ASSIGNMENT VI - SOLUTIONS
Chapter 4, Problem 44.
For the circuit in Fig. 4.105, obtain the Thevenin equivalent as seen from terminals
(a) a-b
(b) b-c
Figure 4.105
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
2Ω
b
(a)
+
−
2Ω
i
VTh
4Ω
24V
RTh
4Ω
5Ω
+
−
b
10V
5Ω
(b)
(b)
For RTh, consider the circuit in Fig. (c).
3Ω
1Ω
4Ω
3Ω
b
24V
2Ω
RTh
5Ω
1Ω
4Ω
vo
+
−
b
+
2Ω
5Ω
2A
c
(c)
VTh
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 51.
Given the circuit in Fig. 4.110, obtain the Norton equivalent as viewed from terminals
(a) a-b
(b) c-d
Figure 4.110
Chapter 4, Solution 51.
(a)
From the circuit in Fig. (a),
RN = 4||(2 + 6||3) = 4||4 = 2 ohms
RTh
4Ω
6Ω
VTh
+
6Ω
3Ω
2Ω
120V
4Ω
+
−
3Ω
6A
2Ω
(b)
(a)
For IN or VTh, consider the circuit in Fig. (b). After some source transformations, the
circuit becomes that shown in Fig. (c).
+ VTh
2Ω
40V
+
−
4Ω
i
2Ω
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
(d)
VTh
12V
+
−
(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 62.
Find the Thevenin equivalent of the circuit in Fig. 4.120.
Figure 4.120
Chapter 4, Solution 62.
Since there are no independent sources, VTh = 0 V
To obtain RTh, consider the circuit below.
0.1io
+
vo
−
1
40 Ω
ix
2
10 Ω
v1
io
VS
+
−
20 Ω
2vo
+ −
At node 2,
ix + 0.1io = (1 – v1)/10, or 10ix + io = 1 – v1
(1)
(v1/20) + 0.1io = [(2vo – v1)/40] + [(1 – v1)/10]
(2)
At node 1,
But io = (v1/20) and vo = 1 – v1, then (2) becomes,
1.1v1/20 = [(2 – 3v1)/40] + [(1 – v1)/10]
2.2v1 = 2 – 3v1 + 4 – 4v1 = 6 – 7v1
or
v1 = 6/9.2
(3)
From (1) and (3),
10ix + v1/20 = 1 – v1
10ix = 1 – v1 – v1/20 = 1 – (21/20)v1 = 1 – (21/20)(6/9.2)
ix = 31.52 mA, RTh = 1/ix = 31.73 ohms.
Chapter 4, Solution 69.
We need the Thevenin equivalent across the resistor R. To find RTh, consider the circuit
below.
22 kΩ v1
+
10 kΩ
vo
−
40 kΩ
0.003vo
30 kΩ
1mA
Assume that all resistances are in k ohms and all currents are in mA.
10||40 = 8, and 8 + 22 = 30
1 + 3vo = (v1/30) + (v1/30) = (v1/15)
15 + 45vo = v1
But vo = (8/30)v1, hence,
15 + 45x(8v1/30) v1, which leads to v1 = 1.3636
RTh = v1/1 = –1.3636 k ohms
RTh being negative indicates an active circuit and if you now make R equal to 1.3636 k
ohms, then the active circuit will actually try to supply infinite power to the resistor. The
correct answer is therefore:
2
2
VTh
⎛
⎞
⎛V ⎞
pR = ⎜
⎟ 1363.6 = ⎜ Th ⎟ 1363.6 = ∞
⎝ − 1363.6 + 1363.6 ⎠
⎝ 0 ⎠
It may still be instructive to find VTh. Consider the circuit below.
10 kΩ vo 22 kΩ
v1
+
100V
+
−
+
40 kΩ
vo
−
0.003vo
30 kΩ
VTh
−
(100 – vo)/10 = (vo/40) + (vo – v1)/22
(1)
[(vo – v1)/22] + 3vo = (v1/30)
(2)
Solving (1) and (2),
v1 = VTh = -243.6 volts
Chapter 4, Problem 74.
For the bridge circuit shown in Fig. 4.132, find the load RL for maximum power transfer
and the maximum power absorbed by the load.
Figure 4.132
Chapter 4, Solution 74.
When RL is removed and Vs is short-circuited,
RTh = R1||R2 + R3||R4 = [R1 R2/( R1 + R2)] + [R3 R4/( R3 + R4)]
RL = RTh = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)/[( R1 + R2)( R3 + R4)]
When RL is removed and we apply the voltage division principle,
Voc = VTh = vR2 – vR4
= ([R2/(R1 + R2)] – [R4/(R3 + R4)])Vs = {[(R2R3) – (R1R4)]/[(R1 + R2)(R3 + R4)]}Vs
pmax = VTh2/(4RTh)
= {[(R2R3) – (R1R4)]2/[(R1 + R2)(R3 + R4)]2}Vs2[( R1 + R2)( R3 + R4)]/[4(a)]
where a = (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)
pmax =
[(R2R3) – (R1R4)]2Vs2/[4(R1 + R2)(R3 + R4) (R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4)]