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This is solution to problems related Electrical Circuit Analysis course. It was given by Prof. Gurnam Kanth at Punjab Engineering College. Its main points are: Frequency, Domain, Matrix, Time, Supernode, Mesh, Node, Constraint, Equation, Current, Source
Typology: Exercises
1 / 14
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Chapter 10, Problem 1.
Determine i in the circuit of Fig. 10.50.
Figure 10.
For Prob. 10.1.
Chapter 10, Solution 1.
We first determine the input impedance.
F j
1 1 1 1 1 1.0101 0.1 1.015 5. 10 0.1 1
o Zin j j j
− ⎛ ⎞ = + (^) ⎜ + + (^) ⎟ = − = < − ⎝ − ⎠
o o o
( ) 1.9704 cos(10 5.653 ) A
o i t = t + = 1.9704cos(10t+5.65˚) A
Figure 10.
For Prob. 10.2.
Chapter 10, Solution 2.
Consider the circuit shown below.
2
V o
–j5 j
4 ∠ 0
o V-
At the main node,
4 40 (10 ) 2 5 4
o o o o
V j j j
o Vo j
Figure 10.
For Prob. 10.4.
Chapter 10, Solution 4.
3
3 6
F j j C (^) j x x
Consider the circuit as shown below.
I 1 2000 V 1 -j
o V j
At node 1,
− −
j j
But
1 1
1 1 1 1 1
V j x j V j V V
1 1
i t 1 ( ) =0 A
Find io in the circuit of Fig. 10.54.
Figure 10.
For Prob. 10.5.
Figure 10.
For Prob. 10.6.
Chapter 10, Solution 6.
Let Vo be the voltage across the current source. Using nodal analysis we get:
20 j 10
where (^) x Vo 20 j 10
Combining these we get:
0 ( 1 j 0. 5 3 )V 60 j 30 20 j 10
20 j 10
o
o o o = → + − = +
2 j 0. 5
orV 2 j 0. 5
60 j 30 V (^) o x 29.11 ∠ –166˚ V.
Figure 10.
For Prob. 10.7.
Chapter 10, Solution 7.
At the main node,
j
40 j 20
196 j 3 40 j 20
91 j 31. 058
j 30
40 j 20
120 15 V o
o
04 j 0. 0233
1885 j 4. 7805 V
o = ∠ −
j 100
1 1 1 2 1
o −
−
or 5. 7955 + j 1. 5529 =(− 0. 025 + j 0. 01 ) V 1 − 0. 025 V 2 (1)
At node 2,
1 2
2 1
1 2 0 3 V ( 1 j 2 )V j 20
From (1) and (2),
or AV B 0
( 5. 7955 j 1. 5529 )
3 ( 1 j 2 )
( 0. 025 j 0. 01 ) 0. 025
2
⎟
Using MATLAB,
V = inv(A)*B
leads to V 1 (^) =− 70. 63 − j 127. 23 , V 2 =− 110. 3 + j 161. 09
1 2 o o 7.^27682.^17 40
Thus,
i (t) 7. 276 cos( 200 t 82. 17 )A
o o = −
Use nodal analysis to find vo in the circuit of Fig. 10.58.
Figure 10.
For Prob. 10.9.
Chapter 10, Problem 10.
Figure 10.
For Prob. 10.10.
Chapter 10, Solution 10.
50 mH j L j 2000 x 50 x 10 j 100 , 2000
3 ⎯⎯→ ω = = ω=
−
j 250 j 2000 x 2 x 10
j C
6
ω
μ ⎯⎯→ −
Consider the frequency-domain equivalent circuit below.
V 1 -j250 V 2
o
2k Ω j100 0.1V 1 4k Ω
At node 1,
1 2
1 1 1 2 36 ( 0. 0005 j 0. 006 )V j 0. 004 V j 250
j 100
At node 2,
1 2
2 1
1 2 0 ( 0. 1 j 0. 004 )V ( 0. 00025 j 0. 004 )V 4000
j 250
Solving (1) and (2) gives
o Vo =V 2 =− 535. 6 +j 893. 5 = 8951. 1 ∠ 93. 43
vo (t) = 8.951 sin(2000t +93.
o ) kV
Chapter 10, Problem 11.
Apply nodal analysis to the circuit in Fig. 10.60 and determine I (^) o.
Figure 10.
For Prob. 10.11.
Chapter 10, Solution 11.
Consider the circuit as shown below.
I o –j5 Ω
j8 Ω
4 ∠ 0
o V 2 I o
