ENE 104 Electric Circuit Theory - staff.kmutt.ac.th - kmutt

ENE 104
Electric Circuit Theory
Lecture 10:
AC Power Circuit Analysis
Week #11 : Dejwoot KHAWPARISUTH
http://webstaff.kmutt.ac.th/~dejwoot.kha/

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Objectives : Ch11
AC Circuit Power Analysis
Page 2
Week #11
• 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.

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AC Circuit Power Analysis
Week #11
Instantaneous Power:
p t = v t i(t)
Page 3
the device is a resistor:
p t = i 2 t R = v2 t
R
the device is entirely inductive:
p t = Li t
di t
dt
= 1 v(t) t
v t ′ d
L −∞
t ′
the device is entirely capacitive:
p t = Cv t
dv t
dt
= 1 i(t) t
i t ′ d
C −∞
t ′

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Instantaneous Power:
AC Circuit Power Analysis
Page 4
Week #11
Consider the series RL circuit,
i t = V 0
R
1 − e−Rt L u(t)
The total power delivered by the source
p t = v t i t = V 0 2
R
1 − e−Rt L u(t)

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Instantaneous Power:
AC Circuit Power Analysis
Page 5
Week #11
Consider the series RL circuit,
i t = V 0
R
1 − e−Rt L u(t)
The power delivered to the resistor is
p R t = i 2 t R = V 0 2
R (1 − e−Rt L ) 2 u(t)

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Instantaneous Power:
AC Circuit Power Analysis
Page 6
Week #11
Consider the series RL circuit,
i t = V 0
R
1 − e−Rt L u(t)
to determine the power absorbed by the inductor, first
Then,
v L (t) = L di(t)
dt
= V 0 e −Rt L u t + LV 0
R
= V 0 e −Rt L u t
p L (t) = v L t i(t) = V 0 2
−Rt L du(t)
1 − e
dt
R e−Rt L (1 − e −Rt L ) 2 u(t)

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Instantaneous Power:
AC Circuit Power Analysis
Page 7
Week #11
Consider the series RL circuit,
p t = p R t + p L (t)

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Instantaneous Power:
AC Circuit Power Analysis
Page 8
Week #11
Power Due to Sinusoidal Excitation:
the response is
Where
i( t)
I cos( t
)
m
I
m
R
2
V
m
2
L
2
tan 1
L
R

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Practice: 11.1
AC Circuit Power Analysis
Page 9
Week #11
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

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Practice: 11.1
AC Circuit Power Analysis
Page 10
Week #11

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Average Power:
AC Circuit Power Analysis
Page 11
Week #11
P
For Periodic Waveforms:
t
2
1
t
1
t
2
t
1
p(
t)
dt
f
( t)
f ( t T)
P
1
1
T
t
1
T
t
1
p(
t)
dt
t
T
x
1
P
x
p(
t)
dt
T
t
x

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Average Power:
AC Circuit Power Analysis
Page 12
Week #11
For Periodic Waveforms:
P
1
nT
t
x
nT
t
x
p(
t)
dt
n
1,2,3,...
P
2
1 nT
nT
nT
2
p(
t)
dt
P
lim
n
1
nT
nT
2
nT
2
p(
t)
dt

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Example:
AC Circuit Power Analysis
Page 13
Week #11
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
p(
t)
v(
t)
i(
t)
i
2
( t)
R
v
2
( t)
R

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Example:
AC Circuit Power Analysis
Page 14
Week #11
Im i(
t)
t,
0 t T
T
Im i( t)
( t T),
T t 2T
T
1 2 2
p(
t)
ImRt
,
2
T
0 t T
1 2
2
p(
t)
ImR(
t T)
,
2
T
T t 2T
P
1
T
T
0
2
I R
2
T
t
dt
1
3
I
m 2
2
m
R

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Average Power:
AC Circuit Power Analysis
Page 15
Week #11
In the Sinusoidal Steady State
v( t)
V
cos( t
)
m
i( t)
I cos( t
)
the instantaneous power is
m
Or
p( t)
VmI
m
cos( t
)cos(
t
)
1
1
p( t)
VmI
m
cos( )
VmI
m
cos(2t
)
2
2
P 1
m
cos( )
2
V I m
By inspection:

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Example:
AC Circuit Power Analysis
Page 16
Week #11
Given the time-domain voltage
V.,
find both the average power and an expression for the
instantaneous power that result when the corresponding
phasor voltage V = 4 0 o V. is applied across an
impedance Z = 2 60 o ohm.
v
4cos
t
/ 6
Sol:

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Example:
AC Circuit Power Analysis
Page 17
Week #11
v
4cos
t
/
6
V. V = 4 0 o V.
Z = 2 60 o ohm.
Sol:
The phasor current is V/Z = 2 -60 o
So the average power is:
P
i
1
VmI
m
cos( )
2
1
(4)(2)cos(60)
2
2cos
t
/ 6 60
2W

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Example:
AC Circuit Power Analysis
Page 18
Week #11
v
4cos
t / 6
i 2cos
t / 6 60
the instantaneous power is:
p(
t)
VmI
m
cos( t
)cos( t
)
1
1
VmI
m
cos( )
VmI
m
cos(2t
)
2
2
t
t
t
8cos( )cos( 60)
2 4cos( 60)
6 6
3

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Example:
AC Circuit Power Analysis
Page 19
Week #11
t
p( t)
2 4cos( 60)
3
v
4cos
t / 6
i 2cos
t / 6 60

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Practice: 11.2
AC Circuit Power Analysis
Page 20
Week #11
115
45
Given the phasor voltage V =
V. across
an impedance Z = 16.26 19.3 o ohm., obtain an
expression for the instantaneous power, and compute
the average power if = 50 rad/s.
2
Sol:
p( t)
VmI
m
cos( t
)cos(
t
)
1
1
p( t)
VmI
m
cos( )
VmI
m
cos(2t
)
2
2

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Practice: 11.2
AC Circuit Power Analysis
Page 21
Week #11

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Average Power:
AC Circuit Power Analysis
Page 22
Week #11
Absorbed by an Ideal Resistor:
Or
P
R
1
V
2
P
R
m
I
m
1
2
cos(0)
I
2
m
R
1
2
2
Vm
2R
V
m
I
m
Absorbed by Purely Reactive Elements:
P x
0

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Example:
AC Circuit Power Analysis
Page 23
Week #11
Find the average power being delivered to an
impedance Z L = 8-j11 ohm. by a current I = 5 20 o A.
Sol:
P
R
1 2
I R
2
m
2
1
2
5 8100
W.

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Practice: 11.3
AC Circuit Power Analysis
Page 24
Week #11
Calculate the average power delivered to the
impedance 6∠25°Ω by the current I = 2 + j5 A.

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Practice: 11.4
AC Circuit Power Analysis
Page 25
Week #11
Compute the average power delivered to each of the
passive elements.
Sol:

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Practice: 11.4
AC Circuit Power Analysis
Page 26
Week #11
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.

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Practice: 11.4
AC Circuit Power Analysis
Page 27
Week #11

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Practice: 11.4
AC Circuit Power Analysis
Page 28
Week #11

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Maximum Power Transfer:
AC Circuit Power Analysis
Page 29
Week #11
An independent voltage source in series with an
impedance Z th delivers a maximum average power
to that load impedance Z L which is the conjugate of
Z th , or Z L =Z * th

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Example 11.5:
AC Circuit Power Analysis
Page 30
Week #11
If we are assured that the voltage source is delivering
maximum average power to the unknown impedance,
what is its value
Sol:
Z
*
Z 500 3
th
j

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Practice: 11.5
AC Circuit Power Analysis
Page 31
Week #11
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

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Practice: 11.5
AC Circuit Power Analysis
Page 32
Week #11

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Average Power:
AC Circuit Power Analysis
Page 33
Week #11
For Nonperiodic Function: for example
i( t)
sin t sin
t
the average power delivered to a 1 ohm resistor:
P
1
lim
1 1
2 2
2
2
(sin
1
2
t
Watt
sin
2
t
2sin
t
sin t)
dt

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Average Power:
AC Circuit Power Analysis
Page 34
Week #11
For
i( t)
Im
1
cos 1t
Im2
cos 2t
...
I
mN
cos t
N
the average power delivered to a resistor R:
P
1
2
I
2
m
2
1
Im2
...
I
2
mN
R

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Practice: 11.6
AC Circuit Power Analysis
Page 35
Week #11
A voltage source v s is connected across a 4-Ω
resistor. Find the average power absorbed by
the resistor if v s equals (a) 8sin200t (b)
8sin200t − 6cos⁡(200t − 45°) V. (c) 8sin200t −
4sin100t V. (d) 8sin200t − 6 cos 200t − 45° −
5sin100t + 4 V.

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Practice: 11.6
AC Circuit Power Analysis
Page 36
Week #11

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Effective Values:
AC Circuit Power Analysis
Page 37
Week #11
If the resistor received the
same average power in part
(a) and (b), then the effective
value of i(t) is equal to I eff ,
and the effective value of v(t)
is equal to V eff

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Effective Values:
AC Circuit Power Analysis
Page 38
Week #11
of a Periodic Waveform:
the average power delivered
to the resistor,
P
1
T
T
0
i
the power delivered by the
direct current is:
2
dt
R
T
T
0
i
2
dt
P
I
2
eff
R

Effective Values:
Page 39
of a Periodic Waveform:
R
T
T
0
i
2
dt
I
2
eff
R
we get
I
eff
1
T
T
0
i
2
dt
ENE 104
note: the effective value is
often called the root-meansquare
value, or simply the
rms value.
AC Circuit Power Analysis
Week #11

Effective Values:
Page 40
of a sinusoidal Waveform:
i( t)
I cos( t
)
m
I
eff
1
T
T
0
i
2
dt
which has a period
ENE 104
to obtain the effective value
I
eff
I
1
T
m
T
0
I
2
2
m
2
0
cos
[
2
1
2
AC Circuit Power Analysis
( t
)
dt
1
2
cos(2t
2
)] dt
Week #11

Effective Values:
Page 41
of a sinusoidal Waveform
i( t)
I cos( t
)
m
I
eff
1
T
T
0
i
2
dt
ENE 104
I
I
eff
eff
I
I
I
1
T
m
m
m
2
T
0
I
2
m
2
2
0
[ t]
4
cos
[
2
0
2
1
2
AC Circuit Power Analysis
( t
)
dt
1
2
cos(2t
2
)] dt
Week #11

ENE 104
Effective Values:
AC Circuit Power Analysis
Page 42
Week #11
Use of RMS Values to
Compute Average Power
I
eff
I
m
2
P
1
2
I
2
m
R
I
2
eff
R
V
2
eff
R
P
1
2
V
m
I
m
cos( )
V
eff
I
eff
cos( )

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Effective Values:
AC Circuit Power Analysis
Page 43
Week #11
with Multiple-Frequency
Circuits
I
eff
I
m
2
P
2 2
I1 eff
I2eff
...
I
2
Neff
R
I
eff
I
I
2 2
2
1 eff 2eff
...
Neff
I

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Effective Values: Example
AC Circuit Power Analysis
Page 44
Week #11
Ex22 Page 384: find the effective value of:
I
eff
1
T
T
0
i
2
dt

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Practice: 11.7
AC Circuit Power Analysis
Page 45
Week #11
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.

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Practice: 11.7
AC Circuit Power Analysis
Page 46
Week #11

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Practice: 11.7
AC Circuit Power Analysis
Page 47
Week #11

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Practice: 11.7
AC Circuit Power Analysis
Page 48
Week #11

Apparent power and Power factor:
Page 49
The sinusoidal voltage
v
V
m
cos( t
)
ENE 104
is applied to the network and the resultant
sinusoidal current is
the average power delivered to the network
P
1
V
2
m
the apparent power:
I
m
the power factor:
cos(
i
eff I eff
I
I eff eff
AC Circuit Power Analysis
m
cos( t
)
)
Veff
Ieff
cos( )
V
PF
V P
Week #11

ENE 104
Example:
AC Circuit Power Analysis
Page 50
Week #11
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
Example:
AC Circuit Power Analysis
Page 51
Week #11
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.
P
V
eff
I
eff
cos( angV angI)
PF
V I eff eff
P
V I eff eff

ENE 104
Example:
AC Circuit Power Analysis
Page 52
Week #11
I 60 0
12
53.13
A rms
S
(2 j1)
(1 j5)
.
P
S
V
eff
I
eff
(60) (12)cos[0
432 W.
cos( angV angI)
( 53.1)]

ENE 104
AC Circuit Power Analysis
Week #11
Example:
the apparent power
supplied by the source:
Page 53
V
eff
I
eff
( 60) (12)
720 VA.
the power factor of the
combined load.
PF
P
V I eff eff
432
(60) (12)
P (12) 2 2
0.6
P (12) 2 1

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Practice: 11.8
AC Circuit Power Analysis
Page 54
Week #11
For the circuit of the figure below, determine the
power factor of the combined loads if Z L = 10Ω.

ENE 104
Practice: 11.8
AC Circuit Power Analysis
Page 55
Week #11

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Complex Power:
AC Circuit Power Analysis
Page 56
Week #11
the average power,
P
V I eff eff
cos( )
express as,
P
V
eff
I
eff
Re
j(
)
j
j
*
e Re V e I e Re V I
eff
eff
eff
eff
the complex power,
S
V
eff
I
V
I
e
* j(
)
eff eff eff
P
jQ
Q
V I eff eff
sin( )

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Power Measurement:
AC Circuit Power Analysis
Page 57
Week #11
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(I 1 +I 2 ) * = V(I 1* +I 2* ) = VI 1* +VI 2
*

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Example:
AC Circuit Power Analysis
Page 58
Week #11
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
Example:
AC Circuit Power Analysis
Page 59
Week #11
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)=36.9 o
50 36.9
Hence, S 1
0.8
50 j37.
5
kVA

ENE 104
AC Circuit Power Analysis
Week #11
Example:
S 1 =
50
36.9
0.8
To achieve a PF of 0.95, the total complex power
must become:
S
50 1
0.95
50
cos
j16.
43
Thus the corrective load is
50
(0.95)
kVA
j37.
5
kVA
Page 60
S 2 =
j21.
07
kVA

ENE 104
Example:
AC Circuit Power Analysis
Page 61
Week #11
S 2 =
j21.
07
kVA
We select a phase angle of 0 o for the voltage
source, and the current drawn by Z 2 is
Or
Therefore,
* S2
j21070
I2 j91.6
A.
V 230
I j91.6
.
2
A
Z
V 230
2. 51
I j91.6
j
2 2

ENE 104
Practice: 11.9
AC Circuit Power Analysis
Page 62
Week #11
Find the complex power
absorbed by the
(a) 1-ohm R
(b) -j10 C
(c) 5+j10 Z
(d) source

ENE 104
Practice: 11.9
AC Circuit Power Analysis
Page 63
Week #11

ENE 104
Practice: 11.9
AC Circuit Power Analysis
Page 64
Week #11

ENE 104
Practice: 11.9
AC Circuit Power Analysis
Page 65
Week #11

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Practice: 11.10
AC Circuit Power Analysis
Page 66
Week #11
A 440-Vrms source supplies power to a load
Z L = 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
Practice: 11.10
AC Circuit Power Analysis
Page 67
Week #11

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Example: Final 2/47
AC Circuit Power Analysis
Page 68
Week #11
จากวงจรตามรูป ให้หา
1) the average power ที่จ่ายโดย แหล่งจ่าย (source)
2) ค่า power factor ที่แหล่งจ่าย (source)
3) ขนาดของ ตัวเก็บประจุ (capacitor) ที่เมื่อน าไปต่อ
ขนานกับ แหล่งจ่าย ท าให้ค่า power factor = 1
4 ohm
120 Vrms 60Hz
j16 ohm
12 ohm

ENE 104
Example: Final 2/47
AC Circuit Power Analysis
Page 69
Week #11

ENE 104
Example: Final 2/47
AC Circuit Power Analysis
Page 70
Week #11
4 ohm
120 Vrms 60Hz
j16 ohm
12 ohm

ENE 104
Example: Final 2/47
AC Circuit Power Analysis
Page 71
Week #11

ENE 104
Ex:
Page 72
5 Ω
1 µF
50 µH
V g
7.5 Ω
Find the average power, the reactive power,
and the apparent power supplied by the voltage
source in the circuit if v g = 50 cos 10 5 t V.

ENE 104
Ex:
Page 73
5 Ω
1 µF
50 µH
V g
7.5 Ω

ENE 104
Summary:
AC Circuit Power Analysis
Page 74
Week #11

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Hw:
AC Circuit Power Analysis
Page 75
Week #11

ENE 104
Reference:
Circuit Analysis and Electrical Engineering
Page 76
W.H. Hayt, Jr., J.E. Kemmerly, S.M. Durbin, Engineering Circuit Analysis, Sixth Edition.
Copyright ©2002 McGraw-Hill. All rights reserved.