Chapter 2 - Alexander`s Fundamentals of Electric circuits Ch.2 Solution PDF

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Alexander`s Fundamentals of Electric circuits Ch.2 Solution...

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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-k resistor is 16 V. Find the current through the resistor. Solution

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 For silicon,   6.4 x10 2 -m.

R

L A



L r2

A   r 2 . Hence,  

r2

 L 6.4 x102 x4 x102   0.033953 R  x 240

r = 184.3 mm

Chapter 2, Solution 4 (a) (b)

i = 40/100 = 400 mA i = 40/250 = 160 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 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 i a , i b , and i c , shown in Fig. 2.72, and asking them to solve for values of i 1 , i 2 , and i 3 . Be careful specify realistic currents. 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 Use KCL to obtain currents i 1, i 2, and i 3 in the circuit shown in Fig. 2.72. Solution 12 A a i1 b

8A i2

i3 12 A

c At node a, At node c, At node d,

9A d

8 = 12 + i 1 9 = 8 + i2 9 = 12 + i 3

i 1 = - 4A i 2 = 1A i 3 = -3A

Chapter 2, Solution 9 At A, 1+6–i 1 = 0 or i 1 = 1+6 = 7 A

At B, –6+i 2 +7 = 0 or i 2 = 6–7 = –1 A At C, 2+i 3 –7 = 0 or i3 = 7–2 = 5 A

Chapter 2, Solution 10 2

–8A

1

4A i2

i1

3

–6A At node 1,

–8–i1 –6 = 0 or i 1 = –8–6 = –14 A

At node 2,

–(–8)+i1 +i 2 –4 = 0 or i 2 = –8–i 1 +4 = –8+14+4 = 10 A

Chapter 2, Solution 11 V1  1  5  0

 

5  2  V2  0

 

V1  6 V V2  3 V

Chapter 2, Solution 12 + 30v –

loop 2 – 50v + + 40v -

+ 20v –

loop 1

+ v1 –

+ v2 –

loop 3

+ v3 –

For loop 1,

–40 –50 +20 + v 1 = 0 or v 1 = 40+50–20 = 70 V

For loop 2,

–20 +30 –v 2 = 0 or v 2 = 30–20 = 10 V

For loop 3,

–v1 +v 2 +v 3 = 0 or v 3 = 70–10 = 60 V

Chapter 2, Solution 13 2A

I2

7A

1

I4

2

3

4 4A

I1 3A

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

 

I4  2  4  2 A

At node 3, 7I4  I3

 

I 3  7 2  5 A

Hence, I 1  12 A,

I 2  10 A,

I 3  5A ,

I 4  2 A

I3

Chapter 2, Solution 14

+ 3V -

+ I3

-

+ V3 -

4V

I4 + 2V -

+ - V4

I2 +

V2 +

+ 5V

I1 -

For mesh 1, V 4  2  5  0

 

V 4  7V

For mesh 2, 4  V3  V4  0

 

V3  4  7  11V

 

V1  V3  3   8V

 

V 2  V 1  2  6V

For mesh 3, 3  V1  V3  0

For mesh 4, V 1 V 2  2  0

Thus, V1  8V ,

V1

V 2  6V ,

V 3  11V ,

V 4  7V

Chapter 2, Solution 15 Calculate v and i x in the circuit of Fig. 2.79. 12  + v

10 V

+ _

+ 16 V – –

ix + _

+ 4V _

Figure 2.79 For Prob. 2.15.

Solution

For loop 1, –10 + v +4 = 0, v = 6 V For loop 2, –4 + 16 + 3i x =0, ix =

–4 A

3 ix

Chapter 2, Solution 16 Determine V o 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 + V o = 0 or V o = 10 – 16(–0.5) = 10+8 = 18 V

Chapter 2, Solution 17 Applying KVL around the entire outside loop we get, –24 + v 1 + 10 + 12 = 0 or v 1 = 2V Applying KVL around the loop containing v2 , the 10-volt source, and the 12-volt source we get, v 2 + 10 + 12 = 0 or v 2 = –22V Applying KVL around the loop containing v3 and the 10-volt source we get, –v 3 + 10 = 0 or v 3 = 10V

Chapter 2, Solution 18

Applying KVL, -30 -10 +8 + I(3+5) = 0 8I = 32

I = 4A

-Vab + 5I + 8 = 0

V ab = 28V

Chapter 2, Solution 19 Applying KVL around the loop, we obtain –(–8) – 12 + 10 + 3i = 0 Power dissipated by the resistor: p 3 = i2R = 4(3) = 12W Power supplied by the sources: p 12V = 12 ((–2)) = –24W p 10V = 10 (–(–2)) = 20W p 8V = (–8)(–2) = 16W

i = –2A

Chapter 2, Solution 20 Determine i o in the circuit of Fig. 2.84. io

54V

22 

+ 

+ –

Figure 2.84 For Prob. 2.20 Solution Applying KVL around the loop, –54 + 22i o + 5i o = 0

i o = 4A

5i o

Chapter 2, Solution 21 Applying KVL, -15 + (1+5+2)I + 2 V x = 0 But V x = 5I, -15 +8I + 10I =0, I = 5/6 Vx = 5I = 25/6 = 4.167 V

Chapter 2, Solution 22 Find V o in the circuit in Fig. 2.86 and the power absorbed by the dependent source. 10  + 10 

V1

Vo  25A

2V o

Figure 2.86 For Prob. 2.22 Solution At the node, KCL requires that [–V o /10]+[–25]+[–2V o ] = 0 or 2.1V o = –25 or V o = –11.905 V The current through the controlled source is i = 2V 0 = –23.81 A and the voltage across it is V 1 = (10+10) i 0 (where i 0 = –V 0 /10) = 20(11.905/10) = 23.81 V. Hence, p dependent source = V 1 (–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!

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. 1

ix + 20A

vx

–

2

3

Applying current division, i x = [2/(2+1+3)]20 = 6.667 and v x = 1x6.667 = 6.667 V ix 

2 ( 6 A)  2 A, 2 1 3

v x  1i x  2V

The current through the 1.2-  resistor is 0.5i x = 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

Chapter 2, Solution 24 (a)

I0 =

Vs R1  R2

V 0    I 0  R3 R4  = 

R 3R 4  Vs  R1  R 2 R 3  R 4

 R 3 R 4 V0  Vs  R1  R2  R3  R4 

(b)

If R1 = R 2 = R 3 = R 4 = R, V0 VS



 R     10 2R 2 4

 = 40

Chapter 2, Solution 25 V 0 = 5 x 10-3 x 10 x 103 = 50V Using current division, I 20 

5 (0.01x50)  0.1 A 5  20

V20 = 20 x 0.1 kV = 2 kV p 20 = I20 V 20 = 0.2 kW

Chapter 2, Problem 26. For the circuit in Fig. 2.90, i o = 3 A. Calculate i x 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 i 16 = i o = 3A, then v = 16x3 = 48 V and i 8 = 48/8 = 6A; i4 = 48/4 = 12A; and i 2 = 48/2 = 24A. Thus, i x = i 8 + i 4 + i 2 + i 16 = 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 I o in the circuit of Fig. 2.91. 8

10V

+ 

Io

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, I o = 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. 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 Find v 1 , v 2 , and v 3 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 (40)  28 V 14  6

v 2 = v3 = Hence,

6 (40)  12 V 14  6

v 1 = 28 V, v 2 = 12 V, vs = 12 V

Chapter 2, Solution 29 All resistors in Fig. 2.93 are 5  each. Find Req .

R eq

Figure 2.93 For Prob. 2.29. Solution

R eq = 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 R eq for the circuit in Fig. 2.94.

25 

180  60 

Re q

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, R eq = 25+48 = 73 Ω.

Chapter 2, Solution 31

Req  3  2 // 4 //1  3 

1  3.5714  1/ 2 1/ 4  1 i 1 = 200/3.5714 = 56 A

v 1 = 0.5714xi 1 = 32 V and i 2 = 32/4 = 8 A i 4 = 32/1 = 32 A; i 5 = 32/2 = 16 A; and i 3 = 32+16 = 48 A

Chapter 2, Solution 32 Find i 1 through i 4 in the circuit in Fig. 2.96. 60 

i2

i4

200 

40 

50  i3

i1

16 A

Figure 2.96 For Prob. 2.32. Solution We first combine resistors in parallel. 40 60 

40x 60 50x 200  40   24  and 50 200  100 250

Using current division principle, 24 40 ( 16)   10A ( 16)   6A, i3  i4  i1  i 2  24  40 64

i1 

200 50 (6)  –4.8 A and i 2  ( 6)  –1.2 A 250 250

i3 

60 40 ( 10)  –6 A and i 4  ( 10)  –4 A 100 100

Chapter 2, Solution 33 Combining the conductance leads to the equivalent circuit below i + v -

9A

6 S 3S 

1S

4S

i

4S

6x3  2S and 2S + 2S = 4S 9

Using current division,

i

1 1

1 2

(9)  6 A, v = 3(1) = 3 V

9A

+ v -

1S

2S

Chapter 2, Solution 34 160//(60 + 80 + 20)= 80 , 160//(28+80 + 52) = 80  R eq = 20+80 = 100 Ω I = 200/100 = 2 A or p = VI = 200x2 = 400 W.

Chapter 2, Solution 35

i 70  +

200V

a

-

+ V1 i1 -

30  Io +

20  i2

b

Vo 5  -

Combining the resistors that are in parallel, 70 30 

i=

70x 30  21 , 100

20 5 

20 x5 4 25

200 8A 21  4

v 1 = 21i = 168 V, vo = 4i = 32 V v v i 1 = 1  2.4 A, i 2 = o  1.6 A 70 20 At node a, KCL must be satisfied i1 = i 2 + I o

2.4 = 1.6 + I o

I o = 0.8 A

Hence, v o = 32 V and I o = 800 mA

Chapter 2, Solution 36 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 i 1 is the current through the 24- resistor and i o is the current through the 50- resistor, using current division gives i 1 = [40/(40+40)]0.2 = 0.1 and i o = [20/(20+80)]0.1 = 0.02 A or v o = 30io = 30x0.02 = 600 mV.

Chapter 2, Solution 37 Applying KVL, -20 + 10 + 10I – 30 = 0, I = 4

10  RI

  R

10  2.5  I

Chapter 2, Solution 38 20//80 = 80x20/100 = 16, 6//12 = 6x12/18 = 4 The circuit is reduced to that shown below. 4

2.5 

60  15 

16 

R eq (4 + 16)//60 = 20x60/80 = 15 R eq = 2.5+15||15 = 2.5+7.5 = 10 Ω and i o = 35/10 = 3.5 A.

Chapter 2, Solution 39 (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,

R eq = [(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, R eq = [(4x12)/16]k = 3 kΩ.

Chapter 2, Solution 40 Req = 8  4 (2  6 3)  8  2  10 

I=

15 15   1.5 A R eq 10

Chapter 2, Solution 41 Let R0 = combination of three 12 resistors in parallel

1 1 1 1    R o 12 12 12

Ro = 4

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

50  30 

60(14  R ) 74  R

or R = 16 

74 + R = 42 + 3R

Chapter 2, Solution 42 5x 20  4 25

(a)

R ab = 5 (8  20 30)  5 (8  12) 

(b)

R ab = 2  4 (5  3) 8  5 10 4  2  4 4  5 2.857  2  2  1.8181  5.818 

Chapter 2, Solution 43 5x 20 400   4  8  12  25 50

(a)

R ab = 5 20  10 40 

(b)

1 1  60  1  60 20 30     10   6  60 20 30 

1

R ab = 80 (10  10) 

80  20  16  100

Chapter 2, Solution 44 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, R ab = [(6x3)/(6+3)] = 2 Ω.

Chapter 2, Solution 45 (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 

Chapter 2, Solution 46 R eq = 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

Chapter 2, Solution 47 5 20 

6 3

5x 20  4 25

6 x3  2 9 8

10  a

b

4

Rab = 10 + 4 + 2 + 8 = 24 

2

Chapter 2, Solution 48 (a)

(b)

Ra =

R 1 R 2  R 2 R 3  R 3 R1

R3 Ra = R b = Rc = 30 



100  100  100  30 10

30 x 20  30x 50  20x 50 3100   103.3 30 30 3100 3100 Rb   155, R c   62 20 50 R a = 103.3 , R b = 155 , R c = 62  Ra 

Chapter 2, Solution 49 (a)

R1 =

R aR c 12 * 12   4 R a R b R c 36

R 1 = R 2 = R3 = 4 

(b)

60x 30  18 60  30  10 60x10 R2   6 100 30 x10  3 R3  100

R1 

R 1 = 18, R 2 = 6, R 3 = 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:

R 30mA

3R 3R

3R

3R R 

30mA R

3RxR 3  R 4R 4

3 3Rx R 3  3 3 2 =R 3R  R  R   3R R  3 4 4 2   3R  R 2 P = I2R R = 889 

800 x 10-3 = (30 x 10-3)2 R

3R

3R/2

Chapter 2, Solution 51

(a)

30 30  15 and 30 20  30x 20 /(50)  12 R ab = 15 (12 12)  15x 24 /(39)  9.231 

a

a 30 

30 

30  30 

b

12 

20  15 

12 

20  b

(b)

Converting the T-subnetwork into its equivalent  network gives R a'b' = 10x20 + 20x5 + 5x10/(5) = 350/(5) = 70  R b'c' = 350/(10) = 35, Ra'c' = 350/(20) = 17.5 

Also

30 70  30x 70 /(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 

10 

20 

a

a 5

b

15 

25 

a’

b’

35 

17.5 

b

70 

c’

15 

c’

Chapter 2, Solution 52 Converting the wye-subnetwork to delta-subnetwork, we obtain the circuit below. 9

3 9

3

9

3

3

3 6

3

3//1 = 3x1/4 = 0.75, 2//1 =2x1/3 = 0.6667. Combining these resistances leads to the circuit below. 3 2.25 

3

2.25

9

3

2

We now convert the wye-subnetwork to the delta-subnetwork. R a = [(2.25x3+2.25x3+2.25x2.25)/3] = 6.188 Ω R b = R c = 18.562/2.25 = 8.25 Ω This leads to the circuit below. 3 6.188 8.25 3

9

8.25 2

R = 9||6.188+8.25||2 = 3.667+1.6098 = 5.277 R eq = 3+3+8.25||5.277 = 9.218 Ω.

Chapter 2, Solution 53 (a)

Converting one  to T yields the equivalent circuit below: 30 

a’

20 

a

60 

b’

b

4

20 

c’

5

80 

10 x 50 40x 50 40x10  5 , R c 'n   20  4, R b 'n  100 100 40  10  50 R ab = 20 + 80 + 20 + (30  4) (60  5)  120  34 65

R a'n =

R ab = 142.32  (b) We combine the resistor in series and in parallel. 30 (30  30) 

30 x 60  20 90

We convert the balanced  s to Ts as shown below: a

30 

30 

a 10 

30  30 

20 

10 

30 

b 30 

10  10 

b R ab = 10 + (10 10) (10  20 10) 10  20 20 40 R ab = 33.33 

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