − j 0.17)
k1 − k 2 = − j 0.0353
179
Copyrighted by Ben M. Chen
vtr (t ) = k1e z1 t + k 2 e z2 t= k1e ( −0.003 +
(
= e −0.003t k1e j 0.17 t + k 2 e − j 0.17 t
j 0.17 ) t
)
+ k 2 e ( −0.003 −
j 0.17 ) t
= e −0.003t {k1 [cos(0.17t ) + j sin (0.17t )] + k2 [cos(0.17t ) − j sin (0.17t )]}
= e −0.003t [(k1 + k 2 ) cos(0.17t ) + j (k1−k 2 )sin (0.17t )]
= e −0.003t [2 cos(0.17t ) + 0.0353 sin (0.17t )] , t ≥ 0
0.0353
2
⎤
⎡
(
)
(
)
= e −0.003t 2 2 + 0.03532 ⎢ 2
cos
0
.
17
t
+
sin
0
.
17
t
⎥
2 2 + 0.03532
⎦
⎣ 2 + 0.03532
[
]
= 2e −0.003t cos1o cos(0.17t ) + sin 1o sin(0.17t ) , t ≥ 0
(
)
= 2e −0.003t cos 0.17t − 1o , t ≥ 0
180
Copyrighted by Ben M. Chen
When an external input is applied to an
− 0.003 t
2
,t≥0
2e
underdamped system, the system will
oscillate. The oscillation will decay expon-
Frequency = 0.17
entially but it may take some time for the
system to reach its steady state position.
t Underdamped systems have large Q
factors and are used in systems such as
0
tune circuit. However, they will be not be
suitable in situations such as car
−
v tr ( t ) = 2 e 0.003 t cos (0.17 t _ 1 ) , t ≥ 0
o
suspensions or instruments with moving
pointers.
Since this is an exponentially decaying
sinusoid, the response vtr(t) and the RLC
circuit are said to be underdamped.
It will take too long for the pointer to
oscillate and settle down to its final position
if the damping system for the pointer is
highly underdamped in nature.
181
Copyrighted by Ben M. Chen
6.13 Critically Damped Response
t ≥0
d v tr ( t )
d 2 v tr ( t )
140
35
dt
dt2
5
1
Q=
=
20 7 7 2
5
20
7
500
7
0
d v tr ( t )
dt
2
v tr ( t )
7
d 2 vtr (t )
dvtr (t )
35
+ 140
+ vtr (t ) = 0, t ≥ 0
2
dt
dt
vtr (t ) = (k1 + k2t ) e z1 t = (k1 + k2t ) e
itr (0) d
= (k1 + k2t ) e
7
dt
182
−
t
35
t =0
−
1
z1 , z2 = −
35
t
35
⇒ vtr (0) = k1 = 2
k
kt ⎞
⎛
= ⎜ k 2 − 1 − 2 ⎟e
35
35 ⎠
⎝
−
t
35
= 0 ⇒ k2 =
t =0
2
35
Copyrighted by Ben M. Chen
183
Copyrighted by Ben M. Chen
Example: The switch in the circuit shown in the following
circuit is closed at time t = 0. Obtain the current i2(t) for t > 0.
10 Ω
A
B
C
t≥0
i2
5Ω
100 V
5Ω
0.01 H
i1
F
E
D
After the switch is closed, the current passing through the source
or the 10Ω resistor is i1 + i2. Applying the KVL to the loops,
ABEFA and ABCDEFA, respectively, we obtain
184
Copyrighted by Ben M. Chen
di1
10(i1 + i2 ) + 5i1 + 0.01 = 100
dt
10(i1 + i2 ) + 5i2 = 100
i2 = (100 − 10i1 ) / 15
di1
+ 833i1 = 3333
dt
i1 (∞ ) = 3333 / 833 = 4.0A
i1 (0 ) = 0
i1 (t ) = α e −833t + 4.0
α = − 4 .0
i1 (t ) = 4.0(1 − e −833t )
i2 (t ) = 4.0 + 2.67e −833t
185
Copyrighted by Ben M. Chen
Chapter 7: Magnetic Circuit
186
Copyrighted by Ben M. Chen
7.1 Magnetic Field and Material
In electrostatic, an electric field is formed by static charges. It is described in
terms of the electric field intensity. The permittivity is a measure of how easy
it is for the field to be established in a medium given the same charges.
Similarly, a magnetic field is formed by moving charges or electric currents.
It is described in terms of the magnetic flux density B, which has a unit of
tesla(T) = (N/A)m. The permeability μ is a measure of how easy it is for a
magnetic field to be formed in a material. The higher the μ, the greater the
B for the same currents.
−7
In free space, μ is μ0 = 4π × 10 H m. The relative permeability μr is
μ
μr =
μ0
187
Most "non-magnetic" materials such as air and wood have
μr ≈ 1. However, "magnetic materials" such as iron and steel
may have μr ≈ 1000.
Copyrighted by Ben M. Chen
7.2 Magnetic Flux ---- Consider the following magnetic system:
Magnetic material
Permeability μ
If μ is large, almost
the entire magnetic
i
field will be
N turns
Cross sectional area A
concentrated inside
Average length l
the material and
there will be no flux
leakage.
The distribution of flux density B or field lines will be
Field lines
form closed paths
Since the field lines form
closed paths and there
i
Total flux Φ
N turns
is no leakage, the total
flux Φ passing through
any cross section of the
material is the same.
188
Copyrighted by Ben M. Chen
Assuming the flux to be uniformly distributed so that the flux density B have
the same value over the entire cross sectional area A:
Total flux Φ
Cross sectional area A
Same flux density B =
Φ
B=
A
189
Φ
A
weber (Wb)
with units tesla (T ) =
m2
Copyrighted by Ben M. Chen
7.3 Ampere's Law
The values of Φ or B can be calculated using Ampere's law:
Line integral of
B
μ
along any closed path = current enclosed by path
Permeability μ
Cross sectional area A
i
Path length l
N turns
Line integral of
B
μ
along dotted path =
⎛ l ⎞
= ⎜ ⎟Φ
μ ⎝ μA ⎠
Bl
= Current enclosed by dotted path Ni
190
⎛ l ⎞
Ni = ⎜ ⎟ Φ
⎝ μA ⎠
Copyrighted by Ben M. Chen
Note that the ratio H = B/μ is called the magnetic field
intensity and Ampere's law is usually stated in terms of
H. Stating the law in terms of H has the advantage that
the effects of magnetic materials, which influence μ, is
not in the main equation. In a certain sense,
characterizes the magnetic field due to only current
distributions. By multiplying H with μ to end up with the
most important flux density B, the effect of the medium is
taken into consideration.
⎛ l ⎞
Ni = ⎜ ⎟ Φ
⎝ μA ⎠
191
Φ NiμA ⎛ N ⎞
H= =
=
=⎜ ⎟i
μ μA lμA ⎝ l ⎠
B
Copyrighted by Ben M. Chen
B
Saturated
H
0
Saturated
192
H=
B
μ
N⎞
⎛
H = ⎜ ⎟ i, which is propotional to the current i.
⎝ l ⎠
Copyrighted by Ben M. Chen
7.4 Magnetic and Electric Circuits --- A more complicated example:
Assuming no flux
i
Total flux Φ
Area A2
Flux density B2 = Φ
A2
leakage and
uniform flux
N turns
distribution, the
field lines, total
flux and flux
Total flux Φ
Area A1
Flux density B1 = Φ
A1
densities are:
Average length l 2
i
Area A2
N turns
Permeability μ2
Ampere’s Law:
B
Line integral of μ
along any closed
path = current
enclosed by path
193
Average length l 1
Area A1
Permeability μ1
Copyrighted by Ben M. Chen
Averagelength
length
A2
A2
Average
l 2 l2Area Area
Flux densityB2
Permeability μ2
Permeability μ2l 2 BFlux density
B2
l2
2
Line integral = μ =
Φ
μ2 A2
2
i
N turns
Average length l
Permeability μ1
Line integral =
Average length 1l1
Area A1
Area A1
Flux densityB
1
),
+ .
),
+ .
l B
l
2 2
μ2
=
2
μ 2 A2
Φ
⎛ l
l ⎞
Ni = ⎜⎜ 1 + 2 ⎟⎟Φ
⎝ μ1 A1 μ 2 A2 ⎠
Same total flux Φ
B1 density
l 1Flux
l1
Permeability
=
Line integralμ=1 μ
ΦB1
μ1 A1
1
Line integral =
l12B12
μ12
Line integral of
194
==
B
μ
l1l2
ΦΦ
μμ12AA12
= Current enclosed by path = Ni
along entire path =
B1l1
μ1
+
B2l2
μ2
⎛ l1
l2 ⎞
= ⎜⎜
+
⎟⎟Φ
⎝ μ1 A1 μ 2 A2 ⎠
Copyrighted by Ben M. Chen
Note that the above process of calculation magnetic flux is the same
as the calculation of current in the following electric circuit:
ℜ1 =
Φ
l1
μ1 A1
Φℜ1
ℜ2 =
Φ
l2
μ2 A2
Φℜ2
Ni
From KVL, the same equation can be obtained:
⎛ l1
l2 ⎞
+
Ni = Φ ℜ1+ Φ ℜ2 = ⎜⎜
⎟⎟Φ
⎝ μ1 A1 μ 2 A2 ⎠
This is not surprising because the two basic laws in electric circuits are
equivalent to the two basic laws in magnetic circuits and the following
quantities are equivalent:
195
Copyrighted by Ben M. Chen
196
Electric circuits
Magnetic circuits
KCL:
total current entering a closed
surface equals total current
leaving the surface
Flux lines form closed path:
total flux entering a closed surface
equals total flux leaving the surface
KVL:
sum of voltages along a closed
path equals zero
Ampere's law:
integral or sum of B μ ("magnetic
voltage drops") along a closed path
equals currents enclosed ("magnetic
voltage sources")
Voltage
Ni (magnetomotive force or mmf)
Current
Φ (flux)
Resistance
ℜ=
l1
(reluctance)
μ1 A1
Copyrighted by Ben M. Chen
Thus, provided there is no flux leakage and uniform distribution of flux across
any cross section, the parallel magnetic circuit with reluctances as indicated:
ℜ2
ℜ4
i
equivalent one to the other
N turns
Φ1
ℜ1
ℜ3
ℜ2
ℜ1
Φ1
ℜ4
Φ2
ℜ6
ℜ7
ℜ5
Φ2
You can use the DC
ℜ6
ℜ7
circuit techniques to
Ni
ℜ3
197
ℜ5
solve this problem.
Copyrighted by Ben M. Chen
7.5 Inductance --- Consider the magnetic circuit:
Permeability μ
i (t )
Area A
N turns
Average length l
Assuming no flux leakage and uniform flux distribution, the reluctance and
the flux linking or enclosed by the winding is
l
ℜ=
μA
198
and
Φ (t ) =
mmf
Ni (t )
=
ℜ
reluctance
Copyrighted by Ben M. Chen
From Faraday's law of induction, a voltage will be induced in the winding
if the flux linking the winding changes as a function of time. This induced
voltage, called the back emf (electromotive force) will attempt to oppose
the change and is given by
d Φ (t ) N 2 di (t )
v (t ) = N
=
dt
ℜ dt
Flux linking winding
Ni ( t)
Φ ( t) =
ℜ
i ( t)
An equivalent inductor:
i (t)
d i (t)
v (t) = L
dt
Voltage induced
d Φ (t ) N22 di(t )
d Φ ( t) = N d i ( t)
vv((t t))== N
N dt
= ℜ dt
ℜ dt
dt
),
+.
N turns
N2
L= ℜ
The inductance
199
Copyrighted by Ben M. Chen
The following summarize the main features of an ideal magnetic
system with no flux leakage:
Num ber of turns
N
Current at tim e t
i (t )
Reluctance
ℜ
MMF at tim e t
Ni (t )
Flux at tim e t
Φ (t ) =
Back em f at tim e t
v (t ) = N
Inductance
Energy stored in
m agnetic field at
tim e t
200
Ni (t )
ℜ
d Φ (t )
dt
N2
L=
ℜ
Li 2 (t ) [Ni (t )]2
ℜ Φ 2 (t )
e (t ) =
=
=
2
2ℜ
2
Copyrighted by Ben M. Chen
7.6 Force --- Consider the magnetic relay:
Air gap permeability μ0
With no flux
0.1 cm
leakage and
2
2 cm
Movable armature
held stationary
by spring
5A
Permeability
4000 μ0
300 turns
uniform flux
distribution
7 cm
(even in air
gaps)
8 cm
3
10
Reluctance of entire magnetic material Total reluctance = ℜ = 8μ + μ =
0
0
=
2(7cm+8− 0.1cm )
0.298m
2.98
=
=
4000 μ 0 (2cm 2 )
4000 μ 0 (2 × 10 −4 m 2 ) 8μ 0
Reluctance of two air gaps
=
201
2(0.1cm )
0 .1
10
=
=
μ0 2cm 2 μ0 (1cm ) μ0
(
)
10.375
μ0
Flux
300(5)
mmf
Φ=
=
10.375 μ0
ℜ
(
)
= 145 4π × 10−7 = 0.182 × 10−3 Wb
Copyrighted by Ben M. Chen
Due to the much smaller permeability, the reluctance of the air gaps is much
larger than that of the entire magnetic material. The inductance and energy
stored in the system are
Inductance:
L=
(no. of turns )2
ℜ
=
(300)2
10.375 μ0
(
ℜΦ
10.375 0.182 × 10
=
2
2 μ0
2
Energy stored in magnetic field =
= 10.9mH
)
−3 2
= 0.137J
To determine the force of attraction f on the armature, suppose the
armature moves in the direction of f by δl so that the total reluctance
changes by δℜ. Also, suppose the current is changed by δi but the flux is
not changed:
202
Copyrighted by Ben M. Chen
2 cm2
0.1 cm − δ l
Work done by armature = f (δl)
5+ δi
Same flux Φ as before
300 turns
Force f
Increase in energy stored in magnetic field
(
ℜ+δ ℜ ) Φ 2
=
2
As there is no change in flux
linkage (which will be the case if
the magnetic system is close to
saturation), there is no back emf
and there is no energy supplied
by the electrical system. Then
203
ℜ Φ2 δ ℜ Φ2
−
=
2
2
From energy conservation:
f (δ l ) = −
δ ℜ Φ2
2
⎛ Φ2 ⎞ δ ℜ
⎟⎟
⇒ f = − ⎜⎜
⎝ 2 ⎠ δl
For our example
⎡ 0.1cm−δ l
0.1cm ⎤
104 δ l
=−
δ ℜ = 2⎢
−
2
2 ⎥
μ0
μ0 2cm ⎦
⎣ μ0 2cm
(
)
(
(
)
⎛ Φ 2 ⎞ δ ℜ 104 Φ 2 104 0.182 × 10−3
⎟⎟
=
=
f = − ⎜⎜
−7
2
2
l
δ
μ
π
2
4
×
10
0
⎠
⎝
(
)
)
2
= 132N
Copyrighted by Ben M. Chen
7.7 Mutual Inductor
Reluctance ℜ
i 1 (t)
The two dots are
associated with the
v 1 (t)
i 2 (t)
N1
v 2 (t)
N2
directions of the
windings. The fields
produced by the two
windings will be
constructive If the
currents going into
the dots have the
same sign.
N1N 2
=
ℜ
204
N12 N 22
•
ℜ ℜ
Primary
winding
Flux Φ (t)
Secondary
winding
Total mmf = N1i1 (t ) + N 2i2 (t )
Total flux = Φ (t ) =
total mmf N1i1 (t ) + N 2i2 (t )
=
ℜ
ℜ
dΦ (t ) ⎛ N12 ⎞ di1 (t ) ⎛ N1 N 2 ⎞ di2 (t )
⎟
= ⎜⎜
+⎜
v1 (t ) = N1
⎟
⎟
ℜ
dt
dt
ℜ
⎝
⎠ dt
⎝
⎠
dΦ (t ) ⎛ N1N 2 ⎞ di1 (t ) ⎛ N 22 ⎞ di2 (t )
⎟⎟
+ ⎜⎜
v2 (t ) = N 2
=⎜
⎟
dt
⎝ ℜ ⎠ dt
⎝ ℜ ⎠ dt
Copyrighted by Ben M. Chen
N12
Inductance of primary winding on its own = L1 =
ℜ
N2 2
Inductance of secondary winding on its own = L2 =
ℜ
di1 (t )
di2 (t )
di1 (t )
di2 (t )
(
)
+ L1L2
= L1
+M
v1 t = L1
dt
dt
dt
dt
v2 (t ) = L1L2
where M =
di1 (t )
di (t )
di (t )
di (t )
+ L2 2 = M 1 + L2 2
dt
dt
dt
dt
L1L2 is called the mutual inductance between the two
windings. Graphically,
v 1 (t) = L1
d i (t)
d i 1 (t)
+M 2
dt
dt
M
i 1 (t)
L1
i 2 (t)
L2
v 2 (t) = M
d i (t)
d i 1 (t)
+ L2 2
dt
dt
N12 ,
N22
L1 =
, M 2 = L1 L2 for no flux leakage and perfect coupling
L2 =
ℜ
ℜ
205
Copyrighted by Ben M. Chen
In an ac environment when the currents i1(t) and i2(t) are given by:
[ (
i1 (t ) = I1 2 cos[ω t + Arg ( I1 )] = Re I1
2e jω t
[ (
[ (
)]
i2 (t ) = Re I 2
2e jω t
[ (
)]
)]
di1 (t )
di2 (t )
d Re I1 2e jω t
d Re I 2 2e jω t
v1 (t ) = L1
+M
= L1
+M
dt
dt
dt
dt
[
)]
(
[
= L1 Re jω I1 2e jω t + M Re jω I 2
2e jω t
)]
[
(
)]
V1 = jω L1 I1 + jω MI 2
[
(
)]
V2 = jω MI1 + jω L2 I 2
= Re ( jω L1 I1 + jω MI 2 ) 2e jω t
v2 (t ) = Re ( jω MI1 + jω L2 I 2 ) 2e jω t
M
I1
V1 = j ω L 1 I 1 + j ω M I 2
206
(
)]
L1
I2
L2
V2 = j ω M I 1 + j ωL 2 I 2
Copyrighted by Ben M. Chen
7.8 Transformer
Now consider connecting a mutual inductor to a load with impedance ZL
N12
L1 =
ℜ
M
I1
V1 = j ω L 1 I 1 + j ω M I 2
N2 2
L2 =
ℜ
L1
I2
L2
M = L1L2
L1L2 I1 + L2 I 2
V2 MI1 + L2 I 2
=
=
V1 L1 I1 + MI 2 L1 I1 + L1L2 I 2
=
L2 ( L1 I1 + L2 I 2 )
L1 ( L1 I1 + L2 I 2 )
=
L2
N 22 ℜ N 2
=
= n, turn ratio
=
2
L1
Ν 1 ℜ N1
V2 = j ω M I 1 + j ωL 2 I 2
ZL I 2
V2 = − Z L I 2 = jω MI1 + jω L2 I 2
− jω MI1 = ( jω L2 + Z L ) I 2
I1
jω L2 + Z L
jω L2
≈−
=−
I2
jω M
jω M
=−
if
207
ZL
L2
L
= − 2 = −n
L1L2
L1
| jω L2 | >> Z L
Copyrighted by Ben M. Chen
I1
Voltages and currents of the primary
1: n
I1
n
and secondary windings of the ideal
transformer with | jω L2 | >> Z L
I1
V1
1: n
nV1
V1
ZL
I1
n
nV1
I1
nV1 = Z L
n
V1 Z L
⇒
= 2
I1 n
I1
Equivalent Load: A load connected
to the secondary of a transformer
can be replaced by an equivalent
load directly connected to the primary
208
V1
ZL
n2
Copyrighted by Ben M. Chen
Annex G.7. A Past Year Exam Paper
Appendix C.4 will be attached to this year’s paper!
209
Copyrighted by Ben M. Chen
Q.1 (a) Using nodal analysis, derive (but DO NOT simplify or solve) the
equations for determining the nodal voltages in the circuit of Fig. 1(a).
10
10
10
8
v2
40
v1
84
20
v3
2
Fig. 1(a)
Numbering the nodes in the circuit by 1, 2 and 3 from left to right, and
applying KCL:
v1 − 84 v1 − v2 v1 − v3 − 10
v v −v v −v
+
= 0 2 + 2 1 + 2 3 = 0 v3 − v2 + 2 + v3 − v1 +10 = 0
+
8
10
10
20 10
40
40
10
210
Copyrighted by Ben M. Chen
(b) Using mesh analysis, derive (but DO NOT solve) the matrix
equation for determining the loop currents in the circuit of Fig. 1(b).
Note that the circuit has a dependent source.
9
15
i1
v
12
i2
6
i3
v
Fig. 1(b)
Relating loop to branch currents and applying KVL:
15 = v = 12(i1 − i2 )
6(i2 − i3 ) + 9i2 + 12(i2 − i1 ) = 0
⇒ − 12i1 + 27i2 − 6i3 = 0
i3 = v
211
− 1⎤ ⎡ v ⎤ ⎡ 0 ⎤
0
⎡1 0
⎢0 − 12 27 − 6⎥ ⎢ i ⎥ ⎢ 0 ⎥
⎢
⎥ ⎢ 1⎥ = ⎢ ⎥
0
0 ⎥ ⎢i2 ⎥ ⎢15⎥
⎢1 0
⎢
⎥ ⎢i ⎥ ⎢ ⎥
−
0
12
12
0
⎣
⎦ ⎣ 3 ⎦ ⎣15⎦
Copyrighted by Ben M. Chen
(c) Determine the Thevenin or Norton
Replacing all independent sources
equivalent circuits as seen from
with their internal resistances, the
terminals A and B of the network of
resistance across A and B is
R = 20 || 20 = 10
Fig. 1(c). What is the maximum power
that can be obtained from these two
terminals?
circuit voltage across A and B is
20
A
120
Using superposition, the open
25
⎛ 20 ⎞
⎛ 20 ⎞
+
25
v AB = 120⎜
⎟
⎜
⎟20
⎝ 20 + 20 ⎠
⎝ 20 + 20 ⎠
= 310
10
A
20
B
310
Fig. 1(c)
212
B
3102
The maximum power p =
4(10)
Copyrighted by Ben M. Chen
Q.2 (a) A 5 kW electric motor is operating at a lagging power factor of 0.5.
If the input voltage is
(
v(t ) = 500 sin ωt + 100
)
determine the apparent power, and find the phasor and sinusoidal
expression for the input current.
Letting V and I to be the voltage and current phasors, the apparent power is
5000
= 10000VA = VI = | V | | I |
0. 5
0
0
0
500
500
(
)
j
10
−
90
−
j
80
where V =
e
=
e
2
2
i (t ) = 20 2 2 cos(ωt − 80 0 − cos −1 0.5)
= 40 cos(ωt − 80 0 − cos −1 0.5)
213
and
I =
10000 10000 2
=
= 20 2
500
V
arg(I ) − arg(V ) = − cos −1 (0.5)
I = 20 2e (
j −800 −cos−1 0.5
)
Copyrighted by Ben M. Chen
(b) In the circuit of Fig. 2(b), the
Using phasor analysis
current i(t) is the excitation and the
voltage v(t) is the response.
Determine the frequency response
of the circuit. Derive (but DO NOT
solve) an equation for finding the
"resonant" frequency at which the
frequency response becomes
1
214
Fig. 2(b)
The phase response is
The resonant frequency is therefore given by
1
0.1
⎞
⎟
⎟
⎟
⎟
⎠
⎛ω ⎞
−1 ⎛ 0.1ω ⎞
arg[H ( f )] = tan −1 ⎜
⎟ − tan ⎜
2 ⎟
−
0
.
1
1
ω
⎝
⎠
⎝
⎠
purely real.
i (t)
⎛
⎜
V
1 ⎜ 0.1 + jω
H( f )= =
I
jω ⎜ 0.1 + jω + 1
⎜
jω
⎝
0.1 + jω
=
1 + j 0.1ω − ω 2
v(t)
⎛ 0.1ω ⎞
tan −1 (10ω ) = tan −1 ⎜
2 ⎟
−
1
ω
⎝
⎠
Copyrighted by Ben M. Chen
(c) A series RLC resonant circuit is to be designed for use in a communication
receiver. Based on measurements using an oscilloscope, the coil that is
available is found to have an inductance of 25.3mH and a resistance of 2 Ω.
Determine the value of the capacitor that will give a resonant frequency of
1.kHz. If a Q factor of 100 is required, will the coil be good enough?
2
f0 =
1
2π LC
2
⎛
⎞ ⎛
⎞
1
1
⎟ =⎜
⎟ = 1μF
⇒ C = ⎜⎜
⎟
⎜
⎟
−3
π
2
L
f
π
2
25
.
3
10
1000
×
⎠
⎝
0 ⎠
⎝
2π 1000(25.3)10 −3
=
= 79.5
Q=
2
R
ω0 L
Since this is less than 100, the coil is not good enough.
215
Copyrighted by Ben M. Chen
Q.3 (a) In the circuit of Fig. 3(a), the switch has been in the position shown
for a long time and is thrown to the other position for time t ≥ 0. Determine
the values of i(t), vC(t), vR(t), vL(t), and di(t)/dt just after the switch has been
moved to the final position?
Taking all the voltages
and currents to be
vR (t)
vL (t)
t =0
R
i (t)
vC (t)
L
C
1
2
constants for t < 0:
i (t ) = C
dvC (t )
=0
dt
vR (t ) = Ri (t ) = 0
vL (t ) = L
di (t )
=0
dt
vC (t ) + vR (t ) + vL (t ) = 1 ⇒ vC (t ) = 1
216
Fig. 3(a)
Applying continuity for i(t) and vC(t):
i (0 ) = 0 vR (0 ) = Ri (0 ) = 0
vC (0 ) = 1
vC (0) + vR (0 ) + vL (0) = 2 ⇒ vL (0) = 1
di (t )
di (t )
vL (0 ) 1
vL (t ) = L
⇒
=
=
dt
dt t =0
L
L
Copyrighted by Ben M. Chen
(b) For vS(t) = cos(t+1), derive
(but DO NOT solve) the
differential equation from which
i(t) can be found in the circuit of
Fig. 3(b). Is this differential
equation sufficient for i(t) to be
determined?
vL (t ) = L
di (t )
dt
dvC (t )
dvL (t )
d 2i (t )
iC (t ) = C
=C
= CL
dt
dt
dt 2
d 2i (t )
iR (t ) = iC (t ) + i (t ) = CL
+ i (t )
2
dt
d 2i (t )
vR (t ) = RiR (t ) = RCL
+ Ri(t )
2
dt
Applying KVL:
R
vS (t)
i (t)
C
Fig. 3(b)
217
L
vS (t ) = vR (t ) + vL (t )
d 2i (t )
di (t )
(
)
= RCL
+
Ri
t
+
L
= cos(t + 1)
2
dt
dt
This is not sufficient for i(t) to be determined.
Copyrighted by Ben M. Chen
(c) The differential equation characterizing the current i(t) in a certain RCL
d 2i (t ) 1 di (t ) i (t )
jt
e
+
+
=
dt 2
CR dt
CL
circuit is
Determine the condition for R, L and C such that the circuit is critically
damped.
The characteristic equation for the transient response is
1
z
+
=0
z +
CR CL
2
z1, 2
1
1
4
−
±
−
2 2
C R CL
= CR
2
Thus, the circuit will be critically damped if
1
4
=
2 2
C R
CL
218
Copyrighted by Ben M. Chen
Q.4 (a) Determine the mean and rms values of the voltage waveform in
Fig. 4(a). If this waveform is applied to a 20 Ω resistor, what is the power
absorbed by the resistor?
Volt
40
0
2
4
6
8
10
second
Fig. 4(a)
One period of the waveform is
⎧20t , 0 ≤ t < 2
v(t ) = ⎨
2≤t <4
⎩0,
⎛2
2
20⎜⎜
20
tdt
2
∫
= ⎝
vm = 0
4
4
2
219
⎞
⎟⎟
⎠ = 10
vms
vrms =
3
⎛
⎞
2
2
400⎜⎜ ⎟⎟
2
(20t ) dt
∫
⎝ 3 ⎠ = 800
0
=
=
4
4
3
800
3
2
v
800 40
=
p = rms =
20
3(20 ) 3
Copyrighted by Ben M. Chen
(b) In the circuit of Fig. 4(b), a transformer is used to couple a loudspeaker
to a amplifier. The loudspeaker is represented by an impedance of value
ZL=.6 + j 2, while the amplifier is represented by a Thevenin equivalent circuit
consisting of a voltage source in series with an impedance of ZS = 3 + j a.
Determine the voltage across the loudspeaker. Hence, find the value of a
such that this voltage is maximized. Will maximum power be delivered to the
loudspeaker under this condition?
ZS
1: 2
ZL
10
Amplifier
Loudspeaker
Fig. 4(b)
220
Copyrighted by Ben M. Chen
If V is the voltage across the loudspeaker, the currents in the primary & secondary
windings are
V
I2 =
ZL
The primary voltage is
2V
I1 = 2 I 2 =
ZL
V
V1 =
2
Applying KVL to the primary circuit:
V=
10 = I1Z S + V1 =
2VZ S V
+
2
ZL
10
20
20(6 + 2 j )
=
=
2Z S 1
⎛ 3 + aj ⎞ 18 + j (4a + 2)
+
⎟⎟
1 + 4⎜⎜
ZL 2
⎝6+2 j ⎠
For the magnitude of this to be maximized, the denominator has to be minimized:
⎡ 20(6 + 2 j ) ⎤
max ⎢
⎥ = min[18 + j (4a + 2 ) ]
⎣ 18 + j (4a + 2 ) ⎦
2
1
a=− =−
4
2
Maximum power will be delivered since power is proportional to |V |2.
221
Copyrighted by Ben M. Chen
Method 2: The given circuit is equivalent to the following one,
3+ja
10
V1
6+j2
4
Then, we have
⎞
⎛
⎟
⎜
1⎞
10(3 + j )
10
⎛3
⎟
⎜
V1 =
⎜ + j ⎟=
⎟
⎜
1⎞ ⎝2
2 ⎠ 9 + j (2a + 1)
⎛3
(
)
3
+
+
+
ja
j
⎜
⎟⎟
⎜
2⎠⎠
⎝2
⎝
⇒ Vload = V2 = nV1 =
20(3 + j )
9 + j (2a + 1)
The rest follows …...
222
Copyrighted by Ben M. Chen