Solutions to Chapter27

8/10/2019 Solutions to Chapter27

1/18

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

95

ANSWERS TO QUESTIONS

*Q27.1 (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).

Q27.2 Geometry and resistivity. In turn, the resistivity of the

material depends on the temperature.

Q27.3 Voltage is a measure of potential difference, not of

current. Surge implies a flowand only charge, incoulombs, can flow through a system. It would also be

correct to say that the victim carried a certain current, in

amperes.

Q27.4 The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electrons

more efficiently.

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 resistance of copper increaseswith 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.

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

(ii) Answer (a).

Q27.8 Because there are so many electrons in a conductor (approximately 1028electrons/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.

*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 constantapplied 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.


2/18


3/18

Current and Resistance 97

P27.4 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= = ( ) ( )= 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

P27.5 I Q

t=

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

N

Q

e= = =

1 20 10

1 60 10 7 50 1

3

19

.

. .

C

C electron 0015

electrons

*P27.6 (a) J I

A= =

( )=

5 0099 5

3 2

..

A

4.00 10 m kA 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

1

2

24 4 2= = =or so rr1 0 800= . cm

P27.7 (a) J I

A= =

( ) =

8 00 10

1 00 102 55

6

3 2

.

..

A

mA m2

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

e d= =

( ) ( )=

v

2 55

1 60 10 3 00 1019 8.

. .

A m

C m s

2

55 31 1010 3. . m

(c) FromI Q

t=

, we have

t Q

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!)

P27.8 I dq

dt=

q dq Idt

tdt= = = ( )

100120

0

1 240

As

s

sin

qq=

=+100

120 20 100C C

1cos cos

220C

= 0 265.


4/18

98 Chapter 27

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 I q

t=

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

V k q

r

e= = ( ) ( )

8 99 10 1 60 10

2 21

9 19. .

.

N m C C2 2

1106 49 10

7

3

=

mV.

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 nis 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 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, vdI

nqA= =

( ) ( ) 5 00

6 02 10 1 60 1028 3 19.

. .

A

m C 44 00 101 30 10

6

4

..

( )=

m m 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

V I

A=

: I

VA= =

( ) ( )

0 900 6 00 10

5 60 10

7

8

. .

.

V m

2

mm m( )( )1 50.

I= 6 43. A

P27.12 (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~

continued on next page


5/18


(c) I V

R=

~ ~10

102

16V

10A

18

I~ ~10

102

7

9V

10A

P27.13 (a) Given M V Ad d= =

where rdmass density,

we obtain: A M

d

=

Taking rrresistivity, 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) V M

d

=

, or

r M

d

2=

Thus, r M

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 I V

R= = = =

1200 500 500

V

240A mA.

P27.15 J E= so = =

= ( )

J

E

6 00 10

1006 00 10

1315 1. .

A m

V m m

2

Section 27.3 A Model for Electrical Conduction

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

(b) The current density is proportional to current according to J I

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.

P27.17

= m

nq2We 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 10 2 47 10192

14

( ) =

s

vdqE

m=

gives 7 84 101 60 10 2 47 10

9 11 10

4

19 14

.. .

. =

( ) ( )

E331

Therefore, E= 0 180. V m


6/18

100 Chapter 27

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 R R T= +[ ]0 1

R R R T

R R

RT

=

= = ( ) =0 0

0

0

35 00 10 25 0 0 125

. . .

*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 5 101 0 5 10

5

13 2

3m

m

=

.

( . ).

TT

C

m

m

+

+

1 5 10

1 5 101 0 4

6

23 2

.

( . ).

10 3TC

simplifying gives 10 = 4.951 5

1+ 0.212 21

2

and 0 = 2.475 71031+ 8.488 3 105

2

These conditions are just sufficient to determine 1and

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 (a) = + ( ) = ( ) +

0 0

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

(b) J E= =

= 0 200

3 15 106 35 10

8

6.

..

V m

mA m2

(c) I JA J d= =

= ( )

( 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( )=

C m s

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

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

))+( )

=( )

=

11 234

1 39

1 002 41 71

T.

.

..


7/18


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 R V

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 R V

=( )

=( )

=

2 2

220

8 8205 49

P

V

W. .

P27.26 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 outputtotal 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

T Qmc

=

= = =

221 15 1J kg C0.015 kg 975 J C.

P27.27 (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


8/18

102 Chapter 27

*P27.28 (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 23.

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

=

PAl

m

1.7 mW W=

=

2 82 10

102 05 3 41

8

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.

P27.29 P= ( )=( )

=I V V

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

= = =

1495 97

V

25.0 mV 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

I V

R

V I

= =( )

( )=

=( ) =( )

149

3370 443

149

VA

V

.

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

P27.31 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. $ . .


9/18


P27.32 The total clock power is

270 10 2 503 6006( )

clocks

J s

clock

s

1 h.

= 2 43 10

12. J h

From e W

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

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

k1000

W sJ=

3 96 10 0 086. $ . =h3600 s$ .0 088

For the incandescent bulb,

Pt= ( )= 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 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.

P27.35 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


10/18

104 Chapter 27

P27.36 You pay the electric company for energy transferred in the amountE 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 weeks7 d

1 wePt

eek

h

1 d

k $

kWh

24

1000

0 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

=

P27.37 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 10 17J s s d d J kWh3.6 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$ . $ ~$

Additional Problems

P27.38 The original stored energy is U Q V Q

Ci i= =1

2

1

2

2

.

(a) When the switch is closed, charge Qdistributes itself over the plates of Cand 3Cin

parallel, presenting equivalent capacitance 4C. Then the final potential difference is

V Q

Cf=

4

for both.

(b) The smaller capacitor then carries charge C V Q

CC

Qf

= =4 4

. The larger capacitor

carries charge 34

3

4C

Q

C

Q= .

(c) The smaller capacitor stores final energy

1

2

1

2 4 32

22 2

C V C

Q

C

Q

Cf( ) =

= . The larger

capacitor possesses energy1

23

4

3

32

2 2

C Q

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


11/18


P27.39 We begin with the differential equation

=

1d

dT

(a) Separating variables,d

dTT

T

0 0

=

ln

0

0

= ( )T T

and

= ( )0

0e T T

.

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


12/18

106 Chapter 27

(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 01 94 10

8

kWh

kWh

3.60 10 J J6

*P27.43 (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) I V

R= = =

4 006 28

..

V

0.637A

(d) J IA

= =( )

= =6 281 00 102 00 10 200

4 28.

..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

*P27.44 (a)E i i= =dV x

dx

V

L

( )

(b) RA

L

d= =

42

(c) I VR

V dL= =

2

4

(d) J I

A

V

L= =

The field and the current are both in thexdirection.

(e) J V

LE= =

P27.45 (a) P=I V

so IV

= =

=P

8 00 10

6673. W

12.0 VA

(b) t U

= =

= 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


13/18


14/18

108 Chapter 27

P27.48 (a) A thin cylindrical shell of radius r, thickness dr, and lengthLcontributes resistance

dR d

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

P27.49 (a) V E= or dV E dx =

V IR E

I dq

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.

P27.50 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 .

*P27.51 The original resistance isRi= rL

i/A

i.

The new length isL = Li+ dL = 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 1

The new resistance is R L

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.

P27.52 From the geometry of the longitudinal section of the resistor shown in

the figure, we see that

b r

y

b a

h

( )=

( )

From this, the radius at a distanceyfrom the base is r a b y

hb= ( ) +

For a disk-shaped element of volume dR dy

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. P27.52


15/18


P27.53 R dx

A

dx

wy= =

where y y y 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

*P27.54 A spherical layer within the shell, with radius r and thickness dr, has resistance

dR dr

r=

4 2

The whole resistance is the absolute value of the quantity

R dR dr

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

=

*P27.55 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.

P27.56 I I e V

k TB=

0 1exp

and R V

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 specified

temperatures.

(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

FIG. P27.56(i)

FIG. P27.53

continued on next page


16/18

110 Chapter 27

(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 4 220 520 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

P27.57 The volume of the gram of gold is given by =mV

V m A= = = =

1019 3 10

5 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

.

. ..4 10 m

m

3

2

( )

= 2 16 102 71 10

11

6

..

P27.58 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

FIG. P27.56(ii)

FIG. P27.56(iii)


17/18


P27.59 (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=

Q V

dx x=

+ + ( )0

22 2

The current is

I dQ

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 .

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

d

A= =

.

The capacitance of the capacitor is C A

d=

0 .

Then RC d

A

A

d=

=

0 0 is a characteristic of the material only.

(b) RC C

=

=

= ( )

0 016 1275 10 8 85 10m 3.78 C . 22

2F N m14 101 79 10

9

15

=

.

ANSWERS TO EVEN PROBLEMS

P27.2 qw/2p

P27.4 3 64. h

P27.6 (a) 99.5 kA/m2 (b) Current is the same, current density is smaller. 5.00 A, 24.9 kA/m2,

0.800 cm

P27.8 0.265 C

P27.10 0.130 mm/s

P27.12 (a) ~1018 (b) ~107 (c) ~100 aA, ~1 GA

P27.14 500 mA

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.


18/18

112 Chapter 27

P27.22 1 71.

P27.24 7.50 W

P27.26 (a) 0.530 (b) 221 J (c) 15.1C

P27.28 (a) 2.05 W (b) 3.41 W. It would not be as safe. If surrounded by thermal insulation, it would getmuch hotter than a copper wire.

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

P27.32 295 metric ton h

P27.34 672 s

P27.36 (a) $1.61 (b) $0.005 82 (c) $0.416

P27.38 (a) Q/4C (b) Q/4 and 3Q/4 (c) Q2/32Cand 3Q2/32C (d) 3Q2/8C

P27.40 8.50 108

s = 27.0 yr

P27.42 (a) 576 and 144 (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) $1.26, energy, 1.94 108$/J

P27.44 (a)E= V/L in thexdirection (b)R = 4rL/pd2 (c)I = Vpd2/4rL (d)J= V/rL(e) See the solution.

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

P27.48 (a) RL

r

r

b

a

= 2

ln (b) =

( )

2 L V

I r rb a

ln

P27.50 No. The fuses should pass no more than 3.87 A.

P27.52 See the solution.



P27.58 50 0. MW

P27.60 (b) 1.79 P

Documents

Solutions to Chapter27