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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: Circuit, Source, Current, Input, Voltage, Y, Transform, Equivalent, Nodal, Analysis, Resistors
Typology: Exercises
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Calculate the current io in the circuit of Fig. 4.69. What does this current become when
the input voltage is raised to 10 V?
Figure 4.
Chapter 4, Solution 1.
i =
i 2
i (^) o 0.1A
Since the resistance remains the same we get i = 10/5 = 2A which leads to
io = (1/2)i = (1/2)2 = 1A.
Figure 4.
Chapter 4, Solution 2.
6 ( 4 + 2 )= 3 Ω, i 1 =i 2 =
i 2
i (^) o = 1 = v (^) o = 2 io = 0.5V
If i (^) s = 1μA, then vo = 0.5 μ V
Use linearity to determine i (^) o in the circuit in Fig. 4.72.
Figure 4.
Chapter 4, Solution 4.
If Io = 1, the voltage across the 6Ω resistor is 6V so that the current through the 3Ω
resistor is 2A.
v i
o 3 6 = 2 Ω, v (^) o= 3(4) = 12V, 1 = =
Hence Is = 3 + 3 = 6A
If Is = 6A Io = 1
Is = 9A Io = 9/6 = 1.5A
For the circuit in Fig. 4.73, assume vo = 1 V, and use linearity to find the actual value
of v (^) o.
Figure 4.
Chapter 4, Solution 5.
If v (^) o= 1V, =
v 3
V (^) s 2 ⎟+ 1 = ⎠
If v (^) s = v (^) o= 1
x 15 = 10
Then vs = 15 vo = 4.5V
Use linearity and the assumption that Vo = 1V to find the actual value of V (^) oin Fig. 4.75.
.
V o
Figure 4.75 For Prob. 4.7.
Chapter 4, Solution 7.
If Vo = 1V, then the current through the 2-Ω and 4-Ω resistors is ½ = 0.5. The voltage
across the 3-Ω resistor is ½ (4 + 2) = 3 V. The total current through the 1-Ω resistor is
0.5 +3/3 = 1.5 A. Hence the source voltage
v (^) s = 1 1.5 x + 3 =4.5 V
If v (^) s = 4.5 ⎯⎯→ 1 V
Then v s = ⎯⎯→ x = = 888.9 mV.
Using superposition, find Vo in the circuit of Fig. 4.76.
Vo
Figure 4.76 For Prob. 4.8.
