Chapter 4, Problem 32.
Use source transformation to find ix in the circuit of Fig. 4.100.
Figure 4.100
Chapter 4, Solution 32.
As shown in Fig. (a), we transform the dependent current source to a voltage source,
5ix
15 Ω 10 Ω
−+
+
60V −
50 Ω 40 Ω
(a)
15 Ω
+
60V −
50 Ω 50 Ω 0.1ix
(b)
ix 15 Ω 25 Ω
+ − 2.5ix
60V −
ix
+
(c)
In Fig. (b), 50||50 = 25 ohms. Applying KVL in Fig. (c),
-60 + 40ix – 2.5ix = 0, or ix = 1.6 A
Chapter 4, Problem 33.
Determine RTh and VTh at terminals 1-2 of each of the circuits of Fig. 4.101.
Figure 4.101
Chapter 4, Solution 33.
(a) RTh = 10||40 = 400/50 = 8 ohms
VTh = (40/(40 + 10))20 = 16 V
(b) RTh = 30||60 = 1800/90 = 20 ohms
2 + (30 – v1)/60 = v1/30, and v1 = VTh
120 + 30 – v1 = 2v1, or v1 = 50 V
VTh = 50 V
Chapter 4, Problem 39.
Obtain the Thevenin equivalent at terminals a-b of the circuit in Fig. 4.106.
1A
10 Ω 16 Ω
c a
10 Ω
8V + 5Ω
_
c b
Figure 4.106 For Prob. 4.39.
Chapter 4, Solution 39.
We obtain RTh using the circuit below.
10 Ω 16
10 Ω 5Ω RTh
20 x5
RTh = 16 + 20 // 5 = 16 + = 20 Ω
25
To find VTh, we use the circuit below.
1A
10 16
V1 V2
+
10 Ω
+
+ V2 5 VTh
8V _
_
_
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 44.
For the circuit in Fig. 4.111, obtain the Thevenin equivalent as seen from terminals
(a) a-b (b) b-c
Figure 4.111
Chapter 4, Solution 44.
(a) For RTh, consider the circuit in Fig. (a).
RTh = 1 + 4||(3 + 2 + 5) = 3.857 ohms
For VTh, consider the circuit in Fig. (b). Applying KVL gives,
10 – 24 + i(3 + 4 + 5 + 2), or i = 1
VTh = 4i = 4 V
3Ω 1Ω a
+
3Ω 1Ω a 4Ω VTh
+
− 24V
b
4Ω RTh +
− 10V
2Ω
b 2Ω
5Ω i 5Ω
(b)
(a)
(b) For RTh, consider the circuit in Fig. (c).
3Ω 1Ω 3Ω 1Ω
4Ω + 4Ω
b 24V − vo b
2Ω
+
RTh 2Ω
5Ω VTh
5Ω 2A
c c
(c) (d)
RTh = 5||(2 + 3 + 4) = 3.214 ohms
To get VTh, consider the circuit in Fig. (d). At the node, KCL gives,
[(24 – vo)/9] + 2 = vo/5, or vo = 15
VTh = vo = 15 V
Chapter 4, Problem 47.
Obtain the Thèvenin and Norton equivalent circuits of the circuit in Fig. 4.114
with respect to terminals a and b.
50 V
Figure 4.114
Chapter 4, Solution 47
Since VTh = Vab = Vx, we apply KCL at the node a and obtain
50 − VTh VTh
= + 2VTh ⎯⎯→ VTh = 250 / 126 = 1.9841 V
12 60
To find RTh, consider the circuit below.
12 Ω Vx a
2Vx
60 Ω
1A
At node a, KCL gives
V V
1 = 2V x + x + x ⎯⎯→ V x = 60 / 126 = 0.4762
60 12
V V
R Th = x = 0.4762Ω, I N = Th = 1.9841 / 0.4762 = 4.167 A
1 R Th
Thus,
VTh = 1.9841 V, Req = RTh = RN = 476.2 mΩ, IN = 4.167 A
Chapter 4, Problem 48.
Determine the Norton equivalent at terminals a-b for the circuit in Fig. 4.115.
4A 8
Figure 4.115
Chapter 4, Solution 48.
To get RTh, consider the circuit in Fig. (a).
10Io 42 Ω
10Io 42 Ω +−
+− Io +
Io + VTh
84 Ω
84 Ω V
4A −
− 1A
(a) (b)
From Fig. (a), Io = 1, 4 – 10 +8 – V = 0, or V = 2
RN = RTh = 2/1 = 2 ohms
To get VTh, consider the circuit in Fig. (b),
Io = 4, VTh = -10Io + 8Io = –8 V
IN = VTh/RTh = -8/2 = -4A
Chapter 4, Problem 51.
Given the circuit in Fig. 4.117, obtain the Norton equivalent as viewed from terminals
(a) a-b (b) c-d
Figure 4.117
Chapter 4, Solution 51.
(a) From the circuit in Fig. (a),
RN = 4||(2 + 6||3) = 4||4 = 2 ohms
RTh VTh
+
6Ω 4Ω 6Ω 4Ω
3Ω 2Ω + 3Ω 6A 2Ω
120V −
(a) (b)
For IN or VTh, consider the circuit in Fig. (b). After some source transformations, the
circuit becomes that shown in Fig. (c).
+ VTh
2Ω 4Ω 2Ω
+ i +
40V − 12V −
(c)
Applying KVL to the circuit in Fig. (c),
-40 + 8i + 12 = 0 which gives i = 7/2
VTh = 4i = 14 therefore IN = VTh/RN = 14/2 = 7 A
(b) To get RN, consider the circuit in Fig. (d).
RN = 2||(4 + 6||3) = 2||6 = 1.5 ohms
6Ω 4Ω 2Ω
i
+
3Ω 2Ω RN +
VTh 12V −
(d) (e)
To get IN, the circuit in Fig. (c) applies except that it needs slight modification as in
Fig. (e).
i = 7/2, VTh = 12 + 2i = 19, IN = VTh/RN = 19/1.5 = 12.667 A
Chapter 4, Problem 53.
Find the Norton equivalent at terminals a-b of the circuit in Fig. 4.119.
Figure 4.119
Chapter 4, Solution 53.
To get RTh, consider the circuit in Fig. (a).
0.25vo 0.25vo
1/2
2Ω a 2Ω a
+ + 1/2 +
6Ω 3Ω vo
1A 2Ω vo vab 1A
− − −
b b
(a) (b)
From Fig. (b),
vo = 2x1 = 2V, -vab + 2x(1/2) +vo = 0
vab = 3V
RN = vab/1 = 3 ohms
To get IN, consider the circuit in Fig. (c).
0.25vo
6Ω 2Ω a
+
+ 3Ω vo
18V Isc = IN
−
−
b
(c)
[(18 – vo)/6] + 0.25vo = (vo/2) + (vo/3) or vo = 4V
But, (vo/2) = 0.25vo + IN, which leads to IN = 1 A
Chapter 4, Problem 59.
Determine the Thevenin and Norton equivalents at terminals a-b of the circuit in Fig.
4.125.
Figure 4.125
Chapter 4, Solution 59.
RTh = (10 + 20)||(50 + 40) 30||90 = 22.5 ohms
To find VTh, consider the circuit below.
i1 i2
10 Ω 20 Ω
+ VTh
8A 50 Ω 40 Ω
i1 = i2 = 8/2 = 4, 10i1 + VTh – 20i2 = 0, or VTh = 20i2 –10i1 = 10i1 = 10x4
VTh = 40V, and IN = VTh/RTh = 40/22.5 = 1.7778 A
Chapter 4, Problem 66.
Find the maximum power that can be delivered to the resistor R in the circuit in Fig.
4.132.
Figure 4.132
Chapter 4, Solution 66.
We first find the Thevenin equivalent at terminals a and b. We find RTh using the circuit
in Fig. (a).
2Ω 10V
− +
3Ω
2Ω a b
+
VTh
3Ω 5Ω
a b
+
−
RTh i −
5Ω 20V +
30V
(a) (b)
RTh = 2||(3 + 5) = 2||8 = 1.6 ohms
By performing source transformation on the given circuit, we obatin the circuit in (b).
We now use this to find VTh.
10i + 30 + 20 + 10 = 0, or i = –6
VTh + 10 + 2i = 0, or VTh = 2 V
p = VTh2/(4RTh) = (2)2/[4(1.6)] = 625 m watts
Chapter 4, Problem 73.
Determine the maximum power that can be delivered to the variable resistor R in
the circuit of Fig. 4.139.
Figure 4.139
Chapter 4, Solution 73
Find the Thevenin’s equivalent circuit across the terminals of R.
10 Ω 25 Ω
RTh
20 Ω 5Ω
RTh = 10 // 20 + 25 // 5 = 325 / 30 = 10.833Ω
10 Ω 25 Ω
+ + VTh -
60 V + +
-
Va Vb
20 Ω 5Ω
- -
20 5
Va = (60) = 40, Vb =(60) = 10
30 30
− Va + VTh + Vb = 0 ⎯
⎯→ VTh = Va − Vb = 40 − 10 = 30 V
2
V 30 2
p max = Th = = 20.77 W
4 RTh 4 x10.833
