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CHAPTER I
APPLICATION OF CIRCUIT LAWS
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Introduction
Generally, we require 3 steps to analyzeAC 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.
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Time domain to Phasor
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Impedance in Frequency Domain
ELEMENT
IMPEDANCE FORMULAPHASOR 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
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Methods of solving
Kirchhoff’s Voltage Law (KVL) Kirchhoff’s Current Law (KCL) Voltage Divider Rule (VDR) Current Divider Rule (CDR) Star / Delta Transformation ( / )
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Kirchhoff’s Voltage Law (KVL)
Algebraic sum of voltage drops around closed loop is zero
V = 0Voltage drop = voltage rise
V = V1+ V2 + … + VN
= I (Z1 + Z2 + … + ZN)
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Kirchhoff’s Current Law (KCL)
Algebraic sum of current at any node is zero
I = 0Current In = Current Out
I = I1+ I2 + … + IN
= V (1 / Z1 + 1 / Z2 + … + 1 / ZN)
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Voltage Divider Rule (VDR)
VZZ
ZV
21
11
VZZ
ZV
21
22
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Current Divider Rule (CDR)
IZZ
ZI
21
21
IZZ
ZI
21
12
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Star / Delta ( / ) Transformation
1
313221a Z
ZZZZZZZ
2
313221b Z
ZZZZZZZ
- Conversion
3
313221C Z
ZZZZZZZ
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Delta / Star ( / Y) Transformation
cba
cb1 ZZZ
ZZZ
cba
ca2 ZZZ
ZZZ
- Y Conversion
cba
ba3 ZZZ
ZZZ
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Example 1Find the input impedance, Zin of the circuit
(ω=50rad/s)
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Solution Example 1
j10Ω)10j(50)(2
1
jω
1Z
31
C
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
j2)Ω(3)10j(50)(10
13
jω
13Z
32
C
j10)Ω(8j(50)(0.2)8jω8Z3 L
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Solution Example 1
The input impedance is
j811
j10)j2)(8(3j10Z||ZZZ 321in
j1.073.22j10
)07.1122.3( j
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Example 2Find the input impedance, Zin of the circuit (ω=10rad/s)
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Solution Example 2
j50)Ω20()10j(10)(2
120
jω
120Z
31
C
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
j25Ω)10j(10)(4
1
jω
1Z
32
C
j20)Ω(50(2))j(1005jω05Z3 L
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Solution Example 2
The input impedance is
32
321321in ZZ
ZZZZ||ZZZ
j20)50j25
j20)j25(50j5020
j73.76)Ω(32.38
j23.7638.21j5020
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Example 3Determine Vo (t)
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Solution example 3
Step 1: Transfer the circuit into frequency domain
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Solution Example 3
0Ω6Z1
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
j100Ωj20j25-
)(-j25)(j20j20||j25Z2
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Solution Example 3
By using voltage divider rule
)15(20j10060
j100V
ZZ
ZV o
s21
20
)15)(2030.96(0.8575 oo
V)96.154cos(15.17)( o ttvo
V15.9617.15 o
Step 3: Convert the result to the time domain
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Example 4Determine Vo (t)
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Solution Example 4
os 7510V
Step 1: Transfer to the frequency domain
j2)j(10)(1/20
1
jω
1
C
j5j(10)(0.5)jω L
Voltage source
0.5H inductor
(1/20)F capacitor
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Solution Example 4
j2Z2
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
j42j510
(10)(j5)j5||10Z1
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Solution Example 4
By using voltage divider rule
)57(10j2-j42
j2-V
ZZ
ZV o
s21
20
V)6010cos(071.7)( o ttvo
V60-071.7 o
Step 3: Convert the result to the time domain
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Example 5Find current I
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Solution Example 5Step 1: Transform the circuit from delta to star connection ( to Y)
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Solution Example 5
j3.2Ω10
j2)4(4
10
j4(8)Zbn
j3.2)Ω(1.610
j2)4(4
10
j4)-8(2Zcn
Calculate new impedances after the transformation
j0.8)Ω(1.610
j2)4(4
8j42j4
j4)-j4(2Zan
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Solution Example 5
The total input impedance is
8)j6(Z||j3)(ZZ12Z cnbnanin
j2.8)(9.6||(j0.2)j0.81.612
Ω4.20413.64j113.6 o
j39.6
j2.8)j0.2(9.6j0.813.6
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Solution Example 5
The desire current is
o
o
4.20413.64
050
Z
VI
A4.204666.3 o
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Example 6Find current I
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Solution Example 6
Step 1: Transform the circuit from
delta to star connection
( to Y)
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Solution Example 6
j2.82)Ω0.24(j68
j5)j3(8-Zbn
j0.72)Ω(0.9668
j4(-j3)Zcn
j
Calculate new impedances after the transformation
j3.76)Ω(0.32j68
j8)4(-5
j3-j58j4
j5)j4(8Zan
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Solution Example 6
j2.82)-9.76j1.76(5.32
j2.82)6j1.76)(9.7(5.32Zcn
The total input impedance is
)01(Z||j2)5(ZZZ bnancnin
j2.82)(9.76||j1.76)(5.32Zcn
j0.40743.744j0.7296.0
)3.802-4.714(j0.3126704.4 o
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Solution Example 6
The desire current is
o
o
3.802714.4
030
Z
VI
A802.3364.6 o