Solucionario serway cap 27

27Current and Resistance

CHAPTER OUTLINE

27.1 Electric Current27.2 Resistance27.3 A Model for Electrical Conduction27.4 Resistance and Temperature27.5 Superconductors27.6 Electrical Power

101

ANSWERS TO QUESTIONS

Q27.1 Voltage is a measure of potential difference, not of current. “Surge” implies a fl ow—and only charge, in coulombs, can fl ow through a system. It would also be correct to say that the victim carried a certain current, in amperes.

Q27.2 Geometry and resistivity. In turn, the resistivity of the material depends on the temperature.

*Q27.3 (i) We require rL /AA = 3rL /A

B. Then A

A/A

B = 1/3,

answer (f ).

(ii) πrA

2/πrB

2 = 1/3 gives rA /r

B = 1/ 3, answer (e).

*Q27.4 Originally, RA

= ρ. Finally, R

A A

Rf = = =ρ ρ( / )

. 3

3 9 9

Answer (b).

Q27.5 The conductor does not follow Ohm’s law, and must have a resistivity that is current-dependent, or more likely temperature-dependent.

Q27.6 The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electrons more effi ciently.

Q27.7 (i) The current density increases, so the drift speed must increase. Answer (a). (ii) Answer (a).

Q27.8 The resistance of copper increases with temperature, while the resistance of silicon decreases with increasing temperature. The conduction electrons are scattered more by vibrating atoms when copper heats up. Silicon’s charge carrier density increases as temperature increases and more atomic electrons are promoted to become conduction electrons.

*Q27.9 In a normal metal, suppose that we could proceed to a limit of zero resistance by lengthening the average time between collisions. The classical model of conduction then suggests that a constant applied voltage would cause constant acceleration of the free electrons. The drift speed and the current would increase steadily in time.

It is not the situation envisioned in the question, but we can actually switch to zero resistance by substituting a superconducting wire for the normal metal. In this case, the drift velocity of electrons is established by vibrations of atoms in the crystal lattice; the maximum current is limited; and it becomes impossible to establish a potential difference across the superconductor.

Q27.10 Because there are so many electrons in a conductor (approximately 1028 electrons/m3) the average velocity of charges is very slow. When you connect a wire to a potential difference, you establish an electric fi eld everywhere in the wire nearly instantaneously, to make electrons start drifting everywhere all at once.

ISMV2_5104_27.indd 101ISMV2_5104_27.indd 101 6/12/07 4:17:56 PM6/12/07 4:17:56 PM

102 Chapter 27

*Q27.11 Action (a) makes the current three times larger. (b) causes no change in current. (c) corresponds to a current 3 times larger. (d) R is 1/4 as large, so current is 4 times larger. (e) R is 2 times larger, so current is half as large. (f) R increases by a small percentage as current has a small decrease. (g) Current decreases by a large factor. The ranking is then d > a > c > b > f > e > g.

*Q27.12 RL

d

L

d

L

dAA

A

B

B

B

B

= = = =ρπ

ρπ

ρπ( / ) ( / ) ( / )2

2

2 2

1

2 22 2 2

RRB

2

P PA A A B BI V V R V R= = = =∆ ∆ ∆( ) / ( ) /2 22 2 Answer (e).

*Q27.13 RL

A

L

ARA

A BB= = =ρ ρ2

2

P PA A A B BI V V R V R= = = =∆ ∆ ∆( ) / ( ) / /2 2 2 2 Answer (f ).

*Q27.14 (i) Bulb (a) must have higher resistance so that it will carry less current and have lower power.(ii) Bulb (b) carries more current.

*Q27.15 One ampere–hour is (1 C/s)(3 600 s) = 3 600 coulombs. The ampere–hour rating is the quantity of charge that the battery can lift though its nominal potential difference. Answer (d).

Q27.16 Choose the voltage of the power supply you will use to drive the heater. Next calculate the

required resistance R as ∆V 2

P. Knowing the resistivity ρ of the material, choose a combination

of wire length and cross-sectional area to makeA

R⎛⎝

⎞⎠ =

⎛⎝⎜

⎞⎠⎟ρ

. You will have to pay for less

material if you make both and A smaller, but if you go too far the wire will have too little surface area to radiate away the energy; then the resistor will melt.

SOLUTIONS TO PROBLEMS

Section 27.1 Electric Current

P27.1 IQ

t= ∆

∆ ∆ ∆Q I t= = ×( )( ) = ×− −30 0 10 40 0 1 20 106 3. . . A s C

NQ

e= = ×

×= ×

−

−

1 20 10

1 60 107 50 1

3

19

.

..

C

C electron0015 electrons


Current and Resistance 103

P27.2 The molar mass of silver = 107 9. g mole and the volume V is

V = ( )( ) = ×( ) ×− −area thickness m2700 10 0 133 104 3. m m3( ) = × −9 31 10 6.

The mass of silver deposited is m VAg3 3kg m m= = ×( ) ×( ) = ×− −ρ 10 5 10 9 31 10 9 78 103 6. . . 22 kg.

And the number of silver atoms deposited is

N = ×( ) ×⎛

⎝⎜⎞⎠

−9 78 106 02 102

23

..

kg atoms

107.9 g ⎟⎟⎛⎝⎜

⎞⎠⎟

= ×

= =

1 0005 45 10

12

23 g

1 kg atoms.

.I

V

R

∆ 006 67 6 67

5 45 10

V

1.80 A C s

Ω

∆ ∆

= =

= = =×

. .

.t

Q

I

Ne

I

223 194

1 60 10

6 671 31 10 3 64

( ) ×( )= × =

−.

.. .

C

C s s hh

P27.3 Q t Idt I et

t( ) = = −( )∫ −

0

0 1τ τ

(a) Q I e Iτ τ τ( ) = −( ) = ( )−0

101 0 632.

(b) Q I e I10 1 0 999 95010

0τ τ τ( ) = −( ) = ( )− .

(c) Q I e I∞( ) = −( ) =−∞0 01τ τ

P27.4 The period of revolution for the sphere is T = 2πω

, and the average current represented by this

revolving charge is Iq

T

q= = ωπ2

.

P27.5 q t t= + +4 5 63

A = ( )⎛⎝⎜

⎞⎠⎟ = × −2 00

1 002 00 10

24.

.. cm

m

100 cm m2 22

(a) Idq

dtt

tt

1 00 12 5 171 00

2

1 00. .

..

s s

s( ) = = +( ) =

==

00 A

(b) JI

A= =

×=−

17 0

2 00 1085 04

.

..

A

m kA m2

2

P27.6 Idq

dt=

q dq Idt

tdt= = = ( ) ⎛

⎝⎜⎞⎠⎟∫ ∫ ∫ 100

120

0

1 240

As

s

sinπ

qq = − ⎛⎝⎜

⎞⎠⎟ −⎡

⎣⎢⎤⎦⎥

= +100

120 20

100 C C

1ππ

cos cos220

Cπ

= 0 265.

P27.7 (a) JI

A= = ×

×( ) =−

−

8 00 10

1 00 102 55

6

3 2

.

..

A

mA m2

π

(b) From J ne d= v , we have nJ

e d

= =×( ) ×( ) =−v

2 55

1 60 10 3 00 1019 8

.

. .

A m

C m s

2

55 31 1010 3. .× −m

(c) From IQ

t= ∆

∆, we have ∆ ∆

tQ

I

N e

I= = =

×( ) ×( )×

−A

C6 02 10 1 60 10

8 00 1

23 19. .

. 001 20 106

10− = ×

As. .

(This is about 382 years!)


104 Chapter 27

*P27.8 (a) JI

A= =

×( )=

−

5 0099 5

3 2

..

A

4.00 10 m kA m2

π (b) Current is the same and current density is smaller. Then I = 5.00 A ,

J J2 1

4 41

4

1

49 95 10 2 49 10= = × = ×. . A/m A/m2 2

A A r r r2 1 22

12

24 4 2= = =or soπ π rr1 0 800= . cm

P27.9 (a) The speed of each deuteron is given by K m= 1

22v

2 00 10 1 60 101

22 1 67 106 19 27. . .×( ) ×( ) = × ×(− − J kg)) v2

and v = ×1 38 107. m s

The time between deuterons passing a stationary point is t in Iq

t=

10 0 10 1 60 106 19. .× = ×− − C s C t or t = × −1 60 10 14. s

So the distance between them is vt = ×( ) ×( ) = ×− −1 38 10 1 60 10 2 21 107 14 7. . . .m s s m

(b) One nucleus will put its nearest neighbor at potential

Vk q

re= =

× ⋅( ) ×( )×

−8 99 10 1 60 10

2 21

9 19. .

.

N m C C2 2

1106 49 107

3−

−= ×m

V.

This is very small compared to the 2 MV accelerating potential, so repulsion within the beam is a small effect.

P27.10 We use I nqA d= v n is the number of charge carriers per unit volume, and is identical to the number of atoms per unit volume. We assume a contribution of 1 free electron per atom in the relationship above. For aluminum, which has a molar mass of 27, we know that Avogadro’s number of atoms, NA, has a mass of 27.0 g. Thus, the mass per atom is

27 0 27 0

6 02 104 49 1023

23. .

..

g g g atom

NA

=×

= × −

Thus, n = =density of aluminum

mass per atom

g cm2 70. 33

g atom4 49 10 23. × −

n = × = ×6 02 10 6 02 1022 28. .atoms cm atoms m3 3

Therefore, vd

I

nqA= =

×( ) ×( )− −

5 00

6 02 10 1 60 1028 3 19

.

. .

A

m C 44 00 101 30 10

64

..

×( ) = ×−−

m m s

2

or, vd = 0 130. mm s

Section 27.2 Resistance

P27.11 ∆V IR=

and RA

= ρ: A = ( ) ⎛

⎝⎜⎞⎠⎟

= × −0 6001 00

6 00 102

2

7..

. mm m

1 000 mm m2

∆VI

A= ρ

: IVA= =

( ) ×( )× ⋅

−

−

∆Ωρ

0 900 6 00 10

5 60 10

7

8

. .

.

V m

2

mm m( )( )1 50.

I = 6 43. A



P27.12 IV

R= = = =∆

Ω120

0 500 500 V

240 A mA.

P27.13 (a) Given M V Ad d= =ρ ρ where rd ≡ mass density,

we obtain: AM

d

=ρ

Taking rr ≡ resistivity, R

A M Mr r

d

r d= = =ρ ρρ

ρ ρ

2

Thus, = =×( )( )

×( ) ×

−

−

MR

r dρ ρ1 00 10 0 500

1 70 10 8 92

3

8

. .

. . 1103( ) = 1 82. m

(b) VM

d

=ρ

, or πρ

rM

d

2 =

Thus, rM

d

= = ××( )( )

−

π ρ π1 00 10

8 92 10 1 82

3

3

.

. . r = × −1 40 10 4. m

The diameter is twice this distance: diameter = 280 mµ

P27.14 (a) Suppose the rubber is 10 cm long and 1 mm in diameter.

RA d

= =⋅( )( )

( )=

−

−

ρ ρπ π

4 4 10 10

102

13 1

3 2~ m m

m

Ω~~1018 Ω

(b) Rd

=× ⋅( )( )

×( )− −

−

4 4 1 7 10 10

2 102

8 3

2

ρπ π

~

. m m

m

Ω22

710~ − Ω

(c) IV

R= −∆

Ω~ ~

1010

216 V

10 A18

I ~ ~10

102

79 V

10 A− Ω

P27.15 J E= σ so σ = = × = × ⋅( )−

− −J

E

6 00 10

1006 00 10

1315 1.

.A m

V m m

2

Ω

Section 27.3 A Model for Electrical Conduction

*P27.16 (a) The density of charge carriers n is set by the material and is unaffected .

(b) The current density is proportional to current according to JI

A=

so it doubles .

(c) For larger current density in J ne d= v the drift speed vd doubles .

(d) The time between collisions τ σ= m

nq2 is unchanged as long as σ does not change due to a

temperature change in the conductor.


106 Chapter 27

P27.17 ρτ

= m

nq2 We take the density of conduction electrons from an Example in the chapter text.

so τρ

= =×( )

×( ) ×( )−

−

m

nq2

31

8 28

9 11 10

1 70 10 8 46 10 1

.

. . ...

60 102 47 10

19 214

×( )= ×

−

− s

vd

qE

m= τ

gives 7 84 101 60 10 2 47 10

9 11 104

19 14

.. .

.× =

×( ) ×( )×

−− −

−

E331

Therefore, E = 0 180. V m

Section 27.4 Resistance and Temperature

P27.18 R R T= + ( )⎡⎣ ⎤⎦0 1 α ∆ gives 140 19 0 1 4 50 10 3 °CΩ Ω ∆= ( ) + ×( )⎡⎣ ⎤⎦

−. . T

Solving, ∆T T= × = −1 42 10 20 03. .°C °C

And the fi nal temperature is T = ×1 44 103. °C

P27.19 (a) ρ ρ α= + −( )⎡⎣ ⎤⎦ = × ⋅( ) + ×− −0 0

81 2 82 10 1 3 90 10T T . . mΩ 33 830 0 3 15 10. .° m( )⎡⎣ ⎤⎦ = × ⋅− Ω

(b) JE= =

× ⋅= ×−ρ

0 200

3 15 106 35 10

86.

..

V m

mA m2

Ω

(c) I JA Jd= =

⎛

⎝⎜

⎞

⎠⎟ = ×( ) ×( −

π π26

4

46 35 10

1 00 10.

.A m

m2 ))⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

=2

449 9. mA

(d) n = ×

×( )⎡⎣

6 02 10

26 98

23.

.

electrons

g 2.70 10 g m6 3 ⎤⎤⎦= ×6 02 1028. electrons m3

vd

J

ne= =

×( )×( )

6 35 10

6 02 10 1

6

28

.

.

A m

electrons m

2

3 ..60 10659

19×( ) =− C m sµ

(e) ∆V E= = ( )( ) = 0 200 2 00 0 400. . . V m m V

*P27.20 We require 103 5 10

1 5 10

1 5 1051

3 2

6

m

mΩ Ω Ω= × ⋅

×+ × ⋅−

−

−.

( . )

.π

mm

m

23 21 5 10π ( . )× −

and for

any ∆T 103 5 10

1 5 101 0 5 10

51

3 23

m

mΩ Ω ∆= × ⋅

×− ×

−

−−.

( . ).

π

TT

°C

m

m⎛⎝⎜

⎞⎠⎟ + × ⋅

×+

−

−

1 5 10

1 5 101 0 4

62

3 2

.

( . ).

Ω π

××⎛⎝⎜

⎞⎠⎟

−10 3 ∆T

°C

simplifying gives 10 = 4.951 5 1 + 0.212 21

2

and 0 = – 2.475 7 × 10–3

1 + 8.488 3 × 10–5

2

These conditions are just suffi cient to determine 1 and

2. The design goal can be met.

We have 2 = 29.167

1 so 10 = 4.951 5

1 + 0.212 21 (29.167

1)

and 1 = 10/11.141 = 0 898 26 21 2. . m m= =



P27.21 R R T= +[ ]0 1 α

R R R T

R R

RT

− =− = = ×( ) =−

0 0

0

0

35 00 10 25 0 0 125

α

α

∆

∆ . . .

P27.22 For aluminum, αE = × − −3 90 10 3 1. °C (Table 27.2)

α = × − −24 0 10 6 1. °C (Table 19.1)

RA

T T

A TR

TE E= =+( ) +( )

+( )=

+(ρ ρ α αα

α 02 0

1 1

1

1∆ ∆∆

∆ ))+( ) = ( )⎛

⎝⎜⎞⎠⎟

=1

1 2341 39

1 002 41 71

α∆Ω Ω

T.

.

..

Section 27.5 Superconductors

Problem 50 in Chapter 43 can be assigned with this section.

Section 27.6 Electrical Power

P27.23 IV

= = =P∆

6005 00

W

120 VA.

and RV

I= = =∆ Ω120

24 0 V

5.00 A .

P27.24 P = = × ×( ) =−I V∆ 500 10 15 10 7 506 3A V W.

*P27.25 The energy that must be added to the water is

Q mc T= = ( )( )( ) = ×∆ 109 4 186 29 0 1 32 107kg J kg°C °C. . JJ

Thus, the power supplied by the heater is

P = = = ××

=W

t

Q

t∆ ∆1 32 10

8 8207. J

25 60 sW

and the resistance is RV= ( )

= ( )=∆ Ω

2 2220

8 8205 49

PV

W. .

*P27.26 (a) efficiencymechanical power output

total pow=

eer input

hp(746 W/1 hp)

(120 V) = =0 900

2 50.

.

II

I = = =1 860

0 9 120

2 070

12017 3

J/s

V)

J/s

J/C. (. A

(b) energy input = Pinput

∆t = (2 070 J/s) 3 (3 600 s) = J2 24 107. ×

(c) cost = 2.24 × 107 J S

1 kWh

k

10

J

W s

h

3 600 s0.93

/⎛⎝

⎞⎠

⎛⎝

⎞⎠ =0 16.

$ 995


108 Chapter 27

P27.27 PP0

2

0

20

2 2140

120= ( )

( )=

⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠ =∆

∆∆∆

V R

V R

V

V11 361.

∆% % % .=−⎛

⎝⎜⎞⎠⎟

( ) = −⎛⎝⎜

⎞⎠⎟

( ) =P P

PPP

0

0 0

100 1 100 1 3661 1 100 36 1−( ) =% . %

P27.28 The battery takes in energy by electric transmission

P ∆ ∆ ∆t V I t= ( ) ( ) = ×( )−2 3 13 5 10 4 23 6003. . .J C C s h

ss

1 hJ⎛

⎝⎞⎠ = 469

It puts out energy by electric transmission

∆ ∆V I t( ) ( ) = ×( ) ⎛⎝

−1 6 18 10 2 43 6003. .J C C s h

s

1 h⎞⎞⎠ = 249 J

(a) effi ciency = = =useful output

total input

J

469 J

2490 530.

(b) The only place for the missing energy to go is into internal energy:

469 249

221

J J

J

int

int

= +

=

∆

∆

E

E

(c) We imagine toasting the battery over a fi re with 221 J of heat input:

Q mc T

TQ

mc

=

= = ° = °

∆

∆ 22115 1

J kg C

0.015 kg 975 JC.

P27.29 P = ( ) = ( )=I V

V

R∆ ∆ 2

500 W

R = ( )( )

=110

50024 2

2 V

W . Ω

(a) RA

= ρ so = =

( ) ×( )× ⋅

=−

−

RA

ρπ24 2 2 50 10

1 50 10

4 2

6

. .

.

m

m

ΩΩ

33 17. m

(b) R R T= +[ ] = + ×( )( )⎡⎣ ⎤⎦ =−0

31 24 2 1 0 400 10 1180α∆ Ω. . 335 6. Ω

P = ( )= ( )

=∆V

R

2 2110

35 6340

.W

P27.30 RA

= =× ⋅( )

×( )−

−

ρπ

1 50 10 25 06

3 2

. .m m

0.200 10 m

Ω== 298 Ω

∆ ΩV IR= = ( )( ) =0 500 298 149. A V

(a) EV= = =∆

1495 97

V

25.0 mV m.

(b) P = ( ) = ( )( ) =∆V I 149 0 500 74 6V A W. .

(c) R R T T= + −( )⎡⎣ ⎤⎦ = + × °( ) °−0 0

31 298 0 400 10 320α 1 CΩ . CC ⎡⎣ ⎤⎦ = 337 Ω

I

V

R

V I

= = ( )( ) =

= ( ) = ( )

∆Ω

∆

149

3370 443

149

VA

V

.

P 00 443 66 1. .A W( ) =



P27.31 (a) ∆ ∆ ∆U q V It V= ( ) = ( ) = ⋅( ) ( )⋅

⎛55 0 12 0

1. .A h V

C

1 A s⎝⎝⎜

⎞⎠⎟

⋅⎛⎝⎜

⎞⎠⎟ ⋅⎛

⎝⎜

⎞⎠⎟

= ⋅ =

1 1

660

J

1 V C

W s

1 J

W h 00 660. kWh

(b) Cost =⎛⎝⎜

⎞⎠⎟ =0 660

0 060 0

13 96.

$ .. kWh

kWh¢

*P27.32 (a) The resistance of 1 m of 12-gauge copper wire is

RA d d

= =( )

= =× ⋅( )−ρ ρ

πρ

π π

2

4 4 1 7 10 12 2

8. m m

0.

Ω

2205 3 m×( ) = ×−

−

105 14 10

2 23. Ω

The rate of internal energy production is P = = = ( ) × =−I V I R∆ Ω2 2 320 5 14 10 2 05A W .. .

(b) PAlAl= =I R

I

d2

2

2

4ρπ

PP

Al

Cu

Al

Cu

= ρρ

PAl

m

1.7 mW W= × ⋅

× ⋅=

−

−

2 82 10

102 05 3 41

8

8

.. .

ΩΩ

Aluminum of the same diameter will get hotter than copper. It would not be as safe. If it is surrounded by thermal insulation, it could get much hotter than a copper wire.

P27.33 The energy taken in by electric transmission for the fl uorescent lamp is

P∆t = ( )⎛⎝

⎞⎠ = ×11 100

3 6003 96 106J s h

s

1 hJ

cos

.

tt JkWh

k

1 000

W s

J= × ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⋅⎛3 96 100 086.

$ .⎝⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

=h

3 600 s$ .0 088

For the incandescent bulb,

P∆t = ( )⎛⎝

⎞⎠ = ×40 100

3 6001 44 107W h

s

1 hJ

cost

.

== ××

⎛⎝⎜

⎞⎠⎟ =1 44 10

0 080 327.

$ .$ .J

3.6 10 J

savin

6

gg = − =$ . $ . $ .0 32 0 088 0 232

P27.34 The total clock power is

270 10 2 503 6006×( )⎛

⎝⎞⎠

⎛⎝clocks

J s

clock

s

1 h. ⎞⎞

⎠ = ×2 43 1012. J h

From eW

Q= out

in

, the power input to the generating plants must be:

Q

t

W t

ein out J h

J∆

∆= = × = ×2 43 10

0 2509 72 10

1212.

.. hh

and the rate of coal consumption is

Rate J hkg coal

33.0 10 J6= ×( )×

⎛⎝

⎞9 72 101 0012.

.⎠⎠ = × =2 95 10 2955. kg coal h metric ton h


110 Chapter 27

P27.35 P = ( ) = ( )( ) =I V∆ 1 70 110 187. A V W

Energy used in a 24-hour day = ( )( ) =0 187 24 0 4 49. . .kW h kWh.

Therefore daily cost kWh$0.060 0

kWh¢ .= ⎛

⎝⎜⎞⎠⎟ = =4 49 0 269 26 9. $ . .

P27.36 P = = ( )( ) =I V∆ 2 120 240.00 A V W

∆Eint kg J kg °C °C k= ( ) ⋅( )( ) =0 500 4 186 77 0 161. . JJ

J

Wsint∆

∆t

E= = × =

P1 61 10

240672

5.

P27.37 At operating temperature,

(a) P = = ( )( ) =I V∆ 1 53 120 184. A V W

(b) Use the change in resistance to fi nd the fi nal operating temperature of the toaster.

R R T= +( )0 1 α∆ 120

1 53

120

1 801 0 400 10 3

. ..= + ×( )⎡

⎣⎤⎦

− ∆T

∆T = 441°C T = + =20 0 441 461. °C °C °C

P27.38 You pay the electric company for energy transferred in the amount E t= P ∆ .

(a) P ∆t = ( )⎛⎝

⎞⎠

⎛⎝

⎞40 286 400

W weeks7 d

1 week

s

1 d ⎠⎠ ⋅⎛⎝

⎞⎠ =1

48 4J

1 W sMJ.

P ∆t = ( )⎛⎝

⎞⎠

⎛⎝

⎞⎠40 2

24

1W weeks

7 d

1 week

h

1 d

k

000013 4

40 2

⎛⎝⎜

⎞⎠⎟

=

= ( )

. kWh

W weeks7 d

1 weP ∆t

eek

h

1 d

k $

kWh⎛⎝

⎞⎠

⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠

24

1 000

0 12.⎟⎟ = $ .1 61

(b) P ∆t = ( )⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

970 31 000

0 1W min

1 h

60 min

k . 220 005 82 0 582

$

kWh¢⎛

⎝⎜⎞⎠⎟ = =$ . .

(c) P ∆t = ( )⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

5 200 41 000

0W 0 min

1 h

60 min

k ..$ .

120 416

$

kWh⎛⎝⎜

⎞⎠⎟ =

P27.39 Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to the dryer is

P ∆t = ( )( )( ) ≈ ×400 600 365 9 1017J s s d d J

kWh

3.6 ××⎛⎝

⎞⎠ ≈

10 JkWh6 20

We suppose that electrically transmitted energy costs on the order of ten cents per kilowatt-hour. Then the cost of using the dryer for a year is on the order of

Cost kWh kWh≈ ( )( ) =20 0 10 2 1$ . $ ~$



Additional Problems

*P27.40 (a) IV

R= ∆

so P = = ( )I V

V

R∆ ∆ 2

RV= ( )

= ( )=∆ Ω

2 2120

25 05

PV

W76

. and R

V= ( )= ( )

=∆ Ω2 2120

100144

PV

W

(b) IV

Q

t t= = = = =P

∆ ∆ ∆25 0

0 2081 00.

..W

120 VA

C

∆t = =1 004 80

..

C

0.208 A s

The charge itself is the same. It comes out at a location that is at lower potential.

(c) P = = =25 01 00

..

WJ∆

∆ ∆U

t t ∆t = =1 00

0 040 0.

. J

25.0 W s

The energy itself is the same. It enters the bulb by electrical transmission and leaves by heat and electromagnetic radiation.

(d) ∆ ∆U t= = ( )( )( ) = ×P 25 0 86 400 30 0 64 8 106. . .J s s d d J

The electric company sells energy .

Cost J$0.070 0

kWh

k

1 000

W= ×⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

64 8 106.⋅⋅⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=s

J

h

3 600 s

Cost per joule

$ .1 26

==×

⎛⎝⎜

⎞⎠⎟

= × −$ .$ .

0 070 01 94 10 8

kWh

kWh

3.60 10 JJ

6

*P27.41 The heater should put out constant power

P = =−( )

=( )( ) −Q

t

mc T T

tf i

∆ ∆0 250 4 186 100 2. kg J °C 00 1

349°C

kg °C 4 min

min

60 sJ s

( )⋅ ( )

⎛⎝

⎞⎠ =

Then its resistance should be described by

P

P= ( ) = ( )( )

=( )

=( ) =∆ ∆ ∆ ∆

V IV V

RR

V2

120

3494

2J C

J s11 3. Ω

Its resistivity at 100°C is given by

ρ ρ α= + −( )⎡⎣ ⎤⎦ = × ⋅( ) + ×− −0 0

6 31 1 50 10 1 0 4 10T T . .mΩ 880 1 55 10 6( )⎡⎣ ⎤⎦ = × ⋅−. mΩ

Then for a wire of circular cross section

RA r d

= = =

= × ⋅( )−

ρ ρπ

ρπ

π

2 2

6

4

41 3 1 55 104

. . mΩ Ωdd

dd

2

2

7 2 82 09 10 4 77 10

= × = ×( )+ −. .m or m

One possible choice is = 0 900. m and d = × −2 07 10 4. m. If and d are made too small,

the surface area will be inadequate to transfer heat into the water fast enough to prevent

overheating of the fi lament. To make the volume less than 0 5. cm ,3 we want and d less

than those described by π d 2

6

40 5 10 = × −. m .3 Substituting d 2 84 77 10= ×( )−. m gives

π4

4 77 10 0 5 108 2 6. .×( ) = ×− −m m ,3 = 3 65. m and d = × −4 18 10 4. m. Thus our answer is:

Any diameter d and length related by d 2 84 77 10= ×( )−. m would have the right resistance. One possibility is length 0.900 m and diameter 0.207 mm, but such a small wire might overheat

rapidly if it were not surrounded by water. The volume can be less than 0 5. cm .3


112 Chapter 27

P27.42 The original stored energy is U Q VQ

Ci i= =1

2

1

2

2

∆ .

(a) When the switch is closed, charge Q distributes itself over the plates of C and 3C in parallel, presenting equivalent capacitance 4C. Then the fi nal potential difference is

∆VQ

Cf=

4 for both.

(b) The smaller capacitor then carries charge C VQ

CC

Qf

∆ = =4 4

. The larger capacitor

carries charge 34

3

4C

Q

C

Q= .

(c) The smaller capacitor stores fi nal energy 1

2

1

2 4 32

22 2

C V CQ

C

Q

Cf∆( ) =

⎛⎝⎜

⎞⎠⎟

= . The larger

capacitor possesses energy 1

23

4

3

32

2 2

CQ

C

Q

C

⎛⎝⎜

⎞⎠⎟

= .

(d) The total fi nal energy is Q

C

Q

C

Q

C

2 2 2

32

3

32 8+ = . The loss of potential energy is the energy

appearing as internal energy in the resistor: Q

C

Q

CE

2 2

2 8= + ∆ int ∆E

Q

Cint = 3

8

2

P27.43 We begin with the differential equation αρ

ρ= 1 d

dT

(a) Separating variables, d

dTT

Tρρ

αρ

ρ

0 0

∫ ∫=

lnρρ

α0

0

⎛

⎝⎜⎞

⎠⎟= −( )T T

and

ρ ρ α= −( )

00e

T T .

(b) From the series expansion e xx ≈ +1 , x <<( )1 , we have

ρ ρ α≈ + −( )⎡⎣ ⎤⎦0 01 T T .

P27.44 We fi nd the drift velocity from I nq A nq rd d= =v v π 2

vd

I

nq r= =

× ×( )− −π π2 28 3 19

1 000

8 46 10 1 60 10

A

m C. . 1102 35 10

200 10

2 24

3

−−

( ) = ×

= = = ××

mm s

m

2.35

.

vv

x

tt

x

1108 50 10 24

8− = × =

m ss 7.0 yr.

*P27.45 From ρ = =( )RA V

I

A

∆ we compute (m) ( ) ( m)

0.540

1.028

1.543

10.4

21.1

31.

R Ω Ωρ ⋅

88

1 41 10

1 50 10

1 50 10

6

6

6

.

.

.

×××

−

−

−

ρ = × ⋅−1 47 10 6. m .Ω With its uncertainty range from 1.41 to 1.50, this average value agrees

with the tabulated value of 1 50 10 6. × ⋅− mΩ in Table 27.2.



P27.46 2 wires → = 100 m

R = ( ) =0 108100 0 036 0

..

300 mm

Ω Ω

(a) ∆ ∆V V IR( ) = ( ) − = − ( )( ) =home line

120 110 0 036 0 116. VV

(b) P = ( ) = ( )( ) =I V∆ 110 116 12 8A V kW.

(c) Pwires A W= = ( ) ( ) =I R2 2110 0 036 0 436. Ω

*P27.47 (a) E i i= − = −

−( )−( ) =dV

dxˆ .

.. ˆ0 4 00

0 500 08 00

V

mV m

(b) RA

= =× ⋅( )( )

×(−

−

ρ

π

4 00 10 0 500

1 00 10

8

4

. .

.

m m

m

Ω

))=

20 637. Ω

(c) IV

R= = =∆

Ω4 00

6 28.

.V

0.637A

(d) JI

A= =

×( )= × =

−

6 28

1 00 102 00 10 200

4 28.

..

A

mA m M2

πAA m2

The fi eld and the current

are both in the x direction.

(e) ρJ E= × ⋅( ) ×( ) = =−4 00 10 2 00 10 8 008 8. . .m A m V m2Ω

*P27.48 (a) E i i= − =dV x

dx

V

L

( ) ˆ ˆ

(b) RA

L

d= =ρ ρ

π 4

2

(c) IV

R

V d

L= =∆ π

ρ

2

4

(d) JI

A

V

L= =

ρ

The fi eld and the current are both in the x direction.

(e) ρ JV

LE= =


114 Chapter 27

P27.49 (a) P = I V∆

so IV

= = × =P∆

8 00 10667

3. W

12.0 VA

(b) ∆ ∆t

U= = ××

= ×P

2 00 102 50 10

73.

.J

8.00 10 Ws3

and ∆ ∆x t= = ( ) ×( ) =v 20 0 2 50 10 50 03. . .m s s km

*P27.50 (a) We begin with RA

T T T T

A= =

+ −( )⎡⎣ ⎤⎦ + ′ −( )⎡⎣ ⎤⎦+ ′

ρ ρ α α 0 0 0 0

0

1 1

1 2αα T T−( )⎡⎣ ⎤⎦0

,

which reduces to RR T T T T

T T=

+ −( )⎡⎣ ⎤⎦ + ′ −( )⎡⎣ ⎤⎦+ ′ −( )⎡

0 0 0

0

1 1

1 2

α αα⎣⎣ ⎤⎦

(b) For copper: ρ081 70 10= × ⋅−. mΩ , α = × − −3 90 10 3 1. °C , and

′ = × − −α 17 0 10 6 1. °C

RA00 0

0

8

3 2

1 70 10 2 00

0 100 101= =

×( )( )×( ) =−

−

ρπ

. .

...08 Ω

The simple formula for R gives:

R = ( ) + ×( ) −( )⎡⎣− −1 08 1 3 90 10 100 20 03 1. . .°C °C °CΩ ⎤⎤⎦ = 1 420. Ω

while the more complicated formula gives:

R =( ) + ×( )( )⎡⎣ ⎤⎦ +− −1 08 1 3 90 10 80 0 1 13 1. . .°C °CΩ 77 0 10 80 0

1 2 17 0 10

6 1

6 1

. .

.

×( )( )⎡⎣+ ×

− −

− −

°C °C

°C(( )( )⎡⎣ ⎤⎦=

80 0

1 418

.

.

°C

Ω

The results agree to three digits. The variation of resistance with temperature is typically a much larger effect than thermal expansion in size.



P27.51 Let a be the temperature coeffi cient at 20.0°C, and ′α be the temperature coeffi cient at 0°C.

Then ρ ρ α= + −( )⎡⎣ ⎤⎦0 1 20 0T . ,°C and ρ ρ α= ′ + ′ −( )⎡⎣ ⎤⎦1 0T °C must both give the correct

resistivity at any temperature T. That is, we must have:

ρ α ρ α0 1 20 0 1 0+ −( )[ ] = ′ + ′ −( )[ ]T T. °C °C (1)

Setting T = 0 in equation (1) yields: ′ = − ( )⎡⎣ ⎤⎦ρ ρ α0 1 20 0. ,°C

and setting T = 20 0. °C in equation (1) gives: ρ ρ α0 1 20 0= ′ + ′( )[ ]. °C

Put ′ρ from the fi rst of these results into the second to obtain:

ρ ρ α α0 0 1 20 0 1 20 0= − ( )[ ] + ′( )[ ]. .°C °C

Therefore 1 20 01

1 20 0+ ′ ( ) =

− ( )αα

..

°C°C

which simplifi es to ′ =− ( )[ ]α α

α1 20 0. °C

From this, the temperature coeffi cient, based on a reference temperature of 0°C, may be

computed for any material. For example, using this, Table 27.2 becomes at 0°C :

Material Temp Coeffi cients at 0°C

Silver 4 1 10 3. × − °C

Copper 4 2 10 3. × − °C

Gold 3 6 10 3. × °− C

Aluminum 4 2 10 3. × °− C

Tungsten 4 9 10 3. × °− C

Iron 5 6 10 3. × °− C

Platinum 4 25 10 3. × °− C

Lead 4 2 10 3. × °− C

Nichrome 0 4 10 3. × °− C

Carbon − × °−0 5 10 3. C

Germanium − × °−24 10 3 C

Silicon − × °−30 10 3 C


116 Chapter 27

P27.52 (a) A thin cylindrical shell of radius r, thickness dr, and length L contributes resistance

dRd

A

dr

r L L

dr

r= =

( ) =⎛⎝⎜

⎞⎠⎟

ρ ρπ

ρπ

2 2

The resistance of the whole annulus is the series summation of the contributions of the thin shells:

RL

dr

r L

r

rr

r

b

aa

b

= =⎛⎝⎜

⎞⎠⎟∫

ρπ

ρπ2 2

ln

(b) In this equation ∆V

I L

r

rb

a

=⎛

⎝⎜

⎞

⎠⎟

ρπ2

ln

we solve for ρ π= ( )2 L V

I r rb a

∆ln

*P27.53 The original resistance is Ri = rL

i/A

i.

The new length is L = Li + d L = L

i(1 + d ).

Constancy of volume implies AL = AiL

i so A =

A L

L

A L

L

Ai i i i

i

i=+

=+( ) ( )1 1δ δ

The new resistance is RL

A

L

AR Ri

ii i= = +

+= + = + +ρ ρ δ

δδ δ δ( )

/ ( )( ) ( )

1

11 1 22 2 ..

The result is exact if the assumptions are precisely true. Our derivation contains no approxima-tion steps where delta is assumed to be small.

P27.54 Each speaker receives 60.0 W of power. Using P = I R2 , we then have

IR

= = =P 60 03 87

..

W

4.00A

Ω

The system is not adequately protected since the fuse should be set to melt at 3.87 A, or lesss .

P27.55 (a) ∆V E= − ⋅ or dV E dx= − ⋅

∆V IR E

Idq

dt

E

R

AE

AE A

dV

dx

= − = − ⋅

= = ⋅ = ⋅ = = − =

ρ ρ

σ σ AAdV

dx

(b) Current fl ows in the direction of decreasing voltage. Energy fl ows by heat in the direction of decreasing temperature.

P27.56 From the geometry of the longitudinal section of the resistor shown in the fi gure, we see that

b r

y

b a

h

−( )= −( )

From this, the radius at a distance y from the base is r a by

hb= −( ) +

For a disk-shaped element of volume dRdy

r= ρ

π 2 : Rdy

a b y h b

h

=−( )( ) +⎡⎣ ⎤⎦

∫ρπ 2

0

Using the integral formula du

au b a au b+( )= −

+( )∫ 2

1, R

h

ab= ρ

π

FIG. P27.56



P27.57 Rdx

A

dx

wy= =∫ ∫ρ ρ

where y y

y y

Lx= + −

12 1

Rw

dx

y y y L x

L

w y yy

y yL

=+ −( )[ ]

=−( )

+ −∫ρ ρ1 2 10 2 1

12ln 11

0

2 1

2

1

Lx

RL

w y y

y

y

L⎡⎣⎢

⎤⎦⎥

=−( )

⎛

⎝⎜

⎞

⎠⎟

ρln

*P27.58 A spherical layer within the shell, with radius r and thickness dr, has resistance

dRdr

r= ρ

π4 2

The whole resistance is the absolute value of the quantity

R dRdr

r

r

rr

r

a

b

r

r

aa

b

a

b

= = =−

= − − +∫∫−ρ

πρπ

ρπ4 4 1 4

12

1 11

4

1 1

r r rb a b

⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟

ρπ

*P27.59 Coat the surfaces of entry and exit with material of much higher conductivity than the bulk mate-rial of the object. The electric potential will be essentially uniform over each of these electrodes. Current will be distributed over the whole area where each electrode is in contact with the resis-tive object.

P27.60 (a) The resistance of the dielectric block is RA

d

A= =ρ

σ

.

The capacitance of the capacitor is CA

d= κ∈0 .

Then RCd

A

A

d= ∈ = ∈

σκ κ

σ0 0 is a characteristic of the material only.

(b) RC C

= ∈ = ∈ =× ⋅ ( ) × −κ

σρκ0 0

16 1275 10 8 85 10m 3.78 CΩ . 22

2F N m14 101 79 10

915

× ⋅= ×− . Ω

P27.61 (a) Think of the device as two capacitors in parallel. The one on the left has κ1 1= ,

A x1 2= +⎛

⎝⎜

⎞⎠⎟ . The equivalent capacitance is

κ κ κ1 0 1 2 0 2 0 0

2 2

∈+

∈=

∈+⎛

⎝⎞⎠ +

∈−⎛

⎝⎞⎠ =

A

d

A

d dx

dx

∈∈+ + −( )0

22 2

dx xκ κ

(b) The charge on the capacitor is Q C V= ∆

QV

dx x=

∈+ + −( )0

22 2

∆ κ κ

The current is

IdQ

dt

dQ

dx

dx

dt

V

d

V

d= = =

∈+ + −( ) = −

∈0 0

20 2 0 2

∆ ∆κ vv κκ −( )1

The negative value indicates that the current drains charge from the capacitor. Positive

current is clockwise∈ −( )0 1

∆V

d

v κ .

FIG. P27.57


118 Chapter 27

P27.62 I Ie V

k TB

=⎛

⎝⎜

⎞

⎠⎟ −

⎡

⎣⎢

⎤

⎦⎥0 1exp

∆

and RV

I= ∆

with I091 00 10= × −. A, e = × −1 60 10 19. C, and kB = × −1 38 10 23. J K

The following includes a partial table of calculated values and a graph for each of the specifi ed temperatures.

(i) For T = 280 K:

∆ ΩV I RV A

( ) ( ) ( )0 400 0 015 6 25 6

0 440 0 081 8 5

. . .

. . .338

0 480 0 429 1 12

0 520 2 25 0 232

0 560 11 8 0

. . .

. . .

. .

..

. . .

047 6

0 600 61 6 0 009 7

(ii) For T = 300 K:

∆ ΩV I RV A

( ) ( ) ( )0 400 0 005 77 3

0 440 0 024 18 1

. . .

. . .

00 480 0 114 4 22

0 520 0 534 0 973

0 560 2 51 0

. . .

. . .

. .

..

. . .

223

0 600 11 8 0 051

(iii) For T = 320 K:

∆ ΩV I RV A( ) ( ) ( )0 400 0 002 0 203

0 440 0 008 4 52 5

. .

. . .

00 480 0 035 7 4

0 520 0 152 3 42

0 560 0 648 0

. . .

. . .

. .

13

..

. . .

864

0 600 2 76 0 217

FIG. P27.62(i)

FIG. P27.62(ii)

FIG. P27.62(iii)



P27.63 The volume of the gram of gold is given by ρ = m

V

Vm

A= =×

= × = ×−

−

ρ10

19 3 105 18 10 2 40

3

38kg

kg mm3

3

.. . 110

2 16 10

2 44 10

3

11

8

m

m

m 2

2

( )= ×

= =× ⋅

−

−

A

RA

.

.ρ Ω ..4 10 m

m

3

2

×( )×

= ×−2 16 102 71 1011

6

.. Ω

P27.64 The resistance of one wire is 0 500

100 50 0.

. .

mi mi

Ω Ω⎛⎝⎜

⎞⎠⎟ ( ) =

The whole wire is at nominal 700 kV away from ground potential, but the potential difference between its two ends is

IR = ( )( ) =1 000 50 0 50 0A kV. .Ω

Then it radiates as heat power P = ( ) = ×( )( ) =∆V I 50 0 10 1 000 50 03. . .V A MW

P27.65 R R T T= + −( )⎡⎣ ⎤⎦0 01 α so T TR

RT

I

I= + −

⎡

⎣⎢

⎤

⎦⎥ = + −⎡

⎣⎢⎤⎦⎥0

00

011

11

α α

In this case, II= 0

10, so T T= + ( ) = + =0

19 20

9

0 004 502 020

α°

°C°C

.

ANSWERS TO EVEN PROBLEMS

P27.2 3 64. h

P27.4 qw /2p

P27.6 0.265 C

P27.8 (a) 99.5 kA/m2 (b) Current is the same, current density is smaller. 5.00 A, 24.9 kA/m2, 0.800 cm

P27.10 0.130 mm/s

P27.12 500 mA

P27.14 (a) ~1018 Ω (b) ~10−7 Ω (c) ~100 aA, ~1 GA

P27.16 (a) no change (b) doubles (c) doubles (d) no change

P27.18 1.44 × 103 °C

P27.20 She can meet the design goal by choosing 1 = 0.898 m and

2 = 26.2 m.

P27.22 1 71. Ω

P27.24 7.50 W


120 Chapter 27

P27.26 (a) 17.3 A (b) 22.4 MJ (c) $0.995

P27.28 (a) 0.530 (b) 221 J (c) 15.1°C

P27.30 (a) 5.97 V/m (b) 74.6 W (c) 66.1 W

P27.32 (a) 2.05 W (b) 3.41 W. It would not be as safe. If surrounded by thermal insulation, it would get much hotter than a copper wire.

P27.34 295 metric ton h

P27.36 672 s

P27.38 (a) $1.61 (b) $0.005 82 (c) $0.416

P27.40 (a) 576 Ω and 144 Ω (b) 4.80 s. The charge itself is the same. It is at a location that is lower in potential. (c) 0.040 0 s. The energy itself is the same. It enters the bulb by electric transmission and leaves by heat and electromagnetic radiation. (d) $1.26, energy, 1.94 × 10−8 $/J

P27.42 (a) Q/4C (b) Q/4 and 3Q/4 (c) Q2/32C and 3Q2/32C (d) 3Q2/8C

P27.44 8.50 × 108 s = 27.0 yr

P27.46 (a) 116 V (b) 12 8. kW (c) 436 W

P27.48 (a) E = V/L in the x direction (b) R = 4rL/pd 2 (c) I = Vpd 2/4rL (d) J = V/rL(e) See the solution.

P27.50 (a) See the solution. (b) 1 418. Ω nearly agrees with 1 420. .Ω

P27.52 (a) RL

r

rb

a

= ρπ2

ln (b) ρ π=( )

2 L V

I r rb a

∆ln

P27.54 No. The fuses should pass no more than 3.87 A.

P27.56 See the solution.


P27.60 (b) 1.79 PΩ


P27.64 50 0. MW


Solucionario serway cap 27

Engineering