current and resistance - uccs home · 106 current and resistance q27.8 the bottom of the rods on...

27

CHAPTER OUTLINE27.1 Electric Current27.2 Resistance27.3 A Model for Electrical Conduction27.4 Resistance and Temperature27.5 Superconductors27.6 Electric Power

Current and Resistance

ANSWERS TO QUESTIONS

Q27.1 Individual vehicles—cars, trucks and motorcycles—wouldcorrespond to charge. The number of vehicles that pass acertain point in a given time would correspond to the current.

Q27.2 Voltage is a measure of potential difference, not of current.“Surge” implies a flow—and only charge, in coulombs, can flowthrough a system. It would also be correct to say that the victimcarried a certain current, in amperes.

Q27.3 Geometry and resistivity. In turn, the resistivity of the materialdepends on the temperature.

Q27.4 Resistance is a physical property of the conductor based on thematerial of which it is made and its size and shape, includingthe locations where current is put in and taken out. Resistivityis a physical property only of the material of which the resistoris made.

Q27.5 The radius of wire B is 3 times the radius of wire A, to make its cross–sectional area 3 times larger.

Q27.6 Not all conductors obey Ohm’s law at all times. For example, consider an experiment in which avariable potential difference is applied across an incandescent light bulb, and the current ismeasured. At very low voltages, the filament follows Ohm’s law nicely. But then long before the

filament begins to glow, the plot of ∆V

I becomes non-linear, because the resistivity is temperature-

dependent.

Q27.7 A conductor is not in electrostatic equilibrium when it is carrying a current, duh! If charges areplaced on an isolated conductor, the electric fields established in the conductor by the charges willcause the charges to move until they are in positions such that there is zero electric field throughoutthe conductor. A conductor carrying a steady current is not an isolated conductor—its ends must beconnected to a source of emf, such as a battery. The battery maintains a potential difference acrossthe conductor and, therefore, an electric field in the conductor. The steady current is due to theresponse of the electrons in the conductor due to this constant electric field.

105

106 Current and Resistance

Q27.8 The bottom of the rods on the Jacob’s Ladder are close enough so that the supplied voltage issufficient to produce dielectric breakdown of the air. The initial spark at the bottom includes a tubeof ionized air molecules. Since this tube containing ions is warmer than the air around it, it is buoyedup by the surrounding air and begins to rise. The ions themselves significantly decrease theresistivity of the air. They significantly lower the dielectric strength of the air, marking longer sparkspossible. Internal resistance in the power supply will typically make its terminal voltage drop, sothat it cannot produce a spark across the bottom ends of the rods. A single “continuous” spark,therefore will rise up, becoming longer and longer, until the potential difference is not large enoughto sustain dielectric breakdown of the air. Once the initial spark stops, another one will form at thebottom, where again, the supplied potential difference is sufficient to break down the air.

Q27.9 The conductor does not follow Ohm’s law, and must have a resistivity that is current-dependent, ormore likely temperature-dependent.

Q27.10 A power supply would correspond to a water pump; a resistor corresponds to a pipe of a certaindiameter, and thus resistance to flow; charge corresponds to the water itself; potential differencecorresponds to difference in height between the ends of a pipe or the ports of a water pump.

Q27.11 The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electronsmore efficiently.

Q27.12 In a metal, the conduction electrons are not strongly bound to individual ion cores. They can movein response to an applied electric field to constitute an electric current. Each metal ion in the latticeof a microcrystal exerts Coulomb forces on its neighbors. When one ion is vibrating rapidly, it can setits neighbors into vibration. This process represents energy moving though the material by heat.

Q27.13 The resistance of copper increases with temperature, while the resistance of silicon decreases withincreasing temperature. The conduction electrons are scattered more by vibrating atoms whencopper heats up. Silicon’s charge carrier density increases as temperature increases and more atomicelectrons are promoted to become conduction electrons.

Q27.14 A current will continue to exist in a superconductor without voltage because there is no resistanceloss.

Q27.15 Superconductors have no resistance when they are below a certain critical temperature. For mostsuperconducting materials, this critical temperature is close to absolute zero. It requires expensiverefrigeration, often using liquid helium. Liquid nitrogen at 77 K is much less expensive. Recentdiscoveries of materials that have higher critical temperatures suggest the possibility of developingsuperconductors that do not require expensive cooling systems.

Q27.16 In a normal metal, suppose that we could proceed to a limit of zero resistance by lengthening theaverage time between collisions. The classical model of conduction then suggests that a constantapplied voltage would cause constant acceleration of the free electrons, and a current steadilyincreasing in time.

On the other hand, we can actually switch to zero resistance by substituting asuperconducting wire for the normal metal. In this case, the drift velocity of electrons is establishedby vibrations of atoms in the crystal lattice; the maximum current is limited; and it becomesimpossible to establish a potential difference across the superconductor.

Q27.17 Because there are so many electrons in a conductor (approximately 1028 electrons m3 ) the averagevelocity of charges is very slow. When you connect a wire to a potential difference, you establish anelectric field everywhere in the wire nearly instantaneously, to make electrons start driftingeverywhere all at once.

Chapter 27 107

Q27.18 Current moving through a wire is analogous to a longitudinal wave moving through the electrons ofthe atoms. The wave speed depends on the speed at which the disturbance in the electric field canbe communicated between neighboring atoms, not on the drift velocities of the electronsthemselves. If you leave a direct-current light bulb on for a reasonably short time, it is likely that nosingle electron will enter one end of the filament and leave at the other end.

Q27.19 More power is delivered to the resistor with the smaller resistance, since P =∆V

R

2

.

Q27.20 The 25 W bulb has a higher resistance. The 100 W bulb carries more current.

Q27.21 One ampere–hour is 3 600 coulombs. The ampere–hour rating is the quantity of charge that thebattery can lift though its nominal potential difference.

Q27.22 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 make A

RFHGIKJ =FHGIKJρ . 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 theenergy; then the resistor will melt.

SOLUTIONS TO PROBLEMS

Section 27.1 Electric Current

P27.1 IQt

=∆∆

∆ ∆Q I t= = × = ×− −30 0 10 40 0 1 20 106 3. . . A s Ce ja f

NQe

= =×

×= ×

−

−1 20 10

1 60 107 50 10

3

1915.

..

C C electron

electrons

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

V = = × × = ×− − −area thickness m m m2 3a fa f e je j700 10 0 133 10 9 31 104 3 6. . .

The mass of silver deposited is m VAg3 3 kg m m kg= = × × = ×− −ρ 10 5 10 9 31 10 9 78 103 6 2. . .e je j .

And the number of silver atoms deposited is

N

IV

R

tQI

NeI

= ××F

HGIKJFHG

IKJ = ×

= = = =

= = =× ×

= × =

−

−

9 78 106 02 10 1 000

5 45 10

12 06 67 6 67

5 45 10 1 60 10

6 671 31 10 3 64

223

23

23 194

..

.

.. .

. .

.. .

kg atoms

107.9 g g

1 kg atoms

V1.80

A C s

C

C s s h

e j

e je j

∆Ω

∆∆


P27.3 Q t Idt I et

ta f e j= = −z −

00 1τ τ

(a) Q I e Iτ τ τa f e j a f= − =−0

101 0 632.

(b) Q I e I10 1 0 999 95010

0τ τ τa f e j b g= − =− .

(c) Q I e I∞ = − =−∞a f e j0 01τ τ

P27.4 (a) Using k er

mvr

e2

2

2

= , we get: vk emre= = ×

262 19 10. m s .

(b) The time for the electron to revolve around the proton once is:

tr

v= =

×

×= ×

−−2 2 5 29 10

2 19 101 52 10

11

616π π .

..

m

m s s

e je j

.

The total charge flow in this time is 1 60 10 19. × − C , so the current is

I =××

= × =−

−−1 60 10

1 05 10 1 0519

163.

. . C

1.52 10 s A mA .

P27.5 The period of revolution for the sphere is T =2πω

, and the average current represented by this

revolving charge is IqT

q= =

ωπ2

.

P27.6 q t t= + +4 5 63

A = FHG

IKJ = × −2 00

1 002 00 10

24.

.. cm

m100 cm

m2 2e j

(a) Idqdt

tt t

1 00 12 5 17 01.00

2

1.00. . s A

s sa f e j= = + =

= =

(b) J = =×

=−IA

17 02 00 10

85 04.

..

A m

kA m22

P27.7 Idqdt

=

q dq Idtt

dt

q

= = = FHG

IKJ

=− F

HGIKJ −

LNM

OQP =

+=

z z z 100120

100120 2

0100

0 265

0

1 240

As

C C120

C

s

a fsin

cos cos .

π

ππ

π

Chapter 27 109

P27.8 (a) JIA

= =×

=−

5 0099 5

3 2.

. A

4.00 10 m kA m2

π e j

(b) J J2 114

= ; I

AI

A2 1

14

=

A A1 214

= so π π4 00 1014

3 222. × =−e j r

r23 32 4 00 10 8 00 10 8 00= × = × =− −. . .e j m mm

P27.9 (a) JIA

= =×

×=

−

−

8 00 10

1 00 102 55

6

3 2.

..

A

m A m2

π e j

(b) From J nevd= , we have nJ

evd= =

× ×= ×

−−2 55

1 60 10 3 00 105 31 10

19 810 3.

. ..

A m

C m s m

2

e je j.

(c) From IQt

=∆∆

, we have ∆∆

tQI

N eI

= = =× ×

×= ×

−

−A

C

A s

6 02 10 1 60 10

8 00 101 20 10

23 19

610

. .

..

e je j.

(This is about 382 years!)

P27.10 (a) The speed of each deuteron is given by K mv=12

2

2 00 10 1 60 1012

2 1 67 106 19 27 2. . .× × = × ×− −e je j e j J kg v and v = ×1 38 107. m s .

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

=

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 me je j .

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

Vk qre= =

× ⋅ ×

×= ×

−

−−

8 99 10 1 60 10

2 21 106 49 10

9 19

73

. .

..

N m C C

m V

2 2e je j.

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


P27.11 We use I nqAvd= n is the number of charge carriers per unit volume, and is identical to the numberof atoms per unit volume. We assume a contribution of 1 free electron per atom in the relationshipabove. 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 06 02 10

4 49 102323. .

..

g g g atom

NA=

×= × − .

Thus, n = =× −

density of aluminummass per atom

g cm g atom

32 704 49 10 23

..

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

Therefore, vI

nqAd = =× × ×

= ×− − −

−5 00

6 02 10 1 60 10 4 00 101 30 10

28 3 19 64.

. . ..

A

m C m m s

2e je je jor, vd = 0 130. mm s .

Section 27.2 Resistance

*P27.12 J EE

= = =× ⋅

⋅FHG

IKJ = ×−σ

ρ0 740

2 44 101

3 03 1087.

..

V m m

A1 V

A m2

ΩΩ

P27.13 IV

R= = = =∆

Ω120

0 500 500 V

240 A mA.

P27.14 (a) Applying its definition, we find the resistance of the rod,

RVI

= =×

=−∆

Ω Ω15 0

103 7503

. V4.00 A

= 3.75 k .

(b) The length of the rod is determined from the definition of resistivity: RA

=ρ

. Solving for

and substituting numerical values for R, A, and the value of ρ given for carbon in Table 27.1,we obtain

= =× ×

× ⋅=

−

−

RAρ

3 75 10 5 00 10

3 50 10536

3 6

5

. .

.

m

m m

2Ω

Ω

e je je j

.

P27.15 ∆V IR=

and RA

=ρ

: A =FHG

IKJ = × −0 600

1 006 00 102

27.

.. mm

m1 000 mm

m2a f

∆VIA

=ρ

: IVA

= =×

× ⋅

−

−

∆

Ωρ

0 900 6 00 10

5 60 10 1 50

7

8

. .

. .

V m

m m

2a fe je ja f

I = 6 43. A

Chapter 27 111

P27.16 JIr

E= = = =π

σπ

σ2 23 00

120. A

0.012 0 m N C

b g b g

σ = ⋅ −55 3 1. Ω ma f ρσ

= = ⋅1

0 018 1. mΩ

P27.17 (a) Given M V Ad d= =ρ ρ where ρd ≡ mass density,

we obtain: AM

d=ρ

. Taking ρ r ≡ resistivity, RA M Mr r

d

r d= = =ρ ρ

ρρ ρ 2

.

Thus, = =×

× ×

−

−

MR

r dρ ρ

1 00 10 0 500

1 70 10 8 92 10

3

8 3

. .

. .

e ja fe je j

= 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

.

. .e ja fr = × −1 40 10 4. m .

The diameter is twice this distance: diameter = 280 mµ .

*P27.18 The volume of the gram of gold is given by ρ =mV

Vm

A

A

RA

= =×

= × = ×

= ×

= =× ⋅ ×

×= ×

−−

−

−

−

ρ

ρ

1019 3 10

5 18 10 2 40 10

2 16 10

2 44 10

2 16 102 71 10

3

38 3

11

8

116

kg kg m

m m

m

m 2.4 10 m

m

33

2

3

2

.. .

.

.

..

e j

e jΩΩ

P27.19 (a) Suppose the rubber is 10 cm long and 1 mm in diameter.

RA d

= =⋅

=−

−

ρ ρπ π

4 4 10 10

10102

13 1

3 218~ ~

m m

m

ΩΩ

e je je j

(b) Rd

=× ⋅

×

− −

−

−4 4 1 7 10 10

2 10102

8 3

2 27ρ

π π~

.~

m m

m

ΩΩ

e je je j

(c) IV

R= −∆

Ω~ ~

1010

216 V

10 A18

I ~ ~10

102

79 V

10 A− Ω


P27.20 The distance between opposite faces of the cube is =FHG

IKJ =

90 02 05

1 3.

. g

10.5 g cm cm3 .

(a) RA

= = = =× ⋅×

= × =−

−−ρ ρ ρ

2

8

271 59 10

107 77 10 777

..

m2.05 m

nΩ

Ω Ω

(b) IV

R= =

××

=−

−∆

Ω1 00 107 77 10

12 95

7..

. V

A

n

n

= ×

= ××F

HGIKJ = ×

10 5107 87

6 02 10

5 86 101 00 10

1 005 86 10

23

226

28

..

.

..

..

g cm g mol

electrons mol

electrons cm cm

mm

3

33

33

e j

e j

I nqvA= and vI

nqA= =

× ×=

−

12 9

5 86 10 1 60 10 0 020 53 28

28 19 2

.

. . ..

C s

m C m m s

3e je jb gµ

P27.21 Originally, RA

=ρ

. Finally, RA A

Rf = = =

ρ ρ3

3 9 9b g

.

P27.22ρ

π

ρ

πAl

Al

Cu

Cur rb g b g2 2=

rr

Al

Cu

Al

Cu= =

××

=−

−ρρ

2 82 101 70 10

1 298

8..

.

P27.23 J E= σ so σ = =×

= × ⋅−

− −JE

6 00 10100

6 00 1013

15 1..

A m V m

m2

Ωa f

P27.24 RA A d

= + =+ρ ρ ρ ρ1 1

1

2 2

2

1 1 2 22

R =× ⋅ + × ⋅

×=

− −

−

4 00 10 0 250 6 00 10 0 400

3 00 10378

3 3

3 2

. . . .

.

m m m m

m

Ω ΩΩ

e ja f e ja fe j

Section 27.3 A Model for Electrical Conduction

P27.25 ρτ

=m

nq2

so τρ

= =×

× × ×= ×

−

− −−m

nq2

31

8 28 1914

9 11 10

1 70 10 8 49 10 1 60 102 47 10

.

. . ..

e je je je j

s

vqEmd = τ

so 7 84 101 60 10 2 47 10

9 11 104

19 14

31.. .

.× =

× ×

×−

− −

−

e j e jE

Therefore, E = 0 181. V m .

Chapter 27 113

P27.26 (a) n is unaffected

(b) JIA

I= ∝

so it doubles .

(c) J nevd=

so vd doubles .

(d) τσ

=mnq2 is unchanged as long as σ does not change due to a temperature change in the

conductor.

P27.27 From Equation 27.17,

τρ

τ

= =×

× × ×= ×

= = × × = × =

−

− −

−

− −

mnq

v

e2

31

19 2 8

14

5 14 8

9 11 10

1 60 10 1 70 102 47 10

8 60 10 2 47 10 2 12 10 21 2

.

. ..

. . . .

8.49 10 s

m s s m nm

28e je j e je je j

Section 27.4 Resistance and Temperature

P27.28 At the low temperature TC we write RV

IR T TC

CC= = + −

∆0 01 αb g where T0 20 0= °. C .

At the high temperature Th , RV

IV

R T Thh

h= = = + −∆ ∆

110 0 A

αb g .

Then∆

∆V

V IC

a f a fa f

e ja fe ja f

1 00 1 3 90 10 38 0

1 3 90 10 108

3

3

. . .

.

A=

+ ×

+ × −

−

−

and IC = FHGIKJ =1 00

1 150 579

1 98...

. A Aa f .

P27.29 R R T= +0 1 α ∆a f gives 140 19 0 1 4 50 10 3 CΩ Ω ∆= + × °−. .a f e j T .

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

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


P27.30 R R R R T T R T Tc n c c n n= + = + − + + −1 10 0α αb g b g

0 0 0= − + −R T T R T Tc c n nα αb g b g so R Rc nn

c= −

αα

R R R

R R R R

R

nn

cn

nn

cc

c

n

n

= − +

= −FHG

IKJ = −

FHG

IKJ

= −× °

− × °

L

NMM

O

QPP

− −

−

−

−

αα

αα

αα

1 1

10 0 10 400 10

0 500 10

1 1

3

3

1

..

. k

C

CΩe je j

Rn = 5 56. kΩ and Rc = 4 44. kΩ

P27.31 (a) ρ ρ α= + − = × ⋅ + × ° = × ⋅− − −0 0

8 3 81 2 82 10 1 3 90 10 30 0 3 15 10T Tb g e j a f. . . . m mΩ Ω

(b) JE

= =× ⋅

= ×−ρ0 200

3 15 106 35 108

6..

. V m

m A m2

Ω

(c) I JA Jd

= =FHGIKJ = ×

×L

NMMM

O

QPPP=

−π π2

64 2

46 35 10

1 00 10

449 9.

.. A m

m mA2e j e j

(d) n =×

×= ×

6 02 10

26 986 02 10

2328.

..

electrons

g 2.70 10 g m electrons m

6 33

e j

vJ

ned = =×

× ×=

−

6 35 10

6 02 10 1 60 10659

6

28 19

.

. .

A m

electrons m C m s

2

3

e je je j

µ

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

P27.32 For aluminum,

α E = × °− −3 90 10 3 1. C (Table 27.1)

α = × °− −24 0 10 6 1. C (Table 19.1)

RA

T T

A TR

TT

E E= =+ +

+=

+

+=

FHG

IKJ =

ρ ρ α α

α

αα

02 0

1 1

1

11

1 2341 39

1 002 41 71

∆ ∆

∆

∆

∆Ω Ω

b g a fa f

b ga f a f.

..

.

Chapter 27 115

P27.33 R R T= +0 1 α

R R R TR R

RT

− =−

= = × =−

0 0

0

0

35 00 10 25 0 0 125

α

α

∆

∆ . . .e j

P27.34 Assuming linear change of resistance with temperature, R R T= +0 1 α∆a fR77

31 00 1 3 92 10 216 0 153 K C = + × − ° =−. . .Ω Ωa f e ja f .

P27.35 ρ ρ α= +0 1 ∆Ta f or ∆TWW

W

W= −FHG

IKJ

11

0αρρ

Require that ρ ρW = 4 0Cu so that ∆TW =

× °

FHG

IKJ

×

×−

FHGG

IKJJ = °−

−

−1

4 50 10

4 1 70 10

5 60 101 47 63

8

8.

.

..

CC

e j.

Therefore, T TW = ° + = °47 6 67 60. .C C .

Section 27.5 Superconductors

Problem 48 in Chapter 43 can be assigned with this section.

Section 27.6 Electric Power

P27.36 IV

= = =P∆

6005 00

W120 V

A.

and RVI

= = =∆

Ω120

24 0 V

5.00 A . .

*P27.37 P = = × × =−I V∆ 500 10 15 10 7 506 3 A V We j .

P27.38 P = = ×0 800 1 500 746 8 95 105. . hp W hp Wb gb gP = I V∆ 8 95 10 2 0005. × = Ib g I = 448 A

P27.39 The heat that must be added to the water is

Q mc T= = ° ° = ×∆ 1 50 4 186 40 0 2 51 105. . . kg J kg C C Jb gb ga f .

Thus, the power supplied by the heater is

P = = =×

=W

tQ

t∆ ∆2 51 10

4195. J

600 s W

and the resistance is RV

= = =∆

Ωa f a f2 2110

41928 9

P V W

. .


*P27.40 The battery takes in energy by electric transmission

P∆ ∆ ∆t V I t= = × FHG

IKJ =

−a f a f e j2 3 13 5 10 4 23 600

4693. . . J C C s h s

1 h J .

It puts out energy by electric transmission

∆ ∆V I ta f a f e j= × FHG

IKJ =

−1 6 18 10 2 43 600

2493. . J C C s h s

1 h J .

(a) efficiency = = =useful output

total input J

469 J249

0 530.

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

469 249

221

J J

Jint

int

= +

=

∆

∆

E

E

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

Q mc T

TQmc

=

= =°

= °

∆

∆221

15 1 J kg C

0.015 kg 975 JC.

P27.41PP0

2

02

0

2 2140120

1 361= =FHGIKJ = FHG

IKJ =

∆

∆

∆∆

V R

V R

VV

a fb g

.

∆% . .=−FHGIKJ = −

FHGIKJ = − =

P PP

PP

0

0 0100% 1 100% 1 361 1 100% 36 1%a f a f a f

P27.42 P = = =I VVR

∆∆a f a f

2

500 W

R = =110500

24 22 V

W

a fa f . Ω

(a) RA

=ρ

so = =×

× ⋅=

−

−RAρ

π24 2 2 50 10

1 50 103 17

4 2

6

. .

..

m

m m

Ω

Ω

a f e j

(b) R R T= + = + × =−0

31 24 2 1 0 400 10 1 180 35 6α∆ Ω Ω. . . e jb g

P = = =∆V

Ra f a f2 2110

35 6340

. W

Chapter 27 117

P27.43 RA

= =× ⋅

×=

−

−

ρ

π

1 50 10 25 0298

6

3 2

. . m m

0.200 10 m

ΩΩ

e je j

∆ ΩV IR= = =0 500 298 149. A Va fa f

(a) EV

= = =∆ 149

5 97 V

25.0 m V m.

(b) P = = =∆V Ia f a fa f149 0 500 74 6 V A W. .

(c) R R T T= + − = + × ° ° =−0 0

31 298 0 400 10 320 337αb g e j 1 C C Ω Ω.

IV

R

V I

= = =

= = =

∆Ω

∆

149337

0 443

149 0 443 66 1

V

A

V A W

a fa f

a f a fa f.

. .P

P27.44 (a) ∆ ∆ ∆U q V It V= = = ⋅⋅

FHGIKJ ⋅FHG

IKJ

⋅FHG

IKJ = ⋅ =a f a f a fa f55 0 12 0

1 1 1660 0 660. . . A h V

C1 A s

J1 V C

W s1 J

W h kWh

(b) Cost = FHG

IKJ =0 660

060 01

3 96¢.$0.

. kWh kWh

P27.45 P = I V∆a f ∆V IR=

P = = =∆V

Ra f a f2 210 0

1200 833

.. W

*P27.46 (a) The resistance of 1 m of 12-gauge copper wire is

RA d d

= = = =× ⋅

×= ×

−

−

−ρ ρ

π

ρπ π2

4 4 1 7 10 1

105 14 102 2

8

2 23

b ge je j.

. m m

0.205 3 m

ΩΩ .

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

(b) PAlAl= =I R

Id

22

24ρπ

PP

Al

Cu

Al

Cu=ρρ

PAl m

1.7 m W W=

× ⋅× ⋅

=−

−2 82 10

102 05 3 41

8

8.

. .ΩΩ

Aluminum of the same diameter will get hotter than copper.


*P27.47 The energy taken in by electric transmission for the fluorescent lamp is

P∆t = FHG

IKJ = ×

= × FHGIKJFHGIKJ

⋅FHGIKJFHG

IKJ =

11 1003 600

3 96 10

3 96 1008

088

6

6

J s h s

1 h J

cost JkWh

k1 000

W sJ

h3 600 s

a f .

.$0.

$0.

For the incandescent bulb,

P∆t = FHG

IKJ = ×

= ××

FHG

IKJ =

= − =

40 1003 600

1 44 10

1 44 1008

32

32 088 232

7

7

W h s

1 h J

cost J3.6 10 J

saving

6

a f .

.$0.

$0.

$0. $0. $0.

P27.48 The total clock power is

270 10 2 503 600

2 43 106 12× FHG

IKJFHG

IKJ = × clocks

J sclock

s1 h

J he j . . .

From eWQ

= out

in, the power input to the generating plants must be:

Qt

W te

in out J h J h

∆∆

= =×

= ×2 43 10

0 2509 72 10

1212.

..

and the rate of coal consumption is

Rate = ××

FHG

IKJ = × =9 72 10

1 002 95 10 29512 5.

.. J h

kg coal33.0 10 J

kg coal h metric ton h6e j .

P27.49 P = = =I V∆a f a fa f1 70 110 187. A V W

Energy used in a 24-hour day = =0 187 24 0 4 49. . . kW h kWha fa f

∴ =FHG

IKJ = =cost kWh

$0.060 0kWh

4 49 269 26 9¢. $0. .

P27.50 P = = =I V∆ 2 120 240.00 A V Wa fa f∆

∆∆

E

tE

int

int

kg J kg C C kJ

J W

s

= ⋅° ° =

= =×

=

0 500 4 186 77 0 161

1 61 10240

6725

. .

.

b gb ga f

P

Chapter 27 119

P27.51 At operating temperature,

(a) P = = =I V∆ 1 53 120 184. A V Wa fa f

(b) Use the change in resistance to find the final operating temperature of the toaster.

R R T= +0 1 α∆a f 1201 53

1201 80

1 0 400 10 3

. ..= + × −e j∆T

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

*P27.52 You pay the electric company for energy transferred in the amount E t= P ∆

(a) P ∆t = FHG

IKJFHG

IKJ ⋅FHG

IKJ =40 2

86 400 148 4 W weeks

7 d1 week

s1 d

J1 W s

MJa f .

P

P

∆

∆

t

t

= FHG

IKJFHGIKJFHGIKJ =

= FHG

IKJFHGIKJFHGIKJFHGIKJ =

40 224

1 00013 4

40 224

1 0000 12

61

W weeks7 d

1 week h

1 dk

kWh

W weeks7 d

1 week h

1 dk $

kWh

a f

a f

.

.$1.

(b) P ∆t = FHG

IKJFHGIKJFHGIKJ = =970 3

1 0000 12

005 82 0 582¢ W min1 h

60 mink $

kWha f .

$0. .

(c) P ∆t = FHG

IKJFHGIKJFHGIKJ =5 200 40 min

1 0000 12

416 W1 h

60 mink $

kWha f .

$0.

P27.53 Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to thedryer is

P ∆t = ≈ ××

FHG

IKJ ≈400 600 365 9 10

1207 J s s d d J

kWh3.6 10 J

kWh6b gb ga f .

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 ≅ =20 10 kWh kWha fb g$0. $2 ~ $1 .


Additional Problems

P27.54 (a) IV

R=∆

so P = =I VVR

∆∆a f2

RV

= = =∆

Ωa f a f2 2120

25 0576

P V W.

and RV

= = =∆

Ωa f a f2 2120

100144

P V W

(b) IV

Qt t

= = = = =P∆ ∆ ∆

25 00 208

1 00..

. W120 V

A C

∆t = =1 00

4 80.

. C

0.208 A s

The bulb takes in charge at high potential and puts out the same amount of charge at lowpotential.

(c) P = = =25 01 00

..

W J∆

∆ ∆Ut t

∆t = =1 00

0 040 0.

. J

25.0 W s

The bulb takes in energy by electrical transmission and puts out the same amount of energyby heat and light.

(d) ∆ ∆U t= = = ×P 25 0 86 400 30 0 64 8 108. . . J s s d d Jb gb ga fThe electric company sells energy .

Cost J$0.070 0

kWhk

1 000W s

Jh

3 600 s

Cost per joulekWh

kWh3.60 10 J

J6

= ×FHG

IKJFHGIKJ

⋅FHGIKJFHG

IKJ =

=×

FHG

IKJ = × −

64 8 10 26

070 094 10

6

8

. $1.

$0.$1.

*P27.55 The original stored energy is U Q VQCi i= =

12

12

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 final potential difference is ∆VQCf = 4

for

both.

(b) The smaller capacitor then carries charge C VQC

CQ

f∆ = =4 4

. The larger capacitor carries

charge 34

34

CQC

Q= .

(c) The smaller capacitor stores final energy 12

12 4 32

2 2 2

C V CQC

QCf∆d i = F

HGIKJ = . The larger

capacitor possesses energy 12

34

332

2 2

CQC

QC

FHGIKJ = .

(d) The total final energy is Q

CQ

CQ

C

2 2 2

32332 8

+ = . The loss of potential energy is the energy

appearing as internal energy in the resistor: Q

CQ

CE

2 2

2 8= + ∆ int ∆E

QCint =

38

2

.

Chapter 27 121

P27.56 We find the drift velocity from I nqv A nqv rd d= = π 2

vI

nq r

vxt

txv

d = =× ×

= ×

= = =××

= × =

− − −

−

−

π π2 28 3 19 2 2

4

3

48

1 000

8 49 10 1 60 10 102 34 10

200 1010

8 54 10 27.0 yr

A

m C m m s

m2.34 m s

s

. ..

.

e j e j

P27.57 We begin with the differential equation αρ

ρ=

1 ddT

.

(a) Separating variables,d

dTT

Tρρ

αρ

ρ

0 0

z z=ln

ρρ

α0

0FHGIKJ = −T Tb g and ρ ρ α= −

00e T Tb g .

(b) From the series expansion e xx ≅ +1 , x << 1a f ,ρ ρ α≅ + −0 01 T Tb g .

P27.58 The resistance of one wire is 0 500

100 50 0.

.

mi mi

ΩΩF

HGIKJ =a f .

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

IR = =1 000 50 0 50 0 A kVb ga f. .Ω .

Then it radiates as heat power P = = × =∆V Ia f e jb g50 0 10 1 000 50 03. . V A MW .

P27.59 ρ = =RA V

IA∆a f (m) ( ) ( m)

0.540

1.028

1.543

10.4

21.1

31.8

R Ω Ωρ ⋅×

×

×

−

−

−

1 41 10

1 50 10

1 50 10

6

6

6

.

.

.

ρ = × ⋅−1 47 10 6. mΩ (in agreement with tabulated value of 1 50 10 6. × ⋅− mΩ in Table 27.1)

P27.60 2 wires → = 100 m

R = =0 108

100 0 036 0.

.

300 m m

ΩΩa f

(a) ∆ ∆V V IRa f a f a fb ghome line V= − = − =120 110 0 036 0 116.

(b) P = = =I V∆a f a fa f110 116 12 8 A V kW.

(c) Pwires A W= = =I R2 2110 0 036 0 436a f b g. Ω


P27.61 (a) E i i= − = −−

−=

dVdx

..

.0 4 00

0 500 08 00

a fa f

V m

V m

(b) RA

= =× ⋅

×=

−

−

ρ

π

4 00 10 0 500

1 00 100 637

8

4 2

. .

..

m m

m

ΩΩ

e ja fe j

(c) IV

R= = =∆

Ω4 00

6 28.

. V

0.637 A

(d) J i i i= =×

= × =−

IA

.

..

6 28

1 00 102 00 10 200

4 28 A

m A m MA m2 2

π e j

(e) ρJ i i E= × ⋅ × = =−4 00 10 2 00 10 8 008 8. . . m A m V m2Ωe je j

P27.62 (a) E i i= − =dV x

dxVL

a f

(b) RA

Ld

= =ρ ρ

π4

2

(c) IV

RV d

L= =∆ π

ρ

2

4

(d) J i i= =IA

VLρ

(e) ρ J i E= =VL

P27.63 R R T T= + −0 01 αb g so T TRR

TII

= + −LNMOQP= + −LNM

OQP0

00

011

11

α α.

In this case, II

= 0

10, so T T= + = °+

°= °0

19 20

90 004 50

2 020αa f

. CC .

P27.64 RVI I I

= = =−

∆ 12 0 6 003 00

. ..a f thus 12 0 36 0 6 00. . .I I− = and I = 6 00. A .

Therefore, R = =12 0

2 00.

. V

6.00 A Ω .

Chapter 27 123

P27.65 (a) P = I V∆

so IV

= =×

=P∆

8 00 10667

3. W12.0 V

A .

(b) ∆∆

tU

= =××

= ×P

2 00 102 50 10

73.

. J

8.00 10 W s3

and ∆ ∆x v t= = × =20 0 2 50 10 50 03. . . m s s kmb ge j .

P27.66 (a) We begin with RA

T T T T

A T T= =

+ − + ′ −

+ ′ −ρ ρ α α

α0 0 0 0

0 0

1 1

1 2

b g b gb g ,

which reduces to RR T T T T

T T=

+ − + ′ −

+ ′ −0 0 0

0

1 1

1 2

α α

α

b g b gb g .

(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 08= =

×

×=

−

−

ρ

π

. .

..

e ja fe j

Ω .

The simple formula for R gives:

R = + × ° ° − ° =− −1 08 1 3 90 10 100 20 0 1 4203 1. . . . C C C Ω Ωa f e ja fwhile the more complicated formula gives:

R =+ × ° ° + × ° °

+ × ° °=

− − − −

− −

1 08 1 3 90 10 80 0 1 17 0 10 80 0

1 2 17 0 10 80 01 418

3 1 6 1

6 1

. . . . .

. ..

C C C C

C C

ΩΩ

a f e ja f e ja fe ja f

.

P27.67 Let α be the temperature coefficient at 20.0°C, and ′α be the temperature coefficient at 0 °C. Thenρ ρ α= + − °0 1 20 0T . Ca f , and ρ ρ α= ′ + ′ − °1 0T Ca f 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 Ca f a f . (1)

Setting T = 0 in equation (1) yields: ′ = − °ρ ρ α0 1 20 0. Ca f ,

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

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

ρ ρ α α0 0 1 20 0 1 20 0= − ° + ′ °. .C Ca f a f .

continued on next page


Therefore 1 20 01

1 20 0+ ′ ° =

− °α

α.

.C

Ca f a f

which simplifies to ′ =− °

αα

α1 20 0. Ca f .

From this, the temperature coefficient, based on a reference temperature of 0°C, may be computedfor any material. For example, using this, Table 27.1 becomes at 0 C° :

Material Temp Coefficients at 0°CSilver 4 1 10 3. × °− CCopper 4 2 10 3. × °− CGold 3 6 10 3. × °− CAluminum 4 2 10 3. × °− CTungsten 4 9 10 3. × °− CIron 5 6 10 3. × °− CPlatinum 4 25 10 3. × °− CLead 4 2 10 3. × °− CNichrome 0 4 10 3. × °− CCarbon − × °−0 5 10 3. CGermanium − × °−24 10 3 CSilicon − × °−30 10 3 C

P27.68 (a) A thin cylindrical shell of radius r, thickness dr, and length L contributes resistance

dRdA

drr L L

drr

= = =FHGIKJ

ρ ρπ

ρπ2 2b g .

The resistance of the whole annulus is the series summation of the contributions of the thinshells:

RL

drr L

rrr

rb

aa

b

= =FHGIKJzρ

πρπ2 2

ln .

(b) In this equation ∆V

I Lrr

b

a=

FHGIKJ

ρπ2

ln

we solve for ρπ

=2 L V

I r rb a

∆lnb g .

Chapter 27 125

P27.69 Each speaker receives 60.0 W of power. Using P = I R2 , we then have

IR

= = =P 60 0

3 87.

. W

4.00 A

Ω.

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

P27.70 ∆V E= − ⋅ or dV E dx= − ⋅

∆V IR E

Idqdt

ER

AE

AE A

dVdx

AdVdx

= − = − ⋅

= =⋅

= ⋅ = = − =ρ ρ

σ σ

Current flows in the direction of decreasing voltage. Energy flows as heat in the direction ofdecreasing temperature.

P27.71 RdxA

dxwy

= =z zρ ρ where y y

y yL

x= +−

12 1

Rw

dxy y y L x

Lw y y

yy y

Lx

RL

w y yyy

L L

=+ −

=−

+−L

NMOQP

=−

FHGIKJ

zρ ρ

ρ

1 2 10 2 11

2 1

0

2 1

2

1

b g b g

b g

ln

lnFIG. P27.71

P27.72 From the geometry of the longitudinal section of the resistor shown in the figure,we see that

b ry

b ah

−=

−a f a f.

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

b= − +a f .

For a disk-shaped element of volume dRdyr

=ρπ 2 : R

dy

a b y h b

h

=− +zρπ a fb g 20

.

Using the integral formula du

au b a au b+= −

+z a f a f21

, Rhab

=ρπ

.

FIG. P27.72

*P27.73 (a) The resistance of the dielectric block is RA

dA

= =ρ

σ.

The capacitance of the capacitor is CA

d=

∈κ 0 .

Then RCdA

Ad

=∈

=∈

σκ κ

σ0 0 is a characteristic of the material only.

(b) RC C

=∈

=∈

=× ⋅ ×

× ⋅= ×

−

−κσ

ρκ0 016 12

91575 10 8 85 10

14 101 79 10

m 3.78 C

F N m

2

2

ΩΩ

a f ..


P27.74 I Ie Vk TB

=FHGIKJ −

LNMM

OQPP0 1exp

∆ and R

VI

=∆

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 :

∆ ΩV I RV Aa f a f a f0 400 0 015 6 25 60 440 0 081 8 5 380 480 0 429 1 120 520 2 25 0 2320 560 11 8 0 047 60 600 61 6 0 009 7

. . .

. . .

. . .

. . .

. . .

. . .

FIG. P27.74(i)

(ii) For T = 300 K :

∆ ΩV I RV Aa f a f a f0 400 0 005 77 30 440 0 024 18 10 480 0 114 4 220 520 0 534 0 9730 560 2 51 0 2230 600 11 8 0 051

. . .

. . .

. . .

. . .

. . .

. . .

FIG. P27.74(ii)

(iii) For T = 320 K :

∆ ΩV I RV Aa f a f a f0 400 0 002 0 2030 440 0 008 4 52 50 480 0 035 7 13 40 520 0 152 3 420 560 0 648 0 8640 600 2 76 0 217

. .

. . .

. . .

. . .

. . .

. . .

FIG. P27.74(iii)

Chapter 27 127

*P27.75 (a) Think of the device as two capacitors in parallel. The one on the left has κ 1 1= ,

A x1 2= +FHG

IKJ . The equivalent capacitance is

κ κ κκ κ1 0 1 2 0 2 0 0 0

2 2 22 2

∈+

∈=∈

+FHGIKJ +

∈−FHGIKJ =

∈+ + −

Ad

Ad d

xd

xd

x xa f .

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

QV

dx x=

∈+ + −0

22 2

∆κ κa f.

The current is

IdQdt

dQdx

dxdt

Vd

vVv

d= = =

∈+ + − = −

∈−0 0

20 2 0 2 1

∆ ∆κ κa f a f.

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

current is clockwise ∈

−0 1∆Vvd

κa f .

ANSWERS TO EVEN PROBLEMS

P27.2 3 64. h P27.32 1 71. Ω

P27.4 (a) see the solution; (b) 1 05. mA P27.34 0 153. Ω

P27.6 (a) 17 0. A ; (b) 85 0. kA m2 P27.36 5 00. A , 24 0. Ω

P27.38 448 AP27.8 (a) 99 5. kA m2 ; (b) 8 00. mm

P27.40 (a) 0.530; (b) 221 J; (c) 15.1°CP27.10 (a) 221 nm ; (b) no; see the solution

P27.42 (a) 3 17. m ; (b) 340 WP27.12 30 3. MA m2

P27.44 (a) 0 660. kWh ; (b) 3 96¢.P27.14 (a) 3.75 kΩ ; (b) 536 m

P27.46 (a) 2.05 W; (b) 3.41 W; noP27.16 0 018 1. mΩ⋅

P27.48 295 metric ton hP27.18 2 71. MΩ

P27.50 672 sP27.20 (a) 777 nΩ ; (b) 3 28. m sµ

P27.52 (a) $1.61; (b) $0.005 82; (c) $0.416P27.22

rr

Al

Cu= 1 29.

P27.54 (a) 576 Ω and 144 Ω ;(b) 4.80 s; The charge is the same. Thecharge-field system is in a lower-energyconfiguration.

P27.24 378 Ω

P27.26 (a) nothing; (b) doubles; (c) doubles;(c) 0.040 0 s; The energy enters by electrictransmission and exits by heat andelectromagnetic radiation;

(d) nothing

P27.28 1 98. A(d) $1.26; energy; 1 94 10 8. × − $ J

P27.30 carbon, 4 44. kΩ ; nichrome, 5 56. kΩ


P27.56 27 0. yr P27.66 (a) see the solution;(b) 1 418. Ω nearly agrees with 1 420. Ω

P27.58 50 0. MW

P27.68 (a) RL

rr

b

a=

ρπ2

ln ; (b) ρπ

=2 L V

I r rb a

∆lnb gP27.60 (a) 116 V ; (b) 12 8. kW ; (c) 436 W

P27.62 (a) Ei

=VL

; (b) RL

d=

42

ρπ

; (c) IV d

L=

πρ

2

4;

P27.70 see the solution

P27.72 see the solution(d) J

i=

VLρ

; (e) see the solutionP27.74 see the solution

P27.64 2 00. Ω

current and resistance - uccs home · 106 current and resistance q27.8 the bottom of the rods on...

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