ECE 35
Homework #3 Solution (Spring 2015, Taur)
All homework problems come from the textbook, “Introduction to Electric Circuits”, by
Svoboda & Dorf, 9
thEdition. Question number in the 8
thedition is listed for reference.
Question Number
Svoboda & Dorf, 8th Edition
Svoboda & Dorf, 9th Edition
1
P 4.3-3
P 4.3-3
2
P 4.3-7
P 4.3-7
3
P 4.3-9
P 4.3-9
4
P 4.3-12
P 4.3-12
5
P 4.4-5
P 4.4-5
6
P 4.4-11
P 4.4-11
7
P 4.4-15
P 4.4-14
8
P 4.5-2
P 4.5-2
9
P 4.5-6
P 4.5-6
10
P 4.6-5
P 4.6-5
11
P 4.6-8
P 4.6-8
12
P 4.7-7
P 4.7-7
13
P 4.7-13
P 4.7-12
14
P 4.11-2
P 4.11-2
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, and express the resistor currents in terms of the node voltages.
Identify the super node corresponding to the 24 V source
Apply KCL to the super node to get
a
a aa a
12
24
24
0.6
196
6
32 V
10
40
40
v
v
v
v
v
The 12 V source supplies
12
12
a24
12
12
32 24
4.8 W
10
10
v
The 24 V source supplies
a32
24
0.6
24
0.6
4.8 W
40
40
v
P 4.3-7
Determine the values of the node voltages, v1
and
v2, in Figure P 4.3-7. Determine the values of the
currents ia and ib.
Figure P 4.3-7
Solution:
Apply KCL at nodes 1 and 2 to get
1 1 1 2
1 2
10
23 3 150 1000 3000 5000
v v v v
v v
2 1 3 3
1 3
10
-4 19 50 4000 5000 2000
v v v v
v v
Solving, e.g. using MATLAB, gives
1
1 1
2
23
3
150
7.06 V and
4.12 V
4 19
50
v
v
v
v
Then b 1 2
7.06 4.12
0.588 mA
5000 5000
v v
i
Apply KCL at the top node to get
1 2
a
10 10 7.06 10 4.12 10
4.41 mA
1000 4000 1000 4000
v v
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
2 1 5 and 4 15
v v v
Apply KCL to the supernode corresponding to the 5 V source to get
1 3 2
1 2 3
15
1.25 0 80 5 2 5
8 20
v v v
v v v
Apply KCL at node 3 to get
1 3 3 3
1 3
15
15 28 150
8 40 12
v v v v
v v
Solving, e.g. using MATLAB, gives
1 1
2 2
3 3
1 1 0 5 22.4
5 2 5 80 27.4
15 0 28 150 17.4
v v
v v
v v
So the node voltages are:
1 22.4 V, 2 27.4 V, 3 17.4 V, and 4 15
P 4.3-12
Determine the values of the node voltages of the circuit shown in Figure P 4.3-12.
Figure P 4.3-12
Solution:
Express the voltage source voltages in terms of the node voltages:
2 1 8 and 3 1 12
v v v v
Apply KVL to the supernode to get
2 1 3
2 1 3
0 2 5 4 0
10 4 5
v v v
v v v
so
1
1
1
164
2 8 5 4 12 0 V
11
v v v v
The node voltages are
1
2
3
5.818 V
2.182 V
6.182 V
v v v
P 4.4-5
Determine the value of the current
ix in the
circuit of Figure P 4.4-5.
Answer: i
x = 2.4 AFigure P 4.4-5
Solution:
First, express the controlling current of the CCVS in terms of the node voltages: 2
2
x
v i
Next, express the controlled voltage in terms of the node voltages:
2
2 2
24
12 3 3 V
2 5
x
v
v i v
P 4.4-11
Determine the power supplied by the
dependent source in the circuit shown in Figure P
4.4-11.
Figure P 4.4-11
Solution:
This circuit contains two ungrounded voltage sources, both incident to node x. In such a circuit it is necessary to merge the supernodes corresponding to the two ungrounded voltage sources into a single supernode. That single supernode separates the two voltage sources and their nodes from the rest of the circuit. It consists of the two resistors and the current source. Apply KCL to this supernode to get
x x
x
20
4 0 10 V
2 10
v v
v
.
The power supplied by the dependent source is
0.1vx
30
30 W.
P 4.4-14
The voltages
v1, v2,
v3, and v4 are
the node voltages corresponding to nodes 1, 2,
3, and 4 in Figure P 4.4-14. Determine the
values of these node voltages.
Solution:
Express the controlling voltage and current of the dependent sources in terms of the node voltages:
a 4
v v and b 3 4 2
v
v
i
R
Express the voltage source voltages in terms of the node voltages:
1 s
v V and v2v3 Ava Av4
Apply KCL to the supernode corresponding to the dependent voltage source
2 1 3 4
s 2 1 2 2 1 3 1 4 1 2 s
1 2
v
v
v
v
I
R v
R v
R v
R v
R R I
R
R
Apply KCL at node 4:
3 4 3 4 4 2
3 4
2 2 3 3
1 1 0
v v v v v R
B B v B v
R R R R
Organizing these equations into matrix form:
1 s
2
2 2 1 1
3 1 2 s
2 4 3
1 0 0 0
0 1 1
0
0 0 1 1 0
v V
A
v
R R R R
v R R I
R v B B R
With the given values:
1 1
2 2
3 3
4 4
1
0
0
0
25
25
0
1
1
5
0
44.4
20
20 10
10
400
8.4
0
0
4
4.667
0
7.2
P 4.5-2
The values of the mesh currents in the
circuit shown in Figure P 4.5-2 are
i1 = 2 A, i2 = 3 A, and i3 = 4 A.
Determine the values of the resistance R and of
the voltages v1 and v2 of the voltage sources.
Answers: R
= 12 Ω, v1 = –4 V, and v2 = –28 V
Figure P 4.5-2
Solution:
The mesh equations are:
Top mesh: 4 (2 3) R(2) 10 (2 4)0 so R = 12 .
Bottom, right mesh: 8 (4 3) 10 (4 2) v2 0
so v2 = 28 V.
Bottom left mesh v1 4 (3 2) 8 (3 4) 0
P 4.5-6
Simplify the circuit shown in
Figure P 4.5-6 by replacing series and
parallel resistors by equivalent resistors.
Next, analyze the simplified circuit by
writing and solving mesh equations.
(a)
Determine the power supplied by
each source.
(b)
Determine the power absorbed by the
30-Ω resistor.
Figure P 4.5-6Solution:
Replace series and parallel resistors with equivalent resistors:
60 300 50
,
40
60
100
and
100
30
80
560
200
so the simplified circuit isThe mesh equations are
1 1 2
200i 50 i i 120
2 1 2
100i 8 50 i i 0 or
1 1
2 2
250
50
12
0.04
50
150
8
0.04
i
i
i
i
The power supplied by the 12 V source is
12
i
1
12 0.04
0.48 W
. The power supplied by the 8 Vsource is
8
i
2
8
0.04
0.32 W
. The power absorbed by the 30 resistor is
2 21 30 0.04 30 0.048 W
P 4.6-5
Determine the value of the voltage measured by the voltmeter in Figure P 4.6-5.
Answer:
8 V
Figure P 4.6-5
Solution:
Label the mesh currents:
Express the current source current in terms of the mesh currents:
3 1 2 1 3 2
i i i i
Supermesh: 6i13i35
i2i3
8 0 6i15i28i38Lower, left mesh: 12 8 5
i2i3
0 5i2 4 5i3Eliminating i1 and i2 from the supermesh equation:
3
3
3 36 i 2 4 5i 8i 8 9i 24
The voltage measured by the meter is: 3
24
3 3 8 V
9
i
P 4.6-8
Determine values of the mesh
currents, i1, i2, and i3, in the circuit shown in
Figure P 4.6-8.
Figure P 4.6-8
Solution:
Use units of V, mA and k
. Express the currents to the supermesh to get
1 3 2
i i
Apply KVL to the supermesh to get
3 3
3
1 2
1 2 34 i i 1 i 3 1 i i 0 i 5i 5i 3
Apply KVL to mesh 2 to get
2 2 3 2 1 1 2 3
2i 4 i i 1 i i 0 1 i 7i 4i 0
Solving, e.g. using MATLAB, gives
1 1
2 2
3 3
1 0 1 2 3
1 5 5 3 1
1 7 4 0 1
i i
i i
i i
P 4.7-7
The currents
i1,
i2 and
i3 are the
mesh currents of the circuit shown in Figure
P 4.7-7. Determine the values of i1, i2, and i3.
Figure P 4.7-7
Solution:
Express va and ib, the controlling voltage and current of the dependent sources, in terms of the mesh
currents
a 5 2 3 and b 2
v i i i i
Next express 20 ib and 3 va, the controlled voltages of the dependent sources, in terms of the mesh
currents
b 2
20 i 20 i
and
3 va 15
i2i3
Apply KVL to the meshes
2 3
2
115 i i 20 i 10 i 0
20 i2
5 i2 i3
20 i2 0
2 3
2 3
10 5 i i 15 i i 0These equations can be written in matrix form
1
2
3
10 35 15 0
0 45 5 0
0 10 10 10
i i i
Solving, e.g. using MATLAB, gives
1 1.25 A, 2 0.125 A, and 3 1.125 A
P 4.7-12
The currents i1, i2, and i3 are the
mesh currents corresponding to meshes 1,
2, and 3 in Figure P 4.7-12. Determine the
values of these mesh currents.
Figure P 4.7-12
Solution:
Express the controlling voltage and current of the dependent sources in terms of the mesh currents:
a 3 1 2
v R i i and ib i3 i2
Express the current source currents in terms of the mesh currents:
2 s
i I and i1 i3 B ib B i
3i2
Consequently
1
1
3 si
B
i
B I
Apply KVL to the supermesh corresponding to the dependent current sourceR i1 3A R i3
1i2
R2
i3i2
R i3
1i2
Vs 0OR
A1
R i3 1
R2
A1
R i3
2 R1R2
i3VsOrganizing these equations into matrix form:
1 s
2 s
3 s
3 2 3 1 2
0 1 0
1 0 1
1 1
i I
B i B I
i V
A R R A R R R
With the given values:
1 1
2 2
3 3
0 1 0 2 0.8276
1 0 4 6 2 A
60 80 50 25 1.7069
P 4.11-2
An old lab report asserts that the
node voltages of the circuit of Figure P
4.10-2 are
va = 4 V, vb = 20 V, and vc = 12 V.
Are these correct?
Solution:
Apply KCL at node a:
2 0
4 2
20 4 4
2 4 0
4 2
b a a
v v v
