Chapter 2, Solution 1
v = iR i = v/R = (16/5) mA = 3.2 mA
Chapter 2, Solution 2
p = v2/R → R = v2/p = 14400/60 = 240 ohms
Chapter 2, Solution 3
R = v/i = 120/(2.5x10-3) = 48k ohms
Chapter 2, Solution 4
(a) i = 3/100 = 30 mA
(b) i = 3/150 = 20 mA
Chapter 2, Solution 5
n = 9; l = 7; b = n + l – 1 = 15
Chapter 2, Solution 6
n = 12; l = 8; b = n + l –1 = 19
Chapter 2, Solution 7
7 elements or 7 branches and 4 nodes, as indicated.
30 V
1 20 Ω 2 3
+++-
+ -
2A 30 Ω 60 Ω 40 Ω 10 Ω
4
Chapter 2, Solution 8
12 A
a
i1
8A b
i3
i2
12 A
c
9A d
At node a, 8 = 12 + i1 i1 = - 4A
At node c, 9 = 8 + i2 i2 = 1A
At node d, 9 = 12 + i3 i3 = -3A
Chapter 2, Solution 9
Applying KCL,
i1 + 1 = 10 + 2 i1 = 11A
1 + i2 = 2 + 3 i2 = 4A
i2 = i3 + 3 i3 = 1A
Chapter 2, Solution 10
2
4A -2A
i2
1 i1 3
3A
At node 1, 4 + 3 = i1 i1 = 7A
At node 3, 3 + i2 = -2 i2 = -5A
Chapter 2, Solution 11
Applying KVL to each loop gives
-8 + v1 + 12 = 0 v1 = 4v
-12 - v2 + 6 = 0 v2 = -6v
10 - 6 - v3 = 0 v3 = 4v
-v4 + 8 - 10 = 0 v4 = -2v
Chapter 2, Solution 12
+ 15v -
loop 2
– 25v + + 10v - + v2 -
+ + +
20v loop 1 v1 loop 3 v3
- - -
For loop 1, -20 -25 +10 + v1 = 0 v1 = 35v
For loop 2, -10 +15 -v2 = 0 v2 = 5v
For loop 3, -v1 +v2 +v3 = 0 v3 = 30v
Chapter 2, Solution 13
2A
I2 7A I4
1 2 3 4
4A
I1
3A I3
At node 2,
3 + 7 + I2 = 0 →
I 2 = −10 A
At node 1,
I1 + I 2 = 2 →
I 1 = 2 − I 2 = 12 A
At node 4,
2 = I4 + 4 →
I 4 = 2 − 4 = −2 A
At node 3,
7 + I4 = I3 →
I3 = 7 − 2 = 5 A
Hence,
I 1 = 12 A, I 2 = −10 A, I 3 = 5 A, I 4 = −2 A
Chapter 2, Solution 14
+ + -
3V V1 I4 V2
- I3 - + 2V - +
- + V3 - + +
4V
I2 - V4 I1 5V
+ -
For mesh 1,
−V4 + 2 + 5 = 0 →
V4 = 7V
For mesh 2,
+4 + V3 + V4 = 0 →
V3 = −4 − 7 = −11V
For mesh 3,
−3 + V1 − V3 = 0 →
V1 = V3 + 3 = −8V
For mesh 4,
−V1 − V2 − 2 = 0 →
V2 = −V1 − 2 = 6V
Thus,
V1 = −8V , V2 = 6V , V3 = −11V , V4 = 7V
Chapter 2, Solution 15
+ +
+ 12V 1 v2
- - 8V + -
v1
- 3 + 2 -
v3 10V
- +
For loop 1,
8 − 12 + v2 = 0 →
v2 = 4V
For loop 2,
− v3 − 8 − 10 = 0 →
v3 = −18V
For loop 3,
− v1 + 12 + v3 = 0 →
v1 = −6V
Thus,
v1 = −6V , v2 = 4V , v3 = −18V
Chapter 2, Solution 16
+ v1 -
+
+ loop 1
6V - 10V v1
-
+- +-
12V loop 2
+ v2 -
Applying KVL around loop 1,
–6 + v1 + v1 – 10 – 12 = 0 v1 = 14V
Applying KVL around loop 2,
12 + 10 – v2 = 0 v2 = 22V
Chapter 2, Solution 17
+ v1 -
loop 1
+ - +
24V - v2 v3 +
- 10V
+ -
loop 2
-+
12V
It is evident that v3 = 10V
Applying KVL to loop 2,
v2 + v3 + 12 = 0 v2 = -22V
Applying KVL to loop 1,
-24 + v1 - v2 = 0 v1 = 2V
Thus,
v1 = 2V, v2 = -22V, v3 = 10V
Chapter 2, Solution 18
Applying KVL,
-30 -10 +8 + I(3+5) = 0
8I = 32 I = 4A
-Vab + 5I + 8 = 0 Vab = 28V
Chapter 2, Solution 19
Applying KVL around the loop, we obtain
-12 + 10 - (-8) + 3i = 0 i = -2A
Power dissipated by the resistor:
p 3Ω = i2R = 4(3) = 12W
Power supplied by the sources:
p12V = 12 (- -2) = 24W
p10V = 10 (-2) = -20W
p8V = (- -2) = -16W
Chapter 2, Solution 20
Applying KVL around the loop,
-36 + 4i0 + 5i0 = 0 i0 = 4A
Chapter 2, Solution 21
Apply KVL to obtain
10 Ω
-45 + 10i - 3V0 + 5i = 0
+ v0 -
But v0 = 10i,
+ -
45V - + 3v0
-45 + 15i - 30i = 0 i = -3A
P3 = i2R = 9 x 5 = 45W
5Ω
Chapter 2, Solution 22
4Ω
+ v0 -
6Ω 10A 2v0
At the node, KCL requires that
v0
+ 10 + 2 v 0 = 0 v0 = –4.444V
4
The current through the controlled source is
i = 2V0 = -8.888A
and the voltage across it is
v0
v = (6 + 4) i0 = 10 = −11.111
4
Hence,
p2 vi = (-8.888)(-11.111) = 98.75 W
Chapter 2, Solution 23
8//12 = 4.8, 3//6 = 2, (4 + 2)//(1.2 + 4.8) = 6//6 = 3
The circuit is reduced to that shown below.
ix 1Ω
+ vx -
6A 2Ω 3Ω
Applying current division,
2
ix = (6 A) = 2 A, v x = 1i x = 2V
2 + 1+ 3
The current through the 1.2- Ω resistor is 0.5ix = 1A. The voltage across the 12- Ω
resistor is 1 x 4.8 = 4.8 V. Hence the power is
v 2 4.8 2
p= = = 1.92W
R 12
Chapter 2, Solution 24
Vs
(a) I0 =
R1 + R2
αV0 R3 R4
V0 = −α I0 (R3 R4 ) = − ⋅
R1 + R 2 R3 + R4
V0 − αR3 R4
=
Vs (R1 + R2 )(R3 + R4 )
(b) If R1 = R2 = R3 = R4 = R,
V0 α R α
= ⋅ = = 10 α = 40
VS 2R 2 4
Chapter 2, Solution 25
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
5
I20 = (0.01x50) = 0.1 A
5 + 20
V20 = 20 x 0.1 kV = 2 kV
p20 = I20 V20 = 0.2 kW
Chapter 2, Solution 26
V0 = 5 x 10-3 x 10 x 103 = 50V
Using current division,
5
I20 = (0.01x50) = 0.1 A
5 + 20
V20 = 20 x 0.1 kV = 2 kV
p20 = I20 V20 = 0.2 kW
Chapter 2, Solution 27
Using current division,
4
i1 = (20) = 8 A
4+6
6
i2 = (20) = 12 A
4+6
Chapter 2, Solution 28
We first combine the two resistors in parallel
15 10 = 6 Ω
We now apply voltage division,
14
v1 = (40) = 20 V
14 + 6
6
v2 = v3 = (40) = 12 V
14 + 6
Hence, v1 = 28 V, v2 = 12 V, vs = 12 V
Chapter 2, Solution 29
The series combination of 6 Ω and 3 Ω resistors is shorted. Hence
i2 = 0 = v2
12
v1 = 12, i1 = = 3A
4
Hence v1 = 12 V, i1 = 3 A, i2 = 0 = v2
Chapter 2, Solution 30
8Ω
i1
i
+
9A
6Ω v 4Ω
-
12
By current division, i = (9) = 6 A
6 + 12
i1 = 9 − 6 = 3A, v = 4i1 = 4 x 3 = 12 V
p6 = 12R = 36 x 6 = 216 W
Chapter 2, Solution 31
The 5 Ω resistor is in series with the combination of 10 (4 + 6) = 5Ω .
Hence by the voltage division principle,
5
v= (20V) = 10 V
5+5
by ohm's law,
v 10
i= = = 1A
4 + 6 4+ 6
pp = i2R = (1)2(4) = 4 W
Chapter 2, Solution 32
We first combine resistors in parallel.
20 x30
20 30 = = 12 Ω
50
10x 40
10 40 = = 8Ω
50
Using current division principle,
8 12
i1 + i 2 = (20) = 8A, i 3 + i 4 = (20) = 12A
8 + 12 20
20
i1 = (8) = 3.2 A
50
30
i2 = (8) = 4.8 A
50
10
i3 = (12) = 2.4A
50
40
i4 = (12) = 9.6 A
50
Chapter 2, Solution 33
Combining the conductance leads to the equivalent circuit below
i 4S i
+ +
9A v 1S 9A v 1S
4S 2S
- -
6x3
6 S 3S = = 25 and 25 + 25 = 4 S
9
Using current division,
1
i= (9) = 6 A, v = 3(1) = 3 V
1
1+
2
Chapter 2, Solution 34
By parallel and series combinations, the circuit is reduced to the one below:
i1 8Ω
10 x15
10 ( 2 + 13 ) = = 6Ω
25
15 x15 +
15 (4 + 6) = = 6Ω 28V + v1
25 - 6Ω
-
12 (6 + 6) = 6Ω
28
Thus i1 = = 2 A and v1 = 6i1 = 12 V
8+6
We now work backward to get i2 and v2.
i1 = 2A 8Ω 6Ω 1A
1A
+ +
+ 6V
28V
- 12V 12 Ω 6Ω
- -
i1 = 2A 8Ω 6Ω 1A 4Ω 0.6A
1A
+ + +
+ 6V 3.6V
28V
- 12V 12 Ω 15 Ω 6Ω
- -
-
13 v
Thus, v2 = (3 ⋅ 6) = 3 ⋅ 12, i2 = 2 = 0.24
15 13
p2 = i2R = (0.24)2 (2) = 0.1152 W
i1 = 2 A, i2 = 0.24 A, v1 = 12 V, v2 = 3.12 V, p2 = 0.1152 W
Chapter 2, Solution 35
i
+
70 Ω V1 30 Ω
+ i1 - I0
50V a b
- +
20 Ω V0 5 Ω
i2 -
Combining the versions in parallel,
70x30 20x 5
70 30 = = 21Ω , 20 15 = =4 Ω
100 25
50
i= =2 A
21 + 4
vi = 21i = 42 V, v0 = 4i = 8 V
v v
i1 = 1 = 0.6 A, i2 = 2 = 0.4 A
70 20
At node a, KCL must be satisfied
i1 = i2 + I0 0.6 = 0.4 + I0 I0 = 0.2 A
Hence v0 = 8 V and I0 = 0.2A
Chapter 2, Solution 36
The 8-Ω resistor is shorted. No current flows through the 1-Ω resistor. Hence v0
is the voltage across the 6Ω resistor.
4 4
I0 = = =1 A
2 + 3 16 4
v0 = I0 (3 6 ) = 2I 0 = 2 V
Chapter 2, Solution 37
Let I = current through the 16Ω resistor. If 4 V is the voltage drop across the 6 R
combination, then 20 - 4 = 16 V in the voltage drop across the 16Ω resistor.
16
Hence, I = = 1 A.
16
20 6R
But I = =1 4= 6R= R = 12 Ω
16 + 6 R 6+R
Chapter 2, Solution 38
Let I0 = current through the 6Ω resistor. Since 6Ω and 3Ω resistors are in parallel.
6I0 = 2 x 3 R0 = 1 A
The total current through the 4Ω resistor = 1 + 2 = 3 A.
Hence
vS = (2 + 4 + 2 3 ) (3 A) = 24 V
vS
I= = 2.4 A
10
Chapter 2, Solution 39
(a) Req = R 0 = 0
R R
(b) Req = R R + R R = + = R
2 2
(c) Req = (R + R ) (R + R ) = 2R 2R = R
1
(d) Req = 3R (R + R R ) = 3R (R + R )
2
3
3Rx R
= 2 =R
3
3R + R
2
R ⋅ 2R
(e) Req = R 2R 3R = 3R
3R
2
3Rx R
= 3R
2
R= 3 = 6R
3 2 11
3R + R
3
Chapter 2, Solution 40
Req = 3 + 4 (2 + 6 3) = 3 + 2 = 5Ω
10 10
I= = = 2A
Re q 5
Chapter 2, Solution 41
Let R0 = combination of three 12Ω resistors in parallel
1 1 1 1
= + + Ro = 4
R o 12 12 12
R eq = 30 + 60 (10 + R 0 + R ) = 30 + 60 (14 + R )
60(14 + R )
50 = 30 + 74 + R = 42 + 3R
74 + R
or R = 16 Ω
Chapter 2, Solution 42
5x 20
(a) Rab = 5 (8 + 20 30) = 5 (8 + 12) = = 4Ω
25
(b) Rab = 2 + 4 (5 + 3) 8 + 5 10 (6 + 4) = 2 + 4 4 + 5 5 = 2 + 2 + 2.5 = 6.5 Ω
Chapter 2, Solution 43
5x 20 400
(a) Rab = 5 20 + 10 40 = + = 4 + 8 = 12 Ω
25 50
−1
1 1 1 60
(b) 60 20 30 = + + = = 10Ω
60 20 30 6
80 + 20
Rab = 80 (10 + 10) = = 16 Ω
100
Chapter 2, Solution 44
(a) Convert T to Y and obtain
20 x 20 + 20 x10 + 10 x 20 800
R1 = = = 80 Ω
10 10
800
R2 = = 40 Ω = R3
20
The circuit becomes that shown below.
R1
a
R3
R2 5Ω
b
R1//0 = 0, R3//5 = 40//5 = 4.444 Ω
Rab = R2 / /(0 + 4.444) = 40 / /4.444 = 4Ω
(b) 30//(20+50) = 30//70 = 21 Ω
Convert the T to Y and obtain
20 x10 + 10 x 40 + 40 x 20 1400
R1 = = = 35Ω
40 40
1400 1400
R2 = = 70 Ω , R3 = = 140 Ω
20 10
The circuit is reduced to Ω shown below.
15 that
11 Ω R1
R2 R3
30 Ω 21 Ω
21 Ω
Combining the resistors in parallel
R1//15 =35//15=10.5, 30//R2=30//70 = 21
leads to the circuit below.
11 Ω 10.5 Ω
21 Ω 140 Ω
21 Ω 21 Ω
Coverting the T to Y leads to the circuit below.
11 Ω 10.5 Ω
R4
R5 R6
21 Ω
21x140 + 140 x 21 + 21x 21 6321
R4 = = = 301Ω = R6
21 21
6321
R5 = = 45.15
140
10.5//301 = 10.15, 301//21 = 19.63
R5//(10.15 +19.63) = 45.15//29.78 = 17.94
Rab = 11 + 17 .94 = 28.94Ω
Chapter 2, Solution 45
(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8
Rab = 5 + 50 + 4.8 = 59.8 Ω
(b) 12 and 60 ohm resistors are in parallel. Hence, 12//60 = 10 ohm. This 10 ohm
and 20 ohm are in series to give 30 ohm. This is in parallel with 30 ohm to give
30//30 = 15 ohm. And 25//(15+10) = 12.5. Thus
Rab = 5 + 12.8 + 15 = 32.5Ω
Chapter 2, Solution 46
30x 70 60 + 20
(a) Rab = 30 70 + 40 + 60 20 = + 40 +
100 80
= 21 + 40 + 15 = 76 Ω
(b) The 10-Ω, 50-Ω, 70-Ω, and 80-Ω resistors are shorted.
20x30
20 30 = = 12Ω
50
40x 60
40 60 = = 24
100
Rab = 8 + 12 + 24 + 6 + 0 + 4 = 54 Ω
Chapter 2, Solution 47
5x 20
5 20 = = 4Ω
25
6x3
6 3= = 2Ω
9
10 Ω 8Ω
a b
2Ω
4Ω
Rab = 10 + 4 + 2 + 8 = 24 Ω
Chapter 2, Solution 48
R 1 R 2 + R 2 R 3 + R 3 R 1 100 + 100 + 100
(a) Ra = = = 30
R3 10
Ra = Rb = Rc = 30 Ω
30x 20 + 30x50 + 20x 50 3100
(b) Ra = = = 103.3Ω
30 30
3100 3100
Rb = = 155Ω, R c = = 62Ω
20 50
Ra = 103.3 Ω, Rb = 155 Ω, Rc = 62 Ω
Chapter 2, Solution 49
RaRc 12 + 12
(a) R1 = = = 4Ω
Ra + Rb + Rc 36
R1 = R2 = R3 = 4 Ω
60x30
(b) R1 = = 18Ω
60 + 30 + 10
60 x10
R2 = = 6Ω
100
30x10
R3 = = 3Ω
100
R1 = 18Ω, R2 = 6Ω, R3 = 3Ω
Chapter 2, Solution 50
Using R ∆ = 3RY = 3R, we obtain the equivalent circuit shown below:
R
30mA 3R
3R 30mA 3R 3R/2
3R
R
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