Lecture Electric circuit theory: AC power analysis - Nguyễn Công Phương

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
AC Power Analysis - sites.google.com/site/ncpdhbkhn

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AC Power Analysis
1.
2.
3.

4.
5.
6.

Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Power Factor Improvement

AC Power Analysis - sites.google.com/site/ncpdhbkhn

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Instantaneous Power (1)
p(t ) = v(t )i (t )

v(t ) = Vm sin(ωt + φv )
i (t ) = I m sin(ωt + φi )
→ p (t ) = Vm I m sin(ωt + 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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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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Average Power (1)
1
P=
T

∫

T

0

p(t ) dt

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(2ωt + 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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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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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
P = Vm I m cos(90o ) = 0
2

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Average Power (4)
• Ex.:
v(t) = 150sin(314t – 30o) V
i(t) = 10sin(314t + 45o) A
Find P?

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AC Power Analysis
1.
2.
3.
4.
5.
6.

Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Power Factor Improvement

AC Power Analysis - sites.google.com/site/ncpdhbkhn

10


Maximum Average Power Transfer (1)
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

–
11


Maximum Average Power Transfer (2)
IL

1 2
PL = I Lm RL
2
Eeq
I Lm =
Zeq + Z L

+

V

ZL

–

Zeq + 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

–
12


Maximum Average Power Transfer (3)

2

Eeq RL
1
PL = ×
2 ( Req + RL )2 + ( X eq + X L )2
2
RL ( X eq + X L )
∂PL
 ∂PL
→
= Eeq
=0
2
2 2
 ∂X = 0 ∂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 = Eeq
=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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Maximum Average Power Transfer (4)

2

Eeq RL
1
PL = ×
2 ( Req + RL )2 + ( X eq + X L )2

 X L = − X eq

 RL = Req

→ PL max =

AC Power Analysis - sites.google.com/site/ncpdhbkhn

Eeq

2

8Req

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Maximum Average Power Transfer (5)
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 = Zeq
AC Power Analysis - sites.google.com/site/ncpdhbkhn

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Maximum Average Power Transfer (6)

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

J

Zeq
+
–

Z1Z3
12(− j16)
Z eq =
=

= 7.68 − j 5.76 Ω
Z1 + Z3 12 − j16

Z3

Z2

→ Z 2 = 7.68 + j 5.76 Ω
AC Power Analysis - sites.google.com/site/ncpdhbkhn

Eeq

I2
Z2

Z 2 = Z*eq
16


Maximum Average Power Transfer (7)

+

Z1
E

–

Determine the load impedance Z2 that
maximize the average power. What is

the maximum average power?

–

→ Eeq

E

J

a
+

Eeq

Eeq

8 Req

Zeq

Z3

I2

J

–

20 − 45o

− 5 60o
12
=
= 54.38 − 140.4 o V
1
1
+
2
12 − j16

= 54.38 V → P2 max =

Z3

Z2

=

+

E
−J
Z1
Eeq = Va =
1
1
+
Z1 Z3

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

AC Power Analysis - sites.google.com/site/ncpdhbkhn

Eeq

P2 max =

Z2

Eeq

2

8 Req
17

Maximum Average Power Transfer (8)

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. Eeq = 54.38 − 140.4 V
o

–

3. P2 max = 48.13W

+

2. Z2 = 7.68 + j5.76 Ω

Z1
E

+

Eeq
–

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Z3
J
18


Ex. 2

Maximum Average Power Transfer (9)

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

− j8 Ω
4Ω

16 0o V

16 0o V

Ic

I1

b
Z

2 30o A

c

+–

− j8 Ω

a
4Ω

I1

2I1
Ic
2 30o A

+

b

Voc
–

c

1. Find the Thevenin equivalent
a. Zeq
b. Eeq
2. Z L = Z*eq
3. Pmax =

Eeq

2

8Req

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Ex. 2

Maximum Average Power Transfer (10)

Determine the load impedance Z that maximize the
average power. What is the maximum average power?

Zeq =

Vopen-circuit
I short-circuit

=

− j8 Ω

Eeq
J eq

(Vc − Vb ) − 16 + j 6I 2 + 4I1 = 0

+–

j6 Ω

2I1

a
4Ω

Ic

I1

2 30o A

Voc = Vb − Vc

→ Voc = −16 + j 6I 2 + 4I1

16 0o V

+b

Voc
–c

I1 = 2 30o

1. Find the Thevenin equivalent
a. Zeq
b. Eeq

I 2 = I c = 2I1 = 2 × 2 30o

2. Z L = Z*eq

→ Voc = −16 + j 6 × 2 × 2 30o + 4 × 2 30o
= 32.53 130.4 V
o

3. Pmax =

Eeq

2

8Req

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Ex. 2

Maximum Average Power Transfer (11)

Determine the load impedance Z that maximize the
average power. What is the maximum average power?

Zeq =

Vopen-circuit

=

I short-circuit

j6 Ω

Eeq

− j8 Ω

+–

j6 Ω

2I1

a

J eq

4Ω

16 0o V

16 0o V

Ic

I1

b
Z

2 30o A

c

+–

− j8 Ω

a
4Ω

I1

2I1
Ic

b

Isc

1. Find the Thevenin equivalent
a. Zeq
b. Eeq
2. Z L = Z*eq

2 30o A

c

3. Pmax =

Eeq

2

8Req

AC Power Analysis - sites.google.com/site/ncpdhbkhn

21


Ex. 2

Maximum Average Power Transfer (12)

Determine the load impedance Z that maximize the
average power. What is the maximum average power?

Zeq =

Vopen-circuit
I short-circuit

=

Eeq

4Ω

I1 − 2 30 + I sc = 0 → I sc = 2 30 − I1
o

j 6I 2 + 4I1 = 16 0

+–

j6 Ω

2I1

a

J eq

o

− j8 Ω

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. Zeq
b. Eeq

→ 3I1 − I 2 = 2 30 o

2. Z L = Z*eq

I 2 − I1 + 2 30 − I c = 0
o

→ I1 = 0.67 − j 0.41 A
→ I sc = 2 30 − (0.67 − j0.41) = 1.76 52.9 A
o

o

3. Pmax =

Eeq

2

8Req

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22


Ex. 2

Maximum Average Power Transfer (13)

Determine the load impedance Z that maximize the
average power. What is the maximum average power?

Zeq =

Vopen-circuit
I short-circuit

=

Eeq
J eq

Voc = 32.53 130.4o V

I sc = 1.76 52.9o A
→ Z eq =

32.53 130.4o
1.76 52.9

o

= 1.67 + j19.0 Ω

→ Z = 1.67 − j19.0 Ω
2

Pmax

32.53
=
= 79.2 W
8 × 1.67

− j8 Ω

+–

j6 Ω

2I1

a
4Ω

16 0o V

Ic

I1

b
Z

2 30o A

c

1. Find the Thevenin equivalent
a. Zeq
b. Eeq
2. Z L = Z*eq
3. Pmax =

Eeq

2

8Req

AC Power Analysis - sites.google.com/site/ncpdhbkhn

23


AC Power Analysis
1.
2.
3.
4.
5.

6.

Instantaneous and Average Power
Maximum Average Power Transfer
RMS Value
Apparent Power and Power Factor
Complex Power
Power Factor Improvement

AC Power Analysis - sites.google.com/site/ncpdhbkhn

24


RMS Value (1)

i(t)
+
–

e(t)
(AC)

R

1
→P=
T

∫

T

0

R T 2
i Rdt = ∫ i dt
T 0
2

→ I eff
I
+
–

E
(DC)

R

1 T 2
=
i dt
∫
0
T

→ P = I 2R
Veff

I is the effective/RMS value of i(t)

X eff = X rms

1
=
T

∫

T

0

1
=
T

∫

T

0

v 2 dt

x 2 dt

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