ECE 2006, Fall 2011
Homework #03 Solution
Problem source: the textbook (4th Edition)
Chapter 3, Problem 39.
Determine the mesh currents i1 and i2 in the circuit shown in Fig. 3.85.
Figure 3.85
Chapter 3, Solution 39
For mesh 1,
− 10 − 2 I x + 10 I 1 − 6 I 2 = 0
But I x = I 1 − I 2 . Hence,
10 = −2I1 + 2I 2 + 10I1 − 6I 2

→ 5 = 4I1 − 2I 2
For mesh 2,
12 + 8 I 2 − 6 I 1 = 0

→ 6 = 3I 1 − 4 I 2
Solving (1) and (2) leads to
I 1 = 0.8 A, I 2 = -0.9A
Chapter 3, Problem 44.
Use mesh analysis to obtain io in the circuit of Fig. 3.90.
Figure 3.90
(1)
(2)
Chapter 3, Solution 44
+
2
4
1
+6
–
5
3
Loop 1 and 2 form a supermesh. For the supermesh,
6i1 + 4i2 - 5i3 + 12 = 0
(1)
For loop 3,
-i1 – 4i2 + 7i3 + 6 = 0
(2)
Also,
i2 = 3 + i1
(3)
Solving (1) to (3), i1 = -3.067, i3 = -1.3333; io = i1 – i3 = -1.7333 A
Chapter 3, Problem 49.
Find vo and io in the circuit of Fig. 3.94.
Figure 3.94
Chapter 3, Solution 49
3
2
1
+1
2
–
2
1
2
+
v0
–
+
or
v0
–
1
+
–
For the supermesh in figure (a),
3i1 + 2i2 – 3i3 + 16 = 0
(1)
At node 0,
i2 – i1 = 2i0 and i0 = -i1 which leads to i2 = -i1
(2)
For loop 3,
-i1 –2i2 + 6i3 = 0 which leads to 6i3 = -i1
(3)
Solving (1) to (3), i1 = (-32/3)A, i2 = (32/3)A, i3 = (16/9)A
i0 = -i1 = 10.667 A, from fig. (b), v0 = i3-3i1 = (16/9) + 32 = 33.78 V.
Chapter 3, Problem 51.
Apply nodal and mesh analysis separately to find vo in the circuit in Fig. 3.96.
Figure 3.96
Chapter 3, Solution 51
5
8
2
1
+
4
+4
–
–2
+
v0
For loop 1,
i1 = 5A
(1)
For loop 2,
-40 + 7i2 – 2i1 – 4i3 = 0 which leads to 50 = 7i2 – 4i3
(2)
For loop 3,
-20 + 12i3 – 4i2 = 0 which leads to 5 = - i2 + 3 i3
(3)
Solving with (2) and (3),
And,
i2 = 10 A, i3 = 5 A
v0 = 4(i2 – i3) = 4(10 – 5) = 20 V.
Chapter 4, Problem 25.
Obtain vo in the circuit of Fig. 4.93 using source transformation.
Figure 4.93
Chapter 4, Solution 25.
Transforming only the current source gives the circuit below.
−+
–
+
+
−
−
+
+−
Applying KVL to the loop gives,
–(4 + 9 + 5 + 2)i + 12 – 18 – 30 – 30 = 0
20i = –66 which leads to i = –3.3
vo = 2i = –6.6 V
Chapter 4, Problem 27.
Apply source transformation to find vx in the circuit of Fig. 4.95.
Figure 4.95
Chapter 4, Solution 27.
Transforming the voltage sources to current sources gives the circuit in Fig. (a).
10||40 = 8 ohms
Transforming the current sources to voltage sources yields the circuit in Fig. (b).
Applying KVL to the loop,
-40 + (8 + 12 + 20)i + 200 = 0 leads to i = -4
vx 12i = -48 V
+ v
+
−
+ v
i
−
−
+
−
Chapter 4, Problem 39.
Obtain the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.106.
1A
10 Ω
16 Ω
 a
10 Ω
8V
5Ω
+
_
 b
Figure 4.106 For Prob. 4.39.
Chapter 4, Solution 39.
We obtain RTh using the circuit below.
10 Ω
16
5Ω
10 Ω
RTh
20 x5
=
20 Ω
25
To find VTh, we use the circuit below.
RTh =+
16 20 // 5 =+
16
1A
10
16
V1
+
10 Ω
8V
V2
+
+
_
V2
_
5
VTh
_
At node 1,
[(V1–8)/10] – 1 + [(V1–V2)/10] = 0 or 2V1 – V2 = 18
(1)
At node 2,
[(V2–V1)/10] + [(V2–0)/5] + 1 = 0 or –V1 + 3V2 = –10
(2)
Adding 3(1) to (2) gives
5V1 = 44 or V1 = 8.8 V
Using (2) we get 3V2 = 8.8 – 10 = –1.2 or V2 = –400 mV.
Finally,
VTh = V2 + (–1)(16) = –0.4 – 16 = –16.4 V
Chapter 4, Problem 41.
Find the Thèvenin and Norton equivalents at terminals a-b of the circuit shown in
Fig. 4.108.
Figure 4.108
Chapter 4, Solution 41
To find RTh, consider the circuit below
14 Ω
a
6Ω
5Ω
b
RTh = 5 //(14 + 6) = 4Ω = R N
Applying source transformation to the 1-A current source, we obtain the circuit below.
6Ω
- 14V +
14 Ω
VTh
a
+
6V
3A
5Ω
b
At node a,
14 + 6 − VTh
V
= 3 + Th
5
6 + 14
IN =

→
VTh = −8 V
VTh
= (−8) / 4 = −2 A
RTh
Thus,
RTh = R N = 4Ω,
VTh = −8V,
I N = −2 A