ENE 104
Electric Circuit Theory
Lecture 10:
AC Power Circuit Analysis
(ENE) Mon, 19 Mar 2012 / (EIE) Wed, 28 Mar 2012
Week #11 : Dejwoot KHAWPARISUTH
http://webstaff.kmutt.ac.th/~dejwoot.kha/
Objectives : Ch11
Page 2
• the instantaneous power
• the average power
• the rms value of a time-varying waveform
• complex power: average and reactive power
• the power factor, how to improve.
ENE 104
AC Circuit Power Analysis
Week #11
Instantaneous Power:
Page 3
𝑝 𝑡 = 𝑣 𝑡 𝑖(𝑡)
the device is a resistor:
𝑝 𝑡 = 𝑖2 𝑡 𝑅 =
𝑣2 𝑡
𝑅
the device is entirely inductive:
𝑑𝑖 𝑡
𝑑𝑡
𝑝 𝑡 = 𝐿𝑖 𝑡
=
𝑡
1
𝑣(𝑡) −∞ 𝑣
𝐿
𝑡′ 𝑑 𝑡′
the device is entirely capacitive:
𝑝 𝑡 = 𝐶𝑣 𝑡
ENE 104
𝑑𝑣 𝑡
𝑑𝑡
=
𝑡
1
𝑖(𝑡) −∞ 𝑖
𝐶
AC Circuit Power Analysis
𝑡′ 𝑑 𝑡′
Week #11
Instantaneous Power:
Page 4
Consider the series RL circuit,
𝑉0
𝑖 𝑡 =
1 − 𝑒 −𝑅𝑡
𝑅
𝐿
𝑢(𝑡)
The total power delivered by the source
𝑝 𝑡 =𝑣 𝑡 𝑖 𝑡 =
ENE 104
𝑉0 2
𝑅
1 − 𝑒 −𝑅𝑡
AC Circuit Power Analysis
𝐿
𝑢(𝑡)
Week #11
Instantaneous Power:
Page 5
Consider the series RL circuit,
𝑉0
𝑖 𝑡 =
1 − 𝑒 −𝑅𝑡
𝑅
𝐿
𝑢(𝑡)
The power delivered to the resistor is
𝑝𝑅 𝑡 = 𝑖 2 𝑡 𝑅 =
ENE 104
𝑉0 2
𝑅
(1 − 𝑒 −𝑅𝑡 𝐿 )2 𝑢(𝑡)
AC Circuit Power Analysis
Week #11
Instantaneous Power:
Page 6
Consider the series RL circuit,
𝑉0
𝑖 𝑡 =
1 − 𝑒 −𝑅𝑡 𝐿 𝑢(𝑡)
𝑅
to determine the power absorbed by the inductor, first
𝑣𝐿 (𝑡) =
=
=
𝑑𝑖(𝑡)
𝐿
𝑑𝑡
𝑉0 𝑒 −𝑅𝑡 𝐿 𝑢
𝑉0 𝑒 −𝑅𝑡 𝐿 𝑢
𝑡 +
𝑡
𝐿𝑉0
𝑅
1−
𝑑𝑢(𝑡)
−𝑅𝑡
𝐿
𝑒
𝑑𝑡
Then,
𝑝𝐿 (𝑡) = 𝑣𝐿 𝑡 𝑖(𝑡) =
ENE 104
𝑉0 2
𝑅
𝑒 −𝑅𝑡 𝐿 (1 − 𝑒 −𝑅𝑡 𝐿 )2 𝑢(𝑡)
AC Circuit Power Analysis
Week #11
Instantaneous Power:
Page 7
Consider the series RL circuit,
𝑝 𝑡 = 𝑝𝑅 𝑡 + 𝑝𝐿 (𝑡)
ENE 104
AC Circuit Power Analysis
Week #11
Instantaneous Power:
Page 8
Power Due to Sinusoidal Excitation:
the response is
i(t )  I m cos(t   )
Where
Im 
ENE 104
Vm
R  L
2
2 2
AC Circuit Power Analysis
   tan
1
L
R
Week #11
Practice: 11.1
Page 9
A current source of 12cos(2000t) A., a 200-ohm
resistor, and a 0.2-H inductor are in parallel.
Assume steady-state conditions exist. At t = 1ms.,
find the power being absorbed by the
(a) resistor
(b) inductor
(c) sinusoidal source
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.1
ENE 104
AC Circuit Power Analysis
Page 10
Week #11
Average Power:
1
P
t2  t1
For Periodic Waveforms:
Page 11
t2
p
(
t
)
dt

t1
f (t )  f (t  T )
1
P1 
T
1
Px 
T
ENE 104
AC Circuit Power Analysis
t1 T
 p(t )dt
t1
t x T
 p(t )dt
tx
Week #11
Average Power:
Page 12
For Periodic Waveforms:
1
P
nT
1
P
nT
ENE 104
t x  nT
 p(t )dt
n  1,2,3,...
tx
nT
2
 p(t )dt
nT
2
1
P  lim
n  nT
AC Circuit Power Analysis
nT
2
 p(t )dt
nT
2
Week #11
Example:
Page 13
find the average power delivered to a resistor R
by the periodic sawtooth current waveform
Im
i(t )  t , 0  t  T
T
Im
i(t )  (t  T ), T  t  2T
T
2
v (t )
p(t )  v(t )i(t )  i (t ) R 
R
2
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 14
Im
i(t )  t , 0  t  T
T
Im
i(t )  (t  T ), T  t  2T
T
1 2 2
p(t )  2 I m Rt , 0  t  T
T
1 2
p(t )  2 I m R(t  T ) 2 , T  t  2T
T
T
2
m
1 I R 2
1 2
P   2 t dt  I m R
T 0 T
3
ENE 104
AC Circuit Power Analysis
Week #11
Average Power:
Page 15
In the Sinusoidal Steady State
v(t )  Vm cos(t   )
i(t )  I m cos(t   )
the instantaneous power is
p(t )  Vm I m cos(t   ) cos(t   )
1
1
Orp(t )  Vm I m cos(   )  Vm I m cos(2t     )
2
2
By inspection:
ENE 104
1
P  Vm I m cos(   )
2
AC Circuit Power Analysis
Week #11
Example:
Page 16
Given the time-domain voltage v  4 cost / 6 V.,
find both the average power and an expression for the
instantaneous power that result when the corresponding
phasor voltage V = 4 0o V. is applied across an
impedance Z = 2 60o ohm.
Sol:
ENE 104
AC Circuit Power Analysis
Week #11
Example:
v  4 cost / 6 V.
Page 17
V = 4 0o V.
Z = 2 60o ohm.
Sol:
The phasor current is V/Z = 2 -60o
i  2 cost / 6  60


1
So the average power is: P  Vm I m cos(   )
2
1
 (4)(2) cos(60)  2W
2
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 18
i  2 cost / 6  60
v  4 cost / 6


the instantaneous power is:
p(t )  Vm I m cos(t   ) cos(t   )
1
1
 Vm I m cos(   )  Vm I m cos(2t     )
2
2
t
t
t
 8 cos( ) cos(  60)  2  4 cos(  60)
6
6
3
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 19
p(t )  2  4 cos(
v  4 cost / 6
ENE 104
t
3
 60)
i  2 cost / 6  60 
AC Circuit Power Analysis
Week #11
Practice: 11.2
Page 20
Given the phasor voltage V = 115 2 45 V. across
an impedance Z = 16.26 19.3o ohm., obtain an
expression for the instantaneous power, and compute
the average power if  = 50 rad/s.
Sol:
p(t )  Vm I m cos(t   ) cos(t   )
1
1
p(t )  Vm I m cos(   )  Vm I m cos(2t     )
2
2
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.2
ENE 104
AC Circuit Power Analysis
Page 21
Week #11
Average Power:
Page 22
Absorbed by an Ideal Resistor:
1
1
PR  Vm I m cos(0)  Vm I m
2
2
Or
2
m
V
1 2
PR  I m R 
2
2R
Absorbed by Purely Reactive Elements:
Px  0
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 23
Find the average power being delivered to an
impedance ZL = 8-j11 ohm. by a current I = 5 20o A.
Sol:
1 2
1 2
PR  I m R  5 8  100 W .
2
2
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.3
Page 24
Calculate the average power delivered to the
impedance 6∠25°Ω by the current 𝐈 = 2 + 𝑗5 A.
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.4
Page 25
Compute the average power delivered to each of the
passive elements.
Sol:
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.4
Page 26
For the circuit in the figure below, compute the
average power delivered to each of the passive
elements. Verify your answer by computing the
power delivered by the two sources.
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.4
ENE 104
AC Circuit Power Analysis
Page 27
Week #11
Practice: 11.4
ENE 104
AC Circuit Power Analysis
Page 28
Week #11
Maximum Power Transfer:
Page 29
An independent voltage source in series with an
impedance Zth delivers a maximum average power
to that load impedance ZL which is the conjugate of
Zth, or ZL=Z*th
ENE 104
AC Circuit Power Analysis
Week #11
Example 11.5:
Page 30
If we are assured that the voltage source is delivering
maximum average power to the unknown impedance,
what is its value?
Sol:
Z?  Zth*  500  j3 
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.5
Page 31
If the 30-mH inductor of Example 11.5 is
replaced with a 10-𝜇𝐹 capacitor, what is the
value of the inductive component of the
unknown impedance 𝐙? if it is known that 𝐙? is
absorbing maximum power?
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.5
ENE 104
AC Circuit Power Analysis
Page 32
Week #11
Average Power:
Page 33
For Nonperiodic Function: for example
i(t )  sin t  sin t
the average power delivered to a 1 ohm resistor:
P  lim
 
1


2
 (sin

2
t  sin t  2 sin t sin t )dt
2
2
1 1
   1 Watt
2 2
ENE 104
AC Circuit Power Analysis
Week #11
Average Power:
Page 34
For
i(t )  I m1 cos 1t  I m2 cos 2t  ...  I mN cos N t
the average power delivered to a resistor R:


1 2
2
P  I m1  I m2 2  ...  I mN
R
2
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.6
Page 35
A voltage source 𝑣𝑠 is connected across a 4-Ω
resistor. Find the average power absorbed by
the resistor if 𝑣𝑠 equals (a) 8𝑠𝑖𝑛200𝑡 (b)
8𝑠𝑖𝑛200𝑡 − 6cos⁡(200𝑡 − 45°) V. (c) 8𝑠𝑖𝑛200𝑡 −
4𝑠𝑖𝑛100𝑡 V. (d) 8𝑠𝑖𝑛200𝑡 − 6 cos 200𝑡 − 45° −
5𝑠𝑖𝑛100𝑡 + 4 V.
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.6
ENE 104
AC Circuit Power Analysis
Page 36
Week #11
Effective Values:
Page 37
If the resistor received the
same average power in part
(a) and (b), then the effective
value of i(t) is equal to Ieff,
and the effective value of v(t)
is equal to Veff
ENE 104
AC Circuit Power Analysis
Week #11
Effective Values:
Page 38
of a Periodic Waveform:
the average power delivered
to the resistor,
1 T 2
R T 2
P   i dt   i dt
T 0
T 0
the power delivered by the
direct current is:
PI R
2
eff
ENE 104
AC Circuit Power Analysis
Week #11
Effective Values:
Page 39
of a Periodic Waveform:
R T 2
2
i dt  I eff R

T 0
we get
I eff
1 T 2

i dt

T 0
note: the effective value is
often called the root-meansquare value, or simply the
rms value.
ENE 104
AC Circuit Power Analysis
Week #11
Effective Values:
Page 40
of a sinusoidal Waveform:
I eff
i(t )  I m cos(t   )
T
which has a period
1 T 2

i dt

T 0
2

to obtain the effective value
I eff
1 T 2
2

I
cos
(t   )dt
m

T 0
 Im
ENE 104

2

2
0

[ 12  12 cos(2t  2 )]dt
AC Circuit Power Analysis
Week #11
Effective Values:
of a sinusoidal Waveform
i(t )  I m cos(t   )
I eff
1

T
 Im
 Im
I eff
ENE 104
Im

2

T
0
Page 41
I eff
1 T 2

i dt

T 0
I cos (t   )dt
2
m
2
  1 1
[ 2  2 cos(2t  2 )]dt

2 0


[t ]0 
4
2
2
AC Circuit Power Analysis
Week #11
Effective Values:
Page 42
Use of RMS Values to
Compute Average Power
I eff
Im

2
2
eff
V
1 2
2
P  I m R  I eff R 
2
R
1
P  Vm I m cos(   )  Veff I eff cos(   )
2
ENE 104
AC Circuit Power Analysis
Week #11
Effective Values:
with Multiple-Frequency
Circuits

Page 43
I eff
Im

2

2
P  I12eff  I 22eff  ... I Neff
R
2
I eff  I12eff  I 22eff  ...  I Neff
ENE 104
AC Circuit Power Analysis
Week #11
Effective Values: Example
Page 44
Ex22 Page 384: find the effective value of:
I eff
ENE 104
1 T 2

i dt

T 0
AC Circuit Power Analysis
Week #11
Practice: 11.7
Page 45
Calculate the effective value of each of the
periodic voltages: (a) 6𝑐𝑜𝑠25𝑡; (b) 6𝑐𝑜𝑠25𝑡 +
4sin⁡(25𝑡 + 30°) V. (c) 6𝑐𝑜𝑠25𝑡 + 5𝑐𝑜𝑠 2 (25𝑡) V.
(d) 6𝑐𝑜𝑠25𝑡 + 5𝑠𝑖𝑛30𝑡 + 4 V.
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.7
ENE 104
AC Circuit Power Analysis
Page 46
Week #11
Practice: 11.7
ENE 104
AC Circuit Power Analysis
Page 47
Week #11
Practice: 11.7
ENE 104
AC Circuit Power Analysis
Page 48
Week #11
Apparent power and Power factor:
Page 49
The sinusoidal voltage
v  Vm cos(t   )
is applied to the network and the resultant
sinusoidal current is
i  I m cos(t   )
the average power delivered to the network
1
P  Vm I m cos(   )  Veff I eff cos(   )
2
the apparent power:
Veff I eff
the power factor:
ENE 104
P
PF 
Veff I eff
AC Circuit Power Analysis
Week #11
Example:
Page 50
Calculate values for the average power delivered
to each of the two loads, the apparent power
supplied by the source, and the power factor of
the combined load.
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 51
Calculate values
for the average
power delivered
to each of the two
loads, the
P  Veff I eff cos(angV  angI )
apparent power
supplied by the
source, and the
Veff I eff
power factor of
P
the combined
PF 
Veff I eff
load.
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 52

60 0

IS 
 12  53.13
(2  j1)  (1  j5)
PS
A.rms
 Veff I eff cos(angV  angI )
 (60)  (12) cos[0  (53.1)]
 432 W .
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 53
the apparent power
supplied by the source:
 Veff I eff  (60)  (12)  720 VA.
2
P

(
12
)

2
the power factor of the
combined load.
P  (12) 1
2
P
432
PF 

 0.6
Veff I eff (60)  (12)
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.8
Page 54
For the circuit of the figure below, determine the
power factor of the combined loads if 𝑍𝐿 = 10Ω.
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.8
ENE 104
AC Circuit Power Analysis
Page 55
Week #11
Complex Power:
Page 56
P  Veff I eff cos(   )
the average power,
express as,

P  Veff I eff Re e
j (  )
 ReV
j
eff
e I eff e
 j
 ReV
eff
I
*
eff

the complex power,
S  Veff I
*
eff
 Veff I eff e
j (  )
 P  jQ
Q  Veff I eff sin(   )
ENE 104
AC Circuit Power Analysis
Week #11
Power Measurement:
Page 57
The complex power
delivered to several
interconnected loads is
the sum of the complex
power delivered to
each of the individual
loads.
S = VI* = V(I1+I2)* = V(I1*+I2*) = VI1*+VI2*
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 58
An industrial consumer
is operating a 50-kW
induction motor at a
lagging PF of 0.8. The
source voltage is 230
Vrms. In order to obtain
lower electrical rates,
the customer wishes to
raise the PF to 0.95
lagging. Specify a
suitable solution.
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 59
A purely reactive load must be added to the
system, in parallel, since the supply voltage must
not change.
The complex power supplied to the motor must
have a real part of 50-kW and an angle of
cos-1(0.8)=36.9o
50 36.9
Hence,
S1 
0.8
 50  j 37.5 kVA
ENE 104
AC Circuit Power Analysis
Week #11
Example:
Page 60
50 36.9
S1 =
 50  j37.5 kVA
0.8
To achieve a PF of 0.95, the total complex power
must become:
50
1
S 
cos (0.95)
0.95
 50  j16.43 kVA
Thus the corrective load is
S2 =  j 21.07
ENE 104
kVA
AC Circuit Power Analysis
Week #11
Example:
S2 =  j 21.07
Page 61
kVA
We select a phase angle of 0o for the voltage
source, and the current drawn by Z2 is
S 2  j 21070
I 

  j91.6
V
230
I 2  j91.6 A.
*
2
Or
Therefore,
ENE 104
A.
V
230
Z2  
  j 2.51 
I 2 j91.6
AC Circuit Power Analysis
Week #11
Practice: 11.9
Page 62
Find the complex power
absorbed by the
(a) 1-ohm R
(b) -j10 C
(c) 5+j10 Z
(d) source
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.9
ENE 104
AC Circuit Power Analysis
Page 63
Week #11
Practice: 11.9
ENE 104
AC Circuit Power Analysis
Page 64
Week #11
Practice: 11.9
ENE 104
AC Circuit Power Analysis
Page 65
Week #11
Practice: 11.10
Page 66
A 440-Vrms source supplies power to a load
𝑍𝐿 = 10 + 𝑗2⁡Ω through a transmission line
having a total resistance of 1.5Ω. Find (a) the
average and apparent power supplied to the
load; (b) the average and apparent power lost in
the transmission line; (c) the average and
apparent power supplied by the source; (d) the
power factor at which the source operates.
ENE 104
AC Circuit Power Analysis
Week #11
Practice: 11.10
ENE 104
AC Circuit Power Analysis
Page 67
Week #11
Example: Final 2/47
Page 68
จากวงจรตามรู ป ให้หา
1) the average power ที่จ่ายโดย แหล่งจ่าย (source)
2) ค่า power factor ที่แหล่งจ่าย (source)
3) ขนาดของ ตัวเก็บประจุ (capacitor) ที่เมื่อนาไปต่อ
ขนานกับ แหล่งจ่าย ทาให้ค่า power factor = 1
4 ohm
120 Vrms 60Hz
ENE 104
j16 ohm
AC Circuit Power Analysis
12 ohm
Week #11
Example: Final 2/47
ENE 104
AC Circuit Power Analysis
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Week #11
Example: Final 2/47
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4 ohm
120 Vrms 60Hz
ENE 104
AC Circuit Power Analysis
j16 ohm
12 ohm
Week #11
Example: Final 2/47
ENE 104
AC Circuit Power Analysis
Page 71
Week #11
Ex:
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5Ω
1 µF
Vg
50 µH
7.5 Ω
Find the average power, the reactive power,
and the apparent power supplied by the voltage
source in the circuit if vg = 50 cos 105t V.
ENE 104
Ex:
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5Ω
1 µF
Vg
ENE 104
50 µH
7.5 Ω
Summary:
ENE 104
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AC Circuit Power Analysis
Week #11
Hw:
ENE 104
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AC Circuit Power Analysis
Week #11
Reference:
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.
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Circuit Analysis and Electrical Engineering
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