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

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 15Ωthat shown below.

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
 3RxR 3
3R R = = R
4R 4
3R (3RxR ) /(4R ) = 3 /(4R )
3
3Rx R
3 3  3 2
3R  R + R  = 3R R =
4 4  2 3
3R + R = R
2
P = I2 R 800 x 10-3 = (30 x 10-3)2 R

R = 889 Ω

Chapter 2, Solution 51

(a) 30 30 = 15Ω and 30 20 = 30 x 20 /(50) = 12Ω

Rab = 15 (12 + 12) = 15x 24 /(39) = 9.31 Ω

a a
20 Ω 12 Ω
30 Ω
30 Ω 30 Ω 15 Ω
30 Ω
12 Ω
20 Ω
b b

(b) Converting the T-subnetwork into its equivalent ∆ network gives

Ra'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70 Ω

Rb'c' = 350/(10) = 35Ω, Ra'c' = 350/(20) = 17.5 Ω

Also 30 70 = 30 x 70 /(100) = 21Ω and 35/(15) = 35x15/(50) = 10.5

Rab = 25 + 17.5 (21 + 10.5) = 25 + 17.5 31.5
Rab = 36.25 Ω
30 Ω
30 Ω

25 Ω a’ 70 Ω b’
25 Ω 10 Ω 20 Ω a
a
15 Ω 17.5 Ω 35 Ω 15 Ω
5Ω
b
b c’ c’
Chapter 2, Solution 52

(a) We first convert from T to ∆ .

100 Ω
a 100 Ω
a
100 Ω 100 Ω 200 Ω
100 Ω 100 Ω
100 Ω
100 Ω R3
R1
100 Ω 200 Ω R2
100 Ω 100 Ω 100 Ω 100 Ω
100 Ω
b 100 Ω
b

100x 200 + 200x 200 + 200 x100 80000

R1 = = = 800Ω
100 100

R2 = R3 = 80000/(200) = 400
100x 400
But 100 400 = = 80Ω
500
We connect the ∆ to Y.

100 Ω Ra
100 Ω
a a
100 Ω 80 Ω 100 Ω
100 Ω Rb
800 Ω 100 Ω

100 Ω 80 Ω 100 Ω
100 Ω Rc
100 Ω
b b
80 x800 64,000 400
Ra = Rc = = = Ω
80 + 80 + 800 960 3
80x80 20
Rb = = Ω
960 3

We convert T to ∆ .
500/3 Ω 500/3 Ω
a a
100 Ω R2’
320/3 Ω
R1’
R3’
100 Ω
500/3 Ω 500/3 Ω
b b
 320 320
100 x100 + 100 x + 100 x
R 1' = 3 3 = 94,000 /(3) = 293.75Ω
320 320 /(3)
3

94,000 /(3)
R '2 = R 13 = = 313.33
100

940 /(3) x500 /(3)

940 /(30) 500 /(3) = = 108.796
1440 /(3)

293.75x 217.6
Rab = 293.75 (2 x108.796) = = 125 Ω
511.36

(b) Converting the Ts to ∆ s, we have the equivalent circuit below.

100 Ω 100 Ω
a
100 Ω 100 Ω
300 Ω 300 Ω
300 Ω 300 Ω
300 Ω 100 Ω 300 Ω 100 Ω

b 100 Ω
100 Ω

100 Ω 100 Ω
a
100 Ω

100 Ω

100 Ω 100 Ω
b

300 100 = 300 x100 /(400) = 75, 300 (75 + 75) = 300 x150 /(450) = 100
Rab = 100 + 100 300 + 100 = 200 + 100 x 300 /(400)
Rab = 2.75 Ω

100 Ω

100 Ω
300 Ω
300 Ω
300 Ω 100 Ω

100 Ω
Chapter 2, Solution 53

(a) Converting one ∆ to T yields the equivalent circuit below:

30 Ω
a’
4Ω
20 Ω 20 Ω
a c’
60 Ω
5Ω 80 Ω
b’
b

40 x10 10 x50 40x50

Ra'n = = 4Ω, R b 'n = = 5Ω, R c 'n = = 20Ω
40 + 10 + 50 100 100
Rab = 20 + 80 + 20 + (30 + 4) (60 + 5) = 120 + 34 65
Rab = 142.32 Ω

(a) We combine the resistor in series and in parallel.

30x 60
30 (30 + 30) = = 20Ω
90

We convert the balanced ∆ s to Ts as shown below:

a a
30 Ω 30 Ω
30 Ω 10 Ω
30 Ω
20 Ω 10 Ω 10 Ω
30 Ω
b
30 Ω
10 Ω

10 Ω 10 Ω 20 Ω
b

Rab = 10 + (10 + 10) (10 + 20 + 10) + 10 = 20 + 20 40

Rab = 33.33 Ω

Chapter 2, Solution 54

(a) Rab = 50 + 100 / /(150 + 100 + 150 ) = 50 + 100 / /400 = 130 Ω

(b) Rab = 60 + 100 / /(150 + 100 + 150 ) = 60 + 100 / /400 = 140 Ω

Chapter 2, Solution 55

We convert the T to ∆ .

I0
a I0
a
24 V 20 Ω 60 Ω
40 Ω 140 Ω
+ 24 V 60 Ω
-
10 Ω 50 Ω + 35 Ω
-
20 Ω 70 Ω
70 Ω
b
Req b
Req
R R + R 2 R 3 + R 3 R 1 20 x 40 + 40 x10 + 10 x 20 1400
Rab = 1 2 = = = 35Ω
R3 40 40
Rac = 1400/(10) = 140Ω, Rbc = 1400/(40) = 35Ω
70 70 = 35 and 140 160 = 140x60/(200) = 42
Req = 35 (35 + 42) = 24.0625Ω
I0 = 24/(Rab) = 0.9774A

Chapter 2, Solution 56

We need to find Req and apply voltage division. We first tranform the Y network to ∆ .
30 Ω 30 Ω

16 Ω 15 Ω 10 Ω 16 Ω 37.5 Ω
a b
+ + 30 Ω 20 Ω
20 Ω
100 V 35 Ω 12 Ω 100 V 35 Ω
45 Ω
- -
c
Req Req

15x10 + 10x12 + 12x15 450

Rab = = = 37.5Ω
12 12
Rac = 450/(10) = 45Ω, Rbc = 450/(15) = 30Ω

Combining the resistors in parallel,

 30||20 = (600/50) = 12 Ω,

37.5||30 = (37.5x30/67.5) = 16.667 Ω

35||45 = (35x45/80) = 19.688 Ω

Req = 19.688||(12 + 16.667) = 11.672Ω

By voltage division,

11.672
v = 100 = 42.18 V
11.672 + 16

Chapter 2, Solution 57

4Ω a 2Ω

27 Ω
1Ω
18 Ω
36 Ω c
e
b
d 7Ω
14 Ω
10 Ω
28 Ω
f

6x12 + 12x8 + 8x 6 216

Rab = = = 18 Ω
12 12
Rac = 216/(8) = 27Ω, Rbc = 36 Ω
4x 2 + 2x8 + 8x 4 56
Rde = = 7Ω
8 8
Ref = 56/(4) = 14Ω, Rdf = 56/(2) = 28 Ω

Combining resistors in parallel,

280 36x 7
10 28 = = 7.368Ω, 36 7 = = 5.868Ω
38 43
27 x 3
27 3 = = 2.7Ω
30
 4Ω
4Ω
1.829 Ω
18 Ω 2.7 Ω
3.977 Ω
0.5964 Ω
5.868 Ω 14 Ω
7.568 Ω
7.568 Ω 14 Ω

18x 2.7 18x 2.7

R an = = = 1.829 Ω
18 + 2.7 + 5.867 26.567
18x5.868
R bn = = 3.977 Ω
26.567
5.868x 2.7
R cn = = 0.5904 Ω
26.567
R eq = 4 + 1.829 + (3.977 + 7.368) (0.5964 + 14)
= 5.829 + 11.346 14.5964 = 12.21 Ω
i = 20/(Req) = 1.64 A

Chapter 2, Solution 58

The resistor of the bulb is 120/(0.75) = 160Ω

40 Ω 2.25 A 1.5 A

+ 90 V - 0.75 A +
+ 160 Ω 80 Ω
VS - 120

Once the 160Ω and 80Ω resistors are in parallel, they have the same voltage 120V.
Hence the current through the 40Ω resistor is

40(0.75 + 1.5) = 2.25 x 40 = 90

Thus

vs = 90 + 120 = 210 V
Chapter 2, Solution 59

Total power p = 30 + 40 + 50 + 120 W = vi

or i = p/(v) = 120/(100) = 1.2 A

Chapter 2, Solution 60

p = iv i = p/(v)
i30W = 30/(100) = 0.3 A
i40W = 40/(100) = 0.4 A
i50W = 50/(100) = 0.5 A

Chapter 2, Solution 61

There are three possibilities

(a) Use R1 and R2:

R = R 1 R 2 = 80 90 = 42.35Ω
p = i2R
i = 1.2A + 5% = 1.2 ± 0.06 = 1.26, 1.14A
p = 67.23W or 55.04W, cost = $1.50

(b) Use R1 and R3:

R = R 1 R 3 = 80 100 = 44.44 Ω
p = I2R = 70.52W or 57.76W, cost = $1.35

(c) Use R2 and R3:

R = R 2 R 3 = 90 100 = 47.37Ω
p = I2R = 75.2W or 61.56W, cost = $1.65

Note that cases (b) and (c) give p that exceed 70W that can be supplied.
Hence case (a) is the right choice, i.e.

R1 and R2

Chapter 2, Solution 62

pA = 110x8 = 880 W, pB = 110x2 = 220 W

Energy cost = $0.06 x 360 x10 x (880 + 220)/1000 = $237.60

Chapter 2, Solution 63

Use eq. (2.61),

Im 2 x10 −3 x100
Rn = Rm = = 0.04Ω
I − Im 5 − 2 x10 −3
In = I - Im = 4.998 A
p = I 2n R = (4.998) 2 (0.04) = 0.9992 ≅ 1 W

Chapter 2, Solution 64

110
When Rx = 0, i x = 10A R= = 11 Ω
10
110
When Rx is maximum, ix = 1A R + Rx = = 110 Ω
1
i.e., Rx = 110 - R = 99 Ω
Thus, R = 11 Ω, Rx = 99 Ω

Chapter 2, Solution 65

Vfs 50
Rn = − Rm = − 1 kΩ = 4 kΩ
I fs 10mA

Chapter 2, Solution 66

1
20 kΩ/V = sensitivity =
I fs
1
i.e., Ifs = kΩ / V = 50 µA
20
V
The intended resistance Rm = fs = 10(20kΩ / V) = 200kΩ
I fs
V 50 V
(a) R n = fs − R m = − 200 kΩ = 800 kΩ
i fs 50 µA
(b) p = I fs2 R n = (50 µA) 2 (800 kΩ) = 2 mW
Chapter 2, Solution 67

(a) By current division,

i0 = 5/(5 + 5) (2 mA) = 1 mA
V0 = (4 kΩ) i0 = 4 x 103 x 10-3 = 4 V

(b) 4k 6k = 2.4kΩ. By current division,

5
i '0 = (2mA) = 1.19 mA
1 + 2.4 + 5
v '0 = (2.4 kΩ)(1.19 mA) = 2.857 V

v 0 − v '0 1.143
(c) % error = x 100% = x100 = 28.57%
v0 4
(d) 4k 30 kΩ = 3.6 kΩ. By current division,
5
i '0 = (2mA) = 1.042mA
1 + 3.6 + 5
v '0 (3.6 kΩ)(1.042 mA) = 3.75V
v − v '0 0.25x100
% error = x100% = = 6.25%
v0 4

Chapter 2, Solution 68

(a) 40 = 24 60Ω
4
i= = 0.1 A
16 + 24
4
(b) i' = = 0.09756 A
16 + 1 + 24
0.1 − 0.09756
(c) % error = x100% = 2.44%
0.1
Chapter 2, Solution 69

With the voltmeter in place,

R2 Rm
V0 = VS
R1 + R S + R 2 R m
where Rm = 100 kΩ without the voltmeter,
R2
V0 = VS
R1 + R 2 + R S

100
(a) When R2 = 1 kΩ, R m R 2 = kΩ
101
100
V0 = 101 (40) = 1.278 V (with)
100
101 + 30
1
V0 = (40) = 1.29 V (without)
1 + 30
1000
(b) When R2 = 10 kΩ, R 2 R m = = 9.091kΩ
110
9.091
V0 = (40) = 9.30 V (with)
9.091 + 30
10
V0 = (40) = 10 V (without)
10 + 30
(c) When R2 = 100 kΩ, R 2 R m = 50kΩ
50
V0 = (40) = 25 V (with)
50 + 30
100
V0 = (40) = 30.77 V (without)
100 + 30

Chapter 2, Solution 70

(a) Using voltage division,

12
va = (25) = 15V
12 + 8
10
vb = (25) = 10V
10 + 15
vab = va − vb = 15 − 10 = 5V
 (b)

+ 8k Ω 15k Ω
25 V
- a b

12k Ω 10k Ω

va = 0 , vb = 10V , vab = va − vb = 0 − 10 = −10V

Chapter 2, Solution 71
R1

iL

Vs +
− RL

Given that vs = 30 V, R1 = 20 Ω, IL = 1 A, find RL.

vs 30
v s = i L ( R1 + R L ) 
→ RL = − R1 = − 20 = 10Ω
iL 1
Chapter 2, Solution 72

The system can be modeled as shown.

12A

9V R R ••• R
-

The n parallel resistors R give a combined resistance of R/n. Thus,

R 12 xR 12 x15
9 = 12 x 
→ n= = = 20
n 9 9

Chapter 2, Solution 73

By the current division principle, the current through the ammeter will be
one-half its previous value when

R = 20 + Rx
65 = 20 + Rx Rx = 45 Ω

Chapter 2, Solution 74

With the switch in high position,

6 = (0.01 + R3 + 0.02) x 5 R3 = 1.17 Ω

At the medium position,

6 = (0.01 + R2 + R3 + 0.02) x 3 R2 + R3 = 1.97

or R2 = 1.97 - 1.17 = 0.8 Ω

At the low position,

6 = (0.01 + R1 + R2 + R3 + 0.02) x 1 R1 + R2 + R3 = 5.97

R1 = 5.97 - 1.97 = 4 Ω
Chapter 2, Solution 75
R 100 Ω
M

+ 12 Ω
VS
-

(a) When Rx = 0, then

t E2 2
Im = Ifs = R2 = − Rm= − 100 = 19.9kΩ
R + Rm I fs 0.1x10 3
I fs
(b) For half-scale deflection, Im = = 0.05mA
2
E E 2
Im = Rx = − (R + R m ) = − 20kΩ = 20 kΩ
R + Rm + Rx Im 0.05x10 −3

Chapter 2, Solution 76

For series connection, R = 2 x 0.4Ω = 0.8Ω

V 2 (120) 2
p= = = 18 kΩ (low)
R 0.8
For parallel connection, R = 1/2 x 0.4Ω = 0.2Ω
V 2 (120) 2
p= = = 72 kW (high)
R 0.2

Chapter 2, Solution 77

(a) 5 Ω = 10 10 = 20 20 20 20
i.e., four 20 Ω resistors in parallel.

(b) 311.8 = 300 + 10 + 1.8 = 300 + 20 20 + 1.8

i.e., one 300Ω resistor in series with 1.8Ω resistor and
a parallel combination of two 20Ω resistors.

(c) 40kΩ = 12kΩ + 28kΩ = 24 24k + 56k 50k

i.e., Two 24kΩ resistors in parallel connected in series with two
50kΩ resistors in parallel.

(d) 42.32kΩ = 42l + 320

= 24k + 28k = 320
= 24k = 56k 56k + 300 + 20
i.e., A series combination of 20Ω resistor, 300Ω resistor, 24kΩ
resistor and a parallel combination of two 56kΩ resistors.
Chapter 2, Solution 78

The equivalent circuit is shown below:

R

VS + +
- V0 (1-α)R
-

(1 − α)R
V0 = VS = (1 − α )R 0 VS
R + (1 − α)R
V0
= (1 − α)R
VS

Chapter 2, Solution 79

Since p = v2/R, the resistance of the sharpener is

R = v2/(p) = 62/(240 x 10-3) = 150Ω
I = p/(v) = 240 mW/(6V) = 40 mA

Since R and Rx are in series, I flows through both.

IRx = Vx = 9 - 6 = 3 V
Rx = 3/(I) = 3/(40 mA) = 3000/(40) = 75 Ω

Chapter 2, Solution 80

The amplifier can be modeled as a voltage source and the loudspeaker as a resistor:

V + V +
- R1 R2
-

Case 1 Case 2

V 2 p2 R1 R1 10
Hence p = , = p2 = p1 = (12) = 30 W
R p1 R 2 R2 4
Chapter 2, Solution 81

Let R1 and R2 be in kΩ.

R eq = R 1 + R 2 5 (1)
V0 5 R2
= (2)
VS 5 R 2 + R 1

5 R1 5R 2
From (1) and (2), 0.05 = 2 = 5 R2 = or R2 = 3.33 kΩ
40 5+ R2
From (1), 40 = R1 + 2 R1 = 38 kΩ

Thus R1 = 38 kΩ, R2 = 3.33 kΩ

Chapter 2, Solution 82

(a) 10 Ω

40 Ω
10 Ω

80 Ω

1 2

R12

50
R12 = 80 + 10 (10 + 40) = 80 + = 88.33 Ω
6

(b)
3
20 Ω
10 Ω

40 Ω
10 Ω R13

80 Ω

R13 = 80 + 10 (10 + 40) + 20 = 100 + 10 50 = 108.33 Ω

(c) 4

20 Ω
10 Ω
R14
40 Ω
10 Ω

80 Ω

R14 = 80 + 0 (10 + 40 + 10) + 20 = 80 + 0 + 20 = 100 Ω

Chapter 2, Solution 83

The voltage across the tube is 2 x 60 mV = 0.06 V, which is negligible

compared with 24 V. Ignoring this voltage amp, we can calculate the
current through the devices.

p1 45mW
I1 = = = 5mA
V1 9V
p 480mW
I2 = 2 = = 20mA
V2 24
60 mA i2 = 20 mA
iR1
R1
+ i1 = 5 mA
24 V
- R2

iR2

By applying KCL, we obtain

I R1 = 60 − 20 = 40 mA and I R 2 = 40 − 5 = 35 mA

15V
Hence, I R1 R1 = 24 - 9 = 15 V R1 = = 375 Ω
40mA
9V
I R 2 R 2 = 9V R2 = = 257.14 Ω
35mA