Solucionario Serway

7/28/2019 Solucionario Serway

1/20

27

Current and Resistance

CHAPTER OUTLINE

27.1 Electric Current

27.2 Resistance

27.3 A Model for Electrical Conduction

27.4 Resistance and Temperature

27.5 Superconductors

27.6 Electrical Power

101

ANSWERS TO QUESTIONS

Q27.1 Voltage is a measure of potential difference, not of

current. Surge implies a flowand only charge, in

coulombs, can flow 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 requirerL/AA

= 3rL/AB

. ThenAA/A

B= 1/3,

answer (f ).

(ii) rA

2/rB

2 = 1/3 gives rA

/rB

= 1/ 3, answer (e).

*Q27.4 Originally, RA

=

. Finally, RA A

Rf = = =

( / ).

3

3 9 9

Answer (b).

Q27.5 The conductor does not follow Ohms 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 efficiently.

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. Silicons 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 field everywhere in the wire nearly instantaneously, to make electrons start

drifting everywhere all at once.


2/20

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.

*Q27.12RL

d

L

d

L

dA

A

A

B

B

B

B

= = = =

( / ) ( / ) ( / )2

2

2 2

1

2 22 2 2RRB

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= = =

22

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 amperehour is (1 C/s)(3 600 s) = 3 600 coulombs. The amperehour 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 resistanceR asV2

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

of wire length and cross-sectional area to make

A

R

=

. You will have to pay for less

material if you make both andA 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


3/20

Current and Resistance 103

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

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= = ( ) ( ) = 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

..

kgatoms

107.9 g

=

= =

1 0005 45 10

12

23g

1 kgatoms.

.I

V

R

006 67 6 67

5 45 10

V

1.80A C s

= =

= = =

. .

.t

Q

I

Ne

I

223 19

41 60 10

6 671 31 10 3 64

( ) ( )= =

.

.. .

C

C ss hh

P27.3 Q t Idt I et

t( ) = = ( ) 0

0 1

(a) Q I e I ( ) = ( ) = ( )

0

1

01 0 632.

(b) Q I e I 10 1 0 999 95010

0 ( ) = ( ) = ( ) .

(c) Q I e I ( ) = ( ) =0 01

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

, and the average current represented by this

revolving charge is Iq

T

q= =

2.

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

A = ( )

= 2 001 00

2 00 10

2

4..

.cmm

100 cm

m2 22

(a) Idq

dtt

tt

1 00 12 5 171 00

2

1 00. .

..

ss

s( ) = = +( ) =

==

00 A

(b) JI

A= =

=

17 0

2 00 1085 0

4

.

..

A

mkA m

2

2

P27.6 Idq

dt=

q dq Idt

tdt= = = ( )

100

120

0

1 240

As

s

sin

qq =

=

+100120 2

0100C C

1

cos cos

220C

= 0 265.

P27.7 (a) J IA

= = ( )

=

8 00 101 00 10

2 556

3 2..

.Am

A m2

(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) FromIQ

t=

, we have

tQ

I

N e

I= = =

( ) ( )

A

C6 02 10 1 60 10

8 00 1

23 19. .

. 001 20 10

6

10

= As. .

(This is about 382 years!)


4/20

104 Chapter 27

*P27.8 (a) JI

A= =

( )=

5 0099 5

3 2

..

A

4.00 10 mkA m2

(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 so rr1 0 800= . cm

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

2

2v

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 tin Iq

t=

10 0 10 1 60 106 19. . = C s C tor 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

V

k q

r

e

= =

( ) ( )

8 99 10 1 60 10

2 21

9 19. .

.

N m C C2 2

110 6 49 1073

= 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 thenumber 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 Avogadros

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 10

23

23. .

..

g gg 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, vdI

nqA= =

( ) ( ) 5 00

6 02 10 1 60 1028 3 19.

. .

A

m C 44 00 101 30 10

6

4

..

( )=

mm s

2

or, vd = 0 130. mm s

Section 27.2 Resistance

P27.11 V IR=

and RA

=

: A = ( )

= 0 6001 00

6 00 102

2

7..

.mmm

1 000 mmm2

VI

A=

: I

VA= =

( ) ( )

0 900 6 00 10

5 60 10

7

8

. .

.

V m

2

mm m( )( )1 50.

I= 6 43. A


5/20


P27.12 IV

R= = = =

1200 500 500

V

240A mA.

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

we obtain: AM

d

=

Takingrr resistivity, R

A M M

r r

d

r d= = =

2

Thus, = =( )( )

( )

MR

r d

1 00 10 0 500

1 70 10 8 92

3

8

. .

. . 1103( ) = 1 82. m

(b) VM

d

=

, or

rM

d

2 =

Thus, rM

d

= =

( )( )

1 00 10

8 92 10 1 82

3

3

.

. . r= 1 40 10 4. m

The diameter is twice this distance: diameter = 280 m

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

RA d

= =( )( )

( )=

4 4 10 10

102

13 1

3 2~

m m

m

~~1018

(b) Rd

= ( )( )

( )

4 4 1 7 10 10

2 102

8 3

2

~

. m m

m

22

710~

(c) IV

R=

~ ~10

102

16V

10A

18

I~ ~10

102

7

9V

10A

P27.15 J E= so = = = ( )

JE

6 00 10100

6 00 1013

15 1. .A mV m

m2

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 = m

nq2is unchanged as long as does not change due to a

temperature change in the conductor.


6/20

106 Chapter 27

P27.17

=m

nq2

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

so

= =( )

( ) ( )

m

nq2

31

8 28

9 11 10

1 70 10 8 46 10 1

.

. . ...

60 102 47 10

19 2

14

( )=

s

vd

qE

m=

gives 7 84 101 60 10 2 47 10

9 11 10

4

19 14

.. .

. =

( ) ( )

E331

Therefore, E= 0 180. V m

Section 27.4 Resistance and Temperature

P27.18 R R T= + ( ) 0 1 gives 140 19 0 1 4 50 103 C = ( ) + ( )

. . T

Solving, T T= = 1 42 10 20 03. .C C

And the final temperature is T= 1 44 103. C

P27.19 (a) = + ( ) = ( ) +

0 0

81 2 82 10 1 3 90 10T T . .m 33 830 0 3 15 10. . m( ) =

(b) JE= =

=

0 200

3 15 106 35 10

8

6.

..

V m

mA m2

(c) I JA Jd= =

= ( )

( 2 64

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

vdJ

ne= =

( )( )

6 35 10

6 02 10 1

6

28

.

.

A m

electrons m

2

3 ..60 10659

19( )=

Cm s

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

*P27.20 We require 103 5 10

1 5 10

1 5 105 13 2

6

m

m

=

+

.

( . )

.

mm

m

23 21 5 10( . )

and for

any T 103 5 10

1 5 10

1 0 5 105

1

3 2

3m

m

=

.

( . )

.

TT

C

m

m

+

+

1 5 10

1 5 10

1 0 46

2

3 2

.

( . )

.

10 3T

Csimplifying gives 10 = 4.951 5

1+ 0.212 21

2

and 0 = 2.475 7 1031+ 8.488 3 105

2

These conditions are just sufficient 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= =


7/20


P27.21 R R T= +[ ]0 1

R R R T

R R

RT

=

= = ( ) =0 0

0

0

35 00 10 25 0 0 125

. . .

P27.22 For aluminum, E =

3 90 10 3 1. C (Table 27.2)

= 24 0 10 6 1. C (Table 19.1)

RA

T T

A TR

TE E= =+( ) +( )

+( )=

+(

02 0

1 1

1

1

))+( )

= ( )

=1

1 2341 39

1 002 41 71

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

600

5 00W

120 VA.

and RV

I= = =

12024 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 kgC 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

=( )

=( )

=

2 2

220

8 8205 49

P

V

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

120

17 3J/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 JS

1 kWh

k

10

J

W s

h

3 600 s0.9

3

/

=

0 16.$ 995


8/20

108 Chapter 27

P27.27P

P0

2

0

2

0

2 2140

120=

( )

( )=

= =

V R

V R

V

V11 361.

% % % .=

( ) =

( ) =P P

P

P

P

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 236003. . .J C C s h

ss

1 hJ

= 469

It puts out energy by electric transmission

V I t( ) ( ) = ( ) 1 6 18 10 2 4

36003. .J C C s hs

1 h

= 249 J

(a) efficiency = = =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 fire with 221 J of heat input:

Q mc T

TQ

mc

=

= =

=

221 15 1J kg C0.015 kg 975 J

C.

P27.29 P = ( ) =( )

=I VV

R

2500 W

R =

( )

( )=

110

50024 2

2V

W.

(a) R

A

=

so = =( ) ( )

=

RA

24 2 2 50 10

1 50 10

4 2

6

. .

.

m

m

33 17. m

(b) R R T= +[ ] = + ( )( ) =

0

31 24 2 1 0 400 10 1180 . . 335 6.

P =( )

=( )

=VR

2 2110

35 6340

.W

P27.30 RA

= = ( )

( )

1 50 10 25 06

3 2

. .m m

0.200 10 m

== 298

V IR= = ( )( ) =0 500 298 149. A V

(a) E V= = =

149 5 97V25.0 m

V m.

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

(c) R R T T= + ( ) = + ( ) 0 031 298 0 400 10 320 1 C . CC = 337

IV

R

V I

= =( )

( )=

= ( ) = ( )

149

3370 443

149

VA

V

.

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


9/20


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

= =( )

= = ( )

2

4 4 1 7 10 12 2

8. m m

0.

2205 3 m( )=

105 14 10

2 2

3.

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

(b) PAlAl= =I R

I

d

22

2

4

P

P

Al

Cu

Al

Cu

=

PAlm

1.7 m

W W=

=

2 82 10

10

2 05 3 418

8

.. .

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 fluorescent lamp is

Pt= ( )= 11 100

3 6003 96 106J s h

s

1 hJ

cos

.

tt JkWh

k

1000

W s

J=

3 96 10

0 086.$ .

=h

3600 s$ .0 088

For the incandescent bulb,

Pt= ( ) = 40 1003 600

1 44 107W hs

1 h J

cost

.

==

=1 44 100 08

0 327.$ .

$ .J3.6 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 10

12. J h

From eW

Q= out

in

, the power input to the generating plants must be:

Q

t

W t

e

in 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 10

1 0012..

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


10/20

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= =

=

P

1 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 find the final operating temperature of the toaster.

R R T= +( )0 1 120

1 53

120

1 801 0 400 10 3

. ..= + ( )

T

T= 441C 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) Pt= ( )

40 2

86 400W weeks

7 d

1 week

s

1 d

=

148 4

J

1 W sMJ.

Pt= ( )

40 2

24

1W weeks

7 d

1 week

h

1 d

k

000013 4

40 2

=

= ( )

. kWh

W weeks 7 d1 we

Pt eekh

1 dk $

kWh 24

10000 12. = $ .1 61

(b) Pt= ( )

970 31000

0 1W min

1 h

60 min

k . 220 005 82 0 582

$

kWh

= =$ . .

(c) Pt= ( )

5 200 41000

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

Pt= ( )( )( ) 400 600 365 9 1017J s s d d J

kWh

3.6

10 J

kWh6

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$ . $ ~$


11/20


Additional Problems

*P27.40 (a) IV

R=

so P = =

( )I V

V

R

2

RV

=( )

=( )

=

2 2

120

25 05

P

V

W76

.and R

V=

( )=

( )=

2 2120

100144

P

V

W

(b) IV

Qt t

= = = = =P

25 0 0 208 1 00. . .W120 V

A C

t= =1 00

4 80.

.C

0.208 As

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= =

1 000 040 0

..

J

25.0 Ws

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

1000

W=

64 8 106.

=s

J

h

3 600 s

Cost per joule

$ .1 26

==

= $ .

$ .0 070 0

1 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

t

f i

0 250 4 186 100 2. kg J C 00 1

349C

kg C 4 min

min

60 sJ s

( )

( )

=

Then its resistance should be described by

P P= ( ) =

( )( )

=

( )

=( )

=

V I

V V

R R

V2

120

349 4

2J C

J s 11 3.

Its resistivity at 100C is given by

= + ( ) = ( ) +

0 0

6 31 1 50 10 1 0 4 10T T . .m 880 1 55 10 6( ) = . m

Then for a wire of circular cross section

RA r d

= = =

= ( )

2 2

6

4

41 3 1 55 104

. . m dd

d

d

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 dare made too small,the surface area will be inadequate to transfer heat into the water fast enough to prevent

overheating of the filament. To make the volume less than 0 5. cm ,3 we want and dless

than those described byd2 6

40 5 10 = . m .3 Substituting d2 84 77 10= ( ). m gives

44 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 dand length related by d2 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


12/20

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 ofCand 3Cin

parallel, presenting equivalent capacitance 4C. Then the final 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 final energy1

2

1

2 4 32

22 2

C V CQ

C

Q

Cf( ) =

= . The larger

capacitor possesses energy1

23

4

3

32

2 2

CQ

C

Q

C

= .

(d) The total final energy isQ

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

=

1 d

dT

(a) Separating variables,d

dTT

T

0 0

=

ln

0

0

= ( )T T

and

= ( )

00e

T T.

(b) From the series expansion e xx +1 , x


13/20


P27.46 2 wires = 100 m

R = ( ) =0 108

100 0 036 0.

.300 m

m

(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.

*P27.47 (a)E i i= =

( )( )

=dV

dx

.

..

0 4 00

0 500 08 00

V

mV m

(b) RA

= = ( )( )

(

4 00 10 0 500

1 00 10

8

4

. .

.

m m

m

))=

20 637.

(c) IV

R= = =

4 006 28

..

V

0.637A

(d) JI

A= =

( )= =

6 28

1 00 102 00 10 200

4 2

8.

..

A

mA m M2

AA m2

The field and the current

are both in thexdirection.

(e) J E= ( ) ( ) = =4 00 10 2 00 10 8 008 8. . .m A m V m2

*P27.48 (a)E i i= =dV x

dx

V

L

( )

(b) R

A

L

d

= =

42

(c) IV

R

V d

L= =

2

4

(d) JI

A

V

L= =

The field and the current are both in thexdirection.

(e) JV

LE= =


14/20

114 Chapter 27

P27.49 (a) P = I V

so IV

= =

=P

8 00 10

6673. W

12.0 VA

(b)

tU

= =

=

P

2 00 102 50 10

73. .

J

8.00 10 Ws

3

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= =

+ ( ) + ( ) +

0 0 0 0

0

1 1

1 2 T T( ) 0,

which reduces to RR T T T T

T T=

+ ( ) + ( ) + ( )

0 0 0

0

1 1

1 2

(b) For copper: 0

81 70 10= . m , = 3 90 10 3 1. C , and

= 17 0 10 6 1. C

R

A0

0 0

0

8

3 2

1 70 10 2 00

0 100 101= = ( )( )

( )=

. .

...08

The simple formula forR gives:

R = ( ) + ( ) ( ) 1 08 1 3 90 10 100 20 03 1. . .C C C = 1 420.

while the more complicated formula gives:

R =( ) + ( )( ) +

1 08 1 3 90 10 80 0 1 13 1. . .C C 77 0 10 80 0

1 2 17 0 10

6 1

6 1

. .

.

( )( )+

C C

C(( )( ) =

80 0

1 418

.

.

C

The results agree to three digits. The variation of resistance with temperature is typically a

much larger effect than thermal expansion in size.


15/20


P27.51 Let abe the temperature coefficient at 20.0C, and be the temperature coefficient at 0C.

Then = + ( ) 0 1 20 0T . ,C and = + ( ) 1 0T C must both give the correctresistivity at any temperature T. That is, we must have:

0 1 20 0 1 0+ ( )[ ] = + ( )[ ]T T. C C (1)

Setting T= 0 in equation (1) yields: = ( ) 0 1 20 0. ,C

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

Put from the first of these results into the second to obtain:

0 0 1 20 0 1 20 0= ( )[ ] + ( )[ ]. .C C

Therefore 1 20 01

1 20 0+ ( ) =

( )

.

.C

C

which simplifies to = ( )[ ]

1 20 0. C

From this, the temperature coefficient, based on a reference temperature of 0C, may be

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

Material Temp Coefficients at 0C

Silver 4 1 10 3. C

Copper 4 2 10 3. C

Gold 3 6 103. C

Aluminum 4 2 103. C

Tungsten 4 9 10

3

.

CIron 5 6 10

3. C

Platinum 4 25 103. C

Lead 4 2 103. C

Nichrome 0 4 10 3. C

Carbon 0 5 10 3. C

Germanium 24 10 3 C

Silicon 30 10 3 C


16/20

116 Chapter 27

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

dRd

A

dr

r L L

dr

r= =

( )=

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

= =

2 2ln

(b) In this equationVI L

r

r

b

a

=

2ln

we solve for

=( )

2 L V

I r rb a

ln

*P27.53 The original resistance isRi=rL

i/A

i.

The new length isL = Li+ dL = L

i(1 + d).

Constancy of volume implies AL = AiL

iso A =

A L

L

A L

L

Ai i i i

i

i=+

=+( ) ( )1 1

The new resistance is RL

A

L

AR Ri

i

i i= =++

= + = + +

( )

/ ( )( ) ( )

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 0

3 87.

.W

4.00A

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

= =

= =

= = = =

AA

dV

dx

(b) Current flows in the direction of decreasing voltage. Energy flows by heat in the direction

of decreasing temperature.

P27.56 From the geometry of the longitudinal section of the resistor shown in

the figure, we see that

b ry

b ah

( ) = ( )

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

hb= ( ) +

For a disk-shaped element of volume dRdy

r=

2: R

dy

a b y h b

h

=( )( ) +

2

0

Using the integral formuladu

au b a au b+( )=

+( ) 2 1 , R

h

ab=

FIG. P27.56


17/20


P27.57 Rdx

A

dx

wy= =

where y yy y

Lx= +

1

2 1

Rw

dx

y y y L x

L

w y yy

y yL

=+ ( )[ ]

=( )

+ 1 2 10 2 1

12ln 11

0

2 1

2

1

Lx

R Lw y y

yy

L

= ( )

ln

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

dRdr

r=

4 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

= = =

= +

4 4 1 4

12

1 11

4

1 1

r r rb a b

=

*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= =

.

The capacitance of the capacitor is CA

d=

0 .

Then RCd

A

A

d=

=

0 0 is a characteristic of the material only.

(b) R C C=

=

= ( )

0 0

16 1275 10 8 85 10m 3.78 C . 222F N m14 10 1 79 109

15

= .

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

A x12

= +

. The equivalent capacitance is

1 0 1 2 0 2 0 0

2 2

+

=

+

+

=

A

d

A

d dx

dx

+ + ( )0

22 2

dx x

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

QV

d

x x=

+ + ( )0

2

2 2

The current is

IdQ

dt

dQ

dx

dx

dt

V

d

V

d= = =

+ + ( ) =

0 02

0 2 0 2

vv

( )1

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

current is clockwise

( )0 1Vd

v .

FIG. P27.57


18/20

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 specifiedtemperatures.

(i) For T= 280 K:

V 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:

V 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:

V 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)


19/20


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

V

Vm

A= =

= =

10

19 3 105 18 10 2 40

3

3

8kg

kg mm

3

3

.. . 110

2 16 10

2 44 10

3

11

8

m

m

m 2

2

( )

=

= =

A

RA

.

. ..4 10 mm

3

2

( )

= 2 16 102 71 10

11

6

..

P27.64 The resistance of one wire is0 500

100 50 0.

. .mi

mi

( ) =

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. .

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 so T TR

RT

I

I= +

= +

0 0

001 1

11

In this case, II

= 010

, so T T= + ( ) = + =01

9 209

0 004 502 020

CC

.

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 (b) ~107 (c) ~100 aA, ~1 GA

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

P27.18 1.44 103C

P27.20 She can meet the design goal by choosing 1

= 0.898 m and 2

= 26.2 m.

P27.22 1 71.

P27.24 7.50 W


20/20

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) 15.1C

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 wouldget 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 and 144 (b) 4.80 s. The charge itself is the same. It is at a location that is lower inpotential. (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, 1.94 108 $/J

P27.42 (a) Q/4C (b) Q/4 and 3Q/4 (c) Q2/32Cand 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 thexdirection (b)R = 4rL/pd2 (c)I = Vpd2/4rL (d)J= V/rL(e) See the solution.

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

P27.52 (a) RL

r

r

b

a

= 2

ln (b) =

( )

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 P


P27.64 50 0. MW

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