252980330 Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual

Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual


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circuits-alexander-5th-edition-solutions-manual

Chapter 2, Solution 1. Design a problem, complete with a solution, to help students to

better understand Ohm’s Law. Use at least two resistors and one voltage source. Hint,

you could use both resistors at once or one at a time, it is up to you. Be creative.

Although there is no correct way to work this problem, this is an example based on the

same kind of problem asked in the third edition.

Problem

The voltage across a 5-kresistor is 16 V. Find the current through the resistor.

Solution

v = iR i = v/R = (16/5) mA = 3.2 mA

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Chapter 2, Solution 2

p = v2/R R = v

2/p = 14400/60 = 240 ohms





For silicon, 6.4x102 -m. A r

2 . Hence,

2 2

R L

L

r 2

L

6.4x10 x4x10

0.033953

A r 2 R x240

r = 184.3 mm






(a) i = 40/100 = 400 mA

(b) i = 40/250 = 160 mA



n = 9; l = 7; b = n + l – 1 = 15



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n = 12; l = 8; b = n + l –1 = 19





6 branches and 4 nodes



Chapter 2, Solution 8. Design a problem, complete with a solution, to help other students

to better understand Kirchhoff’s Current Law. Design the problem by specifying values

of ia, ib, and ic, shown in Fig. 2.72, and asking them to solve for values of i1, i2, and i3.

Be careful specify realistic currents.



Problem

Use KCL to obtain currents i1, i2, and i3 in the circuit shown in Fig. 2.72.

Solution

12 A

a

i1

8 A b

i2

i3

12 A

c 9 A d

At node a, 8 = 12 + i1 i 1 = - 4A

At node c, 9 = 8 + i2 i 2 = 1A At node d, 9 = 12 + i3 i 3 = -3A




At A, 1+6–i1 = 0 or i1 = 1+6 = 7 A

At B, –6+i2+7 = 0 or i2 = 6–7 = –1 A

At C, 2+i3–7 = 0 or i3 = 7–2 = 5 A




2 –8A 4A

i2

1 i1 3

–6A

At node 1, –8–i1–6 = 0 or i1 = –8–6 = –14 A

At node 2, –(–8)+i1+i2–4 = 0 or i2 = –8–i1+4 = –8+14+4 = 10 A




V1 15 0

5 2 V2 0

V1 6 V

V2 3 V




+ 30v –

– 50v +

+ 20v –

loop 2

+ v2 –

+

40v

-

+

loop 1 v1

–

+

loop 3 v3

–

For loop 1, –40 –50 +20 + v1 = 0 or v1 = 40+50–20 = 70 V

For loop 2, –20 +30 –v2 = 0 or v2 = 30–20 = 10 V

For loop 3, –v1 +v2 +v3 = 0 or v3 = 70–10 = 60 V




2A

I2 7A I4

1 2 3 4 4A

I1

3A I3

At node 2,

3 7 I2 0

At node 1,

I1 I2 2

At node 4,

2 I4 4

At node 3,

7 I4 I3

Hence,

I2 10 A

I1 2 I2 12 A

I4 2 4 2 A

I3 7 2 5 A

I1 12 A, I2 10 A, I3 5 A, I4 2 A




+ + -

3V V1 I4 V2

- I3 - + 2V - +

- + V3 - + +

4V

I2 - V4 I1 5V + -

For mesh 1,

V4 2 5 0

For mesh 2,

4 V3 V4 0

For mesh 3,

3 V1 V3 0

For mesh 4,

V1 V2 2 0

Thus,

V4 7V

V3 4 7 11V

V1 V3 3 8V

V2 V1 2 6V

V1 8V , V2 6V , V3 11V , V4 7V




Calculate v and ix in the circuit of Fig. 2.79.

12 + 16 V –

+

10 V _

+ v – ix

+

4 V

_

+ 3 ix

_

Figure 2.79

For Prob. 2.15.

Solution

For loop 1, –10 + v +4 = 0, v = 6 V

For loop 2, –4 + 16 + 3ix =0, ix = –4 A




Determine Vo in the circuit in Fig. 2.80.

16 14

+

10 V +

_

+

Vo _

_

25 V

Figure 2.80

For Prob. 2.16.

Solution

Apply KVL,

–10 + (16+14)I + 25 = 0 or 30I = 10–25 = – or I = –15/30 = –500 mA

Also, –10 + 16I + Vo = 0 or Vo = 10 – 16(–0.5) = 10+8 = 18 V




Applying KVL around the entire outside loop we get,

–24 + v1 + 10 + 12 = 0 or v1 = 2V

Applying KVL around the loop containing v2, the 10-volt source, and the

12-volt source we get,

v2 + 10 + 12 = 0 or v2 = –22V

Applying KVL around the loop containing v3 and the 10-volt source we

get,

–v3 + 10 = 0 or v3 = 10V




Applying KVL,

-30 -10 +8 + I(3+5) = 0

8I = 32 I = 4A

-Vab + 5I + 8 = 0 Vab = 28V




p

Applying KVL around the loop, we obtain

–(–8) – 12 + 10 + 3i = 0 i = –2A

Power dissipated by the resistor:

= i2R = 4(3) = 12W

3

Power supplied by the sources:

p12V = 12 ((–2)) = –24W

p10V = 10 (–(–2)) = 20W

p8V = (–8)(–2) = 16W




Determine io in the circuit of Fig. 2.84.

io 22

54V + + 5io

–

Figure 2.84

For Prob. 2.20

Solution

Applying KVL around the loop,

–54 + 22io + 5io = 0 io = 4A




Applying KVL,

-15 + (1+5+2)I + 2 Vx = 0

But Vx = 5I, -15 +8I + 10I =0, I = 5/6

Vx = 5I = 25/6 = 4.167 V




Find Vo in the circuit in Fig. 2.86 and the power absorbed by the dependent

source.

10 V1

10

+ Vo

25A

2Vo

Figure 2.86

For Prob. 2.22

Solution

At the node, KCL requires that [–Vo/10]+[–25]+[–2Vo] = 0 or 2.1Vo = –25

or Vo = –11.905 V

The current through the controlled source is i = 2V0 = –23.81 A

and the voltage across it is V1 = (10+10) i0 (where i0 = –V0/10) = 20(11.905/10) = 23.81 V.

Hence,

pdependent source = V1(–i) = 23.81x(–(–23.81)) = 566.9 W

Checking, (25–23.81)2(10+10) + (23.81)(–25) + 566.9 = 28.322 – 595.2 + 566.9

= 0.022 which is equal zero since we are using four places of accuracy!




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 1

+ vx –

20A 2 3

Applying current division,

ix = [2/(2+1+3)]20 = 6.667 and vx = 1x6.667 = 6.667 V

i 2

(6 A) 2 A, x

2 1 3

vx 1ix 2V

The current through the 1.2- resistor is 0.5ix = 3.333 A. The voltage across the 12- resistor is 3.333 x 4.8 = 16V. Hence the power absorbed by the 12-ohm

resistor is equal to

(16)2/12 = 21.33 W




3

1

4

3

(a) I0 = Vs

R1 R2

V0 I0 R

R = Vs

R 3R 4

V0 R3 R4

R1 R 2 R 3 R 4

Vs R R2 R R4

(b) If R1 = R2 = R3 = R4 = R,

V0

R

10

= 40

VS 2R 2 4




V0 = 5 x 10-3

x 10 x 103

= 50V

Using current division,

I20 5

5 20

(0.01x50) 0.1 A

V20 = 20 x 0.1 kV = 2 kV

p20 = I20 V20 = 0.2 kW


Chapter 2, Problem 26


For the circuit in Fig. 2.90, io = 3 A. Calculate ix and the total power absorbed by

the entire circuit.

ix 10 io

8 4 2 16

Figure 2.90

For Prob. 2.26.

Solution

If i16= io = 3A, then v = 16x3 = 48 V and i8 = 48/8 = 6A; i4 = 48/4 = 12A; and

i2 = 48/2 = 24A.

Thus,

ix = i8 + i4 + i2 + i16 = 6 + 12 + 24 + 3 = 45 A

p = (45)210 + (6)

28 + (12)

24 + (24)

22 + (3)

216 = 20,250 + 288 + 576 +1152 + 144

= 20250 + 2106 = 22.356 kW.


Chapter 2, Problem 27


Calculate Io in the circuit of Fig. 2.91.

8

+ Io

10V

3 6

Figure 2.91

For Prob. 2.27.

Solution

The 3-ohm resistor is in parallel with the c-ohm resistor and can be replaced by a

[(3x6)/(3+6)] = 2-ohm resistor. Therefore,

Io = 10/(8+2) = 1 A.



Chapter 2, Solution 28 Design a problem, using Fig. 2.92, to help other students better

understand series and parallel circuits.



Problem

Find v1, v2, and v3 in the circuit in Fig. 2.92.

Solution

We first combine the two resistors in parallel

15 10 6

We now apply voltage division,

v1 = 14

14 6

(40) 28 V

v2 = v3 = 6

14 6

(40) 12 V

Hence, v1 = 28 V, v2 = 12 V, vs = 12 V




All resistors in Fig. 2.93 are 5 each. Find Req.

Req

Figure 2.93

For Prob. 2.29.

Solution

Req = 5 + 5||[5+5||(5+5)] = 5 + 5||[5+(5x10/(5+10))] = 5+5||(5+3.333) = 5 +

41.66/13.333

= 8.125 Ω



Chapter 2, Problem 30.

Find Req for the circuit in Fig. 2.94.

25 180

60

Req

60

Figure 2.94

For Prob. 2.30.

Solution

We start by combining the 180-ohm resistor with the 60-ohm resistor which in

turn is in parallel with the 60-ohm resistor or = [60(180+60)/(60+180+60)] = 48.

Thus,

Req = 25+48 = 73 Ω.




Req 3 2 // 4 //1 3 1

1/ 2 1/ 4 1

3.5714

i1 = 200/3.5714 = 56 A

v1 = 0.5714xi1 = 32 V and i2 = 32/4 = 8 A

i4 = 32/1 = 32 A; i5 = 32/2 = 16 A; and i3 = 32+16 = 48 A




Find i1 through i4 in the circuit in Fig. 2.96.

60 i4 i2 200

40 50

i3 i1

16 A

Figure 2.96

For Prob. 2.32.

Solution

We first combine resistors in parallel.

40 60 40x60

100

24 and

50 200 50x200

250

40

Using current division principle,

i1 i2 24

24 40

(16) 6A,i3

i4 40

(16) 10A 64

i 200

(6) –4.8 A and i 1

250 2

50 250

(6) –1.2 A

i3 60

(10) 100

–6 A and i4

40 (10) –4 A

100




Combining the conductance leads to the equivalent circuit below

i 4S i

+

9A v

- 1S 4S

+

9A v 1S 2S -

6S 3S 6x3

2S and 2S + 2S = 4S 9

Using current division,

i 1

1 1

2

(9) 6 A, v = 3(1) = 3 V




160//(60 + 80 + 20)= 80 ,

160//(28+80 + 52) = 80

Req = 20+80 = 100 Ω

I = 200/100 = 2 A or p = VI = 200x2 = 400 W.




-

i

70 +

V1

30

200V +

i1 - Io

a b +

20

i2

Vo 5 -

Combining the resistors that are in parallel,

70 30 70x30 21 ,

100

20 5 20x5 4

25

i = 200

21 4

8 A

v1 = 21i = 168 V, vo = 4i = 32 V v1 vo

i1 = 2.4 A, i2 = 1.6 A

70 20

At node a, KCL must be satisfied

i1 = i2 + Io 2.4 = 1.6 + Io Io = 0.8 A

Hence,

vo = 32 V and Io = 800 mA




20//(30+50) = 16, 24 + 16 = 40, 60//20 = 15

Req = 80+(15+25)40 = 80+20 = 100 Ω

i = 20/100 = 0.2 A

If i1 is the current through the 24-resistor and io is the current through the 50-

resistor, using current division gives

i1 = [40/(40+40)]0.2 = 0.1 and io = [20/(20+80)]0.1 = 0.02 A or

vo = 30io = 30x0.02 = 600 mV.




Applying KVL,

-20 + 10 + 10I – 30 = 0, I = 4

10 RI

R 10

2.5 I




20//80 = 80x20/100 = 16, 6//12 = 6x12/18 = 4

The circuit is reduced to that shown below.

2.5 4

15 16

60

Req

(4 + 16)//60 = 20x60/80 = 15

Req = 2.5+15||15 = 2.5+7.5 = 10 Ω and

io = 35/10 = 3.5 A.




(a) We note that the top 2k-ohm resistor is actually in parallel with the first 1k-ohm

resistor. This can be replaced (2/3)k-ohm resistor. This is now in series with the second

2k-ohm resistor which produces a 2.667k-ohm resistor which is now in parallel with the

second 1k-ohm resistor. This now leads to,

Req = [(1x2.667)/3.667]k = 727.3 Ω.

(b) We note that the two 12k-ohm resistors are in parallel producing a 6k-ohm resistor.

This is in series with the 6k-ohm resistor which results in a 12k-ohm resistor which is in

parallel with the 4k-ohm resistor producing,

Req = [(4x12)/16]k = 3 kΩ.




Req = 8 4 (2 6 3) 8 2 10

I = 15 R eq

15

10

1.5 A




Let R0 = combination of three 12resistors in parallel

1

1

R o 12

1

1 12 12

Ro = 4

R eq 30 60 (10 R 0 R) 30 60 (14 R)

50 30 60(14 R)

74 R 74 + R = 42 + 3R

or R = 16




(a) Rab = 5 (8 20 30) 5 (8 12)

5x20

25

4

(b) Rab = 2 4 (5 3) 8 5 10 4 2 4 4 5 2.857 2 2 1.8181 5.818




1

1

(a) Rab = 5 20 10 40

5x20

25

400

50

4 8 12

(b) 60 20 30 1 1

60 10

60 20 30 6

Rab = 80 (10 10) 80 20

100 16




For the circuits in Fig. 2.108, obtain the equivalent resistance at terminals a-b.

5 20

a

2 3

b

Figure 2.108

For Prob. 2.44

Solution

First we note that the 5 Ω and 20 Ω resistors are in parallel and can be replaced by

a 4 Ω [(5x20)/(5+20)] resistor which in now in series with the 2 Ω resistor and

can be replaced by a 6 Ω resistor in parallel with the 3 Ω resistor thus,

Rab = [(6x3)/(6+3)] = 2 Ω.




(a) 10//40 = 8, 20//30 = 12, 8//12 = 4.8

Rab 5 50 4.8 59.8

(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 32.5




Req = 12 + 5||20 + [1/((1/15)+(1/15)+(1/15))] + 5 + 24||8 = 12 + 4 + 5 + 5 + 6 = 32 Ω

I = 80/32 = 2.5 A




5 20

6 3

5x20 4

25

6x3 2

9

10 8

a b

4 2

Rab = 10 + 4 + 2 + 8 = 24




(a) Ra = R1R 2 R 2 R 3 R 3 R1

100 100 100

30

R 3 10

Ra = Rb = Rc = 30

(b) R 30x 20 30x50 20x50

3100

103.3

a 30

R 3100

155, b 20

30

R 3100

62c 50

(a) R1 =

R a R b R c

36

R1 = R2 = R3 = 4




R1

60 30 10 R

60x10 6

R 30x10

3

Ra = 103.3 , Rb = 155 , Rc = 62

R a R c 12 *12

4




(b)

60x30 18

2 100

3 100

R1 = 18, R2 = 6, R3 = 3



2.50 Design a problem to help other students better understand wye-delta transformations using

Fig. 2.114.

Although there is no correct way to work this problem, this is an example based on the same

kind of problem asked in the third edition.

Problem

What value of R in the circuit of Fig. 2.114 would cause the current source to deliver 800 mW to

the resistors.

Solution

Using R = 3RY = 3R, we obtain the equivalent circuit shown below:

30mA

R 3R

3R 3R

R

30mA 3R

3R/2

3R R 3RxR

3 R

4R 4

3Rx 3

R 3R 3

R 3

R 3R 3

R 2 = R

4 4

2 3R

3 R

2

P = I2R 800 x 10

-3 = (30 x 10

-3)2

R

R = 889




(a) 30 30 15 and 30 20 30x20 /(50) 12

Rab = 15 (12 12) 15x24 /(39) 9.231

a

30

30

30

30

a

20

20

15

12

12

b b

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

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

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

Also 30 70 30x70 /(100) 21 and 35/(15) = 35x15/(50) = 10.5

Rab = 25 + 17.5 (21 10.5) 25 17.5 31.5

Rab = 36.25

30 30

25 70

25 a

10

5

20

15

a a’ b’

17.5 35 15

b b

c’ c’




Converting the wye-subnetwork to delta-subnetwork, we obtain the circuit below.

3 9

9 3 9

3 3

3

3 6

3//1 = 3x1/4 = 0.75, 2//1 =2x1/3 = 0.6667. Combining these resistances leads to the

circuit below. 3

2.25

3

9 2.25

3 2

We now convert the wye-subnetwork to the delta-subnetwork.

Ra = [(2.25x3+2.25x3+2.25x2.25)/3] = 6.188 Ω

Rb = Rc = 18.562/2.25 = 8.25 Ω

This leads to the circuit below.

3

8.25

3

6.188 9

8.25

2



R = 9||6.188+8.25||2 = 3.667+1.6098 = 5.277

Req = 3+3+8.25||5.277 = 9.218 Ω.




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

20 a

b

30 a’ 4

60

5 b’

20

c’

80

Ra'n = 40x10

40 10 50

4, R b'n

10x50

100

5, R c'n

40x50

100

20

Rab = 20 + 80 + 20 + (30 4) (60 5) 120 34 65

Rab = 142.32

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

30 (30 30) 30x60

2090

We convert the balanced s to Ts as shown below:

a

30

b

30

30

30

30

20

30

a

10

10

10 10

10

b

10 20

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

Rab = 33.33




(a) Rab 50 100 / /(150 100 150) 50 100 / /400 130

(b) Rab 60 100 / /(150 100 150) 60 100 / /400 140




We convert the T to .

24 V

+ -

I0

Req

a

20

40

10

20

b

60

50

24 V

+ -

I0

Req

a

35

b

140

70

60

70

Rab =

R1R 2 R 2 R 3 R 3 R1

20x40 40x10 10x20

1400

35

R 3 40 40

Rac = 1400/(10) = 140, Rbc = 1400/(20) = 70

70 70 35 and 140 160 140x60/(200) = 42

Req = 35 (35 42) 24.0625

I0 = 24/(Rab) = 997.4mA




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

30 30

+

100 V

-

Req

16 35

15

12

10

20

+

100 V

-

Req

16 a

35

37.5 b

30

45

c

20

Rab = 15x10 10x12 12x15

12

450

12

37.5

Rac = 450/(10) = 45, Rbc = 450/(15) = 30

Combining the resistors in parallel,

30||20 = (600/50) = 12 ,

37.5||30 = (37.5x30/67.5) = 16.667

35||45 = (35x45/80) = 19.688

Req = 19.688||(12 + 16.667) = 11.672

By voltage division,

v = 11.672

11.672 16

100

= 42.18 V




4 a

18

b

d

2

27

36

7

1

c e

14

10 28

f

Rab = 6x12 12x8 8x6

12

216

12

18

Rac = 216/(8) = 27, Rbc = 36

Rde = 4x2 2x8 8x4

8

56 7

8

Ref = 56/(4) = 14, Rdf = 56/(2) = 28

Combining resistors in parallel,

10 28 280

7.368, 38

27 3 27x3

2.730

36 7 36x7

5.86843

4

7.568

18

5.868

2.7

14

4

1.829

3.977 0.5964

7.568 14

R 18x 2.7

18x 2.7

1.829

an 18 2.7 5.867 26.567

R bn

R cn

18x5.868

3.977

26.567

5.868x2.7

0.5904




26.567



R eq 4 1.829 (3.977 7.368) (0.5964 14)

5.829 11.346 14.5964

i = 20/(Req) = 1.64 A

12.21




The resistance of the bulb is (120)2/60 = 240

40 2 A 1.5 A

+

VS -

+ 90 V - 0.5 A

240

+ 120 V

-

80

Once the 160and 80resistors are in parallel, they have the same voltage

120V. Hence the current through the 40resistor is equal to 2 amps.

40(0.5 + 1.5) = 80 volts.

Thus

vs = 80 + 120 = 200 V.




Three light bulbs are connected in series to a 120-V source as shown in Fig.

2.123. Find the current I through each of the bulbs. Each bulb is rated at 120

volts. How much power is each bulb absorbing? Do they generate much light?

Figure 2.123

For Prob. 2.59.

Solution

Using p = v2/R, we can calculate the resistance of each bulb.

R30W = (120)2/30 = 14,400/30 = 480 Ω

R40W = (120)2/40 = 14,400/40 = 360 Ω

R50W = (120)2/50 = 14,400/50 = 288 Ω

The total resistance of the three bulbs in series is 480+360+288 = 1128 Ω.

The current flowing through each bulb is 120/1128 = 0.10638 A.

p30 = (0.10638)2480 = 0.011317x480 = 5.432 W.

p40 = (0.10638)2360 = 0.011317x360 = 4.074 W.

p50 = (0.10638)2288 = 0.011317x288 = 3.259 W.

Clearly these values are well below the rated powers of each light bulb so we

would not expect very much light from any of them. To work properly, they need

to be connected in parallel.




If the three bulbs of Prob. 2.59 are connected in parallel to the 120-V source,

calculate the current through each bulb.

Solution

Using p = v2/R, we can calculate the resistance of each bulb.

R30W = (120)2/30 = 14,400/30 = 480 Ω

R40W = (120)2/40 = 14,400/40 = 360 Ω

R50W = (120)2/50 = 14,400/50 = 288 Ω

The current flowing through each bulb is 120/R.

i30 = 120/480 = 250 mA.

i40 = 120/360 = 333.3 mA.

i30 = 120/288 = 416.7 mA.

Unlike the light bulbs in 2.59, the lights will glow brightly!




There are three possibilities, but they must also satisfy the current range of 1.2 +

0.06 = 1.26 and 1.2 – 0.06 = 1.14.

(a) Use R1 and R2:

R = R1 R 2 80 90 42.35

p = i2R = 70W

i2

= 70/42.35 = 1.6529 or i = 1.2857 (which is outside our range)

cost = $0.60 + $0.90 = $1.50

(b) Use R1 and R3:

R = R1 R 3 80 100 44.44

i2

= 70/44.44 = 1.5752 or i = 1.2551 (which is within our range),

cost = $1.35

(c) Use R2 and R3:

R = R 2 R 3 90 100 47.37

i2

= 70/47.37 = 1.4777 or i = 1.2156 (which is within our range),

cost = $1.65

Note that cases (b) and (c) satisfy the current range criteria and (b) is the

cheaper of the two, hence the correct choice is:

R1 and R3




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

Energy cost = $0.06 x 365 x10 x (880 + 220)/1000 = $240.90




n

m

Use eq. (2.61),

Rn = Im R 2x10

3 x100

0.04

I Im 5 2x103

In = I - Im = 4.998 A

p = I2 R (4.998)

2 (0.04) 0.9992

1 W




When Rx = 0, i x 10A R = 110 10

11

110

When Rx is maximum, ix = 1A

i.e., Rx = 110 - R = 99

Thus, R = 11 , Rx = 99

R R x 1

110




n

R Vfs

I fs

R m

50

10mA

1 k

4 k




fs

n

n

20 k/V = sensitivity =

1

1

Ifs

i.e., Ifs = k/ V 50 A 20

V The intended resistance Rm = fs

I fs

10(20k/ V) 200k

(a) R Vfs

i fs

R m 50 V

50 A

200 k800 k

(b) p = I2 R (50 A)

2 (800 k) 2 mW




i

v

i

0

0

(a) By current division,

i0 = 5/(5 + 5) (2 mA) = 1 mA

V0 = (4 k) i0 = 4 x 103

x 10-3

= 4 V

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

' 5 0

1 2.4 5

(2mA) 1.19 mA

' (2.4 k)(1.19 mA) 2.857 V

v v'

1.143

(c) % error = 0 0 x 100% = v0

x100 28.57% 4

(d) 4k 36 k3.6 k. By current division,

' 5 0

1 3.6 5

(2mA) 1.042mA

v' (3.6 k)(1.042 mA) 3.75V

% error =

v v' 0 x100%

v0

0.25x100

4

6.25%




(a) 40 24 60

i = 4

16 24

100 mA

(b)

i'

4

16 1 24

97.56 mA

(c) % error = 0.1 0.09756 x100% 2.44%

0.1




V

With the voltmeter in place,

V0 R 2 R m

S

R1 R S R 2 R m

where Rm = 100 kwithout the voltmeter, R 2

V0 VS R1 R 2 R S

(a) When R2 = 1 k, R m

100

R 100

k2 101

V0 = 101

100 30

101

1

(40)

1.278 V (with)

V0 = 1 30

(40) 1.29 V (without)

1000

(b) When R2 = 10 k, R 2

9.091

R m

110 9.091k

V0 = 9.091 30

10

(40) 9.30 V (with)

V0 = 10 30

(40) 10 V (without)

(c) When R2 = 100 k, R 2 R m 50k

V0 50

50 30 100

(40) 25 V (with)

V0 = 100 30

(40) 30.77 V (without)




(a) Using voltage division,

v 12

(25) 15V a

12 8

v 10

(25) 10V b

10 15

vab va vb 15 10 5V

(b) c

+ 8k 15k

25 V

– a b

12k 10k

va = 0; vac = –(8/(8+12))25 = –10V; vcb = (15/(15+10))25 = 15V.

vab = vac + vcb = –10 + 15 = 5V.

vb = –vab = –5V.




V

i 1

R1

iL

+ s − RL

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

v i ( R R )

R

vs R 30

20 10

s L 1 L L 1 L




Converting the delta subnetwork into wye gives the circuit below.

1

⅓

1 ⅓

+

10 V _

⅓ Zin

1

Z 1 (1

1) //(1

1)

1

1 ( 4

) 1 in

3 3 3 3 2 3

Vo Zin

1Zin

(10) 1

11

(10) 5 V




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




With the switch in high position,

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

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

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




Converting delta-subnetworks to wye-subnetworks leads to the circuit below. 1

⅓

1 ⅓

⅓

1 1

1 1

⅓

1 ⅓

1 ⅓

1 (1

1) //(1

1)

1

1 ( 4

) 1 3 3 3 3 2 3 With this combination, the circuit is further reduced to that shown below.

1 1

1

1 1 1

Z 11 (1

1) //(1

1) 11 = 2

ab 3 3 3



1


Zab= 1 + 1 = 2




(a) 5 = 10 10 20 20 20 20

i.e., four 20 resistors in parallel.

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

i.e., one 300resistor in series with 1.8resistor and

a parallel combination of two 20resistors.

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

i.e., Two 24kresistors in parallel connected in series with two

56kresistors in parallel.

(a) 42.32k= 42l + 320

= 24k + 28k = 320

= 24k = 56k 56k 300 20

i.e., A series combination of a 20resistor, 300resistor,

24kresistor, and a parallel combination of two 56k

resistors.




The equivalent circuit is shown below:

R

VS + -

+

V0 (1-)R

-

V0 =

(1 )R V

R (1 )R S

1

V 2

S

V0 1

VS 2




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

R = v2/(p) = 6

2/(240 x 10

-3) = 150

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




2 2 1

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

V + V +

- R1 - R2

Case 1 Case 2

Hence p V

, p2 R1 p

R1 p 10

(12) 30 W

R p1 R 2 R 2 4




Let R1 and R2 be in k.

R eq R1 R 2 5 (1)

V0

VS

5 R 2

5 R 2 R1

(2)

From (1) and (2), 0.05

5 R1 2 = 5 R 2

5R 2

or R2 = 3.333 k

40

From (1), 40 = R1 + 2 R1 = 38 k

5 R 2

Thus,

R1 = 38 k, R2 = 3.333 k




(a) 10

10

40

80

1 2

R12

50 R12 = 80 + 10 (10 40) 80

6

88.33

(b)

10

3

20

40

10 R13

80

1

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

(c) 4

R14

20

10

40

10

80

1

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




The voltage across the fuse should be negligible when compared with 24

V (this can be checked later when we check to see if the fuse rating is

exceeded in the final circuit). We can calculate the current through the

devices.

I1 = p1 45mW

5mA

I2 =

V1

p2

V2

9V

480mW

24

20mA

+

24 V -

Ifuse

R1

R2

iR1

i1 = 5 mA

i2 = 20 mA

iR2

Let R3 represent the resistance of the first device, we can solve for its value from

knowing the voltage across it and the current through it.

R3 = 9/0.005 = 1,800 Ω

This is an interesting problem in that it essentially has two unknowns, R1 and R2 but only

one condition that need to be met and that the voltage across R3 must equal 9 volts.

Since the circuit is powered by a battery we could choose the value of R2 which draws

the least current, R2 = ∞. Thus we can calculate the value of R1 that give 9 volts across

R3.

9 = (24/(R1 + 1800))1800 or R1 = (24/9)1800 – 1800 = 3 kΩ

This value of R1 means that we only have a total of 25 mA flowing out of the battery

through the fuse which means it will not open and produces a voltage drop across it of

0.05V. This is indeed negligible when compared with the 24-volt source.


252980330 Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual

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Transcript of 252980330 Fundamentals of Electric Circuits Alexander 5th Edition Solutions Manual