Nodal Analysis
Chapter 10
10.1 Introduction
•
Step to Analyze AC Circuits:
1. Transform the circuit to the phasor or frequency
domain.
2. Solve the problem using circuit techniques
(nodal analysis, mesh analysis, superposition,
etc.).
3. Transform the resulting phasor to the time
domain.
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10.2 Nodal Analysis
Example 10.1
• Find ix in the circuit of Fig. 10.1 using nodal analysis.
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Fig 10.2
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Example 10.1
V1
V1  V2
• At node 1, 20  V1


10
 j 2.5
j4
or (1  j1.5)V1  j 2.5V2  20
• At node 2,
V1  V2 V2
2I x 

j4
j2
2V1
V1  V2 V2
I x  V1 /  j 2.5, 


 j 2.5
j4
j2
we get 11V1  15V2  0
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Example 10.1
1  j 2.5 j 2.5  V1  20
 


 11

15  V2   0 

1  j 2. 5 j 2. 5

 15  j 5
11
15
1 
20
j 2.5
0
15
 300,  2 
1  j 2.5 20
11
0
 220
1
300
V1 

 18.9718.43 V
 15  j 5
 2  220
V2 

 13.91198.3 V
 15  j 5
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Example 10.1
• The current Ix is given by
V1
18.9718.43
Ix 

 7.59108.4 A
 j 2.5
2.5  90
• Transforming this to the time domain,
ix  7.59 cos(4t  108.4) A
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Example 10.2
• Compute V1 and V2 in the circuit of Fig. 10.4.
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Example 10.2
• By node 1 and 2,
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V1 V2 V2
3


 j 3 j 6 12
or 36  j 4V1  (1  j 2)V2
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Example 10.2
• But a voltage source is connected between nodes 1
and 2, so that
V1  V2  1045
substituting,
36  40135  (1  j 2)V2  V2  31.41  87.18 V
so,
V1  V2  1045  25.87  70.48 V
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10.3 Mesh Analysis
Example 10.3
• Determine current Io in the circuit of Fig. 10.7 using
mesh analysis.
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Example 10.3
• Applying KVL to mesh 1,
(8  j10  j 2)I1  ( j 2)I 2  j10I3  0
• For mesh 2,
(4  j 2  j 2)I 2  ( j 2)I1  ( j 2)I3  2090  0
• For mesh 3, I3 = 5, we get
(8  j8)I1  j 2I 2  j 50
j 2I1  (4  j 4)I 2   j 20  j10
• Put it in matrix form as
j 2   I1   j 50 
8  j8



 j2


I
4

j
4

j
30

 2  

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Example 10.3
• We obtain the determinants
8  j8
j2

 32(1  j )(1  j )  4  68
j2
4  j4
2 
8  j8
j 50
j2
 j 30
 340  j 240  416.17  35.22
 2 416.1735.22
I2 

 6.12  35.22 V

68
• The desired current is
I o  I 2  6.12144.78 A
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Example 10.4
• Solve for Vo in the circuit of Fig. 10.9 using mesh
analysis.
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Example 10.4
For mesh 1, KVL gives
 10  (8  j 2)I1  ( j 2)I 2  8I 3  0
or (8  j 2)I1  j 2I 2  8I 3  10
For mesh 2,
I 2  3
For the supermesh,
(8  j 4)I3  8I 2  (6  j5)I 4  j5I 2  0
Due to the current source between meshes 3 and 4, at
node A,
I 4  I3  4
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Example 10.4
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Example 10.4
(8  j 2)I1  8I 3  10  j 6
 8I1  (14  j )I 3  24  j 35
 8   I1   10  j 6 
8  j 2






I

8
14

j

24

j
35

 3  

we obtain the following determinan ts

8  j2
8
8
14  j
1 
10  j 6
 112  j8  j 28  2  64  50  j 20
8
 24  j 35 14  j
 140  j10  j84  6  192  j 280
 58  j186
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Example 10.4
• Current I1 is obtained as
1  58  j186
I1 

 3.618274.5 A

50  j 20
• The required voltage Vo is
Vo   j 2(I1  I 2 )   j 2(3.618274.5  3)
 7.2134  j 6.568  9.756222.32 V
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10.4 Superposition Theorem
Example 10.6
• Find vo of the circuit of Fig. 10.13 using the
superposition theorem.
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Example 10.6
• We let
vo  v1  v2  v3
• By voltage division,
1
 v1 
(5)  1 V
1 4
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Example 10.6
10 cos 2t  100,   2 red/s
2H

0.1 F

jL  j 4 
1
  j5 
jL
• Let
 j5  4
Z   j5 4 
 2.439  j1.951
4  j5
• By division,
1
10
V2 
(100) 
 2.498  30.79
1  j4  Z
3.439  j 2.049
v2  2.498 cos( 2t  30.79)
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Example 10.6
2 sin 5t  2  90,   5 rad/s
2 H

0.1 F

jL  j10 
1
  j2 
jL
• Let
 j2  4
Z1   j 2 4 
 0.8  j1.6 
4  j2
• By current division,
j10
I1 
(2  90) A
j10  1  Z1
j10
V3  I1  1 
( j 2)  2.32880 V
1.8  j8.4
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Example 10.6
• In the time domain,
v3  2.33 cos(5t  80)  2.33 sin(5t  10) V
v0 (t )  1  2.498 cos( 2t  30.79)  2.33 sin(5t  10)V
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10.5 Source Transformation
Vs  Z s I s
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
Vs
Is 
Zs
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Example 10.7
• Calculus Vx in the circuit of Fig. 10.17 using the
method of source transformation.
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Example 10.7
20  90
Is 
 4  90   j 4 A
5
5(3  j 4)
Z1 
 2.5  j1.25 
8  j4
Vs  I 2 Z1   j 4(2.5  j1.25)  5  j10 V
 Vx 
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(5  j10)  2.519  28 V
10  2.5  j1.25  4  j13
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10.6 Thevenin and Norton
Equivalent Circuits
VTH  Z N I N ,
Z TH  Z N
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Example 10.8
• Obtain the Thevenin equivalent at terminals a-b of
the circuit in Fig. 10.22.
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Example 10.8
 j6  8
Z1   j 6 8 
 2.88  j 3.84 
8  j6
j12  4
Z 2  4 j12 
 3.6  j1.2 
4  j12
ZTH  Z1  Z 2  6.48  j 2.64 
12075
12075
I1 
A, I 2 
A
8  j6
4  j12
VTH  4I 2  ( j 6)I1  0
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Example 10.8
VTH
48075 720(75  90)
 4I 2  j 6I1 

4  j12
8  j6
 37.953.43  72201.87
 28.936  j 24.55
 37.95220.31 V
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Example 10.9
• Find the Thevenin equivalent of the circuit in Fig.
10.25 as seen from terminals a-b.
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Example 10.9
• From KVL at node 1,
15  I o  0.5 o  I o  10 A
 I o (2  j 4)  0.5I o (4  j 3)  VTH  0
or VTH  10(2  j 4)  5(4  j 3)   j 55
Thus, VTH  55  90 V
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10.7 Op Amp AC circuits
1. No current enters either of its input terminals.
2. The voltage across its input terminals is zero.
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Example 10.11
• Determine v0(t) for the o amp circuit in Fig. 10.31(a)
if vs = 3 cos 1000t V.
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Example 10.11
Vs  30,   1000 rad/s
• Apply KCL at node 1,
30  V1
V1 V1  0 V1  Vo



10
 j5
10
20
or 6  (5  j 4)V1  Vo
• At node 2,
V1  0 0  Vo

10
 j10
V1   jVo
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Example 10.11
6   j (5  j 4)Vo  Vo  (3  j 5)Vo
6
Vo 
 1.02959.04
3  j5
• Hence,
vo (t )  1.029 cos(1000t  59.04) V
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10.9 Applications
• 10.9.1 Capacitance Multiplier
Fig. 10.41
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Vi  Vo
Ii 
 jC (Vi  Vo )
1 / jC
Vi  0 0  Vo
R2

or Vo   Vi
R1
R2
R1
 R2 
 R2 
Ii
I i  jC 1  Vi or
 j 1  C
Vi
 R1 
 R1 
Vi
1
Zi  
Ii
jCeq
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 R2 
where Ceq  1  C
 R1 
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