CHAPTER I
APPLICATION OF CIRCUIT LAWS
Introduction
Generally, we require 3 steps to analyze
AC Circuit
 Transform the circuit to the phasor /
frequency domain
 Solve the problem using any technique
 Transform the resulting phasor /
frequency domain to the time domain
expression.
2
Time domain to Phasor
3
Impedance in Frequency Domain
ELEMENT
IMPEDANCE
FORMULA
PHASOR
FORM
RECTANGULAR
FORM
R
ZR
R
R0
R + j0
L
ZL
jωL
XL  90
0 + jXL
C
ZC
1/jωC
XC  -90
0 - jXC
4
Methods of solving





Kirchhoff’s Voltage Law (KVL)
Kirchhoff’s Current Law (KCL)
Voltage Divider Rule (VDR)
Current Divider Rule (CDR)
Star / Delta Transformation ( /  )
5
Kirchhoff’s Voltage Law (KVL)

Algebraic sum of
voltage drops around
closed loop is zero
V=0
Voltage drop = voltage
rise
V = V1+ V2 + … + VN
= I (Z1 + Z2 + … + ZN)
6
Kirchhoff’s Current Law (KCL)

Algebraic sum of
current at any node
is zero
I=0
Current In = Current
Out
I = I 1 + I2 + … + I N
= V (1 / Z1 + 1 / Z2 + … + 1 / ZN)
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Voltage Divider Rule (VDR)
Z1
V1 
V
Z1  Z2
Z2
V2 
V
Z1  Z2
8
Current Divider Rule (CDR)
Z2
I1 
I
Z1  Z2
Z1
I2 
I
Z1  Z2
9
Star / Delta ( /  ) Transformation
 -  Conversion
Z1Z2  Z2 Z3  Z1Z3
Za 
Z1
Zb 
Z1Z2  Z2 Z3  Z1Z3
Z2
Z1Z2  Z2 Z3  Z1Z3
ZC 
Z3
10
Delta / Star ( / Y) Transformation
 - Y Conversion
Z b Zc
Z1 
Za  Z b  Zc
Za Zc
Z2 
Za  Z b  Zc
Za Z b
Z3 
Za  Z b  Zc
11
Example 1
Find the input impedance, Zin of the circuit (ω=50rad/s)
12
Solution Example 1
Let
Z1 = Impedance of the 2mF capacitor
Z2 = Impedance of the 3 resistor in series with the 10mF capacitor
Z3 = Impedance of the 0.2H inductor in series with 8 resistor
Then
1
1

  j10Ω
3
jωC j(50)(2 10 )
1
1
Z2  3 
 3
 (3  j2)Ω
3
jωC
j(50)(10 10 )
Z1 
Z3  8  jωL  8  j(50)(0.2)  (8  j10) Ω
13
Solution Example 1
The input impedance is
Zin  Z1  Z2 || Z3   j10 
(3  j2)(8  j10)
11  j8
  j10  3.22  j1.07
 (3.22  j11.07)
14
Example 2
Find the input impedance, Zin of the circuit (ω=10rad/s)
15
Solution Example 2
Let
Z1 = Impedance of the 2mF capacitor in series with 20 resistor
Z2 = Impedance of the 4mF capacitor
Z3 = Impedance of the 2H inductor in series with 50 resistor
Then
1
1
 20 
 (20  j50) Ω
3
jωC
j(10)(2 10 )
1
1
Z2 

  j25Ω
3
jωC j(10)(4 10 )
Z1  20 
Z3  50  jωL  50  j(10 )(2)  (50  j20) Ω
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Solution Example 2
The input impedance is
Z 2 Z3
Zin  Z1  Z2 || Z3  Z1 
Z 2  Z3
 j25(50  j20)
 20  j50 
 j25  50  j20)
 20  j50  12.38  j23.76
 (32.38  j73.76) Ω
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Example 3
Determine Vo (t)
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Solution example 3
Step 1: Transfer the circuit into frequency domain
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Solution Example 3
Step 2: Solve the circuit using any method
Let
Z1 = Impedance of the 60 resistor
Z2 = Impedance of the parallel combination of the 10mF capacitor and 5H inductor
Then
Z1  60Ω
(-j25)(j20 )
Z2   j25 || j20 
 j100 Ω
- j25  j20
20
Solution Example 3
By using voltage divider rule
Z2
j100
V0 
Vs 
(20  15o )
Z1  Z2
60  j100
 (0.8575 30.96 o )(20  15o )
 17.1515.96 o V
Step 3: Convert the result to the time domain
vo (t )  17.15 cos(4t  15.96 o )V
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Example 4
Determine Vo (t)
22
Solution Example 4
Step 1: Transfer to the frequency domain
Voltage source

0.5H inductor

(1/20)F capacitor

Vs  1075 o
jωL  j(10)(0.5)  j5
1
1

  j2
jωC j(10)(1/20 )
23
Solution Example 4
Step 2: Solve the circuit using any method
Let
Z1 = Impedance of the 0.5H inductor in parallel with the 10 resistor
Z2 = Impedance of the (1/20)F capacitor
Then
(10)(j5)
Z1  10 || j5 
 2  j4
10  j5
Z 2   j2
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Solution Example 4
By using voltage divider rule
Z2
- j2
V0 
Vs 
(1075o )
Z1  Z2
2  j4 - j2
 7.071 - 60 o V
Step 3: Convert the result to the time domain
vo (t )  7.071 cos(10t  60 o )V
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Example 5
Find current I
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Solution Example 5
Step 1: Transform the circuit from delta to star
connection ( to Y)
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Solution Example 5
Calculate new impedances after the transformation
j4(2 - j4)
4(4  j2)
Zan 

 (1.6  j0.8)Ω
j4  2  j4  8
10
Zbn
j4(8) 4(4  j2)


 j3.2Ω
10
10
8(2 - j4) 4(4  j2)
Zcn 

 (1.6  j3.2)Ω
10
10
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Solution Example 5
The total input impedance is
Zin  12  Zan  (Zbn  j3) || (Zcn  j6  8)
 12  1.6  j0.8  (j0.2) || (9.6  j2.8)
j0.2(9.6  j2.8)
 13.6  j0.8 
9.6  j3
 13.6  j1  13.64 4.204 o Ω
29
Solution Example 5
The desire current is
V
500o
I 
Z 13.64 4.204 o
 3.666   4.204 o A
30