Solucionário do livro: Fundamentos de Circuitos Elétricos (Problemas Práticos) [cap. 4] - Sadiku

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Saturday, June 25, 2011 
CHAPTER 4 
 
 
P.P.4.1 
 12  
 
 
 
 
 
 
 
 
By current division, ss2 i6
1i
8124
4i  
 s20 i3
4i8v  
When is = 30 A,  )30(3
4v0 40 V 
When is = 45 A,  )45(3
4v0 60 V 
 
P.P.4.2 
 
 
 
 
 
 
 
 
 
 
 
 
Let vo = 1 volt. Then i = 8
1 and 5.2)812(
8
1v1  
giving vs = 2.5V. 
 
If vs = 40V, then v0 = (40/2.5)(1) = 16 V
i2 
8 
+ 
vo 
 
4 
i1 
iS 
12 
8  
v1 
+ 
vo 
 
5 VS = 30 V 
 
+ 
 
i 
P.P.4.3 Let v0 = v1 + v2, where v1 and v2 are contributions to the 12–V and 5–A 
sources respectively. 
 
 5 3  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
To get v1, consider the circuit in Fig. (a). 
 
 (2 + 3 + 5)i – 12 = 0 or i = 12/(10) = 1.2 A 
 v1 = 2i = 2.4 V 
 
To get v2, consider the circuit in Fig. (b). 
 
Since the resistors are equal (5 = 2 + 3) then the current divides equally and 
i1 = i2 = 5/2 = 2.5 and v2 = 2i2 = 5 V 
 
Thus, 
 v = v1 + v2 = 2.4 + 5 = 7.4 V 
 
 
 
 
 
i 
+ 
v1 
 
12 V 
 
+ 2   
(a) 
5 i2 
+ 
v2 
 
2  
i1 
5 A 
(b) 
3 
 
 
P.P.4.4 Let vx = v1 + v2, where v1 and v2 are due to the 25-V and 5-A sources 
respectively. 
 
 20  v1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
To obtain v1, consider Fig. (a). 
 
 – 0
4
0v
20
25vv1.0 111  or 0.2v1 = 25/20 = 1.25 or v1 = 6.25 V 
 
For v2, consider Fig. (b). 
 
 –5 – 0.1v2 + 0
4
0v
20
0v 22  or 0.2v2 = 5 or v2 = 25 V 
 
vx = v1 + v2 = 31.25 V 
 
 
 
 
 
 
+ 
 0.1v1 4 25 V 
(a) 
5 A 4 
(b) 
20  
0.1v2 
v2 
 
 
P.P.4.5 Let i = i1 + i2 + i3 
 
where i1, i2, and i3 are contributions due to the 8-V, 2-A, and 6-V sources respectively. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
For i1, consider Fig. (a), A5.0826
8i1  
For i2, consider Fig. (b). By current division, 25.0)2(142
2i2  
 
For i3, consider Fig. (c), A375.016
6i3  
Thus, i = i1 + i2 + i3 = 0.5 + 0.25 – 0.375 = 375 mA 
 
P.P.4.6 Combining the 6- and 3- resistors in parallel gives  2
9
3x636 . 
Adding the 1- and 4- resistors in series gives 1 + 4 = 5. Transforming the left 
current source in parallel with the 2- resistor gives the equivalent circuit as shown in 
Fig. (a). 
 
 
 
 
 
6  
8 V 
 
+ 
 
2  8 
i1 
(a) 
2 A 
6 
2 
8  
i2 
(b) 
6 
6 V 
 
+ 
 
2  8 
i3 
(c) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Adding the 10-V and 5-V voltage sources gives a 15-V voltage source. Transforming the 
15-V voltage source in series with the 2- resistor gives the equivalent circuit in Fig. (b). 
Combining the two current sources and the 2- and 5- resistors leads to the circuit in 
Fig. (c). Using circuit division, 
 
 )5.10(
7
7
10
7
10
io 
 = 1.78 A 
 
 
 
10V 
 
+ 
 
2  
7 
 +
(a) 
5  3A 
5 V 
io 
2 7.5A 7 
(b) 
5  3A 
io 
(10/7) 10.5A 7 
(c) 
io 
P.P.4.7 We transform the dependent voltage source as shown in Fig. (a). We combine 
the two current sources in Fig. (a) to obtain Fig. (b). By the current division principle, 
 
  xx i4.0024.015
5i   ix = 7.059 mA 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
P.P.4.8 To find RTh, consider the circuit in Fig. (a). 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

18
4x124)66(R Th 3 
10 24m A 5 
(a) 
ix 
0.4ix 
10 4 – 0.4ix A 5 
(b) 
ix 
6  
4 
(a) 
RTh 
6 
2 A 
6 
4 
(b) 
6  2 A 
+ 
VTh 
 
To find VTh, we use source transformations as shown in Fig. (b) and (c). 
 
 
 
 
 
 
 
 
 
 
 
Using current division in Fig. (c), 
 
  )24(124
4VTh 6 V 
 
  13
6
1R
Vi
Th
Th 1.5 A 
 
P.P.4.9 To find VTh, consider the circuit in Fig. (a). 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
6  6 
24 V 
 
+ 
 
(c) 
4  + 
VTh 
 
6 V 
5  Ix 
4 
 
+ 
 
(a) 
1.5Ix 
i1
i2
i1 i2 
3 
o 
+ 
VTh 
 
b 
a 
1.5Ix 1 V 
 
+ 
 
3 0.5Ix 
5  
(b) 
a 
b 
4 
Ix i 
Ix = i2 
i2 - i1 = 1.5Ix = 1.5i2 i2 = -2i1 (1) 
 
For the supermesh, -6 + 5i1 + 7i2 = 0 (2) 
 
From (1) and (2), i2 = 4/(3)A 
 
VTh = 4i2 = 5.333V 
 
To find RTh, consider the circuit in Fig. (b). Applying KVL around the outer loop, 
 
 I  0I31I5.05 xx  x = -2 
 25.2I
4
1i x  
 
i
1R Th  = 25.2
1 444.4 m 
 
 
P.P.4.10 Since there are no independent sources, VTh = 0 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
To find RTh, consider Fig.(a). Using source transformation, the circuit is transformed to 
that in Fig. (b). Applying KVL, ). 
 
 
15 
(a) 
5  
io 
10  4vx
+ 
+ 
vx 
 
+ 
vo 
 
i 
(b) 
5  
15io 
10  4vx
+ 
+ 
vx 
 
+ 
vo 
 
15 
+ 
– 
 
But vx = -5i. Hence, 30i - 20i + 15io = 0 10i = -15io 
vo = (15i + 15io) = 15(-1.5io + io) = -7.5io 
RTh = vo/(io) = –7.5 It needs to be noted that this negative resistance indicates we 
must have an active source (a dependent source). 
 
P.P.4.11 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
From Fig. (a), RN =  6)33( 3  
 
From Fig. (b), IN =  )45(2
1 4.5A 
 
3  
6 
(a) 
RN 
3 
5A 
3 
(b) 
3  4A 
 
IN 
 
P.P.4.12 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
To get RN consider the circuit in Fig. (a). Applying KVL, 01v2i6 xx  
But vx = 1, 6ix = 3 ix = 0.5 
 15.05.0
2
vii xx  
 
i
1RR ThN 1 
 
To find IN, consider the circuit in Fig. (b). Because the 2 resistor is shorted, vx = 0 and 
the dependent source is inactive. Hence, IN = isc = 10A. 
 
P.P.4.13 We first need to find RTh and VTh. To find RTh, we consider the circuit in 
Fig. (a). 
 
 
 
 
 
 
 
 
 
 
2 
(a) 
6  
2vx
+ 
+ 
vx 
 
+ 
vx 
 
1V 
 
+ 
 ix
i
2 
(b) 
6  10 A
2vx
+ 
+ 
vx 
 
Isc 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Applying KCL at the top node gives 
 
 
2
v
1
vv3
4
v1 ooxo  
 
But vx = -vo. Hence 
 
 
2
v
v4
4
v1 o
o
o  vo = 1/(19) 
 
38
9
4
19
11
4
v1
i o 

 
 RTh = 1/i = 38/(9) = 4.222 
 
To find VTh, consider the circuit in Fig. (b), 
 
 -9 + 2io + io + 3vx = 0 
 
But vx = 2io. Hence, 
 
 9 = 3io + 6io = 9io io = 1A 
 
 VTh = 9 – 2io = 7V 
 
 RL = RTh = 4.222  
 
 maxP  )222.4(4
49
R4
v
L
2
Th 2.901 W 
 
 
2  
4  
1 V 
 
+ 
 
(a) 
1  
3vx 
+ 
 
i 
v0 +  vx 
9 V 
 
+ 
 
io
1  + 
VTh 
 +  3vx 
2 
+  vx 4  
(b) 
 
P.P.4.14 We will use PSpice to find Voc and Isc which then can be used to 
find VTh and Rth. 
 
Clearly Isc = 12 A 
 
 
 
 
Clearly VTh = Ioc = 5.333 volts. RTh = Voc/Isc = 5.333/12 = 444.4 mΩ. 
 
 
 
 
 
P.P.4.15 The schematic is the same as that in Fig. 4.56 except that the 1-k resistor is 
replaced by 2-k resistor. The plot of the power absorbed by RL is shown in the figure 
below. From the plot, it is clear that the maximum power occurs when RL = 2k and it 
is 125 W.P.P.4.16 VTh = 9V,    5.2
8
20)19(
V
R
VvR
L
L
LocTh 
 
 
 
 
 
 
 
 
 
 )9(5.210
10VL 7.2 V 
 
 
 
 
+ 
VL 
 
9 V 
 
+ 
 
2.5 
10 
 
P.P.4.17 R1 = R3 = 1k, R2 = 3.2k 
  22
1
3
x RRR
R
R 3.2 k 
 
P.P.4.18 We first find RTh and VTh. To get RTh, consider the circuit in Fig. (a). 
 
 
100
40x60
50
30x2040603020R Th  
 = 12 + 24 = 36 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
To find VTh, we use Fig. (b). Using voltage division, 
 
 ,6.9)16(
100
60v1  4.6)16(50
20v2  
 
But v0vvv Th21  Th = v1 - v2 = 9.6 - 6.4 = 32V 
 
 4.16.3
2.3
RR
V
I
mTh
Th
G 64mA 
30  
40 60 
20 
+ 
VTh 
+

+

10 V 
v2 
v1 
+

(b) 
a 
b 
30 
40 60  
20  
(a) 
RTh 
b 
a 
	P.P.4.1 
	P.P.4.2
	P.P.4.9 To find VTh, consider the circuit in Fig. (a).
	P.P.4.10 Since there are no independent sources, VTh = 0