P4.2-2, 7
P4.3-2, 3, 6, 9
P4.4-2, 5, 8, 18
P4.5-2, 4, 5
P4.6-2, 4, 8
P4.7-2, 4, 9, 15
P4.8-2
P 4.2-2 Determine the node voltages for the circuit of Figure
P 4.2-2.
Answer: v
1
= 2 V, v
2
= 30 V, and v
3
= 24 V
Figure P 4.2-2
Solution:
KCL at node 1:
KCL at node 2: v v
−
1
20
KCL at node 3: v
−
20
1 2 v
2
10
+ v
− v
2 3 v
5
1
1 v
− v
2 3
10 v
0
3
15
⇒
⇒ −
5
⇒ − v v
3
1
1 v
−
+
2
3 v
+ v
2
5
2
= − v
−
3
2
20
= v
3
30
=
40
Solving gives v
1
= 2 V, v
2
= 30 V and v
3
= 24 V.
P 4.2-7 The node voltages in the circuit shown in Figure P 4.2-7 are v a
= 7 V and v b
= 10 V. Determine values of the current source current, i s
, and the resistance, R .
Figure P 4.2-7

Solution
Apply KCL at node a to get
2
= v
R a + v
4 a + v a
−
2 v b
7
R
7 7 10
4 2
7
R
1
4
⇒
R 4
Apply KCL at node b to get i s
+ v a
−
2 v b = v b + v
8 8 b i s
7 10
2
=
10
+
10
8 8
⇒ i s
=
4 A
P 4.3-2 The voltages v a
, v b
, v c
, and v d
in Figure P 4.3-2 are the node voltages corresponding to nodes a, b, c, and d. The current i is the current in a short circuit connected between nodes b and c. Determine the values of v a
, v b
, v c
, and v d
and of i .
Answer: v a
= –12 V, v b
= v c
= 4 V, v d
= –4 V, i = 2 mA
Figure P 4.3-2
Solution:
Express the branch voltage of each voltage source in terms of its node voltages to get: v a
= −
12 V, v b
= v c
= v d
+
8
KCL at node b : v b
− v a
=
0.002
+ i
⇒ v b
(
12
)
=
0.002
+ i
⇒ v b
+
12 8 4000 i
4000 4000
KCL at the supernode corresponding to the 8 V source:

0.001
= v d
4000
+ i
⇒
4
= v d
+
4000 i so v b
4 4 v d
⇒ ( v d
8
)
Consequently v b
= v c
= v d
8 4 V and i
=
4
− v d
4000
4 4 v d
=
2 mA
⇒ v d
= −
4 V
Figure P4.3-3
P4.3-3.
Determine the values of the power supplied by each of the sources in the circuit shown in Figure
P4.3-3.
Solution: First, label the node voltages. Next, express the resistor currents in terms of the node voltages.
Identify the supernode corresponding to the 24 V source

Apply KCL to the supernode to get
12
−
( v a
−
24
)
+
0.6
=
10
The 12 V source supplies v a
−
24
+
40 v a
40
⇒
196
=
6 v a
⇒
12


12
−
( v a
−
24
)
10

 =
12


12
− (
32 24
)
10


=
4.8 W v a
=
32 V
The 24 V source supplies
The current source supplies
24

−
0.6
+ v a
40


=
24 
−
0.6
+
32
40
0.6
v a
=
=
4.8 W
( ) =
19.2 W
P 4.3-6 The voltmeter in the circuit of Figure P
4.3-6 measures a node voltage. The value of that node voltage depends on the value of the resistance
R .
(a) Determine the value of the resistance R that will cause the voltage measured by the voltmeter to be 4 V.
(b) Determine the voltage measured by the voltmeter when R
= 1.2 kΩ = 1200 Ω.
Answer:
(a) 6 kΩ (b) 2 V
Figure P 4.3-6

Solution:
Label the voltage measured by the meter. Notice that this is a node voltage.
Write a node equation at the node at which the node voltage is measured.
−
12
− v m
6000
+ v
R m +
0.002
+ v m
−
3000
8
That is
3
+
6000
R v R m
= ⇒ =
16 v
6000 m
−
3
(a) The voltage measured by the meter will be 4 volts when R = 6 k
Ω
.
(b) The voltage measured by the meter will be 2 volts when R = 1.2 k
Ω
.
P 4.3-9 Determine the values of the node voltages of the circuit shown in Figure P 4.3-9.
Figure P 4.3-9
Solution:
Express the voltage source voltages as functions of the node voltages to get v
2
− =
1
5 and v
4
=
15
Apply KCL to the supernode corresponding to the 5 V source to get

1.25
= v
1
− v
3 + v
8
2
−
20
15
Apply KCL at node 3 to get
=
0
⇒
80
=
5 v
1
+
2 v
2
−
5 v
3 v
1
−
8 v
3 = v
3 + v
3
−
40 12
15
⇒ − v
1
+
28 v
3
=
150
Solving, e.g. using MATLAB, gives




−
5
1 1
2
−
0
5
−
15 0 28



 v
  v
 
=




5
80
150


⇒ v
  v
=

22.4


 27.4


 17.4

So the node voltages are: v
1
=
22.4 V, v
2
=
27.4 V, v
3
=
17.4 V, and v
4
=
15
P 4.4-2 Find i b
for the circuit shown in Figure P 4.4-2.
Answer: i b
= –12 mA
Solution:
Figure P 4.4-2
Write and solve a node equation: v a
−
6
+ v a
+ v a
−
4 v
1000 2000 3000 a
=
0
⇒ i b
= v a
−
4
3000 v a
= − v a
=
(checked using LNAP 8/13/02)

P 4.4-5 Determine the value of the current i x
in the circuit of
Figure P 4.4-5.
Answer: i x
= 2.4 A
Figure P 4.4-5
Solution:
First, express the controlling current of the CCVS in terms of the node voltages: i x
= v
2
2
Next, express the controlled voltage in terms of the node voltages:
12
− v
2
=
3 i x
=
3 v
2
2 so i x
= 12/5 A = 2.4 A.
⇒ v
2
=
24
V
5
P 4.4-8 Determine the value of the power supplied by the dependent source in Figure P 4.4-8.
Figure P 4.4-8

Solution:
Label the node voltages.
First, v
2
= 10 V due to the independent voltage source. Next, express the controlling current of the dependent source in terms of the node voltages: i a
= v
3
− v
2 = v
16
3
−
16
10
Now the controlled voltage of the dependent source can be expressed as v
1
− v
3
=
8 i a
=
8

 v
3
−
10
16


⇒ v
1
=
3
2 v
3
−
5
Apply KCL to the supernode corresponding to the dependent source to get v
1
−
4 v
2 + v
1 + v
3
−
12 16 v
2 + v
8
3 =
0
Multiplying by 48 and using v
2
= 10 V gives
16 v
1
+
9 v
3
=
150
Substituting the earlier expression for v
1
16
3
2 v
3
−
5
+
9 v
3
=
150
⇒ v
3
=
6.970 V
Then v
1
= 5.455 V and i a
= -0.1894 A. Applying KCL at node 2 gives v
1 i
10
− v
1 ⇒
12 i b
= − +
12 4
So i
b
=
−
0.6817 A. v
1
( )
Finally, the power supplied by the dependent source is p
=
( ) i b
=
8
( −
0.1894
) ( −
0.6817
) =
1.033 W

P4.4-18
The voltages v
2
, v
3
and v for the circuit shown in Figure P4.4-18 are:
4 v
2
=
16 V, v
3
=
8 V and v
4
=
6 V
Determine the values of the following:
(a) The gain, A , of the VCVS
(b) The resistance R
5
(c) The currents i b
and i c
(d) The power received by resistor R
4
Figure P4.4-18
Solution:
Given the node voltages v
2
=
16 V, v
3
=
8 V and v
4
=
6 V
R
5

 v
3
− v
4
15
A
=
A v v a


= v
4 a =
⇒
=
4
V
V
R
5
= =
45
Ω
, i b
=
12
=
2 A p
=
4
and i c
=
40 16
−
16
12 12
=
0.6667 A v a
2
2
2
= =
15 15
0.2667 W

P 4.5-2 The values of the mesh currents in the circuit shown in Figure P 4.5-2 are i
1
= 2 A, i
2
= 3 A, and i
3
= 4 A.
Determine the values of the resistance R and of the voltages v
1
and v
2
of the voltage sources.
Answers: R
= 12 Ω, v
1
= –4 V, and v
2
= –28 V
Solution:
The mesh equations are:
Top mesh:
− +
R
+ − =
0 so R = 12
Ω
.
Bottom, right mesh: 8 (4 3) 10 (4 2) v
2
=
0 so v
2
=
−
28 V.
Bottom left mesh v
1
4 (3 2) 8 (3 4) 0 so v
1
=
−
4 V.
Figure P 4.5-2
(checked using LNAP 8/14/02)

P 4.5-4 Determine the mesh currents, i a
and i b
, in the circuit shown in
Figure P 4.5-4.
Figure P 4.5-4
Solution:
KVL loop 1:
25 i a
− + i a
+
75 i a
+ + i
− i
= −
−
100( i a
− i b
− +
KVL loop 2:
− i
+ i
= − i b
+ i b
Solving these equations: i a
6.5 mA , b
9.3 mA i a
−
+ + i b
)
= i b
0
=
(checked using LNAP 8/14/02)

P 4.5-5 Find the current i for the circuit of Figure P 4.5-5.
Hint: A short circuit can be treated as a 0-V voltage source.
Figure P 4.5-5
Solution:
P 4.6-2 Find v c
for the circuit shown in
Figure P 4.6-2.
Mesh Equations: i
1
+
2 ( i
1
− i
2
)
+
10
=
0 mesh 2 : 2( i
2
− i
1
)
+
4 ( i
2
− i
3
)
=
0 mesh 3 : 10 4 ( i
3
− i
2
)
+ i
=
0
Solving: i
= i
2 i
5
17
= −
0.294 A
(checked using LNAP 8/14/02)
Answer: v c
= 15 V
Figure P 4.6-2
Solution:
Mesh currents: mesh a: i a
=
−
0.25 A mesh b: i b
=
−
0.4 A
Ohm’s Law: v c
=
100( i a
− i b
) = 100(0.15) =15 V

P 4.6-4 Find v c
for the circuit shown in
Figure P 4.6-4.
Solution:
Express the current source current in terms of the mesh currents: i b
0.02
a
−
Apply KVL to the supermesh:
250 i a
+
100 ( i a
− + =
0
∴ i a
.02 A
= −
20 mA v c
=
100( i a
− = −
V
Figure P 4.6-4
(checked using LNAP 8/14/02)

P 4.6-8 Determine values of the mesh currents, i
1
, i
2
, and i
3
, in the circuit shown in Figure P 4.6-8.
Figure P 4.6-8
Solution: Use units of V, mA and k
Ω
. Express the currents to the supermesh to get i
1 i
3
2
Apply KVL to the supermesh to get
4
( i
3
− i
3
)
+ ( ) i
3
− + ( ) ( i
1
− i
2
)
=
0
⇒ i
1
−
5 i
2
+
5 i
3
=
3
Apply KVL to mesh 2 to get
2 i
2
+
4
( i
2
− i
3
)
+ ( ) ( i
2
− i
1
)
=
0
⇒ ( ) i
1
+
7 i
2
−
4 i
3
=
0
Solving, e.g. using MATLAB, gives




1 0
−
1
1
−
5 5


−
1 7
−
4

 i
  i
 
=
 
⇒ i
  i
1
=
 
(checked: LNAP 6/21/04)

P4.7-2 Determine the values of the power supplied by the voltage source and by the CCCS in the circuit shown Figure P4.7-2
Figure P4.7-2
Solution: First, label the mesh currents, taking advantage of the current sources. Next, express the resistor currents in terms of the mesh currents:
Apply KVL to the left mesh: 4000 i a
+
2000 6
The 2 A voltage source supplies
The CCCS supplies
( ) ( )
2 i a
=
(
( ) (
×
×
3
)(
2 0
⇒ i a
1
8
−
3
)
=
0.25 mW
× −
3
)
2 =
0.125 mA
0.9375 10
−
3 =
0.9375 mW
P 4.7-4 Determine the mesh current i a
for the circuit shown in Figure P 4.7-4.
Answer: i a
= –24 mA
Figure P 4.7-4
Solution: Express the controlling voltage of the dependent source as a function of the mesh current: v b
=
100 (.006
− i a
)
Apply KVL to the right mesh:
−
100 (.006
− i a
+ [ − i a
] + i a
=
0
⇒ i a
= −
24 mA
(checked using LNAP 8/14/02)

P 4.7-9 Determine the value of the resistance R in the circuit shown in Figure P 4.7-9.
Figure P 4.7-9
Solution:
Notice that i b and 0.5 mA are the mesh currents. Apply KCL at the top node of the dependent source to get i b
+
0.5 10
−
3 =
4 i b
⇒ i b
=
1
mA
6
Apply KVL to the supermesh corresponding to the dependent source to get
−
5000 i b
+ (
10000
+
R
)
(
−
3
)
−
25
=
0
−
5000
1
6
×
10
−
3 + (
10000
+
R
)
(
× −
3
)
=
25
R
=
125
6
−
3
=
41.67 k
Ω
(checked: LNAP 6/21/04)
P4.7-15 Determine the values of the mesh currents i
1
and i
2
for the circuit shown in Figure P4.7-15.
Figure P4.7-15
Solution: Expressing the dependent source currents in terms of the mesh currents we get: i
1
=
4 i a
=
4
( i
2
+
1
)
⇒
4 i
1
4 i
2
Apply KVL to mesh 2 to get
2 i
2
+
2
( i
2
1
) ( i
1
− i
2
)
=
0 2 2 i
1
+
6 i
2

Solving these equations using MATLAB we get i
1
=
−
8 A and i
2
=
−
3 A
P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, v s
and i s
, and one output v o
. The output is related to the inputs by the equation v o
= ai s
+ bv s where a and b are constants to be determined. Determine the values a and b by
(a) writing and solving mesh equations and
(b) writing and solving node equations.
Solution:
(a)
Figure P 4.8-2
Apply KVL to meshes 1 and 2: v
32 i
1 v s s
+
96
30 i
2
+
120
( i
(
1
− i s
)
)
=
0 i
2
− i s
=
0
150 i
2
= +
120 i s
− v s v o
= i
2
=
4 i
− v s s
5 150
30 i
2
=
24 i s
−
1 s
5 v
So a = 24 and b = -.02.
(b)
Apply KCL to the supernode corresponding to the voltage source to get

So a = 24 and b = -0.2.
So v a
−
( v s
96
+ v o
)
+ v a
−
32 v o = v s
+ v o + v o
120 30 i s
= v s
+ v o + v o = v s + v o
120 30 120 24
Then v o
=
24 i s
−
1 s
5 v
(checked: LNAP 5/24/04)