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Chapter 3, Solution 1
Using Fig. 3.50, design a problem to help other students to better understandnodal analysis.
R R1 2
Figure 3.50For Prob. 3.1 and Prob. 3.39.
Solution
Given R 1 = 4 k Ω , R 2 = 2 k Ω , and R 3 = 2 k Ω , determine the value of I x usingnodal analysis.
Let the node voltage in the top middle of the circuit be designated as V x .
[(V x –12)/4k] + [(V x –0)/2k] + [(V x –9)/2k] = 0 or (multiply this by 4 k)
(1+2+2)V x = 12+18 = 30 or V x = 30/5 = 6 volts and
Ix = 6/(2k) = 3 mA .
12 V+
I x
+9 V R 3
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Chapter 3, Solution 2
At node 1,
2
vv6
5
v
10
v 2111
60 = - 8v 1 + 5v 2 (1)
At node 2,
2vv
634
v 212
36 = - 2v 1 + 3v 2 (2)
Solving (1) and (2),
v1 = 0 V , v 2 = 12 V
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Chapter 3, Solution 3
Applying KCL to the upper node,
60
v20
30
v
20
v
10
v8 0oo0 = 0 or v 0 = –60 V
i1 =10v 0 –6 A , i2 =
20v 0 –3 A ,
i3 =30v 0 –2 A , i 4 =
60v 0 1 A .
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Chapter 3, Solution 4
At node 1,
–6 – 3 + v 1 /(20) + v 1/(10) = 0 or v 1 = 9(200/30) = 60 V
At node 2,3 – 2 + v 2 /(10) + v 2 /(5) = 0 or v 2 = –1(1600/80) = –20 V
i1 = v 1 /(20) = 3 A , i 2 = v 1 /(10) = 6 A ,i3 = v 2 /(40) = –500 mA , i 4 = v 2 /(40) = –500 mA .
i1
10 40 2 A20
3Av v1 2
i i
6 A
2 3 i4
40
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Chapter 3, Solution 6.
Solve for V using nodal analysis.1
10
10
Figure 3.55For Prob. 3.6.
Step 1. The first thing to do is to select a reference node and to identify all the
unknown nodes. We select the bottom of the circuit as the reference node. The onlyunknown node is the one connecting all the resistors together and we will call that nodeV 1 . The other two nodes are at the top of each source. Relative to the reference, the oneat the top of the 10-volt source is –10 V. At the top of the 20-volt source is +20 V.
Step 2. Setup the nodal equation (there is only one since there is only oneunknown).
Step 3. Simplify and solve.
orV 1 = –2 V .
The answer can be checked by calculating all the currents and see if they add up to zero.The top two currents on the left flow right to left and are 0.8 A and 1.6 A respectively.The current flowing up through the 10-ohm resistor is 0.2 A. The current flowing right toleft through the 10-ohm resistor is 2.2 A. Summing all the currents flowing out of the
node, V 1 , we get, +0.8+1.6 –0.2–2.2 = 0. The answer checks.
+
V 1
–
1010 V–+
+5 20 V
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Chapter 3, Solution 7
0V2.020
0V10
0V2 x
xx
0.35V x = 2 or V x = 5.714 V .
Substituting into the original equation for a check we get,
0.5714 + 0.2857 + 1.1428 = 1.9999 checks!
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Chapter 3, Solution 820i v i6 1 1 3
i1 + i 2 + i 3 = 0 020
v5v20
0)60v(10v 0111
But10
v5
2v so that 2v 1 + v 1 – 60 + v 1 – 2v 1 = 0
or v 1 = 60/2 = 30 V, therefore v o = 2v 1 /5 = 12 V .
5V 0
+V 0
– – 20
i2
4
60V
–
+
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Chapter 3, Solution 9
Let V 1 be the unknown node voltage to the right of the 250- Ω resistor. Let the groundreference be placed at the bottom of the 50- Ω resistor. This leads to the following nodalequation:
0I300V5V1572V3
getwegsimplifyin
0150
0I60V50
0V250
24V
b111
b111
But250
V24I 1 b . Substituting this into the nodal equation leads to
or V08.100V2.24 1 1 = 4.165 V.
Thus, I b = (24 – 4.165)/250 = 79.34 mA .
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Chapter 3, Solution 10
v 1 v 2 v 3
At node 1. [(v 1 –0)/8] + [(v 1 –v 3)/1] + 4 = 0At node 2. –4 + [(v 2 –0)/2] + 2i o = 0
At node 3. –2i o + [(v 3 –0)/4] + [(v 3 –v 1)/1] = 0
Finally, we need a constraint equation, i o = v 1 /8
This produces,1.125v 1 – v 3 = 4 (1)
0.25v1 + 0.5v
2 = 4 (2)
–1.25v 1 + 1.25v 3 = 0 or v 1 = v 3 (3)
Substituting (3) into (1) we get (1.125–1)v 1 = 4 or v 1 = 4/0.125 = 32 volts. This leads to,
io = 32/8 = 4 amps .
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Chapter 3, Solution 11
Find V o and the power absorbed by all the resistors in the circuit of Fig. 3.60.
12 6
Figure 3.60For Prob. 3.11.
Solution
At the top node, KCL produces 06
)24(V12
0V12
60V ooo
(1/3)V o = 1 or V o = 3 V .
P 12 = (3–60) 2/1 = 293.9 W (this is for the 12 Ω resistor in series with the 60 Vsource)
P 12Ω = (V o)2/12 = 9/12 = 750 mW (this is for the 12 Ω resistor connecting V o toground)
P 4Ω = (3–(–24)) 2/6 = 121.5 W .
+ _60 V
–+ 24 V
V o
12
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Chapter 3, Solution 12
There are two unknown nodes, as shown in the circuit below.
20
At node 1,
010
VV20
0V20
40V o111 or
(0.05+0.05+.1)V 1 – 0.1V o = 0.2V 1 – 0.1V o = 2 (1)
At node o,
010
0VI4
10VV o
x1o and I x = V 1/20
–0.1V 1 – 0.2V 1 + 0.2V o = –0.3V 1 + 0.2V o = 0 or (2)
V 1 = (2/3)V o (3)
Substituting (3) into (1),
0.2(2/3)V o – 0.1V o = 0.03333V o = 2 or
V o = 60 V .
10
+ _
V
20 10 40 V
V1 o
I x
4 I x
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Chapter 3, Solution 13
Calculate v 1 and v 2 in the circuit of Fig. 3.62 using nodal analysis.
10 V
15 A
Figure 3.62For Prob. 3.13.
Solution
At node number 2, [((v 2 + 10) – 0)/10] + [(v 2 –0)/4] – 15 = 0 or(0.1+0.25)v 2 = 0.35v 2 = –1+15 = 14 or
v 2 = 40 volts .
Next, I = [(v 2 + 10) – 0]/10 = (40 + 10)/10 = 5 amps and
v 1 = 8x5 = 40 volts .
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Chapter 3, Solution 14
Using nodal analysis, find v o in the circuit of Fig. 3.63.
12.5 A
Figure 3.63For Prob. 3.14.
Solution
At node 1,
[(v 1 –100)/1] + [(v 1 –v o)/2] + 12.5 = 0 or 3v 1 – v o = 200–25 = 175 (1)
(2)
dding 4x(1) to 3x(2) yields,
At node o,
[(v o –v 1)/2] – 12.5 + [(v o –0)/4] + [(v o+50)/8] = 0 or –4v 1 + 7v o = 50
A
+
v o
12.5 A
v 1 v 0
4
21 – 50 V
8
– +100 V
+
v o 4
2
8
1 –50 V
100 V –
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(1) + 3(2) = –4v o + 21v o = 700 + 150 or 17v o = 850 or
v o = 50 V .
hecking, we get v 1 = (175+v o)/3 = 75 V.
At node 1,
75–100)/1] + [(75–50)/2] + 12.5 = –25 + 12.5 + 12.5 = 0!
At node o,
[(50–75)/2] + [(50–0)/4] + [(50+50)/8] – 12.5 = –12.5 + 12.5 + 12.5 – 12.5 = 0!
4
C
[(
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Chapter 3, Solution 155 A
8
Nodes 1 and 2 form a supernode so that v 1 = v 2 + 10 (1)
At the supernode, 2 + 6v 1 + 5v 2 = 3 (v 3 - v 2) 2 + 6v 1 + 8v 2 = 3v 3 (2)
At node 3, 2 + 4 = 3 (v 3 - v 2) v 3 = v 2 + 2 (3)
Substituting (1) and (3) into (2),
2 + 6v 2 + 60 + 8v 2 = 3v 2 + 6 v 2 =1156
v 1 = v 2 + 10 =1154
i0 = 6v i = 29.45 A
P 65 =
61154
GvR v 2
21
21 144.6 W
P 55 =
51156
Gv2
22 129.6 W
P 35 = 12 W 3)2(Gvv 223L
–+40 V
–+20 V
0 v
v 1 1 2
4
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Chapter 3, Solution 16
At the supernode,
2 = v 1 + 2 (v 1 - v 3) + 8(v 2 – v 3) + 4v 2 , which leads to 2 = 3v 1 + 12v 2 - 10v 3 (1)But
v 1 = v 2 + 2v 0 and v 0 = v 2 .
Hencev 1 = 3v 2 (2)v 3 = 13V (3)
Substituting (2) and (3) with (1) gives,
v1 = 18.858 V , v
2 = 6.286 V , v
3 = 13 V
2 A
v 3v 2v
1
8 S
4 S1 S
i0
2 S
+ 13 V –+
v 0 –
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Chapter 3, Solution 17
At node 1,2
vv8v
4v60 2111
120 = 7v 1 - 4v 2 (1)
At node 2, 3i 0 + 02
vv
10
v60 212
But i 0 = .4
v60 1
Hence
0
2vv
10v60
4v603 2121 1020 = 5v 1 + 12v 2 (2)
Solving (1) and (2) gives v 1 = 53.08 V. Hence i 0 =4
v60 1 1.73 A
60 V
–+60 V
4
10
2
8
3i 0
i0
v 1
v 2
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Chapter 3, Solution 18
At node 2, in Fig. (a),2
vv
2
vv 3212
–15 = 0 or –0.5v 1 + v 2 – 0.5v 3 = 15 (1)
At the supernode,8v
4v
2vv
2vv 313212 = 0 and (v 1 /4) – 15 + (v 3/8) = 0 or
2v 1+v 3 = 120 (2)
From Fig. (b), – v 1 – 30 + v 3 = 0 or v 3 = v 1 + 30 (3)
Solving (1) to (3), we obtain,v 1 = 30 V , v 2 = 60 V = v 3
(b)
15A
v 3
84
22
v 2v 1
– +
30 V
+v 1
–
+v 3
–
(a)
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Chapter 3, Solution 19
At node 1,
32112131 4716
48235 V V V
V V V V V (1)
At node 2,
32132221 270
428V V V
V V V V V (2)
At node 3,
32132313 724360
428
123 V V V
V V V V V (3)
From (1) to (3),
B AV
V
V
V
36
0
16
724
271
417
3
2
1
Using MATLAB,
V267.12 V,933.4 V,10
267.12
933.4
10
3211 V V V B AV
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Chapter 3, Solution 20
Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence
040414
321321 V V V
V V V (1)
.V 1 . V 2 2 V 3
4 1 4
Between nodes 1 and 3,12012 1331
V V V V (2)Similarly, between nodes 1 and 2,
(3)iV V 221
But . Combining this with (2) and (3) gives4/3V i
. (4)2/6 12 V V
Solving (1), (2), and (4) leads to
V15V,5.4 V,3 321 V V V
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Chapter 3, Solution 21
4 k
Let v 3 be the voltage between the 2k resistor and the voltage-controlled voltage source.At node 1,
2000vv
4000vv
10x3 31213
12 = 3v 1 - v 2 - 2v 3 (1)
At node 2,
1v
2vv
4vv 23121 3v 1 - 5v 2 - 2v 3 = 0 (2)
Note that v 0 = v 2 . We now apply KVL in Fig. (b)
- v 3 - 3v 2 + v 2 = 0 v 3 = - 2v 2 (3)
From (1) to (3),
v 1 = 1 V , v 2 = 3 V
(b)
+v 3
–
+v 2
–
3v 0
3 mA 1 k
v 32 k
3v 0v v1 2
+v 0
–
(a)
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Chapter 3, Solution 22
At node 1,8
vv3
4v
2v12 0110
24 = 7v 1 - v 2 (1)
At node 2, 3 +1
v5v8
vv 2221
But, v 1 = 12 - v 1
Hence, 24 + v 1 - v 2 = 8 (v 2 + 60 + 5v 1) = 4 V
456 = 41v 1 - 9v 2 (2)
Solving (1) and (2),
v 1 = - 10.91 V , v 2 = - 100.36 V
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Chapter 3, Solution 23
We apply nodal analysis to the circuit shown below.
At node o,
30V25.0V25.104
)VV2(V2
0V1
30V1o
1oooo (1)
At node 1,
48V4V50316
0V4
V)VV2(o1
1o1o (2)
From (1), V 1 = 5V o – 120. Substituting this into (2) yields29V o = 648 or V o = 22.34 V .
1
2
2 V o
+ –
3 A
+ _30 V
4 V Vo 1
+
V 16 o
_
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Chapter 3, Solution 24
Consider the circuit below.
8
_
4V125.0V125.108
VV4
10V
41411 (1)
4V25.0V75.004
VV2
0V4 32
322 (2)
2V75.0V25.0022
0V4
VV32
323 (3)
2V125.1V125.001
0V8
VV2 41
414 (4)
2
2
4
4
V
125.100125.0
075.025.00
025.075.00
125.000125.1
Now we can use MATLAB to solve for the unknown node voltages.
>> Y=[1.125,0,0,-0.125;0,0.75,-0.25,0;0,-0.25,0.75,0;-0.125,0,0,1.125]
Y =
1
4
2 1
+ V o
4 A 2 A
V V2 3V V1 4
2
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1.1250 0 0 -0.12500 0.7500 -0.2500 00 -0.2500 0.7500 0
-0.1250 0 0 1.1250
>> I=[4,-4,-2,2]'
I =
4-4-22
>> V=inv(Y)*I
V =3.8000
-7.0000-5.00002.2000
V o = V 1 – V 4 = 3.8 – 2.2 = 1.6 V .
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Chapter 3, Solution 25
Consider the circuit shown below.
20
At node 1.1 2 1 4
1 24 80 2
1 20
V V V V V V
41 20 V (1)
At node 2,2 31 2 2
1 20 80 98 8
1 8 10
V V V V V V V 3
V (2)
At node 3,2 3 3 3 4
2 3 40 2 5 2
10 20 10
V V V V V V V V
(3)
At node 4,3 41 4 4
1 30 3 6 11
20 10 30
V V V V V V V V 4 (4)
Putting (1) to (4) in matrix form gives:
1
2
3
4
80 21 20 0 1
0 80 98 8 0
0 0 2 5 2
0 3 0 6 11
V
V
V
V
B = A V V = A -1 B
Using MATLAB leads to
V 1 = 25.52 V , V 2 = 22.05 V , V 3 = 14.842 V , V 4 = 15.055 V
10
4
1
8
2 1 3
4
10
30
20
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Chapter 3, Solution 26
At node 1,
32121311 24745
5103
2015 V V V V V V V V (1)
At node 2,
554
532221
V V V I V V o
(2)
But10
31 V V
I o . Hence, (2) becomes
321 31570 V V V (3)At node 3,
321
32331 V11V6V37005
VV
15
V10
10
VV3 (4)
Putting (1), (3), and (4) in matrix form produces
BAV
70
0
45
V
V
V
1163
3157
247
3
2
1
Using MATLAB leads to
89.2
78.2
19.7
BAV 1
Thus,V 1 = –7.19V ; V 2 = –2.78V ; V 3 = 2.89V .
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Chapter 3, Solution 27
At node 1,
2 = 2v 1 + v 1 – v 2 + (v 1 – v 3)4 + 3i 0 , i 0 = 4v 2 . Hence,
2 – v (2)
2
2 = 7v 1 + 11v 2 – 4v 3 (1)At node 2,
v 1 – v 2 = 4v 2 + v 2 – v 3 0 = – v 1 + 6v 3
At node 3,v3 = 4 + v 2 – v 3 + 12v 2 + 4(v 1 – v 3)
or – 4 = 4v 1 + 13v 2 – 7v 3 (3)
In matrix form,
4
0
2
v
v
v
7134
161
4117
3
2
1
,1767134
161
4117
1107134
160
4112
1
,66
744
101
427
2 286
4134
061
2117
3
v 1 = ,V625.01761101 v 2 = V375.0
176662
v 3 = .V625.11762863
v 1 = 625 mV , v 2 = 375 mV , v 3 = 1.625 V .
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Chapter 3, Solution 28
At node c,
d cbcbccd V V V
V V V V V 21150
5410 (1)
At node b,
c ba b bc ba V2V4V90
8V
4VV
8V90V (2)
At node a,
d ba baad a V4V2V7600
8V90V
16V
4V60V (3)
At node d,
d cacd d d a V8V2V5300
10VV
20V
4V60V (4)
In matrix form, (1) to (4) become
BAV
300
60
90
0
V
V
V
V
8205
4027
0241
21150
d
c
b
a
We use MATLAB to invert A and obtain
75.43389.1
56.20
56.10
BAV 1
Thus,V a = –10.56 V ; V b = 20.56 V ; V c = 1.389 V ; VC d = –43.75 V .
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Chapter 3, Solution 29
At node 1,
42121141 45025 V V V V V V V V (1)At node 2,
32132221 4700)(42 V V V V V V V V (2)At node 3,
4324332 546)(46 V V V V V V V (3)At node 4,
43144143 5232 V V V V V V V V (4)In matrix form, (1) to (4) become
B AV
V
V
V
V
2
6
0
5
5101
1540
0471
1014
4
3
2
1
Using MATLAB,
7076.0
309.2
209.1
7708.0
1 B AV
i.e.V7076.0 V,309.2 V,209.1 V,7708.0 4321
V V V V
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Chapter 3, Solution 30
At node 1,
[(v 1 –80)/10]+[(v 1 –4v o)/20]+[(v 1 –(v o –96))/40] = 0 or(0.1+0.05+0.025)v 1 – (0.2+0.025)v o =0.175v 1 – 0.225v o = 8–2.4 = 5.6 (1)
At node 2,
–2 i o + [((v o –96)–v 1)/40] + [(v o –0)/80] = 0 and i o = [(v 1 –(v o –96))/40] –2[(v 1 –(v o –96))/40] + [((v o –96)–v 1)/40] + [(v o –0)/80] = 0 –3[(v 1 –(v o –96))/40] + [(v o –0)/80] = 0 or –0.0.075v 1 + (0.075+0.0125)v o = 7.2 = –0.075v
1 + 0.0875v
o = 7.2 (2)
Using (1) and (2) we get,
2.7
6.5
0015625.0175.0075.0
225.00875.0
2.7
6.5
016875.00153125.0175.0075.0
225.00875.0
2.7
6.5
0875.0075.0
225.0175.0
1
1
o
o
v
v
or v
v
v1 = –313.6–1036.8 = –1350.4
vo = –268.8–806.4 = –1.0752 kV
and i o = [(v 1 –(v o –96))/40] = [(–1350.4–(–1075.2–96))/40] = –4.48 amps .
v 1
10
20
80
40
1
v 0
i 0
2i 0
– +
2
– + +
–
96 V
4v80 V 0
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Chapter 3, Solution 31
1
At the supernode,
1 + 2v 0 =1
vv
1
v
4
v 3121
(1)
But v o = v 1 – v 3 . Hence (1) becomes,
4 = -3v 1 + 4v 2 +4v 3 (2)
At node 3,
2v o +2
v10vv
4v 3
313
or 20 = 4v 1 + 0v 2 – v 3 (3)
At the supernode, v 2 = v 1 + 4i o . But i o =4
v 3 . Hence,
v 2 = v 1 + v 3 (4)
Solving (2) to (4) leads to,
v 1 = 4.97V , v 2 = 4.85V , v 3 = –0.12V .
i0
4
v 2v 1
1 A – +10 V
+ v – 0
4
22v v0 3
1
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Chapter 3, Solution 32
v 35 k 10 V 20 V
We have a supernode as shown in figure (a). It is evident that v 2 = 12 V, Applying KVLto loops 1and 2 in figure (b), we obtain,
-v 1 – 10 + 12 = 0 or v 1 = 2 and -12 + 20 + v 3 = 0 or v 3 = -8 V
Thus, v 1 = 2 V , v 2 = 12 V , v 3 = -8V .
(b)
v 1v 2
+ – – +
(a)
4 mA
10 k
+v 1
+
–
v 3 –
– +12 V
loop 1loop 2
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Chapter 3, Solution 33
(a) This is a planar circuit. It can be redrawn as shown below.
5
13
4
(b) This is a planar circuit. It can be redrawn as shown below.
– +12 V
5
4
3
2
1
2 A
2
6
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Chapter 3, Solution 34
(a) This is a planar circuit because it can be redrawn as shown below,
7
231
6
(b) This is a non-planar circuit.
510 V – +
4
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Chapter 3, Solution 35
Assume that i 1 and i 2 are in mA. We apply mesh analysis. For mesh 1,
-30 + 20 + 7i 1 – 5i 2 = 0 or 7i 1 – 5i 2 = 10 (1)For mesh 2,
-20 + 9i 2 – 5i 1 = 0 or -5i 1 + 9i 2 = 20 (2)
Solving (1) and (2), we obtain, i 2 = 5.
v0 = 4i 2 = 20 volts .
5 k
i1
i2+v 0
–
– +30 V 20 V – +
4 k
2 k
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Chapter 3, Solution 36
10 V 4
+ – i
Applying mesh analysis gives,
10I 1 – 6I 2 = 12 and –6I 1 + 8I 2 = –10
or5
611
53
34
IIor
56
II
4335
56
2
1
2
1
I1 = (24–15)/11 = 0.8182 and I 2 = (18–25)/11 = –0.6364
i1 = –I 1 = –818.2 mA ; i 2 = I 1 – I 2 = 0.8182+0.6364 = 1.4546 A ; andi3 = I 2 = –636.4 mA .
I 1 I 2
– +12 V 6 2
1 i i2 3
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Chapter 3, Solution 37
Applying mesh analysis to loops 1 and 2, we get,
30i 1 – 20i 2 + 60 = 0 which leads to i 2 = 1.5i 1 + 3 (1)
–20i 1 + 40i 2 – 60 + 5 v 0 = 0 (2)But, v 0 = –4i 1 (3)
Using (1), (2), and (3) we get –20i 1 + 60i 1 + 120 – 60 – 20i 1 = 0 or
20i 1 = –60 or i 1 = –3 amps and i 2 = 7.5 amps.
Therefore, we get,v 0 = –4i 1 = 12 volts .
20
i1
i2+v 0
– + –
– + 60 V
6
5v 0 4
20
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Chapter 3, Solution 38
Consider the circuit below with the mesh currents.
4 3
+ _
22.5V + _
Io
1
2 2
10 A
I4I31 60 V
I2I11
4
5 A
I1 = –5 A (1)
1(I 2 –I 1) + 2(I 2 –I 4) + 22.5 + 4I 2 = 07I 2 – I 4 = –27.5 (2)
–60 + 4I 3 + 3I 4 + 1I 4 + 2(I 4 –I 2) + 2(I 3 – I 1) = 0 (super mesh) –2I 2 + 6 I 3 + 6I 4 = +60 – 10 = 50 (3)
But, we need one more equation, so we use the constraint equation –I 3 + I 4 = 10. Thisnow gives us three equations with three unknowns.
1050
5.27
II
I
110662
107
4
3
2
We can now use MATLAB to solve the problem.
>> Z=[7,0,-1;-2,6,6;0,-1,0]
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Z =
7 0 -1-2 6 60 -1 0
>> V=[–27.5,50,10]'
V = –27.5
5010
>> I=inv(Z)*V
I = –1.3750 –10.0000
17.8750Io = I 1 – I 2 = –5 – 1.375 = –6.375 A .
Check using the super mesh (equation (3)):
–2I 2 + 6 I 3 + 6I 4 = 2.75 – 60 + 107.25 = 50!
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Chapter 3, Solution 40
Assume all currents are in mA and apply mesh analysis for mesh 1.
–56 + 12i 1 – 6i 2 – 4i 3 = 0 or 6i 1 – 3i 2 – 2i 3 = 28 (1)
for mesh 2, –6i 1 + 14i 2 – 2i 3 = 0 or –3i 1 + 7i 2 – i 3 = 0 (2)
for mesh 3,
–4i 1 – 2i 2 + 10i 3 = 0 or –2i 1 – i 2 + 5i 3 = 0 (3)
Solving (1), (2), and (3) using MATLAB, we obtain,
i o = i 1 = 8 mA .
4 k
– 56V i1
i3
2 k
6 k
4 k
6 k
i 2
2 k
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Chapter 3, Solution 41
10
i 1 6 V
For loop 1,6 = 12i 1 – 2i 2 3 = 6i 1 – i 2 (1)
For loop 2,-8 = – 2i 1 +7i 2 – i 3 (2)
For loop 3,
-8 + 6 + 6i 3 – i 2 = 0 2 = – i 2 + 6i 3 (3)
We put (1), (2), and (3) in matrix form,
2
8
3
i
i
i
610
172
016
3
2
1
,234610172
016
240620182
036
2
5
i3
i2
i3
+ –
– + 8 V
2
1
4
ii2
0
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38
210
872
316
3
At node 0, i + i 2 = i 3 or i = i 3 – i 2 =234
2403823 = 1.188 A
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Chapter 3, Solution 42
Although there are many ways to work this problem, this is an example based on the same kindof problem asked in the third edition.
Problem
Determine the mesh currents in the circuit of Fig. 3.88.
Figure 3.88
Solution
For mesh 1,(1)2121 3050120305012 I I I I
For mesh 2,(2)321312 40100308040301008 I I I I I I
For mesh 3,
(3)3223 50406040506 I I I I Putting eqs. (1) to (3) in matrix form, we get
B AI
I
I
I
6
8
12
50400
4010030
03050
3
2
1
Using Matlab,
44.0
40.0
48.01 B A I
i.e. I 1 = 480 mA , I 2 = 400 mA , I 3 = 440 mA
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Chapter 3, Solution 43
20
For loop 1,
80 = 70i 1 – 20i 2 – 30i 3 8 = 7i 1 – 2i 2 – 3i 3 (1)
For loop 2,
80 = 70i 2 – 20i 1 – 30i 3 8 = -2i 1 + 7i 2 – 3i 3 (2)
For loop 3,
0 = -30i1 – 30i
2 + 90i
3 0 = i
1 + i
2 – 3i
3
(3)
olving (1) to (3), we obtain i 3 = 16/9
Io = i 3 = 16/9 = 1.7778 A
V ab = 30i 3 = 53.33 V .
S
a
i1
i2
i3
– +80 V
– 3080 V +
V ab 3020
– 30
b20
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Chapter 3, Solution 44
90 V
+
Loop 1 and 2 form a supermesh. For the supermesh,
6i 1 + 4i 2 – 5i 3 + 180 = 0 (1)
For loop 3, –i 1 – 4i 2 + 7i 3 + 90 = 0 (2)
Also, i 2 = 45 + i 1 (3)
Solving (1) to (3), i 1 = –46, i 3 = –20; i o = i 1 – i 3 = –26 A
– 180V
45 A
i1
i2i3 4
5
2
1
ii1 2
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Chapter 3, Solution 45
4 8
i
For loop 1, 30 = 5i 1 – 3i 2 – 2i 3 (1)
For loop 2, 10i 2 - 3i 1 – 6i 4 = 0 (2)
For the supermesh, 6i 3 + 14i 4 – 2i 1 – 6i 2 = 0 (3)
But i 4 – i 3 = 4 which leads to i 4 = i 3 + 4 (4)
Solving (1) to (4) by elimination gives i = i 1 = 8.561 A .
1i1 i2
3 i4
–
62
30V
3
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Chapter 3, Solution 46
For loop 1,12811081112)(8312 2121211
iiiiiii (1)For loop 2,
0214826)(8 21212 oo viiviii But ,13iv
o
21121 706148 iiiii (2)Substituting (2) into (1),
1739.012877 222 iii A and 217.17 21
ii A
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Chapter 3, Solution 47
First, transform the current sources as shown below.
- 6V +
2
I3 V 1 8 V 2 4 V 3
4 8
I1 2 I 2 + +
20V 12V- -
For mesh 1,
321321 47100821420 I I I I I I (1)For mesh 2,
321312 2760421412 I I I I I I (2)For mesh 3,
321123 7243084146 I I I I I I (3)Putting (1) to (3) in matrix form, we obtain
B AI
I
I
I
3
6
10
724
271
417
3
2
1
Using MATLAB,
8667.1,0333.0 ,5.2
8667.1
0333.0
2
3211 I I I B A I
But
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111 420
420
I V V
I 10 V
)(2 212 I I V 4.933 V
Also,
23
32 812
8
12 I V
V I 12.267 V .
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Chapter 3, Solution 48
We apply mesh analysis and let the mesh currents be in mA. 3k
I4
4k 2k 5k
Io
1k I 3 -I1 I 2 3 V
+ +6 V + 10k
- 4 V-
For mesh 1,
421421 I4II520I4II586 (1)For mesh 2,
43214312 I2I10I13I40I2I10II134 (2)For mesh 3,
432423 I5I15I1030I5I10I153 (3)
For mesh 4, 014524 4321 I I I I (4)
Putting (1) to (4) in matrix form gives
BAI
0
3
4
2
I
I
I
I
14524
515100
210131
4015
4
3
2
1
Using MATLAB,
3
896.3
044.4
608.3
BAI 1 0.148
The current through the 10k resistor is I o= I 2 – I 3 = 148 mA .
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Chapter 3, Solution 49
3
i3
For the supermesh in figure (a),
3i 1 + 2i 2 – 3i 3 + 27 = 0 (1)
At node 0, i 2 – i 1 = 2i 0 and i 0 = –i 1 which leads to i 2 = –i 1 (2)
For loop 3, –i 1 –2i 2 + 6i 3 = 0 which leads to 6i 3 = –i 1 (3)
Solving (1) to (3), i 1 = (–54/3)A, i 2 = (54/3)A, i 3 = (27/9)A
i0 = –i 1 = 18 A , from fig. (b), v 0 = i 3 –3i 1 = (27/9) + 54 = 57 V .
–+
1
i1
+v 0 or
+v 0
– i2
227V
2
(b)
i1
2 i1 i2
21
27 V –+
2i 0
i20
(a)
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Chapter 3, Solution 52
For mesh 1,
2(i 1 – i 2) + 4(i 1 – i 3) – 12 = 0 which leads to 3i 1 – i 2 – 2i 3 = 6 (1)
(2)
or the independent current source, i 3 = 3 + i 2 (3)
olving (1), (2), and (3), we obtain,
i1 = 3.5 A , i 2 = -0.5 A , i 3 = 2.5 A .
For the supermesh, 2(i 2 – i 1) + 8i 2 + 2v 0 + 4(i 3 – i 1) = 0 But v 0 = 2(i 1 – i 2) which leads to -i 1 + 3i 2 + 2i 3 = 0
F S
– +V S
–2V 0
+iv
0 2
– 3A
i1
i3
8
4
2i2
i3
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Chapter 3, Solution 53
Applying mesh analysis leads to; –12 + 4kI 1 – 3kI 2 – 1kI 3 = 0 (1)
–3kI 1 + 7kI 2 – 4kI 4 = 0
–3kI 1 + 7kI 2 = –12 (2) –1kI 1 + 15kI 3 – 8kI 4 – 6kI 5 = 0 –1kI 1 + 15kI 3 – 6k = –24 (3)
I4 = –3mA (4) –6kI 3 – 8kI 4 + 16kI 5 = 0 –6kI 3 + 16kI 5 = –24 (5)
Putting these in matrix form (having substituted I 4 = 3mA in the above),
24
24
12
12
I
I
I
I
k
16600
61501
0073
0134
5
3
2
1
ZI = V
Using MATLAB,
>> Z = [4,-3,-1,0;-3,7,0,0;-1,0,15,-6;0,0,-6,16]
Z =
4 -3 -1 0-3 7 0 0-1 0 15 -60 0 -6 16
>> V = [12,-12,-24,-24]'
V =
12
-12-24-24
We obtain,
>> I = inv(Z)*V
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I =1.6196 mA
–1.0202 mA–2.461 mA
3 mA
–2.423 mA
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Chapter 3, Solution 54
Let the mesh currents be in mA. For mesh 1,
2121 22021012 I I I I (1)
For mesh 2,(2)321312 3100310 I I I I I I
For mesh 3,
3223 2120212 I I I I (3)Putting (1) to (3) in matrix form leads to
B AI
I
I
I
12
10
2
210
131
012
3
2
1
Using MATLAB,
mA25.10,mA5.8 ,mA25.5
25.10
5.8
25.5
3211 I I I B A I
I 1 = 5.25 mA , I 2 = 8.5 mA , and I 3 = 10.25 mA .
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Chapter 3, Solution 55Chapter 3, Solution 5510 V
It is evident that I 1 = 4 (1)It is evident that I For mesh 4, 12(I 4 – I 1) + 4(I 4 – I 3) – 8 = 0For mesh 4, 12(I (2)(2)
For the supermesh 6(I 2 – I 1) + 10 + 2I 3 + 4(I 3 – I 4) = 0For the supermesh 6(I or -3I 1 + 3I 2 + 3I 3 – 2I 4 = -5 (3)or -3I At node c, I 2 = I 3 + 1 (4)At node c, I Solving (1), (2), (3), and (4) yields, I 1 = 4A, I 2 = 3A, I 3 = 2A, and I 4 = 4ASolving (1), (2), (3), and (4) yields, I
At node b, i 1 = I 2 – I 1 = -1A At node b, i
At node a, i 2 = 4 – I 4 = 0A At node a, i
At node 0, i 3 = I 4 – I 3 = 2A At node 0, i
ddI 1I
1 = 4 (1)
4 – I 1) + 4(I 4 – I 3) – 8 = 0
2 – I 1) + 10 + 2I 3 + 4(I 3 – I 4) = 01 + 3I 2 + 3I 3 – 2I 4 = -5 (3)
2 = I 3 + 1 (4)
1 = 4A, I 2 = 3A, I 3 = 2A, and I 4 = 4A
1 = I 2 – I 1 = -1A
2 = 4 – I 4 = 0A
3 = I 4 – I 3 = 2A
1
I
I 3
I 4
4 A
+ –
+I 2b c
1Ai1
61A
2i2
i3
I 34A 412
a 0I 4
8 V
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Chapter 3, Solution 56+ v – 1
For loop 1, 12 = 4i 1 – 2i 2 – 2i 3 which leads to 6 = 2i 1 – i 2 – i 3 (1)
For loop 2, 0 = 6i 2 –2i 1 – 2 i 3 which leads to 0 = -i 1 + 3i 2 – i 3 (2)For loop 3, 0 = 6i 3 – 2i 1 – 2i 2 which leads to 0 = -i 1 – i 2 + 3i 3 (3)
In matrix form (1), (2), and (3) become,
0
0
6
i
i
i
311
131
112
3
2
1
= ,8
311
131112
2 = 24
301
131162
3 = 24
011
031
612
, therefore i 2 = i 3 = 24/8 = 3A,
v1 = 2i 2 = 6 volts , v = 2i 3 = 6 volts
– +12 V
i1
i2
i3
2
22
+2 2v 2
–
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Chapter 3, Solution 57
Assume R is in kilo-ohms.V306090V90V,V60mA15xk 4V 212
Current through R is
R )15(R 3
330R iViR 3
3i R 1,oR
This leads to R = 90/15 = 6 k Ω .
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Chapter 3, Solution 5830
i2
For loop 1, 120 + 40i 1 – 10i 2 = 0, which leads to -12 = 4i 1 – i 2 (1)
For loop 2, 50i 2 – 10i 1 – 10i 3 = 0, which leads to -i 1 + 5i 2 – i 3 = 0 (2)
For loop 3, -120 – 10i 2 + 40i 3 = 0, which leads to 12 = -i 2 + 4i 3 (3)
Solving (1), (2), and (3), we get, i 1 = -3A , i 2 = 0, and i 3 = 3A
– +
120 V
30
i1 i3
10 1030
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Chapter 3, Solution 59
For loop 1, -80 + 30i 1 – 20i 2 + 4v 0 = 0, where v 0 = 80i 3
(3)
From (1), (2), and (3),
=
or 4 = 1.5i 1 – i 2 + 16i 3 (1)
For the supermesh, 60i 2 – 20i 1 – 96 + 80i 3 – 4 v 0 = 0, where v 0 = 80i 3
or 4.8 = -i 1 + 3i 2 – 12i 3 (2)
Also, 2I 0 = i 3 – i 2 and I 0 = i 2 , hence, 3i 2 = i 3
130
12313223
0
8.48
3
2
1
i
i
i
,5
130
1231
3223
2 = ,4.22
100
128.41
3283
3 = 2.67
030
8.431
823
I0 = i 2 = 2 / = -28/5 = –4.48 A
v 0 = 8i 3 = (-84/5)80 = –1.0752 kvolts
i1
i2
i3
40– +
96 V
4v 0 – –
+80V
+v 0
– 80
20
10
I
i
0
2I 0
i2 3
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Chapter 3, Solution 60
0.5 i0
At node 1, [(v 1 –0)/1] + [(v 1 –56)/4] + 0.5[(v 1 –0)/1] = 0 or 1.75v 1 = 14 or v 1 = 8 VAt node 2, [(v 2 –56)/8] – 0.5[8/1] + [(v 2 –0)/2] = 0 or 0.625v 2 = 11 or v 2 = 17.6 V
P1 = (v 1)2/1 = 64 watts , P 2 = (v 2)2/2 = 154.88 watts ,
P4 = (56 – v 1)2/4 = 576 watts , P 8 = (56 – v 2)2/8 = 1.84.32 watts .
v 2
– +56 V
856 V4v 1
1 2
i0
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Chapter 3, Solution 61
20 10
At node 1, i s = (v 1/30) + ((v 1 – v 2)/20) which leads to 60i s = 5v 1 – 3v 2
s
(1)
But v 2 = -5v 0 and v 0 = v 1 which leads to v 2 = -5v 1 Hence, 60i s = 5v 1 + 15v 1 = 20v 1 which leads to v 1 = 3i s, v 2 = -15i
i0 = v 2 /50 = -15i s/50 which leads to i 0 /i s = -15/50 = –0.3
+v 0
–
v 2
–
v 1
i0
i s 30 40
5v 0
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Chapter 3, Solution 62
4 k
We have a supermesh. Let all R be in k , i in mA, and v in volts.
For the supermesh, -100 +4i 1 + 8i 2 + 2i 3 + 40 = 0 or 30 = 2i 1 + 4i 2 + i 3 (1)
At node A, i 1 + 4 = i 2 (2)
At node B, i 2 = 2i 1 + i 3 (3)
Solving (1), (2), and (3), we get i 1 = 2 mA , i 2 = 6 mA , and i 3 = 2 mA .
i1 i2 i3 – +100V
B8 k 2 k
–
A
40 V
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Chapter 3, Solution 63
10 A
For the supermesh, -50 + 10i 1 + 5i 2 + 4i x = 0, but i x = i 1 . Hence,
2 (2)
50 = 14i 1 + 5i 2 (1)
At node A, i 1 + 3 + (v x /4) = i 2 , but v x = 2(i 1 – i 2), hence, i 1 + 2 = i Solving (1) and (2) gives i 1 = 2.105 A and i 2 = 4.105 A
v x = 2(i 1 – i 2) = –4 volts and i x = i 2 – 2 = 2.105 amp
– +50 V
4i x–
i1 i2
5
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Chapter 3, Solution 64
i
For mesh 2, 20i 2 – 10i 1 + 4i 0 = 0 (1)
But at node A, i o = i 1 – i 2 so that (1) becomes i 1 = (16/6)i 2 (2)
or the supermesh, –250 + 50i 1 + 10(i 1 – i 2) – 4i 0 + 40i 3 = 0
r 28i 1 – 3i 2 + 20i 3 = 125
e B,
becomes i 3 = 5 + (2/3)i 2 (5)
Solving (1) to
v 0 = 10i 2 = 2.941 volts , i 0 = i 1 – i 2 = (5/3)i 2 = 490.2mA .
F o (3)
At nod i 3 + 0.2v 0 = 2 + i 1 (4)
But, v 0 = 10i 2 so that (4)
(5), i 2 = 0.2941 A,
40i1
i2
i3
– 250V
–4i 0
50 iA 10
0.2V 0
1 2
10
5 A
B
i0 + v 0
i i1 3
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Chapter 3, Solution 65
For mesh 1, –12 + 12I 1 – 6I 2 – I 4 = 0 or
421 61212 I I I (1)For mesh 2,
–6I 1 + 16I 2 – 8I 3 – I 4 – I 5 = 0 (2)For mesh 3,
–8I 2 + 15I 3 – I 5 – 9 = 0 or9 = –8I 2 + 15I 3 – I 5 (3)
For mesh 4, –I 1 – I 2 + 7I 4 – 2I 5 – 6 = 0 or
(4)6 = –I 1 – I 2 + 7I 4 – 2I 5 For mesh 5,
–I 2 – I 3 – 2I 4 + 8I 5 – 10 = 0 or 5432 8210 I I I I (5)
Casting (1) to (5) in matrix form gives
BAI
10
6
9
0
12
I
I
I
I
I
82110
27011
101580
118166
010612
5
4
3
2
1
Using MATLAB we input:
Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]and V=[12;0;9;6;10]
This leads to>> Z=[12,-6,0,-1,0;-6,16,-8,-1,-1;0,-8,15,0,-1;-1,-1,0,7,-2;0,-1,-1,-2,8]
Z =
12 -6 0 -1 0-6 16 -8 -1 -10 -8 15 0 -1
-1 -1 0 7 -20 -1 -1 -2 8
>> V=[12;0;9;6;10]
V =
12
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096
10
>> I=inv(Z)*V
I =
2.17011.99121.81192.09422.2489
Thus,
I = [2.17, 1.9912, 1.8119, 2.094, 2.249] A .
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Chapter 3, Solution 66
The mesh equations are obtained as follows.
1 2 3 412 24 30 4 6 2 0I I I I
or30I 1 – 4I 2 – 6I 3 – 2I 4 = –12 (1)
1 2 4 524 40 4 30 2 6 0I I I I
or –4I 1 + 30I 2 – 2I 4 – 6I 5 = –16 (2)
–6I 1 + 18I 3 – 4I 4 = 30 (3)
–2I 1 – 2I 2 – 4I 3 + 12I 4 –4I 5 = 0 (4)
–6I 2 – 4I 4 + 18I 5 = –32 (5)
Putting (1) to (5) in matrix form
32
0
30
16
12
I
184060
412422
041806
620304
026430
ZI = V
Using MATLAB,
>> Z = [30,-4,-6,-2,0;-4,30,0,-2,-6;-6,0,18,-4,0;-2,-2,-4,12,-4;
0,-6,0,-4,18]Z =
30 -4 -6 -2 0-4 30 0 -2 -6-6 0 18 -4 0-2 -2 -4 12 -4
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0 -6 0 -4 18
>> V = [-12,-16,30,0,-32]'
V =
-12-16300
-32
>> I = inv(Z)*V
I =
-0.2779 A-1.0488 A1.4682 A-0.4761 A-2.2332 A
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Chapter 3, Solution 67
Consider the circuit below.
V 3
+ V o -
15
0
V35
V
5.05.00
5.095.025.0
025.035.0 o
Since we actually have four unknowns and only three equations, we need a constraintequation.
Vo = V
2 – V
3
Substituting this back into the matrix equation, the first equation becomes,
0.35V 1 – 3.25V 2 + 3V 3 = –5
This now results in the following matrix equation,
V 2
10 5 10 A 3 V o
V 1 2
5 A
4
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15
0
5
V
5.05.00
5.095.025.0
325.335.0
Now we can use MATLAB to solve for V.
>> Y=[0.35,-3.25,3;-0.25,0.95,-0.5;0,-0.5,0.5]
Y =
0.3500 -3.2500 3.0000-0.2500 0.9500 -0.5000
0 -0.5000 0.5000
>> I=[–5,0,15]'
I =
–5015
>> V=inv(Y)*I
V = –410.5262 –194.7368 –164.7368
V o = V 2 – V 3 = –77.89 + 65.89 = –30 V .
Let us now do a quick check at node 1.
–3(–30) + 0.1(–410.5) + 0.25(–410.5+194.74) + 5 =90–41.05–102.62+48.68+5 = 0.01; essentially zero considering theaccuracy we are using. The answer checks.
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Chapter 3, Solution 68
Although there are many ways to work this problem, this is an example based on thesame kind of problem asked in the third edition.
Problem
Find the voltage V o in the circuit of Fig. 3.112.
3 A
40 20
_
+
V o
25 10
+ _4A 24 V
Figure 3.112For Prob. 3.68.
Solution
Consider the circuit below. There are two non-reference nodes.
3 A
V 1 V o
40 20
_
+
V o
25 10
+ _4 A 24 V
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04.2
7
25/243
34V
19.01.0
1.0125.0
Using MATLAB, we get,
>> Y=[0.125,-0.1;-0.1,0.19]
Y =
0.1250 -0.1000-0.1000 0.1900
>> I=[7,-2.04]'
I =
7.0000-2.0400
>> V=inv(Y)*I
V =
81.890932.3636
Thus, V o = 32.36 V .We can perform a simple check at node V o ,
3 + 0.1(32.36–81.89) + 0.05(32.36) + 0.04(32.36–24) =3 – 4.953 + 1.618 + 0.3344 = – 0.0004; answer checks!
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Chapter 3, Solution 70
7I4
20I4V
50
03
x
x
With two equations and three unknowns, we need a constraint equation,
Ix = 2V 1 , thus the matrix equation becomes,
7
20V
58
05
This results in V 1 = 20/(–5) = –4 V andV 2 = [–8(–4) – 7]/5 = [32 – 7]/5 = 5 V .
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Chapter 3, Solution 71
015
30
I915174
549
We can now use MATLAB solve for our currents.
>> R=[9,–4,–5;–4,7,–1;–5,–1,9]
R =
9 –4 –5 –4 7 –1 –5 –1 9
>> V=[30,–15,0]'
V =
30 –15
0
>> I=inv(R)*V
I =
6.255 A1.9599 A3.694 A
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Chapter 3, Solution 72
R 11 = 5 + 2 = 7, R 22 = 2 + 4 = 6, R 33 = 1 + 4 = 5, R 44 = 1 + 4 = 5,R 12 = -2, R 13 = 0 = R 14 , R 21 = -2, R 23 = -4, R 24 = 0, R 31 = 0,R 32 = -4, R 34 = -1, R 41 = 0 = R 42 , R 43 = -1, we note that R ij = R ji for
all i not equal to j.
v1 = 8, v 2 = 4, v 3 = -10, and v 4 = -4
Hence the mesh-current equations are:
4
10
4
8
5100
1540
0462
0027
4
3
2
1
i
i
i
i
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Chapter 3, Solution 73
R 11 = 2 + 3 +4 = 9, R 22 = 3 + 5 = 8, R 33 = 1+1 + 4 = 6, R 44 = 1 + 1 = 2,R 12 = -3, R 13 = -4, R 14 = 0, R 23 = 0, R 24 = 0, R 34 = -1
v 1 = 6, v 2 = 4, v 3 = 2, and v 4 = -3 Hence,
3
2
4
6
2100
1604
0083
0439
4
3
2
1
i
i
i
i
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Chapter 3, Solution 74
R 11 = R 1 + R 4 + R 6 , R 22 = R 2 + R 4 + R 5 , R 33 = R 6 + R 7 + R 8 ,,R 44 = R 3 + R 5 + R 8 , R 12 = -R 4 , R 13 = -R 6 , R 14 = 0, R 23 = 0
R 24 = -R 5 , R 34 = -R 8 , again, we note that R ij = R ji for all i not equal to j.
The input voltage vector is =
4
3
2
1
V
V
V
V
4
3
2
1
4
3
2
1
85385
88766
55424
64641
0
0
0
0R
V
V
V
V
i
i
i
i
R R R R R
R R R R R
R R R R R
R R R R
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Chapter 3, Solution 75
* Schematics Netlist *
R_R4 $N_0002 $N_0001 30R_R2 $N_0001 $N_0003 10
R_R1 $N_0005 $N_0004 30R_R3 $N_0003 $N_0004 10R_R5 $N_0006 $N_0004 30V_V4 $N_0003 0 120Vv_V3 $N_0005 $N_0001 0v_V2 0 $N_0006 0v_V1 0 $N_0002 0
i3
i2i1
Clearly, i 1 = –3 amps , i 2 = 0 amps , and i 3 = 3 amps , which agrees with the answers inProblem 3.44.
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Chapter 3, Solution 76
* Schematics Netlist *
I_I2 0 $N_0001 DC 4AR_R1 $N_0002 $N_0001 0.25
R_R3 $N_0003 $N_0001 1R_R2 $N_0002 $N_0003 1F_F1 $N_0002 $N_0001 VF_F1 3VF_F1 $N_0003 $N_0004 0VR_R4 0 $N_0002 0.5R_R6 0 $N_0001 0.5I_I1 0 $N_0002 DC 2AR_R5 0 $N_0004 0.25
Clearly, v 1 = 625 mVolts , v 2 = 375 mVolts , and v 3 = 1.625 volts , which agrees withthe solution obtained in Problem 3.27.
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Chapter 3, Solution 78
The schematic is shown below. When the circuit is saved and simulated the nodevoltages are displayed on the pseudocomponents as shown. Thus,
,V15V,5.4 V,3 321
V V V
.
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Chapter 3, Solution 79
The schematic is shown below. When the circuit is saved and simulated, we obtain thenode voltages as displayed. Thus,
V a = –10.556 volts ; V b = 20.56 volts ; V c = 1.3889 volts ; and V d = –43.75 volts .
R1
20
R2
8
R3
10
R4
5
R5
4
R6
4 R7
16
R8
8
V1
90VdcV2
60Vdc
20.56 V
1.3889 V
–43.75 V
–10.556 V
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Chapter 3, Solution 80
* Schematics Netlist *
H_H1 $N_0002 $N_0003 VH_H1 6VH_H1 0 $N_0001 0V
I_I1 $N_0004 $N_0005 DC 8AV_V1 $N_0002 0 20VR_R4 0 $N_0003 4R_R1 $N_0005 $N_0003 10R_R2 $N_0003 $N_0002 12R_R5 0 $N_0004 1R_R3 $N_0004 $N_0001 2
Clearly, v 1 = 84 volts , v 2 = 4 volts , v 3 = 20 volts , and v 4 = -5.333 volts
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Chapter 3, Solution 81
Clearly, v 1 = 26.67 volts , v 2 = 6.667 volts , v 3 = 173.33 volts , and v 4 = –46.67 volts which agrees with the results of Example 3.4.
This is the netlist for this circuit.
* Schematics Netlist *
R_R1 0 $N_0001 2R_R2 $N_0003 $N_0002 6R_R3 0 $N_0002 4R_R4 0 $N_0004 1R_R5 $N_0001 $N_0004 3I_I1 0 $N_0003 DC 10AV_V1 $N_0001 $N_0003 20VE_E1 $N_0002 $N_0004 $N_0001 $N_0004 3
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Chapter 3, Solution 82
This network corresponds to the Netlist.This network corresponds to the Netlist.
+ v 0 – + v
4 4
3 k3 k
0 –
2 k
4 k 8 k
6 k
2i 0
+4A
–+100V
2 30 3v1
0
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Chapter 3, Solution 83
The circuit is shown below.
20 70
When the circuit is saved and simulated, we obtain v 2 = –12.5 volts
0
2 A 30 – +
1 2 3
20 V 50
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Chapter 3, Solution 85
The amplifier acts as a source.
R s
+V s R L
-
For maximum power transfer,
9s L R R
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Chapter 3, Solution 87
v 1 = 500(v s)/(500 + 2000) = v s/5
v 0 = -400(60v 1)/(400 + 2000) = -40v 1 = -40(v s /5) = -8v s,
Therefore, v 0/v s = –8
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Chapter 3, Solution 88
Let v 1 be the potential at the top end of the 100-ohm resistor.
(v s – v 1)/200 = v 1/100 + (v 1 – 10 -3v 0)/2000 (1)
For the right loop, v 0 = -40i 0(10,000) = -40(v 1 – 10 -3)10,000/2000,
or, v 0 = -200v 1 + 0.2v 0 = -4x10 -3v 0 (2)
Substituting (2) into (1) gives, (v s + 0.004v 1)/2 = -0.004v 0 + (-0.004v 1 – 0.001v 0)/20
This leads to 0.125v 0 = 10v s or (v 0/v s) = 10/0.125 = –80
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Chapter 3, Solution 89
Consider the circuit below.
For the left loop, applying KVL gives
–2.25 – 0.7 + 10 5IB + V BE = 0 but V BE = 0.7 V means 10 5IB = 2.25 or
IB = 22.5 µA .
For the right loop, –V CE + 15 – I Cx10 3 = 0. Addition ally, I C = βIB = 100x22.5x10 –6 =2.25 mA. Therefore,
V CE = 15–2.25x10 –3x10 3 = 12.75 V .
+ _
1 k
+ _
2.25 V
0.7 V 100 k
C 15 V | |
+ I C
_V CE
E
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Chapter 3, Solution 90
For loop 1, -v s + 10k(I B) + V BE + I E (500) = 0 = -v s + 0.7 + 10,000I B + 500(1 + )I B
which leads to v s + 0.7 = 10,000I B + 500(151)I B = 85,500I B
But, v 0 = 500I E = 500x151I B = 4 which leads to I B = 5.298x10 -5
Therefore, v s = 0.7 + 85,500I B = 5.23 volts
+V 0
–
1 k
500-+
18V -+
v s i1
i2
I10 k B
+V CE +
– V BE –
I E
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Chapter 3, Solution 91
We first determine the Thevenin equivalent for the input circuit.
R Th = 6||2 = 6x2/8 = 1.5 k and VTh = 2(3)/(2+6) = 0.75 volts
For loop 1, -0.75 + 1.5kIB + VBE + 400IE = 0 = -0.75 + 0.7 + 1500I B + 400(1 + )IB
IB = 0.05/81,900 = 0.61 A
v0 = 400IE = 400(1 + )IB = 49 mV
For loop 2, -400IE – VCE – 5kIC + 9 = 0, but, IC = IB and IE = (1 + )IB
VCE = 9 – 5k IB – 400(1 + )IB = 9 – 0.659 = 8.641 volts
+V 0 –
5 k
400-+
9 V -+
I
i1
i2
C
I1.5 k B +V CE + – V BE
– 0.75 V
I E
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Chapter 3, Solution 92
Although there are many ways to work this problem, this is an example based on the same kindof problem asked in the third edition.
ProblemFind I B and V C for the circuit in Fig. 3.128. Let = 100, V BE = 0.7V.
Figure 3.128
Solution
I 5 k 1
I1 = I B + I C = (1 + )I B and I E = I B + I C = I 1
+
V 0 –
4 k -+
12V
10 k
V
+V CE
– +
V BE –
I B
I E
C
I C
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Applying KVL around the outer loop,
4kI E + V BE + 10kI B + 5kI 1 = 12
12 – 0.7 = 5k(1 + )I B + 10kI B + 4k(1 + )I B = 919kI B
IB = 11.3/919k = 12.296 A
Also, 12 = 5kI 1 + V C which leads to V C = 12 – 5k(101)I B = 5.791 volts
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Chapter 3, Solution 93
From (b), -v 1 + 2i – 3v 0 + v 2 = 0 which leads to i = (v 1 + 3v 0 – v 2)/2
2
2
At node 1 in (a), ((24 – v 1)/4) = (v 1 /2) + ((v 1 +3v 0 – v 2)/2) + ((v 1 – v 2)/1), where v 0 = v
or 24 = 9v 1 which leads to v 1 = 2.667 volts
At node 2, ((v 1 – v 2)/1) + ((v 1 + 3v 0 – v 2)/2) = (v 2 /8) + v 2/4, v 0 = v
v 2 = 4v 1 = 10.66 volts
Now we can solve for the currents, i 1 = v 1 /2 = 1.333 A , i 2 = 1.333 A , and
i3 = 2.6667 A.
+v 0
–
i3v 1 v 2i
(b) (a)
2 3v 0
– +24V
1
3v4 0
i i 21 i2 + +2 8 v 1 v 2 4
– –