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Chapter 6 AC Circuits
Difference between AC and DC
• Although DC can deliver power, it is not able to transmit information
efficiently.
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Difference between AC and DC
• In an AC circuit, voltages and currents not only are functions of time,
but their polarities also alternate from time to time.
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Prevalent AC waves
• Sine wave
• Square wave
• Pulse wave
• Triangular wave
• Sawtooth wave
• Irregulate wave
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Sine wave
• Sine wave is the most popular ac signal. A sine wave is described by
i( t )  I P sin(t  )  I P sin( 2ft  )  I P sin(
2
t  )
T
• A few parameters are required to specify a sine wave
• IP is the peak value (maximum amplitude) of i(t).
•  is the angular frequency of i(t).
• f is the frequency of i(t).
• T is the period (cycle time) of i(t).
•  is the phase angle of i(t).
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Sine wave
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Sine wave
• Amplitude
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Sine wave
• Peak value is the maximum amplitude.
• Peak‐to‐peak value = 2 × maximum amplitude
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Sine wave
• Effective value (root‐mean‐square (rms) value)
• When an ac current flows through a resistor, the instantaneous power
consumed by the resistor is p(t) = i2(t)R.
• The energy dissipated at the resistor in one cycle will be
w ( t )  0T i 2 ( t )Rdt[J]
• The average power consumed by the resistor becomes
1
1
Pave  0T i 2 ( t )Rdt  [ 0T i 2 ( t )dt ]R[ W ]
T
T
• The effective ac current is the equivalent dc current that leads to the same
power consumption at the resistor.
1
2
Pave  I 2 R  I eff
R  [ 0T i 2 ( t )dt ]R[ W ]
T
Ieff =
1 T 2
i (t)dt = i 2 (t)之平均值[A]
T 0
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Sine wave
• When 𝑖 𝑡
𝐼 cos 𝜔𝑡
1
𝑇
𝐼 cos
𝑡,
2𝜋
𝑡
𝑇
𝑑𝑡
𝐼 cos
𝐼
𝐼
2
𝐼
𝐼
𝐼
𝐼
𝐼
𝐼
2
2
0.707 𝐼
2𝐼
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Sine wave
• Usually we use the rms value to represent an ac voltage or ac current.
For example, the commonly used wall socket in Taiwan provides 110 V
ac voltage. The peak value of the ac voltage is 110 V × 1.414 = 155.54 V.
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Sine wave
• For a sine wave, the average value of one cycle must be zero. The
average value of a positive cycle of an ac current is
𝐼
2
𝑇
/
𝐼 sin
2𝜋
𝑡 𝑑𝑡
𝑇
2𝐼
𝜋
0.637 𝐼
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Sine wave
• Form factor (FF) is defined as
rms value
average value
• For an ideal sine wave, its FF = 1.11.
𝐹𝐹
• Example 6‐1: Find VP, VP-P, and Vavg of a 120 V sine wave.
Ans: Vrms = 120 V. Thus,
VP = 1.414×120V = 169.68[V]
VP-P = 2VP = 2×169.68V = 339.36[V]
Vave = 0.637VP = 0.637×169.68V = 108.09[V]
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Sine wave
• Frequency and period (cycle time)
• Frequency represents the number of oscillation in a second. It is usually
denoted as f and its unit is Hz.
• Angular frequency f It represents the angle a phasor rotate in a second.
The unit of  is rad/sec.
• The period or cycle time is the required time that a sine wave oscillates once.
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Sine wave
• TaiPower company delivers 110 V and 220 V 60 Hz ac electricity to the
power outlet at your home.
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Sine wave
• Example 6‐2: It takes 25 ms for a sine wave to oscillate twice. What is
its frequency?
Ans: The cycle time is 25/2 = 12.5 ms.
f
1
1

 80[ Hz ]
T 12.5ms
• Example 6‐3: Find the cycle time of a sine wave having the following
frequencies. (a) 100 MHz, (b) 40 cycles in 5 seconds, (c) 500 kHz.
Ans: (a) 1/100106 = 10 ns.
(b) 5/40 = 125 ms.
(c) 1/500103 = 2 s.
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Sine wave
• Phase angle:
• Recall: In a capacitor, the current response leads the voltage response. On the
other hand, in an inductor, the voltage response leads the current response.
• When the applied signal source is an ac source, the current response and the
voltage response have the same frequency but possess a time lead or lag.
• The time difference between two sine waves that possess the same oscillating
frequency can be mathematically treated to different phase angles. For
example:
𝑠 𝑡
𝐴 cos 𝜔𝑡, 𝑠 𝑡
𝐴 cos 𝜔 𝑡 𝑡
𝐴 cos 𝜔𝑡 𝜔𝑡 .
Let 𝜃 𝜔𝑡 , 𝑠 𝑡
𝐴 cos 𝜔𝑡 𝜃 .  is called the phase angle of 𝑠 . We
also say that 𝑠 and 𝑠 have phase difference  .
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Sine wave
• Example:
A leads B by /2 or B lags behind
A by /2.
In phase
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Sine wave
• Example 6‐4:
B leads A by   or A lags behind
B by  .
A leads B by   or B lags behind
A by  .
Phase can be seen as the normalized time.
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Square wave
• A square wave is a periodic wave that has a high pulse and a low
pulse appearing alternatively. Ideally, the high pulse width is equal to ,
the low pulse width.
• For a square wave, the peak value and the rms value are the same.
positive edge
(leading edge)
negative edge
(tailing edge)
VP = 10V, VP-P = 20V = 2VP . If its frequency is 1 kHz,
then its period is 1/f = 1ms.
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Square wave
• Duty cycle: 𝑑𝑢𝑡𝑦 𝑐𝑦𝑐𝑙𝑒 %
• Average value:
𝑉
where 𝑉
ℎ𝑖𝑔ℎ 𝑝𝑢𝑙𝑠𝑒 𝑤𝑖𝑑𝑡ℎ
𝑐𝑦𝑐𝑙𝑒 𝑡𝑖𝑚𝑒
100%
𝑉
𝑑𝑢𝑡𝑦 𝑐𝑦𝑐𝑙𝑒 𝑉
is the low value of the square wave.
Vavg = －10V + (50% × 20V) =－10V + 10V = 0[V]
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Square wave
• Another example:
Vavg = 2V + (50% × 16V) = 2V + 8V = 10[V]
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Square wave
• For a real square wave, the positive edge and the negative edge cannot be
infinitely sharp. A non‐zero rise time for the positive edge and a non‐zero
fall time for the negative edge can be observed in a square wave.
• The rise time (TR) is the time required for a
square wave travelling from 10% to 90% of its
full rising transition.
• The fall time (TF) is the time required for a
square wave travelling from 90% to 10% of its
full falling transition.
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Square wave
• For a real square wave, we employ the width of the high pulse
between its 50% full rising transition and its 50% full falling transition
as the width of the square wave.
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Rectangular wave
• A rectangular wave is also called a pulse wave.
• The square wave is a special case of a more general rectangular wave.
The square wave is a rectangular wave with a 50% duty cycle.
Positive pulse wave
Negative pulse wave
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Rectangular wave
• In pulse wave, we use pulse repetition frequency (PRF) instead of just
frequency. We use pulse repetition time (PRT) instead of cycle time as well.
• For example:
≈
PRF = 1kHz
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Rectangular wave
• Similar to the square wave, duty cycle
𝑝𝑢𝑙𝑠𝑒 𝑤𝑖𝑑𝑡ℎ
𝑝𝑢𝑙𝑠𝑒 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑜𝑛 𝑡𝑖𝑚𝑒
𝑑𝑢𝑡𝑦 𝑐𝑦𝑐𝑙𝑒 %
• Average value: 𝑉
𝑉
𝑑𝑢𝑡𝑦 𝑐𝑦𝑐𝑙𝑒
100%
𝑉
• Example 6‐5: A pulse wave has 1 s pulse width and 3300 pulses/sec
PRF. If its peak value is 20 kV and its Vbaseline is 0 V, find the duty cycle
and average voltage of the pulse wave.
Ans: 𝑃𝑅𝑇
303 s. Duty cycle = 106  3300 = 0.33%.
𝑉
0
3.3
10
20
10
66 V
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Triangular wave
• The slope of the rising waveform is opposite
to the slope of the falling waveform a
triangular wave.
• The average value of a triangular wave is
50.5% of its peak value.
• The rms value of a triangular wave is 62.4% of
its peak value.
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Sawtooth wave
• It looks like
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The other waveforms
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Background Mathematics
• Sine wave is the most prevalent and important signal for ac circuits.
Its relation to the circular motion is shown in the following figure. A
phase vector (phasor) rotates at an angular frequency  around a
circle of radius A. The projection of the phasor is a sine wave with
amplitude A and angular frequency .
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Background Mathematics
• Hence, all operations of sine waves can be equivalently carried out by
using phasors in mathematics.
• Definition of a phasor
• A phasor is a complex number representing a sinusoidal function whose
amplitude, angular frequency, and initial phase angle are time‐invariant.
𝑅𝑒 𝐴 · 𝑒
·𝑒
𝐴 cos 𝜔𝑡 𝜃
𝑗
1. The term 𝐴𝑒
is a phasor.
• In a linear circuit, if the source is a sinusoidal function, then any
response of the circuit is also a sinusoidal function with the same
frequency but with different amplitude and phase angle. Thus, we can
ignore the angular frequency term 𝑒
of the phasor by simply using
𝐴𝑒 or 𝐴∠𝜃.
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Background mathematics
• Review complex numbers.
• Representation in Cartesian coordinates
• +,  , .
• 𝑧
• Conjugate
• Representation in polar coordinates
• Employing phasors (complex numbers) can simplify the math
operations in time domain.
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Background Mathematics
• Time‐domain waveform v(t) = Vocos(t +) = 𝑅𝑒 𝑉 · 𝑒 · 𝑒
. The
corresponding phasor becomes 𝑉 𝑒 .
• The former can be observed by oscilloscope. Depended on the signal
type, the latter can be observed by the spectrum analyzer, network
analyzer, or the vector signal analyzer.
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Background Mathematics
• Example 6‐6: Connect two ac voltage sources v1(t) = 15cos (377t + 45o) [V]
and v2(t) = 15cos (377t + 15o) [V] in series. Calculate the output ac voltage.
Ans: In time domain,
v1(t) = 15cos(377t + 45o)V
= [15cos45ocos377t－15sin45osin377t][V]
v2(t) = 15cos(377t +15o)V
= [15cos15ocos377t－15sin15osin377t][V]
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Background Mathematics
Hence,
vT(t) = v1(t) + v2(t)
= 15(cos15o + cos45o)cos377t
－15(sin15o + sin45o)sin377t
= 15(1.673cos377t－0.966sin377t)
= 15[ (1.673) 2  (0.966) 2 ]cos[377t
0.966
+ tan 1 (
)]
1.673
= 15 [1.932cos(377t + 30o)][V]
= 29.98cos(377t + 30o)[V]
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Background Mathematics
Using phasors,
o
V1  15e j45  10.61  j10.61[V]
o
V2  15e j15  14.49  j3.88[V]
o
VT  V1  V2  25.10  j14.49  29.98e j30 [V]
Taking it back to the time domain, we obtain
vT(t) = 29.98cos(377t + 30o)[V]
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Impedance
• In a dc circuit, the resistance across a device can be obtained by
dividing the voltage across the device by the current flowing through
it.
• In an ac circuit, the voltage and current are time functions and we
introduce the phasors that can be used to represent them. When we
divide the voltage phasor by the current phasor, we will get a complex
number that is called the impedance.
Z
V
[ ]
I
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Impedance
• For example, if v(t) = VPsin(t + )[V] and i(t) = IPsin(t + )[A], the
corresponding phasors are V = VP∠[V]及I = IP∠[A].
𝑉 ∠𝛼
𝑍∠ 𝛼 𝛽
𝑍 ∠𝜃
𝑍
𝐼 ∠𝛽
• It is noticed that the impedance is a complex number, but it is not a
phasor because there is no corresponding time function for an
impedance.
• Impedance is a combination of an ac resistance R() (real part) and
an ac reactance X() (imaginary part).
Z() = R() + jX() []
• Reactance is contributed by inductors and/or capacitors.
• Inductive reactance (caused by inductors) if X() > 0.
• capacitive reactance (caused by capacitors) if X() < 0.
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Impedance
• We can write
𝑍
𝑅 𝜔
𝑋 𝜔
𝑋 𝜔
𝜃 tan
𝑅 𝜔
𝑅
𝑍 cos 𝜃
𝑋
𝑍 sin 𝜃
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Impedance
• For resistors, 𝑣 𝑡
𝑖 𝑡 𝑅. In phasor form, 𝑉
𝐼𝑅. Therefore, 𝑍
• For capacitor, 𝑖 𝑡
𝐶
𝑉𝑒
Thus, 𝐼
𝑗𝜔𝐶𝑉 and 𝑍
. In phasor form, 𝑉
𝑅.
𝑗𝜔𝑉.
.
.
Because
𝑗, 𝑍
𝑗𝑋 𝜔
and 𝑋 𝜔
0. That is why
we call a reactance capacitive when its value is less than 0.
• For inductors, 𝑣 𝑡
𝐿
. In phasor form, 𝐼
𝐼 𝑒
.
𝑗𝜔𝐼.
𝑗𝑋 𝜔
𝑗𝜔𝐿. 𝑋 𝜔
𝜔𝐿 0. That is
Thus, 𝑉 𝑗𝜔𝐿𝐼 and 𝑍
why we call a reactance inductive when its value is greater than 0.
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Impedance
• For resistors, 𝑍
𝑅 and X = 0. V and I are in phase.
𝜔𝐶𝑉 𝑒
• For capacitors, 𝐼 𝑗𝜔𝐶𝑉 𝜔𝑉 𝐶𝑒 / 𝑒
∘
Current leads voltage by 90 .
/
.
• For capacitors, 𝑍
. This implies that capacitors are open
when  = 0 (dc) and are short when  = ∞.
/ .
𝜔𝐿𝐼 𝑒
• For inductors, 𝑉 𝑗𝜔𝐿𝐼 𝜔𝐼 𝐿𝑒 / 𝑒
Voltage leads current by 90∘ .
• For inductors, 𝑍
𝑗𝜔𝐿. This implies that inductors are short
when  = 0 (dc) and are open when  = ∞.
• 𝑍 𝑅 𝑗𝑋. 𝑋 0, 𝑍 is resistive. 𝑋 0, 𝑍 is inductive. 𝑋 0, 𝑍 is
capacitive.
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Impedance
• Example 6‐7: A 5  resistor and a 3  inductive reactor are connected
in series. What is the total impedance?
Ans: Z = 5 + j3 = 5.8330.96o[]
• Example 6‐8: Find the total impedance of the following circuit.
Ans:
Z = 4 + j(8－11) = 4－j3 = 5－36.87o[]
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Admittance
• The inverse of impedance is admittance.
Y
1
 G  jB[S]
Z
• G is called the (ac) conductance and B is called the susceptance. Their
unit are both Siemens (S).
• Capacitive susceptance:
𝐵
1
𝑋
𝜔𝐶
𝐵
1
𝑋
1
𝜔𝐿
• Inductive susceptance:
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Admittance
• In general, G and B should be calculated as follows.
1
1
1
R  jX R  jX
 G  jB 



Z
R  jX R  jX R  jX R 2  X 2
R
X
[S]
 2
j 2
2
R X
R  X2
Y
G
X
R
[S]
[S] , B  2
2
R X
R  X2
2
• In a similar manner,
1
1
1
G  jB G  jB
 R  jX 



Y
G  jB G  jB G  jB G 2  B2
G
B
 2
j 2
[]
2
G B
G  B2
G
B
R 2
[ ] , X  2
[ ]
2
G B
G  B2
Z
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Analysis of ac circuits
• The analysis of ac circuits is exactly the same as that of dc circuits. The
only difference is that in ac circuit analysis, we employ complex
numbers, whereas in dc circuit analysis, we only employ real numbers.
• Lets consider the basic RL, RC, and RLC circuits first. The following
circuit is an RL circuit, now Z  R  jL  Z [] .
V = ZI = (R + jL)I = RI + jLI = VR + VL[V]
Z  R 2  (L) 2 []
L
  tan 1
R
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Analysis of ac circuits
• Since V  ZI  IZ[V ] , when we use I as the reference, we can draw
We observe that V leads I by 
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Analysis of ac circuits
• The following is a parallel RL circuit.
IR 
V
[A]
R
IL 
I  IR  IL  V(
Z
V
1
RL (L  jR )


 Z []
I (1 / R )  (1 / jL)
R 2  ( L ) 2
𝑍
𝜔𝑅𝐿
𝑅
𝜔𝐿
  tan 1
V
[A]
jL
1
1
)[A ]

R jL
We observe that
V leads I by 
R
L
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Analysis of ac circuits
• In a series‐connected RC circuit, the overall impedance is
ZR
1
1 2
 R2  (
) []
j C
C
where    tan 1 ( 1 )
RC
• The current leads voltage by .
49
Analysis of ac circuits
• In a parallel‐connected RC circuit, the overall impedance can be
driven as follows.
V
[A] ,
R
V
 jCV[A]
1 / jC
1
V
I  I R  I C  V (  jC)[A ] , Z  
R
I
IR 
IC 
1
[]
1 2
2
( )  (  C)
R
where  = tan-1(RC)
• Again, the current leads voltage
by .
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Analysis of ac circuits
• In a series‐connected RLC circuit, the overall impedance is
Z  R  jL 
1
1
 R  j(L 
)  Z []
jC
C
where
Z  R 2  (L 
1 2
) [ ] ,
C
  tan 1
L  (1 / C)
R
• When L > 1/C, the impedance is inductive. When
L<1/C, the impedance is capacitive.
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Analysis of ac circuits
• The frequency 𝑓
is called the resonance frequency.
• Below 𝑓 , the series RLC circuit is capacitive.
• Above 𝑓 , the series RLC circuit is inductive.
• At 𝑓 , the series RLC circuit is at resonance and is resistive.
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Analysis of ac circuits
• The following circuit is a parallel RLC circuit, its admittance is
1
1
Y  G  j C 
 G  j(C 
)[S]
L
jL
• Its resonance frequency is also
1
𝑓
2𝜋 𝐿𝐶
• Below 𝑓 , the parallel RLC circuit is inductive
and the voltage leads the current by 
where 𝜃 tan 𝑅
𝜔𝐶 .
• Above 𝑓 , the parallel RLC circuit is capacitive
and the voltage lags behind the current by 
where 𝜃 tan 𝑅 𝜔𝐶
.
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Analysis of ac circuits
• Example 6‐9: The voltage across a load is 10cos(120t+12o)V and the
current flowing through the load is 2.5cos(120t－37o)A. Find the
reactance of the circuit.
Ans: Using phasors to represent the voltage and current, we can write
V = 10∠12o[V] and I = 2.5∠－37o[A]
Z
V
1012o [V ]
 449o []  2.62  j3.02[]

o
I 2.5  37 [A ]
Hence, the reactance is 3.02 . The load is inductive.
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Analysis of ac circuits
• Example 6‐10: In a series RC circuit, R = 10  and C = 0.01 F. At what
frequency does the voltage and current phasors possess a phase
difference of 12.5∘ .
Ans: For a series RC circuit, the phase difference between the voltage
and current phasors is
   tan 1
1
 12.5o
RC
1
 tan 12.5o
RC

1
 4.51  107 [ rad / s]
o
10  10 F  tan 12.5
8
55
Analysis of ac circuits
• Example 6‐11: In a series RLC circuit, R = 10 , L = 2 H, and C = 10 nF.
Find its resonance frequency.
Ans:
o 
1

LC
1
 7.701  106 [rad / s]
2H  10nF
o 7.701 106 rad / s

fo 
 1125[kHz]
2
2
56
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Analysis of ac circuits
• Example 6‐12: In a parallel RC circuit, R = 50 , C = 470 F. What will
the impedance be at  = 377 rad/sec?
Ans: For a parallel RC circuit,
Z
R
[ ]
1  jCR
At  = 377 rad/sec,
50
50

1  j(377rad / s)(50)(470F) 1  j8.86
50(1  j8.86)
50  j443
 0.629  j75.57[]


(1  j8.86)(1  j8.86) 1  j78.5
Z
57
Analysis of ac circuits
• Example 6‐13: In a series RLC circuit, R = 3 , L= 6, and 1/C = 2.
The voltage source is V = 10∠0o. Find the current I and the voltages
across each component.
Abs: Z = R + j[ L－(1/  C)] = 3 + j[6－2] = (3 + j4)[] = 5∠53.1o[]
V 100o
I 
 2  53.1o [A]
o
Z 553.1
VR = IR = (2∠－53.1o)(3) = 6∠－53.1o|[V]
VL = IZL = (2∠－53.1o)(6∠90o) = 12∠36.9o[V]
VC = IZC = (2∠－53.1o)(2∠－90o) = 4∠－143.1o[V]
58
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Analysis of ac circuits
• Example 6‐14: Find the current is(t) in the following circuit.
Ans: The angular frequency is  = 100.
ZC 
Z 2 Z3 
1
1

  j100[]
jC j100rad / s  100  10  6 F
Z 2 Z3
200  ( j100)
2  104   90o


 40  j80[]
Z 2  Z3
200  j100
223.6  26.57 o
Z  Z1  ( Z2 Z3 )  50  40  j80  90  j80[]
IS 
V
10V
100o V


 0.08341.6o [A]
Z (90  j80) 120  41.6o 
Thus, iS(t) = 0.083cos(100t + 41.6o)[A].
59
Analysis of ac circuits
• Example 6‐15: Find the input impedance of the following circuit when
 = 50 rad/s.
Ans:
1
1

  j10[]
jC j50rad / s  ( 2  10  3 ) F
1
1
Z 2  3 
 3 
 (3  j2)[]
j C
j50rad / s  (10  10  3 )F
and Z3 = 8 + jL = 8 + j50rad/s × 0.2H = (8 + j10)[]
Z Z
(3  j2)(8  j10)
Zin  Z1  2 3   j10 
Z 2  Z3
11  j8
Z1 
  j10  3.22  j1.07  3.22  j11.07[]
60
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Analysis of ac circuits
• Example 6‐16: Use the voltage divider method to find
the voltage at node a.
Ans:  = 6 rad/s.
𝑉
4
2 𝑗
2 𝑗
2 𝑗
𝑗2
2 𝑗
4
𝑗2
2
2
𝑉
2 2∠45∘
𝑣 𝑡
2 2 sin 6𝑡
45∘
𝑣 𝑡
2 2 sin 6𝑡
135∘
5
4𝑗
5
4𝑗
2
2𝑗
2 2 sin 6𝑡
225∘
61
Analysis of ac circuits
• Example 6‐17: Find the current flowing through branch bd in the
following circuit.
𝐼
𝐼
0 where
Ans: 𝐼
I1 
Vb  Va Vb  V1

[A ]
Zab
Zab
I2 
Vb  Vc Vb  V2

[A ]
Z bc
Zbc
I3 
Vb  Vd Vb  0 Vb


[A]
Z bd
Z bd
Z bd
Thus,
Vb  V1 Vb  V2 Vb


0
Zab
Zbc
Zbd
Vb (
1
1
1
V
V


) 1  2
Zab Z bc Z bd
Zab Z bc
62
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Analysis of ac circuits
Therefore,
 1
1
1  100o 10  60o
 


[V]

Vb 
1  j1
 1  j1 1  j2 1  j1  1  j1
 1
1  100o
10  60o
2 1
1 1
[V]

Vb   j     j     j  
2 
5 2
2 5
245o
2  45o
 2
Vb [(0.5  j0.5)  (0.2  j0.4)  (0.5  j0.5)]  5  j5  6.81  j1.83[V]
Vb (1.2  j0.4)  11.81  j6.83[V ]
Vb (1.263  18.4o )  13.65  30o [V]
13.65  30o
Vb 
 10.8  11.6o [V]
o
1.263  18.4
10.8  11.6o
I2 
 4.82  75.1o [A]
o
563.5
63
Analysis of ac circuits
• Example 6‐18: Using mesh current method to
find 𝑖 𝑡 and 𝑖 𝑡 .
Ans:  = 2 rad/s.
XL1 = L1 = 2 × 1 = 2 []
XL2 = L2 = 2 × (1/2) = 1 []
1
1
X C1 

 2[]
C1 2  (1 / 4)
The mesh current method results in
(4 + j2)I1－I2 = 18∠0o
－I1 + (2－j1)I2 = 0
64
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Analysis of ac circuits
Applying the Cramer rule to solve the above equations, we can obtain
180o
I1 
1
(4  j2) 180o
0
(2  j1)
 4.47  26.6o [A]
(4  j2)
1
1
(2  j1)
I2 
(2  j1)
(4  j2)
0
1
1
(2  j1)
 20o [A]
Hence,
i1(t) = 4.47sin(2t－26.6o)[A]
i2(t) = 2sin2t[A]
65
Analysis of ac circuits
• Example 6‐19: Using the mesh current method
to find the output voltage Vo.
Ans: Applying the KVL to two loops, we write
I1－(1－j)I2 = 10∠20o
－(1－j)I1 + (2 + j)I2 = 0
Applying the Cramer rule to solve I2.
1
1020o
 (1  j1)
0
I2 
 3.92  81.3o [A]
1
 (1  j)
 (1  j) (2  j)
Hence,
1
VO  (3.92  81.3o A )  ( )  1.96  81.3o [V ]
2
66
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Analysis of ac circuits
• Example 6‐20: Find the Thévenin equivalent circuit looked to the left
from ZL.
Ans: ZTh  Z1 Z2  (5)( j20)  (5)( j20)  5  j20
5  j20
5  j20 5  j20
2000  j500

 4.71  j1.176[]
425
VTh 

Z2
j20
VS 
 100
Z1  Z2
5  j20
( j2200)(5  j20) 44000  j11000

(5  j20)(5  j20)
425
 103.2  j25.88  106.714o [V]
67
Analysis of ac circuits
• Example 6‐21: Find the Thévenin equivalent circuit looked to the left
from ab.
Ans: VTh  10I X   j5  5 2I X   45o []
5  j5
j50IX + 10IX = 200－40IX
𝐼
VTh
2 2∠
45∘
= 20∠－90o[V]
ZTh 
This makes the current in the right portion
circuit only dependent on the Ix.
(5)( j5)
 2.5  j2.5[]
5  j5
68
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Analysis of ac circuits
• Example 6‐22: Use the superposition principle to solve i(t).
Ans:  = 2 rad/sec. Short the 10∠0o V
voltage source first.
Z1  5 
 j10(4  j2)
5(4  j8)
5
 5  5  10[]
 j10  (4  j2)
4  j8
The current produced by the 20∠0o V
voltage source is
200o V
I1 
 20o [A]
10
Next, short the 20∠0o V voltage
source
Z2  4  j2 
 j10(5)
 j10
 4  j2 
 8[]
5  j10
1  j2
69
Analysis of ac circuits
The current produced by the 10∠0o V voltage
source becomes
I3 
and
I2  
100o
 1.250o [A ]
8
 j10
j10
5
I3 
 1.250o 
153.4o [A ]
5  j10
5  j10
2
I  I1  I 2  20o 
5
153.4o  1.1226.6o [A ]
2
Therefore,
i(t) = 1.12sin(2t + 26.6o)[A]
70
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AC Power
• In a dc circuit, resistors consume electrical power. In an ac circuit, the
existence of reactance make the situation more complicated. In an ac
circuit, some power is not dissipated by the elements. It is stored
temporarily in reactive elements and will be returned back to the
circuit or source later.
• Recall that in a linear ac circuit, when the source provides a sinusoidal
voltage or current to excite the circuit, all node voltages and branch
currents are also sinusoidal functions of the same frequency but with
different amplitude and/or phase.
71
AC Power
• Suppose the voltage across an element and the current through the
element are
v( t )  VP cos(t  V )[V]  2V cos(t  V )[V]
A]
i( t )  I P cos(t  I )[V ]  2I cos(t  I )[V
respectively, Then p( t )  v( t )i( t )  VP I P cos(t  V ) cos(t  I )
VP I P
VI
cos   P P cos(2t  V  I )
2
2
VI
VI
 P P cos   P P cos(2t  V  I  I  I )
2
2
VI
VI
 P P cos   P P cos(2t    2I )[ W ]
2
2

where  = V－ I.
72
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AC Power
• If we take the current as the reference, then I = 0. We can write
p( t ) 
VP I P
VI
cos   P P cos( 2t  )[ W ]
2
2
• Let V and I be the rms values of the voltage and current, respectively.
Then the instantaneous power p(t) = VIcos + VIcos(2t + )[W].
• The first term is the average power that is consumed by the resistive
part of the element.
• The second term is time‐varying and its frequency is twice the
frequency of the exciting source.
73
AC Power
• Rewrite
p(t) = VIcos + VIcos2tcos  VIsinsin2t
= VI(cos + coscos2t  sinsin2t)
= VIcos(1 + cos2t)  VIsin(sin2t)
= P(1 + cos2t)  Qsin2t[W]
where P = VIcos[W] , Q = VIsin[VAR]
• P is the average power (or real power) that is the real consumed
power by the element. The power company charge its customer
based on how much total P the customer uses.
74
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AC Power
• Power factor
• Since P = VIcos[W], we see that when the voltage and the current are
orthogonal to each other, the real power is zero. On the other hand, when
they are in phase, the real power is maximized.
• We define the power factor as
PF  cos  
P P

VI S
• The term S = VI is called the apparent power. The unit of S is VA
instead of W for the sake of indicating that S is not a real power.
• Q = VI sin[VAR] is called the reactive power whose unit is VAR
(volt‐ampere reactive). When  = 90 ͦ, Q = VI and P = 0. That is, no
real power is delivered to the load and the power will eventually
returns to the source.
75
AC Power
• In a circuit, when  = 0o, the apparent power turns out to be real
power. However, when  = 90o, the apparent power turns out to be
reactive power and the load receives no real power.
• For a resistor, the apparent power is equal to the real power
dissipated by it and the power factor (PF) is 1.
76
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AC Power
• For an ideal inductor, when iL(t) = IPsint[A],
vL (t )  L
di L
 LI P cos t[ V ]
dt
• The instantaneous power becomes
p(t)=i L (t)v L (t)
1
=ωLI 2P cosωtsinωt= ωLI P2 sin(2ωt)[W]
2
• Thus, no real power will be consumed by the inductor and we call this
situation lossless. At one moment it receives power from the source
but at the next moment ( later) it returns the power back to the
source. When 𝜔 2𝜋 60, the cycle time of p(t) is 1/120 second.
Some heavy electric equipment might produce 120 Hz hum because
power flows back and forth at 120 Hz rate.
77
AC Power
• An inductor can store energy in its magnetic field (magnetic energy).
• The stored energy 𝑊 𝑡 is
𝐿𝐼 1 cos 2𝜔𝑡
1
𝑊 𝑡
𝑝 𝜏 𝑑𝜏
𝜔𝐿𝐼 sin 2𝜔𝜏 𝑑𝜏
2
4
78
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AC Power
1
• The peak stored energy is WmP  LI2P  LI2 [J ] where I is the rms
2
current.
• The average stored energy is
𝑊 𝑡 𝑑𝑡
.
• Example 6‐23: An ideal 0.1 H inductor can store 5 Joule of average
energy. When the ac source frequency is 30 Hz, find the maximum
power delivered to the inductor.
Ans: The rms current is I 
will be
Pmax 
2Wm
2  5J

 10[A] . The maximum power
L
0.1H
LI 2P
 LI 2  2  30Hz  0.1H  (10A) 2  1885[ W ]
2
79
AC Power
• A capacitor can store electric energy. When the applied voltage across
an ideal capacitor is vC(t) = Vpsin t[V], the current will be
𝑖 𝑡
𝐶
𝑑𝑣 𝑡
𝑑𝑡
𝜔𝐶𝑉 cos 𝜔𝑡
and the instentaneous power will be
𝑝 𝑡
𝜔𝐶𝑉 sin 𝜔𝑡 cos 𝜔𝑡
𝜔𝐶𝑉
sin 2𝜔𝑡
2
• Similar to an ideal inductor, the average power is zero. This is also a
lossless case.
• The energy stored in the capacitor is
𝑊 𝑡
𝑣𝐶
𝑑𝑣
𝑑𝜏
𝑑𝜏
𝐶𝑉 sin 𝜔𝑡
2
𝐶𝑉
4
1
cos 2𝜔𝑡
and
,
• The maximum and average stored energy are
respectively, where V is the rms voltage across the capacitor.
80
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AC Power
• Example 6‐24: Driven by a 60 Hz ac source, a 120 F capacitor can store
14 joules average power. Find the maximum current that drives the
capacitor.
Ans: The rms voltage across the capacitor is
V
2WC
2 14J
 483[V ]

C
120 10  6 F
Then the rms current value will be
I
V
 CV  2  60Hz  120  10  6 F  483V  21.9[A ]
XC
Hence,
I P  2I  2  21.9A  30.9[ A ]
81
AC Power
• In summary
82
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AC Power
• Example 6‐25: Find the PF, the average power, the reactive power, and
the maximum stored energy of the following RL circuit.
Ans: Z = 10 + (j2 × 60Hz × 0.04H) = 10 + j15.1
= 18.1∠56.4o[]
I
V
120V

 6.63  56.4o [A ]
o
Z 18.156.4 
PF = cos  = cos56.4o = 0.553
P = VIcos =120V × 6.63A × 0.553 = 440[W] (average power)
Q = VIsin = 120V × 6.63A × sin(56.4o) = 663[VAR] (reactive power)
The maximum stored power is
𝑊
𝐿𝐼
0.04
6.63
1.76 Joules
83
Complex power
• Again, we can use complex numbers (phasors) to calculate powers to
simplify the involved mathematics.
• Recall that the real power delivered to the load is
𝑉𝐼
cos 𝜃 𝑉𝐼 cos 𝜃
𝑃
2
• For example, the real power of a series RL circuit is V
times I cos  that is the projection of the current phasor
on the V phasor.
• We also have the reactive power
Q  V I sin [ VAR ]
• We introduce the complex power S that is expressed as
S = VI* = P + jQ[VA]
84
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Complex power
• The reason why we use conjugate of I rather than I itself is we can
eliminate the phase difference between the voltage and the current.
Only the phase angle in Z counts in calculating the complex power.
• Let V =V∠V [V] and Z = Z∠ []. Then
I
V
 I  V  [A ]
Z
and
I*  I   V  [A]
• Therefore, the complex power observed at the load is
S  VI*  ( V V )  ( I   V  )
 V I   V I (cos   jsin )[VA ]
• We see that the magnitude of the complex power is equal to the
apparent power.
85
Complex power
• In summary,
S  P 2  Q 2 [VA ]
PF 
P

S
P
P2  Q2
Q  S 1  (PF) 2 [ VAR ]
• At the load, when the current lags
behind the voltage across the load, the
load is inductive and we adopt Q < 0.
• When the current leads the voltage
across the load, the load is capacitive
and we adopt Q > 0.
86
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Complex power
• Example 6‐26: A motor is driven by a 226 V source. It can deliver 3
hps of power (hp = horse power). The input current of the motor is
15.6 A and it consumes 2920 W real power. Find the PF and the
reactive power of the motor and draw the corresponding phasor
portrait. Assume the input current lags behind the voltage across the
motor. (1 hp 746 W)
Ans: 𝑃𝐹
.
 = cos-1(0.828) = 34.10o
𝑄
3525.6
2920
.
0.828
1975.72 [VAR]
87
Complex power
• Example 6‐27: Applying a 50∠0oV voltage to a load with Z=10∠15o. Find
the corresponding complex power, apparent power, real power, reactive
power, and the PF.
o
Ans: I  V  500 oV  5  15o [A]
Z
1015 
The complex power is S = VI* = (50∠0o)(5∠15o) = 250∠15o[VA].
The apparent power S = 250 VA.
The real power is P = Scos = (250)(cos15o) = 250 × (0.966) = 241.5[W].
The reactive power is Q = Ssin = (250)(sin15o) = 250 × (0.259) = 64.7[VAR].
The PF = cos15o = 0.966.
88
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Complex power
• What does a power company care about?
• The generator generates apparent power and the power company delivers
the apparent power to its customer through distribution network. If the load’s
PF is small, then the real power that the load receives is low. The
consequence is the customer pay less fee to the power company.
• Since the load is generally inductive, to increase the PF of the load, the power
company can correct it by using a proper shunt capacitor.
89
Complex power
• Example 6‐28: An inductive load is driven by a 460 V, 60 Hz input
voltage. Its PF is known to be 0.75 and it consumes 12 kW real power.
Find a proper shunt capacitor to increase the PF to 0.9.
Ans: The rms current is
Iload 
P
12kW

 34.8[ A ]
V  PF 460V  0.75
and its phase angle is  = cos－1(0.75) = 41.4o. Thus,
Iload = 34.8∠－41.4o = (26.1－j23)[A].
For PF = 0.9, the current phase angle should be corrected to
* = cos－1(0.9) = －25.9o. The shunt capacitor should absorb an
IC current and 23.0－IC = 26.1 × tan25.9o = 12.6.
C
IC
10.4A

 59.8[F]
V 2  60Hz  460V
90
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2024/2/20
Maximum transfer of power
• Consider a Thévenin equivalent circuit and a load. 𝐼
complex power at the load is 𝐼 𝑍 and is equal to
𝑉
𝑅
𝑍
𝑗𝑋
𝑍
𝑉
𝑅
𝑅
𝑅
. The
𝑗𝑋
𝑋
𝑋
• The real power is maximized when
𝑋
𝑋
and when
𝑑
𝑅
𝑅 𝑅
𝑑𝑅
𝑍
0
𝑅
𝑍∗ ,
𝑍
𝑅
𝑍
,𝜃
𝜃
This is called the impedance
match.
The effect is to eliminate the reactive power.
91
Maximum transfer of power
• For example, when the source impedance is inductive, the load
impedance should be a capacitive one that can cancel the imaginary
part of the source impedance and the vice versa.
• When the maximum power transfer occurs,
Z = 2RTh = 2RL[]
IL 
VTh
V
 Th [A ]
Z
2R Th
𝑃 ,,
𝑉
4𝑅
92
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2024/2/20
Maximum transfer of power
• Example 6‐29: Find ZL to have a maximum power transfer. Also find
the maximum power consumed at the load.
Ans: First find the Thévenin equivalent
impedance.
ZTh 
2( j4)
2( j10)

 3.55  15.26o  3.42  j0.93[]
2  j4 2  ( j10)
To achieve the maximum power transfer, we should have
Z L  Z*Th  3.5515.26 o  3.42  j0.93[]
The Thévenin equivalent voltage and the corresponding current are
VTh 
44.74V
j4
 6.54[A]
 500o  44.7426.57 o [V ] and I 
3.42  3.42
2  j4
The maximum power at the load is
Pmax = I2R = (6.54A)2 × 3.42 = 146.28[W]
93
Maximum transfer of power
• Example 6‐30: Find ZL to have a maximum power transfer. Also find
the maximum power consumed at the load.
Ans: The source impedance is
ZTh 
(500)( j2000)
 485.0714.04o  470.58  j117.68[]
500  j2000
The matched load impedance should be
RL = 470.58[] ，X = XC = 117.68[]
ZL = 470.58－j117.68[]
The maximum power at load is
Pmax 
2
VTh
(120V) 2

 7.65[ W ]
4R Th 4  470.58
94
47