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Solutions to FE Problems
Chapter 3
3FE-1 Find Vo in the circuit in Fig. 3PFE-1.
The correct answer is a.
A
6
2
1
I1
I

12V
2
Vo

KCL at node A: I1  I  I 2
I  I1  I 2
KVL around the left loop:
12  2I1  1( I )  2I
12  2I1  1( I1  I 2 )  2( I1  I 2 )
12  5I1  3I 2
KVL around the right loop:
6  6I 2  2I  1( I )
6  6I 2  3( I1  I 2 )
6  9I 2  3I1
Two equations and two unknowns:
12  5I1  3I 2
6  3I1  9I 2
Therefore, I1  3.5 A and I 2  1.833A
V0  2I  2( I1  I 2 )  2(3.5  1.833)
V0  3.33V
I2
6V
3FE-2 Determine the power dissipated in the 6-ohm resistor in the network in
Fig. 3PFE-2.
The correct answer is d.
V1
V2
4
I1
12V
V1  12V
KCL at node 2:
V2  V1 V2 V2


 2I1
4
6 12
V  V2
I1  1
4
V2  V1 V2 V2
 V  V2 


 2 1

4
6 12
 4 
3V2  3V1  2V2  V2  6V1  6V2
12V2  9V1
12V2  9(12)
V2  9V
V 2 92
P6  2 
 13.5W
6
6
6
text
2I 1
12
3FE-3 Find the current Ix in the 4-ohm resistor in the circuit in Fig. 3PFE-3.
The correct answer is b.
12V
3
I1
6
2A

Vx
I2
4

I2  I x  I3
I x  I2  I3
I1  2  I 2
I 2  I1  2
12  6I1  4I x
12  6 I1  4( I 2  I 3 )
12  6I1  4I 2  4I 3
KVL around the right loop:
2Vx  4( I x )  3I 3  0
 2Vx  4( I 2  I 3 )  3I 3
 2(4I x )  4( I 3  I 2 )  3I 3
 8( I 2  I 3 )  4I 3  4I 2  3I 3
8I 2  8I 3  4 I 3  4 I 2  3I 3  0
4I 2  I 3  0
I 3  4I 2
12  6I1  4I 2  4(4I 2 )
12  6I1  12I 2
Two equations and two unknowns:
 I1  I 2  2
6I1  12I 2  12
Therefore, I1  6 A and I 2  4 A .
I 3  4(4)  16 A
I x  4  16
+
-
Ix
Vx  4I x
I3
2V x
I x  12 A
3FE-4 Determine the voltage Vo in the circuit in Fig. 3PFE-4.
The correct answer is a.
12V
4
I1
4
2
I3
I2
Ix
KCL: I1  I x  I 2
I x  I1  I 2
KVL around the left loop:
12  4I1  2I x
12  4I1  2( I1  I 2 )
12  6I1  2I 2
KVL around the outer loop:
12  4 I1  4 I 2  4 I 3
I 2  2I x  I 3
2I x  I 2  I 3
2( I1  I 2 )  I 2  I 3
I 3  2I1  3I 2
12  4I1  4I 2  4(2I1  3I 2 )
12  4I1  16I 2
Two equations and two unknowns:
12  6I1  2I 2
12  4I1  16I 2
Therefore, I1  2.45 A and I 2  1.36 A .
I 3  2(2.45)  3(1.36)
I 3  0.82 A
V o 4I 3  4(0.82)
V o 3.28V
text
2I x

4
Vo

3FE-5 What is the voltage V1 in the circuit in Fig. 3PFE-5?
The correct answer is c.
I1
V1
1
V2
I2
V3
2
10V
3
15V
8A
4A
KCL at the supernode: 4  I1  8  I 2
I1  I 2  4
V1 V2  V3

 4
4
2
V1  2V2  2V3  16
V3  15V
V1  2V2  2(15)  16
V1  2V2  14
V2 V1  10
Two equations and two unknowns:
 V1  V2  10
V1  2V2  14
Therefore, V1  2V and V2  8V .