Solutions manual for fundamentals of electric circuits 6th edition by alexander ibsn 0078028221 PDF

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Solutions Manual for Fundamentals of Electric Circuits 6th Edition by Alexander IBSN 0078028221 Full Download: http://downloadlink.org/product/solutions-manual-for-fundamentals-of-electric-circuits-6th-edition-by-alexande

Solution 2.1 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-kΩ resistor 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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F ll ll h

t

i t

td

l

d l

t S l ti

M

l T tB

k it

d

l

dli k

Solution 2.2 p = v2/R →

R = v2/p = 14400/60 = 240 ohms

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Solution 2.3 For silicon, ρ = 6.4 x10 2 Ω-m. A = π r 2 . Hence, R=

ρL A

=

ρL πr2

ρL πR

 → r2 =

6.4 x102 x4 x10−2 = π x 240

= 0.033953

r = 184.3 mm

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Solution 2.4 (a) (b)

i = 40/100 = 400 mA i = 40/250 = 160 mA

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Solution 2.5 n = 9; l = 7; b = n + l – 1 = 15

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Solution 2.6 n = 8; l = 8; b = n + l –1 = 15

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Solution 2.7 6 branches and 4 nodes

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Solution 2.8 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 one of the many possible solutions. Note that the solution process must follow the same basic steps. Problem Use KCL to obtain currents i 1, i 2, and i 3 in the circuit shown in Fig. 2.72 given that i a = 2 amps, i b = 3 amps, and i c = 4 amps.

Solution

ib

a ia

i2

b

i1 c

i3 ic d At node a, At node b, At node c,

 i a  i 2  i b = 0 or i 2 =  2  3 =  5 amps i b + i 1 + i c = 0 or i 1 =  3  4 =  7 amps i 2 + i 3  i 1 = 0 or i 3 =  7 + 5 =  2 amps

We can use node d as a check, i a  i 3  i c = 2 + 2  4 = 0 which is as expected.

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Solution 2.9 Find

i 1, i 2 , and i 3 in Fig. 2.73. –4 A i2

1A –3 A

–6 A

B

A

i3

2A

i1

–2 A

C

Figure 2.73 For Prob. 2.9. Solution Step 1.

We can apply Kirchhoff’s current law to solve for the unknown currents.

Summing all of the currents flowing out of nodes A, B, and C we get, at A, 1 + (–6) + i 1 = 0; at B, –(–6) + i 2 + 2 = 0; and at C, (–2) + i 3 – 2 = 0. Step 2.

We now can solve for the unknown currents, i 1 = –1 + 6 = 5 amps; i 2 = –6 – 2 = –8 amps; and i 3 = 2 + 2 = 4 amps.

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Solution 2.10 2

–8A

1

4A i2

i1

3

–6A At node 1,

–8–i 1 –6 = 0 or i 1 = –8–6 = –14 A

At node 2,

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

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Solution 2.11 −V 1 +1 +5 =0 −5 +2 +V2

=0

 → V 1 6=V  → V2

3=V

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Solution 2.12 + 30v –

loop 2 – 50v + + 40v -

+ v2 –

+ 20v –

loop 1

+ v1 –

loop 3

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 +v2 +v3 = 0 or v3 = 70–10 = 60 V

+ v3 –

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Solution 2.13 2A

I2

7A

1

I4

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 = 12A

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 = 12A,

I 2 = − 10A ,

I 3 = 5A ,

I 4 = − 2A

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Solution 2.14

+ 3V -

+

V1

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 +V 3 +V 4 = 0

 →

V 3 = −4 − 7 = −11V

 →

V1 = V 3 + 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 ,

V2 = 6V ,

V 3 = −11V ,

V 4 = 7V

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Solution 2.15 Calculate v and i x in the circuit of Fig. 2.79. 12 Ω + v

10 V

+ 16 V – –

+ _

ix +

+

4V

_

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 + 3i x =0, i x =

–4 A

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Solution 2.16

Determine V o in the circuit in Fig. 2.80. 16 Ω

14 Ω



+

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

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Problem 2.17 Obtain v 1 through v 3 in the circuit in Fig. 2.81.

Figure 2.81 For Prob. 2.17.

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Solution 2.18 Find I and V in the circuit of Fig. 2.82. I + 3 amp

20 Ω

10 Ω

4 amp

20 Ω

V

–2 amp −

Figure 2.82 For Prob. 2.18. Solution. Step 1. We can make use of both Kirchhoff’s KVL and KCL. KVL tells us that the voltage across all the elements of this circuit is the same in every case. Ohm’s Law tells us that the current in each resistor is equal to V/R. Finally we can use KCL to find I. Applying KCL and summing all the current flowing out of the top node and setting it to zero we get, –3 + [V/20] + [V/10] + 4 + [V/20] – [–2] = 0. Finally at the node to the left of I we can write the following node equation which will give us I, –3 + [V/20] + [V/10] + I = 0. Step 2. [0.05+0.1+0.05]V = 0.2V = 3–4–2 = –3 or V = –15 volts. I = 3–V[0.05+0.1] = 3–[–15]0.15 = 5.25 amps.

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Solution 2.19 Applying KVL around the loop, we obtain –(–8) – 12 + 10 + 3i = 0

i = –2A

Power dissipated by the resistor: p 3Ω = i2R = 4(3) = 12W Power supplied by the sources: p 12V = 12 ((–2)) = –24W p 10V = 10 (–(–2)) = 20W p 8V = (–8)(–2) = 16W

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Solution 2.20 Determine i o in the circuit of Fig. 2.84.

io

54V

22 Ω

+ −

+ –

5io

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

i o = 4A

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Solution 2.21 Applying KVL, -15 + (1+5+2)I + 2 V x = 0 But V x = 5I, -15 +8I + 10I =0, I = 5/6 V x = 5I = 25/6 = 4.167 V

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Solution 2.22 Find V o in the circuit in Fig. 2.86 and the power absorbed by the dependent source. 10 Ω + 10 Ω

Vo

V1 − 25A

2Vo

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!

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Solution 2.23 In the circuit shown in Fig. 2.87, determine vx and the power absorbed by the 60-Ω resistor. 5Ω 6Ω +

vx

− 20 Ω

10 Ω

60 amp

40 Ω

60 Ω

30 Ω

15 Ω

Figure 2.87 For Prob. 2.23. Step 1. Although we could directly use Kirchhoff’s current law to solve this, it will be easier if we reduce the circuit first. The reduced circuit looks like this, 5Ω −

+ vx 60 amp

R1 R2 R3 R4 R5

10 Ω

R5

= 40x60/(40+60) = 6 + R1 = 15x30/(15+30) = 20+R 3 = R 2 R 4 /(R 2 +R 4 )

Letting V 10 = v x + V R5 and using Kirchhoff’s current law, we get 60 + V 10 /10 + V 10 /(5+R 5 ) = 0 60 + V 10 /10 + V 10 /20 = 0 V 10 = –60x20/3 = –400 volts Copyright © 2017 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

We could have also used current division to find the current through the 5 Ω resistor, however, i5 = V 10 /(5+R 5 ) and v x = 5i 5 Calculating the power delivered to the 60-ohm resistor requires that we find the voltage across the resistor. V R5 = V 10 – v x ; using voltage division we get V 60 = [V R5 /(6+R 1 )]R 1 . Finally P 60 = (V 60 )2/60. Step 2. R1 R2 R3 R4 R5

= 40x60/(40+60) = 2400/100 = 24; = 6 + R 1 = 6+24 = 30; = 15x30/(15+30) 450/45 = 10; = 20+R 3 = 20+10 = 30; = R 2 R 4 /(R 2 +R 4 ) = 30x30/(30+30) = 15.

Now, we have 60 + (V 10 /10) + (V 10 /(20)) = 0 or V 10 = –60x20/3 = –400 and i 10 = –400/20 = –20 and v x = 5i 5 = 5(–20) = –100 volts. V R5 = V 10 – v x = –400 – (–100) = –300; using voltage division we get V 60 = [V R5 /(6+R 1 )]R1 . = [–300/30]24 = –240. Finally, P 60 = (V 60 )2/60 = (–240)2/60 = 960 watts.

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Solution 2.24 (a)

I0 =

Vs R1 + R2

V0 = −α I 0 (R3 R4 ) = −

α Vs R 3R 4 ⋅ R1 + R 2 R 3 + R 4

V0 − αR3 R4 = Vs (R1 + R2 )( R3 + R4 ) (b)

If R 1 = R 2 = R 3 = R 4 = R,

V0 α R α = ⋅ = = 10 VS 2R 2 4

α = 40

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Solution 2.25 V 0 = 5 x 10-3 x 10 x 103 = 50V Using current division, I 20 =

5 (0.01x 50) = 0.1 A 5 + 20

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

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Solution 2.26 For the circuit in Fig. 2.90, i o = 5 A. Calculate i x and the total power absorbed by the entire circuit.

ix

25 Ω

io

20 Ω

10 Ω

5Ω

40 Ω

Figure 2.90 For Prob. 2.26. Solution Step 1. V 40 = 40i o and we can combine the four resistors in parallel to find the equivalent resistance and we get (1/R eq ) = (1/20) + (1/10) + (1/5) + (1/40). This leads to i x = V 40 /R eq and P = (i x )2(25+Req ). Step 2. V 40 = 40x5 = 200 volts and (1/R eq ) = (1/20) + (1/10) + (1/5) + (1/40) = 0.05 + 0.1 + 0.2 + 0.025 = 0.375 or R eq = 2.667 Ω and i x = 200/R eq = 75 amps. P = (75)2(25+2.667) = 155.62 kW.

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Solution 2.27 Calculate I o in the circuit of Fig. 2.91. 8Ω

10V

+ −

Io

3Ω

6Ω

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.

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Solution 2.28 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 Ω We now apply voltage division, v1 =

14 ( 40) = 28 V 14 + 6

v2 = v3 = Hence,

6 ( 40) = 12 V 14 + 6

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

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Solution 2.29 All resistors (R) in Fig. 2.93 are 10 Ω each. Find R eq .

R

R Req R

R

R

R R

Figure 2.93 For Prob. 2.29. Solution Step 1.

All we need to do is to combine all the resistors in series and in parallel.  R(R)  R(R) R(R + R)  +     R + R  R + R R + R + R  R eq = which can be derived by inspection. We  R(R)   R(R) R(R + R)  + +  R + R   R + R R+ R+ R      will look at a simpler approach after we get the answer.

Step 2.

R eq =

5[( 5 + 6.667 )] 58.335 = 3.5 Ω. = 5+ 5+ 6.667 16.667

10 Ω

10 Ω Req 10 Ω

10 Ω

R1

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Checking we get, R 1 = 10(20)/(10+20) = 6.667 Ω. R2

Req 10 Ω

10 Ω

6.667 Ω

We get R 2 = 10(10)/(10+10) = 5 Ω.

R eq 10 Ω

10 Ω

11.667 Ω

Finally we get (noting that 10 in parallel with 10 gives us 5 Ω, R eq = 5(11.667)/(5+11.667) = 3.5 Ω.

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Solution 2.30 Find R eq for the circuit in Fig. 2.94.

25 Ω

180 Ω 60 Ω

Req

60 Ω

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.

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Solution 2.31 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

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Solution 2.32

Find i 1 through i 4 in the circuit in Fig. 2.96. 60 Ω

i4

i2

40 Ω

200 Ω 50 Ω

i3

i1 16 A

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

40x 60 50 x 200 = 24 Ω and 50 200 = = 40 Ω 100 250

Using current division principle, 24 40 i1 + i 2 = (−16) = −6A, i 3 + i 4 = (− 16) = − 10A 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 ...