Chapter 27 - Current and Resistance

P27.1 Δ=

ΔQIt

( )( )− −Δ = Δ = × = ×6 330.0 10 A 40.0 s 1.20 10 CQ I t

−

−

×= = = ×

×

315

19

1.20 10 C 7.50 10 electrons1.60 10 C electron

QNe

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

=2T , and the average current represented

by this revolving charge is ωπ

= =2

q qI

T.

P27.6 =dq

Idt

( ) π

ππ π

⎛ ⎞= = = ⎜ ⎟⎝ ⎠

− +⎡ ⎤⎛ ⎞= − = =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

∫ ∫ ∫1 240 s

0

120100 A sin

s

100 C 100 Ccos cos0 0.265 C120 2 120

tq dq Idt dt

q

P27.9 (a) The speed of each deuteron is given by = 212

K mv

( )( ) ( )− −× × = × ×6 19 2712.00 10 1.60 10 J 2 1.67 10 kg2

v2 and

= × 71.38 10 m sv

The time between deuterons passing a stationary point is t in =q

It

− −× = ×6 110.0 10 C s 1.60 10 C t9 or −= × 141.60 10 st

So the distance between them is ( ) ( )− −= × × = ×7 141.38 10 m s 1.60 10 s 2.21 10 mvt 7 .

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

( )( )−

−−

× ⋅ ×= = = ×

×

9 2 2 193

7

8.99 10 N m C 1.60 10 C6.49 10 V

2.21 10 mek q

Vr

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

P27.13 (a) Given ρ ρ= = ld dM V A where ρ ≡d mass

density,

we obtain: ρ

=ld

MA Taking ρ ≡r resistivity,

ρ ρ ρ ρρ

= = =l l

l

2r r r d

d

RA M M

l

Thus, ( )( )

( )( )ρ ρ

−

−

×= =

× ×l

3

8 3

1.00 10 0.5001.70 10 8.92 10r d

MR =l 1.82 m

(b) ρ

=d

MV , or πρ

=l2d

Mr

Thus, ( )( )π ρ π

−×= =

×l

3

3

1.00 108.92 10 1.82d

Mr −= × 41.40 10 mr

The diameter is twice this distance: diameter μ= 280 m

P27.17 ρτ

= 2

mnq

We take the density of conduction electrons from an Example in the

chapter text.

so ( )

( ) ( )( )τ

ρ

−−

− −

×= = = ×

× × ×

3114

22 8 28 19

9.11 102.47 10 s

1.70 10 8.46 10 1.60 10mnq

τ=dqE

vm

gives ( ) ( )− −

−−

× ×× =

×

19 144

31

1.60 10 2.47 107.84 10

9.11 10E

Therefore, = 0.180 V mE

P27.18 gives ( )α= + Δ⎡ ⎤⎣ ⎦0 1R R T ( ) ( )−⎡ ⎤Ω = Ω + × ° Δ⎣ ⎦

3140 19.0 1 4.50 10 C T

Solving, Δ = × ° = − °31.42 10 C 20.0 CT T

And the final temperature is = × °31.44 10 CT

P27.21 [ ]α= +0 1R R T

( )α

α −

− = Δ−

= Δ = × =

0 0

30

0

5.00 10 25.0 0.125

R R R TR R T

R

P27.22 For aluminum, 33.90 10 CEα

1− −= × ° (Table 27.2)

α − −= × °624.0 10 C 1 (Table 19.1)

( ) ( )( )

( )( )

( )ρ α α αρ

αα+ Δ + Δ + Δ ⎛ ⎞

= = = = Ω = Ω⎜ ⎟+ Δ+ Δ ⎝ ⎠

ll 002

1 1 1 1.391.234 1.711 1.002 41

E ET T TR R

A TA T

*P27.26 (a) = = = 2.50 hp(746 W/ 1 hp)

0.900 (120 V)

mechanical power outputefficiency

total power input I

= = =1 860 J/ s 2 070 J/ s 17.3 A0.9(120 V) 120 J/ C

I

(b) energy input = Pinput Δt = (2 070 J/s) 3 (3 600 s) = 2.24 × 107 J

(c) cost = 2.24 × 107 J / ⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠3

S 0.16 k J h1 kWh W s 3 600 s10

= $ 0.995

P27.27 ( )( )

⎛ ⎞Δ Δ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟Δ ⎝ ⎠Δ ⎝ ⎠

PP

22 2

20 00

140 1.361120

V R VVV R

( ) ( ) ( )⎛ ⎞ ⎛ ⎞−

Δ = = − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

P P PP P

0

0 0

% 100% 1 100% 1.361 1 100% 36.1%

P27.31

(a)

( ) ( ) ( )( ) ⋅⎛ ⎞⎛ ⎞⎛ ⎞Δ = Δ = Δ = ⋅ = ⋅ =⎜ ⎟⎜ ⎟⎜ ⎟⋅ ⋅⎝ ⎠⎝ ⎠⎝ ⎠

1 C 1 J 1 W s55.0 A h 12.0 V 660 W h 0.660 kWh1 A s 1 V C 1 J

U q V It V

(b) Cost ⎛ ⎞= =⎜ ⎟⎝ ⎠

$0.060 00.660 kWh 3.96¢

1 kWh

P27.35 ( ) ( )( )1.70 A 110 V 187 WI V= Δ = =P

Energy used in a 24-hour day ( )( )= =0.187 kW 24.0 h 4.49 kWh .

Therefore daily ⎛ ⎞= =⎜ ⎟⎝ ⎠

$0.060 0cost 4.49 kWh $0.269 26.9¢

kWh= .

P27.38 You pay the electric company for energy transferred in the amount . = ΔPE t

(a) ( ) ⎛ ⎞⎛ ⎞ ⎛ ⎞Δ = =⎜ ⎟⎜ ⎟⎜ ⎟ ⋅⎝ ⎠⎝ ⎠⎝ ⎠P

86 400 s7 d 1 J40 W 2 weeks 48.4 MJ1 week 1 d 1 W s

t

( )

( )

⎛ ⎞⎛ ⎞⎛ ⎞Δ = =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞Δ = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎝ ⎠

P

P

7 d 24 h k40 W 2 weeks 13.4 kWh1 week 1 d 1000

7 d 24 h k 0.12 $40 W 2 weeks $1.611 week 1 d 1000 kWh

t

t

(b) ( )⎛ ⎞⎛ ⎞ ⎛ ⎞Δ = = =⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

P 1 h k 0.12 $970 W 3 min $0.005 82 0.582¢60 min 1000 kWh

t

(c) ( )⎛ ⎞⎛ ⎞ ⎛ ⎞Δ = =⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

P 1 h k 0.12 $5 200 W 40 min $0.41660 min 1000 kWh

t

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

resistance

( )

ρ ρ ρπ π

⎛ ⎞= = = ⎜ ⎟

⎝ ⎠

l

2 2d drdRA r L L

drr

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

ρ ρπ π

⎛ ⎞= = ⎜ ⎟

⎝ ⎠∫ ln

2 2

b

c

rb

ar

rdrRL r L r

(b) In this equation ρπ

⎛ ⎞Δ= ⎜ ⎟

⎝ ⎠ln

2b

a

rVI L r

we solve for ( )

πρ

Δ=

2ln b a

L VI r r

P27.54 Each speaker receives 60.0 W of power. Using , we then have =P 2I R

= = =Ω

P 60.0 W 3.87 A4.00

IR

The system is not adequately protected since

the fuse should be set to melt at 3.87 A, or less .

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

we see that

( ) ( )− −=

b r b ay h

From this, the radius at a distance y from the base is ( )= − +y

r a bh

b

For a disk-shaped element of volumeρπ

= 2

dydR

r:

( )( )ρπ

=⎡ ⎤− +⎣ ⎦∫ 20

h dyR

a b y h b

Using the integral formula( ) ( )

= −++∫ 2

1dua au bau b

, ρπ

=hRab

FIG. P27.56

P27.57 ρ ρ= =∫ ∫

dx dxRA wy

where −= + 2 1

1y y

y y xL

FIG. P27.57

( ) ( )

( )

ρ ρ

ρ

−⎡ ⎤= = +⎢ ⎥−⎡ ⎤+ − ⎣ ⎦⎣ ⎦

⎛ ⎞= ⎜ ⎟− ⎝ ⎠

∫ 2 11

2 10 1 2 1 0

2

2 1 1

ln

ln

LL y ydx LR yw w y yy y y L x

yLRw y y y

xL

)

P27.65 so (α= + −⎡ ⎤⎣ ⎦0 01R R T Tα α⎡ ⎤ ⎡ ⎤= + − = + −⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

00 0

0

1 11 1IRT T TR I

In this case, = 0

10II , so ( )

α= + = °+ = °

°01 99 20 2 020 C

0.004 50 CT T