20 Chapter

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Chapter 20 I

Practice 20A, p. 739

Givens 1. R1 = 6.75 Ω R2 = 15.3 Ω R3 = 21.6 Ω ∆V = 12.0 V 2. R1 = 4.0 Ω R2 = 8.0 Ω R3 = 12.0 Ω ∆V = 24.0 V

Solutions a. Req = R1 + R2 + R3 Req = 6.75 Ω + 15.3 Ω + 21.6 Ω = 43.6 Ω ∆V 12.0 V b. I =  =  = 0.275 A Req 43.6 Ω a. Req = R1 + R2 + R3 = 4.0 Ω + 8.0 Ω + 12.0 Ω = 24.0 Ω ∆V 24.0 V b. I =  =  = 1.00 A Req 24.0 Ω c. I = 1.00 A

3. I = 0.50 A R1 = 2.0 Ω R2 = 4.0 Ω R3 = 5.0 Ω R4 = 7.0 Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved.

4. ∆V = 9.00 V R1 = 7.25 Ω R2 = 4.03 Ω

∆V1 = IR1 = (0.50 A)(2.0 Ω) = 1.0 V ∆V2 = IR2 = (0.50 A)(4.0 Ω) = 2.0 V ∆V3 = IR3 = (0.50 A)(5.0 Ω) = 2.5 V ∆V4 = IR4 = (0.50 A)(7.0 Ω) = 3.5 V a. Req = R1 + R2 = 7.25 Ω + 4.03 Ω = 11.28 Ω ∆V 9.00 V I =  =  = 0.798 A Req 11.28 Ω b. ∆V1 = IR1 = (0.798 A)(7.25 Ω) = 5.79 V ∆V2 = IR2 = (0.798 A)(4.03 Ω) = 3.22 V

5. R1 = 7.0 Ω ∆V = 4.5 V I = 0.60 A

∆V Req = R1 + R2 =  I ∆V 4.5 V R2 =  − R1 =  − 7.0 Ω I 0.60 A R2 = 7.5 Ω − 7.0 Ω = 0.5 Ω

6. ∆V = 115 V I = 1.70 A

∆V 115 V a. Req =  =  = 67.6 Ω I 1.70 A

R = 1.50 Ω

b. NR = Req

Req 67.6 Ω N =  =  = 45 bulbs R 1.50 Ω

Section One—Pupil’s Edition Solutions

I Ch. 20–1

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Practice 20B, p. 744

Givens

Solutions

1. ∆V = 9.0 V R1 = 2.0 Ω

I

R2 = 4.0 Ω R3 = 5.0 Ω R4 = 7.0 Ω

∆V 9.0 V I1 =  =  = 4.5 A R1 2.0 Ω ∆V 9.0 V I2 =  =  = 2.2 A R2 4.0 Ω ∆V 9.0 V I3 =  =  = 1.8 A R3 5.0 Ω ∆V 9.0 V I4 =  =  = 1.3 A R4 7.0 Ω

2. Req = 2.00 Ω



−1

−1

 

1 1 1 1 1 Parallel: Req =  +  +  +  +  R R R R R

5 =  R

R = 5Req = 5(2.00 Ω) = 10.0 Ω Series: Req = 5R = 5(10.0 Ω) = 50.0 Ω 3. R1 = 4.0 Ω R2 = 8.0 Ω R3 = 12.0 Ω ∆V = 24.0 V



−1

 

1 1 1 a. Req =  +  +  R1 R2 R3



−1



1 1 1 =  +  +  4.0 Ω 8.0 Ω 12.0 Ω −1

 

1 1 1 Req = 0.25  + 0.12  + 0.0833  Ω Ω Ω

−1



1 = 0.45  Ω

Req = 2.2 Ω ∆V 24.0 V b. I1 =  =  = 6.0 A R1 4.0 Ω

∆V 24.0 V I3 =  =  = 2.00 A R3 12.0 Ω 4. R1 = 18.0 Ω R2 = 9.00 Ω R3 = 6.00 Ω I2 = 4.00 A



−1

 

1 1 1 a. Req =  +  +  R1 R2 R3





−1

 

1 1 1 Req = 0.0555  + 0.111  + 0.167  Ω Ω Ω Req = 2.99 Ω

b. ∆V = I2 R2 = (4.00 A)(9.00 Ω) = 36.0 V ∆V 36.0 V c. I1 =  =  = 2.00 A R1 18.0 Ω ∆V 36.0 V I3 =  =  = 6.00 A R3 6.00 Ω

I Ch. 20–2

Holt Physics Solution Manual

−1

1 1 1 =  +  +  18.0 Ω 9.00 Ω 6.00 Ω

−1



1 = 0.334  Ω

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∆V 24.0 V I2 =  =  = 3.0 A R2 8.0 Ω

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Givens 4. R1 = 2.0 Ω R2 = 4.0 Ω ∆V = 12 V

Solutions a. Req = R1 + R2 = 2.0 Ω + 4.0 Ω = 6.0 Ω ∆V 12 V I1 = I2 = I =  =  = 2.0 A Req 6.0 Ω

I

∆V1 = I1R1 = (2.0 A)(2.0 Ω) = 4.0 V ∆V2 = I2 R2 = (2.0 A)(4.0 Ω) = 8.0 V

R1 = 2.0 Ω R2 = 4.0 Ω ∆V = 12 V



−1

 

1 1 b. Req =  +  R1 R2



−1



1 1 =  +  2.0 Ω 4.0 Ω −1

−1

 = 0.75 Ω

1 1 Req = 0.50  + 0.25  Ω Ω

1

= 1.3 Ω

∆V1 = ∆V2 = ∆V = 12 V 12 V ∆V I1 = 1 =  = 6.0 A 2.0 Ω R1 12 V ∆V I2 = 2 =  = 3.0 A 4.0 Ω R2 R1 = 4.0 Ω

c. Req = R1 + R2 = 4.0 Ω + 12.0 Ω = 16.0 Ω

R2 = 12.0 Ω

∆V 4.0 V I1 = I2 = I =  =  = 0.25 A Req 16.0 Ω

∆V = 4.0 V

∆V1 = I1R1 = (0.25 A)(4.0 Ω) = 1.0 V ∆V2 = I2 R2 = (0.25 A)(12.0 Ω) = 3.0 V

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R1 = 4.0 Ω R2 = 12.0 Ω ∆V = 4.0 V

−1

1 1   +  =  4.0 Ω 12.0 Ω −1

Req

−1

    1 1 1 = 0.25  + 0.0833  = 0.33   Ω  Ω Ω

1 1 d. Req =  +  R1 R2

−1

= 3.0 Ω

∆V1 = ∆V2 = ∆V = 4.0 V 4.0 V ∆V I1 = 1 =  = 1.0 A 4.0 Ω R1 4.0 V ∆V I2 = 2 =  = 0.33 A 12.0 Ω R2

Section One—Pupil’s Edition Solutions

I Ch. 20–3

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Givens

Solutions

5. R1 = 150 Ω R2 = 180 Ω ∆V = 12 V

a. Req = R1 + R2 = 150 Ω + 180 Ω = 330 Ω ∆V 12 V I1 = I2 = I =  =  = 0.036 A Req 330 Ω ∆V1 = I1R1 = (0.036 A)(150 Ω) = 5.4 V

I

∆V2 = I2 R2 = (0.036 A)(180 Ω) = 6.5 V

R1 = 150 Ω R2 = 180 Ω ∆V = 12 V

b. ∆V1 = ∆V2 = ∆V = 12 V ∆V 12 V I1 = 1 =  = 0.080 A R1 150 Ω ∆V 12 V I2 = 2 =  = 0.067 A R2 180 Ω

6. I = 0.20 A ∆V = 120.0 V N = 35 bulbs

∆V 120.0 V Req =  =  = 600 Ω I 0.20 A Req 600 Ω R =  =  = 17 Ω N 35

Practice 20C, p. 748 1. Ra = 25.0 Ω Rb = 3.0 Ω Rc = 40.0 Ω

−1

−1

    1 1 1 = 0.33  + 0.0250  = 0.36   Ω  Ω Ω

1 1 a. Rbc =  +  Rb Rc

1 1 =  +  3.0 Ω 40.0 Ω −1

Rbc

−1

Rbc = 2.8 Ω

Ra = 12.0 Ω Rb = 35.0 Ω Rc = 25.0 Ω

−1

−1

 +    =  35.0 Ω 25.0 Ω 1 1 1 = 0.0286  + 0.0400  = 0.0686   Ω  Ω Ω

1 1 b. Rbc =  +  Rb Rc

1

1

−1

Rbc

Rbc = 14.6 Ω Req = Ra + Rbc = 12.0 Ω + 14.6 Ω = 26.6 Ω

I Ch. 20–4

Holt Physics Solution Manual

−1

Copyright © by Holt, Rinehart and Winston. All rights reserved.

Req = Ra + Rbc = 25.0 Ω + 2.8 Ω = 27.8 Ω

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Print Givens Ra = 15.0 Ω Rb = 28.0 Ω Rc = 12.0 Ω

Solutions −1

−1

    1 1 1 = 0.0357  + 0.0833  = 0.0119   Ω  Ω Ω

1 1 c. Rbc =  +  Rb Rc

1 1 =  +  28.0 Ω 12.0 Ω −1

Rbc

−1

I

Rbc = 8.40 Ω Req = Ra + Rbc = 15.0 Ω + 8.40 Ω = 23.4 Ω 2. Ra = 25.0 Ω Rb = 3.0 Ω Rc = 40.0 Ω Rd = 15.0 Ω Re = 18.0 Ω

−1

−1

    1 1 1 = 0.0400  + 0.33  = 0.37   Ω Ω  Ω

1 1 a. Rab =  +  Ra Rb

1 1 =  +  25.0 Ω 3.0 Ω −1

Rab

−1

Rab = 2.7 Ω −1

−1

 +    =  15.0 Ω 18.0 Ω 1 1 1 = 0.0667  + 0.0556  = 0.1223   Ω Ω  Ω

1 1 Rde =  +  Rd Re

1

1

−1

Rde

−1

Rde = 8.177 Ω Req = Rab + Rc + Rde = 2.7 Ω + 40.0 Ω + 8.177 Ω = 50.9 Ω

Ra = 12.0 Ω Rb = 35.0 Ω Rc = 25.0 Ω Rd = 50.0 Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved.

Re = 45.0 Ω

−1

−1

    1 1 1 = 0.0833  + 0.0286  = 0.1119   Ω  Ω Ω

1 1 b. Rab =  +  Ra Rb

1 1 =  +  12.0 Ω 35.0 Ω −1

Rab

−1

Rab = 8.937 Ω −1

1

1

−1

Rde

−1

 +    =  50.0 Ω 45.0 Ω 1 1 1 = 0.0200  + 0.0222  = 0.0422   Ω Ω  Ω

1 1 Rde =  +  Rd Re

−1

Rde = 23.7 Ω Req = Rab + Rc + Rde = 8.937 Ω + 25.0 Ω + 23.7 Ω = 57.6 Ω

Section One—Pupil’s Edition Solutions

I Ch. 20–5

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Practice 20D, p. 751

I

Givens

Solutions

Ra = 5.0 Ω

Rab = Ra + Rb = 5.0 Ω + 7.0 Ω = 12.0 Ω

Rb = 7.0 Ω Rc = 4.0 Ω

1 1 Rabc =  +  Rab Rc

Rd = 4.0 Ω

Rabc

Re = 4.0 Ω Rf = 2.0 Ω

−1

−1

 +    =  12.0 Ω 4.0 Ω 1 1 1 = 0.0833  + 0.25  = 0.33  = 3.0 Ω  Ω Ω  Ω 1 1 1 1 =  +  =  +  R R  4.0 Ω 4.0 Ω 1 1 1 = 0.25  + 0.25  = 0.50  = 2.0 Ω  Ω Ω  Ω 1

1

−1

−1

Rde

d

∆V = 14.0 V

−1

−1

e

−1

Rde

−1

Req = Rabc + Rde + Rf = 3.0 Ω + 2.0 Ω + 2.0 Ω = 7.0 Ω ∆V 14.0 V I =  =  = 2.0 A Req 7.0 Ω ∆Vabc = IRabc = (2.0 A)(3.0 Ω) = 6.0 V 6.0 V ∆Vabc Iab =  =  = 0.50 A 12.0 Ω Rab Ra: Ia = Iab = 0.50 A ∆Va = IaRa = (0.50 A)(5.0 Ω) = 2.5 V Rb : Ib = Iab = 0.50 A ∆Vb = Ib Rb = (0.50 A)(7.0 Ω) = 3.5 V Rc : ∆Vc = ∆Vabc = 6.0 V 6.0 V ∆V Ic = c =  = 1.5 A 4.0 Ω Rc Rd: ∆Vd = ∆Vde = 4.0 V 4.0 V ∆V Id = d =  = 1.0 A 4.0 Ω Rd Re : ∆Ve = ∆Vde = 4.0 V 4.0 V ∆V Ie = e =  = 1.0 A 4.0 Ω Re Rf : If = I = 2.0 A ∆Vf = If Rf = (2.0 A)(2.0 Ω) = 4.0 V

I Ch. 20–6

Holt Physics Solution Manual

Copyright © by Holt, Rinehart and Winston. All rights reserved.

∆Vde = IRde = (2.0 A)(2.0 Ω) = 4.0 V

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Givens 1. R1 = 5.0 Ω R2 = 5.0 Ω R3 = 5.0 Ω R4 = 5.0 Ω R5 = 1.5 Ω

Solutions R23 = R2 + R3 = 5.0 Ω + 5.0 Ω = 10.0 Ω −1

−1

 +    =  10.0 Ω 5.0 Ω 1 1 1 = 0.100  + 0.20  = 0.30   Ω Ω  Ω

1 1 R234 =  +  R23 R4

1

1

−1

R234

−1

I = 3.3 Ω

Req = R1 + R234 + R5 = 5.0 Ω + 3.3 Ω + 1.5 Ω = 9.8 Ω

R1 18.0 V

R2 R4

R5 2. Req = 9.8 Ω ∆V = 18.0 V 3. I5 = 1.8 A

R3

∆V 18.0 V I5 = I =  =  = 1.8 A Req 9.8 Ω ∆V5 = I5 R5 = (1.8 A)(1.5 Ω) = 2.7 V

R5 = 1.5 Ω 5. ∆V = 120 V RT = 16.9 Ω RM = 8.0 Ω RP = 10.0 Ω RC = 0.01 Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved.

6. ∆V = 120 V

−1

1 1 1 =  +  +  16.9 Ω 8.0 Ω 10.0 Ω −1

RTMP

Req = 3.6 Ω RTMP = 3.6 Ω

∆V T = ∆V TMP = IRTMP = (33 A)(3.6 Ω) = 120 V

RT = 16.9 Ω

∆V 120 V IT = T =  = 7.1 A RT 16.9 Ω

∆V = 120.0 V N = 35 bulbs n = 3 strands

8. ∆V = 120.0 V

Req, strand = NR = (35)(15.0 Ω) = 525 Ω −1

−1

Req

−1

  = 525Ω + 525Ω + 525Ω 1 1 1 1 = 0.0019  + 0.0019  + 0.0019  = 0.0057  = 175 Ω  Ω   Ω Ω Ω

1 1 1 Req =  +  +  Req, strand Req, strand Req, strand

1

1

1

−1

∆Vstrand = ∆V = 120.0 V

Req, strand = 520 Ω

∆V rand 120.0 V Istrand = st =  = 0.23 A Req, strand 520 Ω

R = 15 Ω

∆V = IstrandR = (0.23 A)(15 Ω) = 3.4 V

Req = 170 Ω

−1

Req = RTMP + RC = 3.6 Ω + 0.01 Ω = 3.6 Ω ∆V 120 V I =  =  = 33 A Req 3.6 Ω

7. R = 15.0 Ω

−1

    1 1 1 1 = 0.0592  + 0.12  + 0.100  = 0.28  = 3.6 Ω  Ω Ω Ω  Ω

1 1 1 RTMP =  +  +  RT RM RP

Section One—Pupil’s Edition Solutions

I Ch. 20–7

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Givens

Solutions

9. Req, strand = 520 Ω R = 15 Ω

∆V rand 120.0 V Istrand = st =  = 0.23 A Req, strand 520 Ω

n = 2 strands

∆V = IstrandR = (0.23 A)(15 Ω) = 3.4 V

∆V = 120.0 V

I

∆Vstrand = ∆V = 120.0 V

N = 35 bulbs

Chapter Review and Assess, pp. 754–759 16. R = 0.15 Ω

Req = 5R = 5(0.15 Ω) = 0.75 Ω

17. R1 = 4.0 Ω

a. Req = R1 + R2 + R3 = 4.0 Ω + 8.0 Ω + 12 Ω = 24 Ω

R2 = 8.0 Ω R3 = 12 Ω ∆V = 24 V 18. R1 = 4.0 Ω R2 = 8.0 Ω R3 = 12 Ω ∆V = 24 V

∆V 24 V b. I =  =  = 1.0 A Req 24 Ω

−1

1 1 1 =  +  +  4.0 Ω 8.0 Ω 12 Ω −1

Req

−1

    1 1 1 1 = 0.25  + 0.12  + 0.083  = 0.45  = 2.2 Ω  Ω Ω Ω  Ω

1 1 1 a. Req =  +  +  R1 R2 R3

−1

∆V 24 V b. I =  =  = 11 A Req 2.2 Ω

R2 = 9.00 Ω R3 = 6.00 Ω ∆V = 12 V

−1

1

1

1

−1

Req

−1

 +  +    =  18.0 Ω 9.00 Ω 6.00 Ω 1 1 1 1 = 0.0556  + 0.111  + 0.167  = 0.334   Ω Ω Ω  Ω

1 1 1 a. Req =  +  +  R1 R2 R3

−1

= 2.99 Ω

∆V 12 V b. I =  =  = 4.0 A Req 2.99 Ω 23. R1 = 12 Ω R2 = 18 Ω R3 = 9.0 Ω R4 = 6.0 Ω ∆V = 30.0 V 24. R1 = 7.0 Ω R2 = 7.0 Ω R3 = 7.0 Ω R4 = 7.0 Ω R5 = 1.5 Ω

1

1

1

−1

R234

−1

−1

= 2.9 Ω

Req = R1 + R234 = 12 Ω + 2.9 Ω = 15 Ω R34 = R3 + R4 = 7.0 Ω + 7.0 Ω = 14.0 Ω



−1

 

1 1 R234 =  +  R2 R34

−1

 

1 1 =  +  7.0 Ω 14.0 Ω

Holt Physics Solution Manual

−1



1 1 = 0.14  + 0.0714  Ω Ω

Req = R1 + R234 + R5 = 7.0 Ω + 4.8 Ω + 1.5 Ω = 13.3 Ω

∆V = 12.0 V

I Ch. 20–8

−1

  = 18Ω + 9.0Ω + 6.0Ω 1 1 1 1 = 0.056  + 0.11  + 0.17  = 0.34   Ω Ω Ω  Ω

1 1 1 R234 =  +  +  R2 R3 R4

= 4.8 Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved.

19. R1 = 18.0 Ω

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Solutions

25. R1 = 6.0 Ω

Current:

R2 = 9.0 Ω

∆V12 = IR12 = (1.8 A)(3.6 Ω) = 6.5 V

R3 = 3.0 Ω

∆V12 6.5 V I1 =  =  = 1.1 A 6.0 Ω R1

∆V = 12 V

I

∆V12 6.5 V I2 =  =  = 0.72 A 9.0 Ω R2 −1

−1

    1 1 1 = 0.17  + 0.11  = 0.28   Ω Ω  Ω

1 1 R12 =  +  R1 R2

1 1 =  +  6.0 Ω 9.0 Ω −1

R12

−1

= 3.6 Ω

Req = R12 + R3 = 3.6 Ω + 3.0 Ω = 6.6 Ω ∆V 12 V I =  =  = 1.8 A Req 6.6 Ω I3 = 1.8 A Potential difference: ∆V1 = ∆V2 = ∆V12 = 6.5 V ∆V3 = I3R3 = (1.8 A)(3.0 Ω) = 5.4 V 26. R1 = 3.0 Ω R2 = 3.0 Ω

−1

 

−1

 

1 1 =  +  6.0 Ω 6.0 Ω

R234 = R2 + R34 = 3.0 Ω + 2.9 Ω = 5.9 Ω

R4 = 6.0 Ω R5 = 4.0 Ω

1 1 R56 =  +  R5 R6

R6 = 12.0 Ω

R56

∆V = 18.0 V

−1

−1



1 = 0.34  Ω

= 2.9 Ω

−1

    1 1 1 = 0.25  + 0.0833  = 0.33   Ω  Ω Ω 1 1 =  +  4.0 Ω 12.0 Ω

−1

 

1 1 = 0.17  + 0.17  Ω Ω

R3 = 6.0 Ω

R7 = 2.0 Ω Copyright © by Holt, Rinehart and Winston. All rights reserved.



1 1 a. R34 =  +  R3 R4

−1

−1

= 3.0 Ω

R567 = R56 + R7 = 3.0 Ω + 2.0 Ω = 5.0 Ω −1

1 1 =  +  5.9 Ω 5.0 Ω −1

R234567

−1

    1 1 1 = 0.17  + 0.20  = 0.37  = 2.7 Ω  Ω Ω  Ω

1 1 R234567 =  +  R234 R567

−1

Req = R1 + R234567 = 3.0 Ω + 2.7 Ω = 5.7 Ω ∆V 18.0 V I =  =  = 3.2 A Req 5.7 Ω ∆V234567 = IR234567 = (3.2 A)(2.7 Ω) = 8.6 V ∆V234567 8.6 V  =  = 1.7 A I7 = I567 =  R567 5.0 Ω

Section One—Pupil’s Edition Solutions

I Ch. 20–9

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Givens

Solutions b. ∆V7 = I7R7 = (1.7 A)(2.0 Ω) = 3.4 V c. ∆V56 = I567R56 = (1.7 A)(3.0 Ω) = 5.1 V ∆V6 = ∆V56 = 5.1 V

I

5.1 V ∆V d. I6 = 6 =  = 0.42 A 12.0 Ω R6 27. R1 = 8.0 Ω R2 = 6.0 Ω ∆V2 = 12 V

28. R1 = 9.0 Ω

Req = R1 + R2 = 8.0 Ω + 6.0 Ω = 14.0 Ω 12 V ∆V I2 = 2 =  = 2.0 A 6.0 Ω R2 ∆V = I2Req = (2.0 A)(14.0 Ω) = 28 V ∆V1 = I1R1 = (0.25 A)(9.0 Ω) = 2.2 V

R2 = 6.0 Ω I1 = 0.25 A 29. R1 = 9.0 Ω

Req = R1 + R2 = 9.0 Ω + 6.0 Ω = 15.0 Ω

R2 = 6.0 Ω

∆V = I1Req = (0.25 A)(15.0 Ω) = 3.8 V

I1 = 0.25 A

R2 = 6.0 Ω

∆V 12 V I = 2 =  = 2.0 A R2 6.0 Ω

∆V2 = 12 V

Req = R1 + R2 = 9.0 Ω + 6.0 Ω = 15.0 Ω

30. R1 = 9.0 Ω

∆V = IReq = (2.0 A)(15.0 Ω) = 3.0 × 101 V

R2 = 9.00 Ω R3 = 6.00 Ω I2 = 4.00 A

a. Req = R1 + R2 + R3 = 18.0 Ω + 9.00 Ω + 6.00 Ω = 33.0 Ω b. I = I2 = 4.00 A ∆V = IReq = (4.00 A)(33.0 Ω) = 132 V c. I1 = I3 = I2 = 4.00 A

33. R1 = 90.0 Ω

R12 = R1 + R2 = 90.0 Ω + 10.0 Ω = 100.0 Ω

R2 = 10.0 Ω

R34 = R3 + R4 = 10.0 Ω + 90.0 Ω = 100.0 Ω

R3 = 10.0 Ω R4 = 90.0 Ω

1 1 R1234 =  +  R12 R34

Req = 60.0 Ω

R1234

−1

   1 = 0.020000  = 50.00 Ω  Ω

−1

 

1 1 =  +  100.0 Ω 100.0 Ω

−1

Req = R + R1234 R = Req − R1234 = 60.0 Ω − 50.00 Ω = 10.0 Ω

I Ch. 20–10 Holt Physics Solution Manual

−1



1 1 = 0.01000  + 0.01000  Ω Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved.

31. R1 = 18.0 Ω

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Solutions

34. Req = 150.0 Ω

N Req, string =  R

∆V = 120.0 V N = 25

−1

−1

  25 =  R

R =  25

R R 2R Req =  +  =  = 150.0 Ω 25 25 25

I

25(150.0 Ω) R =  = 1875 Ω 2 35. R = 6.0 Ω

The following equations represent the circuits as listed. (a) Req = 2R = 2(6.0 Ω) = 12.0 Ω −1

−1

  

2 (b) Req =  R

−1

2 =  6.0 Ω

−1

  

3 (c) Req =  R

3 =  6.0 Ω −1

= 3.0 Ω

= 2.0 Ω −1

   = 6.0Ω +  12.0 Ω 1 1 1 = 0.17  + 0.0833  = 0.25   Ω Ω  Ω

1 1 (d) Req =  +  R 2R

1

1

−1

Req

−1



2 (e) Req =  R 36. ∆V = 9.0 V R1 = 4.5 Ω R2 = 3.0 Ω

−1

= 4.0 Ω

+ R = 3.0 Ω + 6.0 Ω = 9.0 Ω



−1

−1

−1

 = 3.0Ω + 2.0Ω = 0.83 Ω

1 1 a. R23 =  +  R2 R3

1

1

1

= 1.2 Ω

Req = R1 + R23 = 4.5 Ω + 1.2 Ω = 5.7 Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved.

R3 = 2.0 Ω ∆V 9.0 V b. I =  =  = 1.6 A Req 5.7 Ω c. I1 = I = 1.6 A ∆V23 = IR23 = (1.6 A)(1.2 Ω) = 1.9 V ∆V23 1.9 V I2 =  =  = 0.63 A R2 3.0 Ω ∆V23 1.9 V I3 =  =  = 0.95 A R3 2.0 Ω d. ∆V1 = I1R1 = (1.6 A)(4.5 Ω) = 7.2 V ∆V2 = ∆V3 = ∆V23 = 1.9 V

Section One—Pupil’s Edition Solutions

I Ch. 20–11

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Givens

Solutions

37. R1 = 18.0 Ω

Req = R1 + R2 = 18.0 Ω + 6.0 Ω = 24.0 Ω

R2 = 6.0 Ω ∆V = 18.0 V

∆V 18.0 V I1 = I2 = I =  =  = 0.750 A Req 24.0 Ω ∆V1 = I1 R1 = (0.750 A)(18.0 Ω) = 13.5 V

I

∆V2 = I2 R2 = (0.750 A)(6.0 Ω) = 4.5 V 38. R1 = 30.0 Ω R2 = 15.0 Ω R3 = 5.00 Ω ∆V = 30.0 V

−1

−1

 +    =  30.0 Ω 15.0 Ω 1 1 1 = 0.0333  + 0.0667  = 0.1000   Ω Ω  Ω

1 1 b. R12 =  +  R1 R2

1

1

−1

R12

−1

= 10.00 Ω

Req = R12 + R3 = 10.00 Ω + 5.00 Ω = 15.00 Ω ∆V 30.0 V c. I3 = I =  =  = 2.00 A Req 15.00 Ω ∆V12 = IR12 = (2.00 A)(10.00 Ω) = 20.0 V ∆V12 20.0 V I1 =  =  = 0.667 A R1 30.0 Ω ∆V12 20.0 V I2 =  =  = 1.33 A R2 15.0 Ω d. ∆V1 = ∆V2 = ∆V12 = 20.0 V ∆V3 = I3 R3 = (2.00 A)(5.00 Ω) = 10.0 A 39. R2 = 12 Ω ∆V = 12 V I1 = 3.0 A 40. R1 = 18.0 Ω



−1

 

R2 = 6.0 Ω

1 1 Req =  +  R1 R2

∆V = 18.0 V

Req = 4.3 Ω ∆V1 = ∆V2 = ∆V = 18.0 V ∆V 18.0 V I1 = 1 =  = 1.00 A R1 18.0 Ω ∆V 18.0 V I2 = 2 =  = 3.0 A R2 6.0 Ω

I Ch. 20–12

Holt Physics Solution Manual

−1

 

1 1 =  +  18.0 Ω 6.0 Ω

−1

 

1 1 = 0.0556  + 0.17  Ω Ω

−1



1 = 0.23  Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved.

∆V 12 V R1 =  =  = 4.0 Ω I1 3.0 A

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Solutions

41. R1 = 90.0 Ω

Switch open:

R2 = 10.0 Ω

R12 = R1 + R2 = 90.0 Ω + 10.0 Ω = 100.0 Ω

R3 = 10.0 Ω

R34 = R3 + R4 = 10.0 Ω + 90.0 Ω = 100.0 Ω

R4 = 90.0 Ω Req = 2Req,S



−1

 

1 1 R1234 =  +  R12 R34

−1

 

1 1 =  +  100.0 Ω 100.0 Ω

−1



1 = 0.02000  Ω

= 50.00 Ω

I

Req = R + R1234 = R + 50.00 Ω Switch closed: −1

−1

    1 1 1 = 0.0111  + 0.100  = 0.111  = 9.01 Ω  Ω Ω  Ω 1 1 1 1 =  +  =  +  = 9.01 Ω R R  10.0 Ω 90.0 Ω

1 1 R13 =  +  R1 R3

1 1 =  +  90.0 Ω 10.0 Ω −1

R13

−1

R24

2

−1

−1

4

Req,S = R + R13 + R24 = R + 9.01 Ω + 9.01 Ω = R + 18.02 Ω Req = 2Req,S R + 50.00 Ω = 2(R + 18.02 Ω) = 2R + 36.04 Ω 2R − R = 50.00 Ω − 36.04 Ω R = 13.96 Ω 42. R = 20.0 Ω

a. Two resistors in series with two parallel resistors: −1



2 Req = R + R +  R



−1



2 = 20.0 Ω + 20.0 Ω +  20.0 Ω

= 50.0 Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved.

b. Four parallel resistors: −1



4 Req =  R

R 20.0 Ω =  =  = 5.00 Ω 4 4

Section One—Pupil’s Edition Solutions

I Ch. 20–13

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Givens

Solutions

43. ∆V = 12.0 V

a. R12 = R1 + R2 = 30.0 Ω + 50.0 Ω = 80.0 Ω

R1 = 30.0 Ω

I

R3 = 90.0 Ω

R123

R4 = 20.0 Ω

−1

−1

    1 1 1 = 0.0125  + 0.0111  = 0.0236   Ω  Ω Ω

R2 = 50.0 Ω

1 1 R123 =  +  R12 R3

1 1 =  +  80.0 Ω 90.0 Ω −1

−1

= 42.4 Ω

Req = R123 + R4 = 42.4 Ω + 20.0 Ω = 62.4 Ω ∆V 12.0 V b. I =  =  = 0.192 A Req 62.4 Ω c. ∆V123 = IR123 = (0.192 A)(42.4 Ω) = 8.14 V ∆V123 8.14 V I12 =  =  = 0.102 A R12 80.0 Ω I1 = I12 = 0.102 A

d. ∆V2 = I12 R2 = (0.102 A)(50.0 Ω) = 5.10 V (∆V2 )2 (5.10 V)2 P2 =  =  = 0.520 W R2 50.0 Ω e. ∆V4 = IR4 = (0.192 A)(20.0 Ω) = 3.84 V (∆V4 )2 (3.84 V)2 P4 =  =  = 0.737 W R4 20.0 Ω series:

∆VA = 4.0 V (series)

∆V = ∆VA + ∆VB

IB = 2.0 A (parallel)

∆VB = ∆V − ∆VA = 6.0 V − 4.0 V = 2.0 V 2.0 V ∆V IB = B =  = 0.67 A 3.0 Ω RB IA = IB = 0.67 A 4.0 V ∆V RA = A =  = 6.0 Ω 0.67 A IA parallel: ∆VA = ∆VB = 6.0 V ∆V 6.0 V RB = B =  = 3.0 Ω IB 2.0 A

I Ch. 20–14

Holt Physics Solution Manual

Copyright © by Holt, Rinehart and Winston. All rights reserved.

44. ∆V = 6.0 V

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Solutions

46. R1 = 5.0 Ω

a. R789 = R7 + R8 + R9 = 3.0 Ω + 4.0 Ω + 3.0 Ω = 10.0 Ω

R2 = 10.0 Ω



−1

 

−1

 



R3 = 4.0 Ω R4 = 3.0 Ω

R456789 = R4 + R5789 + R6 = 3.0 Ω + 5.00 Ω + 2.0 Ω = 10.0 Ω

R5 = 10.0 Ω R6 = 2.0 Ω R7 = 3.0 Ω

R2456789



1 1 =  +  10.0 Ω 10.0 Ω

−1

1 1 R5789 =  +  R5 R789

−1

 

1 1 =  +  R2 R456789

1 = 0.200  Ω

−1

= 5.00 Ω

 

1 1 =  +  10.0 Ω 10.0 Ω

−1



1 = 0.200  Ω

I = 5.00 Ω

Req = R1 + R2456789 + R3 = 5.0 Ω + 5.00 Ω + 4.0 Ω = 14.0 Ω

R8 = 4.0 Ω R9 = 3.0 Ω ∆V = 28 V

∆V 28 V b. I =  =  = 2.0 A Req 14.0 Ω I1 = I = 2.0 A

47. P = 4.00 W R1 = 3.0 Ω R2 = 10.0 Ω R3 = 5.0 Ω R4 = 4.0 Ω R5 = 3.0 Ω

−1

−1

    1 1 1 = 0.100  + 0.20  = 0.30   Ω Ω  Ω

1 1 a. R23 =  +  R2 R3

1 1 =  +  10.0 Ω 5.0 Ω −1

R23

−1

= 3.3 Ω

R234 = R23 + R4 = 3.3 Ω + 4.0 Ω = 7.3 Ω −1

1 1 =  +  7.3 Ω 3.0 Ω −1

R2345

−1

    1 1 1 = 0.14  + 0.33  = 0.47   Ω Ω  Ω

1 1 R2345 =  +  R234 R5

−1

= 2.1 Ω

Req = R1 + R2345 = 3.0 Ω + 2.1 Ω = 5.1 Ω

Copyright © by Holt, Rinehart and Winston. All rights reserved.







b. ∆V = P R eq  = (4 .0 0W )( 5. 1Ω ) = 4.5 V 48. PT = 1200 W

P = I∆V

PC = 1200 W

PT + PC = I∆V

∆V = 120 V

2(1200 W) I =  = 20 A 120 V

Imax = 15 A

no, because 20 A > 15 A 49. PH = 1300 W PT = 1100 W PG = 1500 W ∆V = 120 V

P 1300 W  =  = 11 A a. heater: I = H ∆V 120 V PT 1100 W toaster: I =  =  = 9.2 A ∆V 120 V PG 1500 W grill: I =  =  = 12 A ∆V 120 V b. yes; Itot = 11 A + 9.2 A + 12 A = 32.2 A

Section One—Pupil’s Edition Solutions

I Ch. 20–15