Nguyễn Công Phương
Electric Circuit Theory
AC Power Analysis
Contents
I. Basic Elements Of Electrical Circuits
II. Basic Laws
III. Electrical Circuit Analysis
IV. Circuit Theorems
V. Active Circuits
VI. Capacitor And Inductor
VII. First Order Circuits
VIII.Second Order Circuits
IX. Sinusoidal Steady State Analysis
X. AC Power Analysis
XI. Three-phase Circuits
XII. Magnetically Coupled Circuits
XIII.Frequency Response
XIV.The Laplace Transform
XV. Two-port Networks
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2
http://electricalacademia.com/induction-motor/electric-motornameplate-details-explained-induction-motor-nameplate/
http://poqynamekyxoqep.oramanageability.com/understandinginduction-motor-nameplate-information-47374dan8099.html
https://w ww.cpsc.gov/Recalls/2010/marley-engineered-products-recalls-baseboard-heaters-sold-at-grainger-due-to-fire
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3
AC Power Analysis
1.
2.
3.
4.
5.
6.
7.
8.
Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Improvement
Average Power and RMS Value of Periodic Signals
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4
Instantaneous Power (1)
p(t ) = v(t )i (t )
v(t ) = Vm sin(ωt + φv )
i (t )
+
v (t )
−
i (t ) = I m sin(ωt + φi )
→ p(t ) = Vm I m sin(ω t + φv )sin(ωt + φi )
Vm I m
=
[cos(φv − φi ) − cos(2ωt + φv + φi )]
2
Vm I m
Vm I m
=
cos(φv − φi ) −
cos(2ωt + φv + φi )
2
2
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5
Instantaneous Power (2)
Vm I m
Vm I m
p(t ) =
cos(φv − φi ) −
cos(2ωt + φv + φi )
2
2
p(t)
Vm I m
2
0
Vm I m
cos(φv − φi )
2
t
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6
Average Power (1)
1 T
P =  p (t )dt
T 0
Vm I m
Vm I m
p (t ) =
cos(φv − φi ) −
cos(2ωt + φv + φi )
2
2
1
1
→ P = Vm I m cos(φv − φi )
2
T

T
0
1
1
dt − Vm I m
2
T

T
0
cos(2ω t + φv + φi ) dt
The average of a sinusoid over its period is zero
1
→ P = Vm I m cos(φv − φi )
2
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7
Average Power (2)
V = Vm φv
I = I m φi
→ VI* = Vm I m φv − φi
→ I* = I m
− φi
VI* = Vm I m φv − φi = Vm I m cos(φv − φi ) + jVm I m sin(φv − φi )
1
P = Vm I m cos(φv − φi )
2
1
*
→ P = Re( VI )
2
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8
Average Power (3)
1
1
*
P = Re( VI ) = Vm I m cos(φv − φi )
2
2
1
1
1 2
P = Vm I m cos(0) = Vm I m = I m R
2
2
2
φv = φi :
φv − φi = ±90 :
o
1
o
P = Vm I m cos(90 ) = 0
2
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9
Average Power (4)
Ex.
v(t) = 150sin(314t – 30o) V, i(t) = 10sin(314t + 45o) A. Find P?
1
1
P = Vm Im cos(ϕu − ϕi ) = 150 ×10 cos(−30o − 45o ) = 194.11 W
2
2
1
P = Re(VI * )
2
V = Vm ϕu = 150 − 30o
I = I m ϕ i = 10 45o → I* = 10 − 45o
(
)(
)
VI * = 150 − 30o 10 − 45 o = 1500 − 75o = 388.23 − j1448.9 VA
1
1
P = Re{388.23 − j1448.9} = 388.23 = 191.11W
2
2
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10
AC Power Analysis
1.
2.
3.
4.
5.
6.
7.
8.
Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Improvement
Average Power and RMS Value of Periodic Signals
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11
Maximum Average Power Transfer (1)
http://www.chegg.com/homework-help/questions-and-answers/use-maximum-power-transfer-theoremdetermine-increase-power-delivered-loudspeaker-resultin-q6983635
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12
Maximum Average Power Transfer (2)
1 2
PL = I Lm RL
2
Eeq
Eeq
IL =
→ I Lm =
Z eq + Z L
Z eq + Z L
IL
+
V
ZL
–
Z eq = Req + jX eq
Z L = RL + jX L
Zeq
→ Z eq + Z L = Req + jX eq + RL + jX L
+
–
= ( Req + RL ) + j ( X eq + X L ) Eeq
→ Z eq + Z L = ( Req + RL ) 2 + ( X eq + X L ) 2
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IL
+
V
ZL
–
13
Maximum Average Power Transfer (3)
IL
1 2
PL = I Lm RL
2
E eq
I Lm =
Z eq + Z L
+
V
ZL
–
Z eq + Z L = ( Req + RL ) 2 + ( X eq + X L ) 2
2
Eeq RL
1
→ PL = ×
2 ( Req + RL ) 2 + ( X eq + X L )2
–
Eeq
+
 ∂PL
 ∂R = 0
PL is maximum if:  L
 ∂PL = 0
 ∂X L
Zeq
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IL
+
V
ZL
–
14
Maximum Average Power Transfer (4)
2
Eeq RL
1
PL = ×
2 ( Req + RL ) 2 + ( X eq + X L )2
RL ( X eq + X L )
2
∂PL
 ∂PL
→
= E eq
=0
=
0
2
2
2
 ∂X
∂X L
[( Req + RL ) + ( X eq + X L ) ]
L

2
2
2 ( Req + RL ) + ( X eq + X L ) − 2 RL ( Req + RL )
∂
P
∂
P
 L = 0 → L = E eq
=0
2
2 2
∂RL
2[( Req + RL ) + ( X eq + X L ) ]
 ∂RL
 X L = − X eq
→
2
2
R
=
R
+
(
X
+
X
)
 L
eq
eq
L
ZL = Z
*
eq
 X L = − X eq
→
 RL = Req
For maximum average power transfer, the load impedance must
be equal to the complex conjugate of the equivalent impedance
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15
Maximum Average Power Transfer (5)
2
Eeq RL
1
PL = ×
2 ( Req + RL )2 + ( X eq + X L )2
 X L = − X eq

 RL = Req
→ PL max =
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E eq
2
8 Req
16
Maximum Average Power Transfer (6)
For maximum average power transfer, the load impedance must
be equal to the complex conjugate of the equivalent impedance
If ZL = RL ?
ZL = Z
*
eq
→ XL = 0
∂PL
= 0 → RL = Req2 + ( X eq + X L )2
∂RL
→ RL = Req2 + X eq2 = Z eq
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17
Maximum Average Power Transfer (7)
Determine the load impedance Z2 that
maximize the average power. What is
the maximum average power?
Z1
Z3
Z1
+
E = 20 − 45o V; J = 5 60 o A
Z 1 = 12 Ω ; Z 3 = − j16 Ω
–
Ex. 1
E
Zeq
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–
→ Z 2 = 7.68 + j5.76 Ω
J
Zeq
+
Z1Z3
12( − j16)
Zeq =
=
= 7.68 − j5.76 Ω
Z1 + Z3 12 − j16
Z3
Z2
Eeq
I2
Z2
Z 2 = Z*eq
18
Maximum Average Power Transfer (8)
+
Z1
E
–
Determine the load impedance Z2 that
maximize the average power. What is
the maximum average power?
E
J
a
+
Eeq
Eeq
8Req
Zeq
Z3
I2
J
–
–
20 − 45o
− 5 60o
12
=
= 54.38 − 140.4o V
1
1
+
2
12 − j16
→ Eeq = 54.38V → P2max =
Z3
Z2
+
E
−J
Z
Eeq = Va = 1
1
1
+
Z1 Z 3
Z1
+
E = 20 − 45o V; J = 5 60 o A
Z 1 = 12 Ω ; Z 3 = − j16 Ω
–
Ex. 1
2
54.38
=
= 48.13W
8 × 7.68
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Eeq
P2 max =
Z2
Eeq
2
8Req
19
Maximum Average Power Transfer (9)
3. Pmax =
–
2. Z L = Z
*
eq
Eeq
Z1
+
1. Find the Thevenin equivalent
a. Zeq
b. Eeq
Z2
E
Z3
J
2
Z1
8Req
Zeq
Z3
1a . Z eq = 7.68 − j5.76 Ω
a
1b . E eq = 54.38 − 140.4 V
o
–
3. P2 max = 48.13W
+
2. Z2 = 7.68 + j 5.76 Ω
Z1
E
+
Eeq
–
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Z3
J
20
Ex. 2
Maximum Average Power Transfer (10)
Determine the load impedance Z that maximize the
average power. What is the maximum average power?
Z eq =
Vopen-circuit
+–
j6 Ω
2I1
a
I short-circuit
j6 Ω
− j8Ω
4Ω
16 0o V
Ic
I1
16 0o V
b
Z
2 30o A
c
+–
− j8 Ω
2I1
a
4Ω
I1
Ic
2 30o A
+
b
Voc
–
c
1. Find the Thevenin equivalent
a. Z eq
b. Eeq
*
2. Z L = Z eq
3. Pmax =
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E eq
2
8Req
21
Ex. 2
Maximum Average Power Transfer (11)
Determine the load impedance Z that maximize the
average power. What is the maximum average power?
Z eq =
Vopen-circuit
I short-circuit
=
− j8 Ω
Eeq
J eq
(Vc − Vb ) −16 + j 6 I2 + 4I1 = 0
+–
j6 Ω
2I1
a
4Ω
16 0o V
Ic
I1
2 30o A
Voc = Vb − Vc
+b
Voc
–c
I1 = 2 30o
1. Find the Thevenin equivalent
a. Z eq
b. Eeq
I 2 = Ic = 2I1 = 2 × 2 30 o
*
2. Z L = Z eq
→ Voc = −16 + j 6 I 2 + 4I1
→ Voc = − 16 + j 6 × 2 × 2 30 + 4 × 2 30
o
= −21.07 + j 24.78 V
o
3. Pmax =
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E eq
2
8Req
22
Ex. 2
Maximum Average Power Transfer (12)
Determine the load impedance Z that maximize the
average power. What is the maximum average power?
Z eq =
Vopen-circuit
I short-circuit
=
j6 Ω
Eeq
− j8 Ω
+–
j6 Ω
2I1
a
J eq
4Ω
16 0o V
Ic
I1
16 0o V
b
Z
2 30o A
c
+–
− j8 Ω
2I1
a
4Ω
I1
Ic
b
Isc
2 30o A
1. Find the Thevenin equivalent
a. Z eq
b. Eeq
*
2. Z L = Z eq
c
3. Pmax =
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E eq
2
8Req
23
Ex. 2
Maximum Average Power Transfer (13)
Determine the load impedance Z that maximize the
average power. What is the maximum average power?
Z eq =
Vopen-circuit
I short-circuit
=
Eeq
I1 − 2 30 + I sc = 0 → Isc = 2 30 − I1
o
j6I 2 + 4I1 = 16 0
+–
j6 Ω
2I1
a
J eq
o
− j8 Ω
4Ω
16 0o V
Ic
I1
b
Isc
2 30o A
c
o
→ I 2 − I1 + 2 30o − 2I1 = 0
1. Find the Thevenin equivalent
a. Z eq
b. Eeq
→ 3I1 − I2 = 2 30o
*
2. Z L = Z eq
I2 − I1 + 2 30 − I c = 0
o
→ I1 = 0.67 − j 0.41 A
→ Isc = 2 30 − (0.67 − j0.41) = 1.06 + j1.41 A
o
3. Pmax =
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E eq
2
8Req
24
Maximum Average Power Transfer (14)
Ex. 2
Determine the load impedance Z that maximize the
average power. What is the maximum average power?
Z eq =
Vopen-circuit
I short-circuit
=
Eeq
J eq
Voc = −21.07 + j 24.78 V
I sc = 1.06 + j1.41 A
→ Z eq =
−21.07 + j 24.78
= 4.00 + j18.00 Ω
1.06 + j1.41
→ Z = 4.00 − j18.00 Ω
21.07 + 24.78
=
= 33.06 W
8× 4
2
Pmax
2
− j8 Ω
+–
j6 Ω
2I1
a
4Ω
16 0o V
Ic
I1
b
Z
2 30o A
c
1. Find the Thevenin equivalent
a. Z eq
b. Eeq
*
2. Z L = Z eq
3. Pmax =
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E eq
2
8Req
25
AC Power Analysis
1.
2.
3.
4.
5.
6.
7.
8.
Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Improvement
Average Power and RMS Value of Periodic Signals
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26
RMS Value (1)
i(t)
+
–
e(t)
(AC)
R
1 T 2
R T 2
→ P =  i Rdt =  i dt
T 0
T 0
→ I eff
I
+
–
E
(DC)
R
1
=
T

T 2
0
i dt
→ P = I 2R
1 T 2
Veff =
v dt

0
T
I is the effective/RMS value of i(t)
X eff = X rms
1 T 2
=
x dt

T 0
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27
RMS Value (2)
I rms
1
=
T

T
0
i 2dt
→ I rms
i (t ) = I m sin ω t
1 T 2
1
=
i dt =

T 0
T

T
0
[ I m sin ωt ]2 dt
1 T 2 1 − cos 2ωt
=
Im
dt

0
T
2
I m2
=
2T
I rms
Im
=
2

T
0
Im
dt =
2
Vrms
Vm
=
2
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28
RMS Value (3)
I
1
P = Vm I m cos(φv − φi )
2
+
V
−
Vrms
Vm
=
2
P = Vrms I rms cos(φv − φi )
I rms
Im
=
2
1
P = Re(VI* )
2
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29
Ex.
RMS Value (4)
v(t) = 150sin(314t – 30o) V, i(t) = 10sin(314t + 45o) A. Find Vrms & Irms?
Vrms
Vm 150
=
=
= 106.07 V
2
2
I rms =
Im
10
=
= 7.07 A
2
2
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30
AC Power Analysis
1.
2.
3.
4.
5.
6.
7.
8.
Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Improvement
Average Power and RMS Value of Periodic Signals
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31
Apparent Power (1)
I
+
S = Vrms I rms
V
−
(in volt- ampere, VA)
https://www.amazon.com/
Ventilated-TransformerEnclosure-NameplateDetails/dp/B07G3DNTXN
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32
Ex.
Apparent Power (2)
v(t) = 150sin(314t – 30o) V, i(t) = 10sin(314t + 45o) A. Find S?
S = Vrms I rms =
Vm I m 150 10
×
=
×
= 750 VA
2
2
2
2
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33
Power Factor (1)
P = Vrms I rms cos(φv − φi )
I
+
V
−
S = Vrms I rms
P
pf = = cos(φv − φi )
S
• φv − φi = 0 → pf = 1 → P = S = Vrms I rms
• φv − φi = ±90o → pf = 0 → P = 0
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34
Ex.
Power Factor (2)
v(t) = 150sin(314t – 30o) V, i(t) = 10sin(314t + 45o) A. Find pf?
I
+
V
−
pf = cos( − 30o − 45o ) = 0.2588
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35
AC Power Analysis
1.
2.
3.
4.
5.
6.
7.
8.
Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Improvement
Average Power and RMS Value of Periodic Signals
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36
Complex Power (1)
I
+
Z
V
1 *
S = VI
2
V = Vm φv
–
I = I m φi → I = I m
*
− φi
1
1
1
→ S = Vm I m φv − φi = Vm I m cos(φv − φi ) + j Vm I m sin(φv − φi )
2
2
2
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37
Complex Power (2)
1 *
S = VI
2
V = ZI
(
)(
)
*
2

1  V  1 VV* 1 Vm φv Vm − φv
Vrms
S = V   =
=
= *
*
*

2 Z 2 Z
2
Z
Z
→
1
1
1 2

*
2
S
=
ZII
=
Z
I
φ
I
−
φ
=
Z
I
=
Z
I
m
i
m
i
rms

2
2
2 m
Z = R + jX
(
2
→ S = ( R + jX ) I rms
)(
)
2

P
=
Re(
S
)
=
RI

rms
2
2
= RI rms + jXI rms → 
2
Q
=
Im(
S
)
=
XI

rms
Reactive power (in volt-ampere reactive, VAR)
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38
Complex Power (3)
I
+
Z
V
–
2
1 *
1
V
2
S = VI = P + jQ = Vm I m φv − φi = Vrms I rms φv − φi = ZI rms
= rms*
2
2
Z
S = S = Vrms I rms = P 2 + Q 2
2
P = Re(S) = Vrms I rms cos(φv − φi ) = S cos(φv − φi ) = RI rms
2
Q = Im(S) = S sin(φv − φi ) = XI rms
pf =
P
= cos(φv − φi )
S
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39
Complex Power (4)
Ex.
v(t) = 150sin(314t – 30o) V, i(t) = 10sin(314t + 45o) A.
I
+
V
−
V = 150 − 30o V, I = 10 45o A
S=
(
)(
)
1 * 1
VI = 150 − 30o 10 − 45o = 750 − 75o VA
2
2
S = S = 750 VA
P = S cos(ϕu − ϕi ) = 750cos(−75o ) = 194.11 W
Q = S sin(ϕu − ϕi ) = 750sin(−75o ) = −724.44 VAR
pf = cos(ϕu − ϕi ) = cos( −75o ) = 0.26
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40
AC Power Analysis
1.
2.
3.
4.
5.
6.
7.
8.
Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Improvement
Average Power and RMS Value of Periodic Signals
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41
Conservation of AC Power (1)
M
S
i =1
N
source ,i
=  S load ,i
i =1
N
M
 Psource ,i =  Pload ,i
 i =1
i =1
→M
N
 Q

source , i =  Qload , i
 i =1
i =1
M
S
i =1
N
source , i
≠  S load ,i
i =1
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42
Ex.
Conservation of AC Power (2)
Z1 = 10Ω; Z 2 = j 20Ω; Z 3 = 5 − j10Ω;
I3
I2
E3
–
Find currents?
I1
+
E1 = 30 V; E3 = 45 15o V; J = 2 − 30o A;
Z1
– +
Z2
E1
J
Z3
I1 = 4.09 75.2o A, I 2 = 2.20 26.4o A, I 3 = 6.16 39.6o A
S Z 1 = Z1 I12rms = 10(4.09)2 / 2 = 83.64 VA
→
S
load
1
1
S E 1 = E1I1* = 30.4.09 − 75, 2o = 61.35 − 75, 2o VA
2
2
1
S E 3 = E3I *3 = 138.60 − 24.6o VA
→
2
1
S J = U J J * = ( −Z 2 I 2 ) J* = 36.65 − j 24.35 VA
2
S
source
S Z 2 = Z 2 I 22rms = j 20(2.20)2 / 2 = j 48.40 VA
= 178.50 − j141.33 VA
S Z 3 = Z3 I32rms = (5 − j10)(6.162 ) / 2 = 94.86 − j189.73 VA
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= 178.34 − j141.36 VA
43
AC Power Analysis
1.
2.
3.
4.
5.
6.
7.
8.
Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Improvement
Average Power and RMS Value of Periodic Signals
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44
Power Factor Improvement (1)
pf = cos φ
R
+
–
E
pf 2 > pf1 → φ2 < φ1
IL
IC
L
φ1
I
+
–
E
R
IL
L
φ2
E
I
C
IC
IL
φ2 < φ1 → C = ? ( P = const)
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45
Power Factor Improvement (2)
Q1
Q1 = P tan φ1 , Q2 = P tan φ2
S1
∆Q = Q1 − Q2
φ
φ1 2
2
∆Q
Erms
2
∆Q =
= ωCErms → C =
2
ω
E
XC
rms
P
I
E
–
tan φ1 − tan φ2
=P
2
ω E rms
Q2
R
+
Q1 − Q2 P tan φ1 − P tan φ2
C=
=
2
2
ω E rms
ω Erms
ΔQ
IL
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C
L
IC
46
Ex
Power Factor Improvement (3)
A load is connected to a 220 V (rms), 50 Hz power line. This load absorbs a power of
1000kW. Its power factor is 0.8. Find the capacitor required to raise the pf to 0.9?
tan φ1 − tan φ2
C=P
2
ω E rms
pf1 = 0.8 → cos φ1 = 0.8 → φ1 = 36.9o → tan φ1 = 0.75
pf 2 = 0.9 → cosφ2 = 0.9 → φ2 = 25.8o → tan φ2 = 0.48
0.75 − 0.48
→ C = 1000 × 10
= 0.0178 F
2
314 × (220)
3
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47
AC Power Analysis
1.
2.
3.
4.
5.
6.
7.
8.
Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Conservation of AC Power
Power Factor Improvement
Average Power and RMS Value of Periodic Signals
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48
Average Power and RMS Value
of Periodic Signals (1)
∞
v (t ) = Vdc + Vn sin( nω0t − θn )
n=1
∞
i (t ) = I dc +  I n sin(mω0t − φm )
m =1
1 ∞
P = Vdc I dc + Vn I n cos(θn − φn )
2 n=1
Vrms
1 ∞ 2
= V + Vn
2 n =1
I rms
1 ∞ 2
= I +  Im
2 m=1
2
dc
2
dc
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49
L
e1
j
R2
C
+
R1
+
–
e1 = 10sin10t V; j = 4sin(50t + 30o) V; e2 = 6 V
(DC); L = 1 H; R1 = 1 Ω; R2 = 5 Ω; C = 0.01 F;
find the RMS value of the voltage across R1?
–
Ex.
Average Power and RMS Value
of Periodic Signals (2)
e2
vR1 = −1 + 1.06sin(10t − 58o ) + 4.14 sin(50t + 32o ) V
Vrms
∞
1
= Vdc2 + Vn2
2 n=1
1
= ( −1) + (1.06 2 + 4.14 2 ) = 3.18 V
2
2
dc
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50