109

Chapter 17

Current and Resistance

Problem Solutions

17.1 The charge that moves past the cross section is Q I t , and the number of electrons

is

3

20

19

80.0 10 C s 10.0 min 60.0 s min3.00 10 electrons

1.60 10 C

I tQn

e e

The negatively charged electrons move in the d irection opposite

to the conventional current flow.

17.2 (a) From Example 17.2 in the textbook, the density of charge carriers (electrons) in a

copper wire is 28 38.46 10 electrons mn . With 2 and A r q e , the drift speed of

electrons in this wire is

5

22 28 -3 19 3

3.70 C s5.57 10 m s

8.46 10 m 1.60 10 C 1.25 10 md

I Iv

n q A ne r

(b) The drift speed is smaller because more electrons are being conducted. To create the

same current, therefore, the drift speed need not be as great.

110 CHAPTER 17

17.3 The period of the electron in its orbit is 2T r v , and the current represented by the

orbiting electron is

6 19

3

11

2

2.19 10 m s 1.60 10 C1.05 10 C s 1.05 mA

2 5.29 10 m

e v eQI

t T r

17.4 If N is the number of protons, each with charge e, that hit the target in time t , the

average current in the beam is I Q t Ne t , giving

6

16

19

125 10 C s 23.0 s1.80 10 protons

1.60 10 C proton

I tN

e

17.5 (a) The carrier density is determined by the physical characteristics of the wire, not the

current in the wire. Hence, n is unaffected .

(b) The drift velocity of the electrons is dv I nqA . Thus, the drift velocity is doubled

when the current is doubled .

17.6 The mass of a single gold atom is

-22 -25

atom 23

197 g mol3.27 10 g 3.27 10 kg

6.02 10 atoms molA

Mm

N

The number of atoms deposited , and hence the number of ions moving to the negative

electrode, is

3

21

-25

atom

3.25 10 kg9.93 10

3.27 10 kg

mn

m

Thus, the current in the cell is

21 199.93 10 1.60 10 C0.159 A 159 mA

2.78 h 3 600 s 1 h

Q neI

t t

Current and Resistance 111

17.7 The drift speed of electrons in the line is 2 4

d

I Iv

nqA n e d, or

-4

228 3 19

4 1000 A2.3 10 m s

8.5 10 m 1.60 10 C 0.020 mdv

The time to travel the length of the 200-km line is then

3

-4 7

200 10 m 1 yr27 yr

2.34 10 m s 3.156 10 sd

Lt

v

17.8 Assuming that, on average, each aluminum atom contributes three electrons, the density

of charge carriers is three times the number of atoms per cubic meter. This is

33

3

A

A

Ndensityn

mass per atom M N M,

or

23 3 6 3 3

29 33 6.02 10 mol 2.7 g cm 10 cm 1 m

1.8 10 m26.98 g mol

n

The drift speed of the electrons in the wire is then

5

29 3 19 6 2

5.0 C s4.3 10 m s

1.8 10 m 1.60 10 C 4.0 10 md

Iv

n e A

17.9 (a) Using the periodic table on the inside back cover of the textbook, we find

3 3

Fe 55.85 g mol 55.85 g mol 1 kg 10 g 55.85 10 kg molM

(b) From Table 9.3, the density of iron is 3 3

Fe 7.86 10 kg m , so the molar density is

3 3

5 3Fe

3FeFe

7.86 10 kg mmolar density 1.41 10 mol m

55.85 10 kg molM

(c) The density of iron atoms is

23 5 28

3 3

density of atoms molar density

atoms mol atoms6.02 10 1.41 10 8.49 10

mol m m

AN

112 CHAPTER 17

(d) With two conduction electrons per iron atom, the density of charge carriers is

28 29 3

3

charge carriers atom density of atoms

electrons atoms2 8.49 10 1.70 10 electrons m

atom m

n

(e) With a current of 30.0 AI and cross-sectional area 6 25.00 10 mA , the drift

speed of the conduction electrons in this wire is

4

29 3 19 6 3

30.0 C s2.21 10 m s

1.70 10 m 1.60 10 C 5.00 10 md

Iv

nqA

17.10 From Ohm’ s law, 21.20 10 V

8.89 13.5 A

VR

I

17.11 6

maxmax80 10 AV I R R

Thus, if 54.0 10 R , max

32 VV

and if 2000 R , max

0.16 VV

17.12 The volume of the copper is

3

7 3

3 3

1.00 10 kg1.12 10 m

8.92 10 kg m

mV

density

Since, V A L , this gives 7 31.12 10 mA L [1]

(a) From L

RA

, we find that

8

81.70 10 m3.40 10 m

0.500 A L L L

R

Inserting this expression for A into Equation [1] gives

8 2 7 33.40 10 m 1.12 10 mL , which yields 1.82 mL

Current and Resistance 113

(b) From Equation [1], 2 7 31.12 10 m

4

dA

L, or

7 3 7 3

4

4 1.12 10 m 4 1.12 10 m

1.82 m

2.80 10 m 0.280 mm

dL

17.13 (a) From Ohm’ s law, 21.20 10 V

13.0 9.25 A

VR

I

(b) Using L

RA

and data from Table 17.1, the required length is found to be

22 3

8

13.0 0.791 10 m17.0 m

150 10 m

R rRAL

17.14

8

2 23

4 1.7 10 m 15 m0.31

4 1.024 10 m

L LR

A d

17.15 (a) 12 V

30 0.40 A

VR

I

(b) From, L

RA

,

22

4

30 0.40 10 m

4.7 10 m3.2 m

R A

L

17.16 Using L

RA

and data from Table 17.1, we have Cu AlCu Al2 2

Cu Al

L L

r r, which reduces to

2

Al Al

2

CuCu

r

r and yields

8

Al Al

8

Cu Cu

2.82 10 m1.29

1.70 10 m

r

r

114 CHAPTER 17

17.17 The resistance is 9.11 V

0.253 36.0 A

VR

I, so the resistivity of the metal is

22 3

84 0.253 2.00 10 m

1.59 10 m4 50.0 m

R dR A

L L

Thus, the metal is seen to be silver .

17.18 With d ifferent orientations of the block, three d ifferent values of the ratio L A are

possible. These are:

1

10 cm 1 1

20 cm 40 cm 80 cm 0.80 m

L

A,

2

20 cm 1 1

10 cm 40 cm 20 cm 0.20 m

L

A,

and 3

40 cm 1 1

10 cm 20 cm 5.0 cm 0.050 m

L

A

(a) 8

max 8

min min

6.0 V 0.80 m2.8 10 A

1.7 10 m

V VI

R L A

(b) 7

min 8

max max

6.0 V 0.050 m1.8 10 A

1.7 10 m

V VI

R L A

17.19 The volume of material, 2

0 0 0V AL r L , in the wire is constant. Thus, as the wire is

stretched to decrease its rad ius, the length increases such that 2 2

0 0f fr L r L giving

2 2

20 00 0 0 0

0

4.0 160.25

f

f

r rL L L L L

r r

The new resistance is then

20 0

02 2 2

00

1616 4 256 256 1.00 256

4

f f

f

f f

L L L LR R

A r rr

Current and Resistance 115

17.20 (a) From Ohm’ s law, 3 6 5500 10 A 1.0 10 5 10 VV IR

(b) Rubber-soled shoes and rubber gloves can increase the resistance to current and

help reduce the likelihood of a serious shock.

17.21 If a conductor of length L has a uniform electric field E maintained within it, the

potential d ifference between the ends of the conductor is V EL . But, from Ohm’ s

law, the relation between the potential d ifference across a conductor and the current

through it is where V IR R L A . Combining these relations, we obtain

V EL IR I L A or E I A J

17.22 Using 0 01R R T T with 0 6.00 R at

0 20.0 °CT and 133.8 10 °Csilver

(from Table 17.1 in the textbook), the resistance at 34.0 CT is

136.00 1 3.8 10 °C 34.0°C 20.0 C 6.32 R

17.23 From Ohm’ s law, i i f fV I R I R , so the current in Antarctica is

0 0

0 0

13

13

1

1

1 3.90 10 C 58.0 C 20.0 C1.00 A 1.98 A

1 3.90 10 C 88.0 C 20.0 C

iif i i

f f

R T TRI I I

R R T T

17.24 (a) Given: Aluminum wire with 133.90 10 °C (See Table 17.1 in textbook), and

0 030.0 at 20.0 CR T . If 46.2 R at temperature T, solving

0 01R R T T gives the final temperature as

0

0 13

1 46.2 30.0 120.0°C 158°C

3.90 10 °C

R RT T

(b) The expansion of the cross-sectional area contributes slightly more than the

expansion of the length of the wire, so the answer would be slightly reduced.

116 CHAPTER 17

17.25 For tungsten, the temperature coefficient of resistivity is 134.5 10 °C . Thus, if

0 015 at 20°CR T , and 160 R at the operating temperature of the filament,

solving 0 01R R T T for the operating temperature gives

0 3

0 13

1 160 15 120°C 2.2 10 °C

4.5 10 °C

R RT T

17.26 For aluminum, the resistivity at room temperature is 8

0 2.82 10 m and the

temperature coefficient of resistivity is 13

Al 3.9 10 °C . Thus, if at some

temperature, the aluminum has a resistivity of

8 8

0 Cu3 3 1.7 10 m 5.1 10 m

solving 0 Al 01 T T for that temperature gives

8

8

0 2

0 13Al

5.1 10 m1

1 2.82 10 m20°C 2.3 10 °C

3.9 10 °CT T

17.27 At 80°C,

3 1

0 0

5.0 V

1 200 1 0.5 10 C 80 C 20 C

V VI

R R T T

or 22.6 10 A 26 mAI

17.28 If 41.0 at 20 C and 41.4 at 29.0 CR T R T , then 0 01R R T T gives

the temperature coefficient of resistivity of the material making up this wire as

130

0 0

41.4 41.0 1.1 10 °C

41.0 29.0 C 20°C

R R

R T T

Current and Resistance 117

17.29 (a) The resistance at 20.0°C is

8

0 23

1.7 10 m 34.5 m3.0

0.25 10 m

LR

A

and the current will be 0

9.0 V3.0 A

3.0

VI

R

(b) At 30.0°C,

0 0

13

1

3.0 1 3.9 10 C 30.0 C 20.0 C 3.1

R R T T

Thus, the current is 9.0 V

2.9 A3.1

VI

R

17.30 The resistance of the heating element when at its operating temperature is

2 2120 V

13.7 1050 W

VR

P

From 00 0 01 1

LR R T T T T

A, the cross-sectional area is

00

8

13

7 2

1

150 10 m 4.00 m1 0.40 10 C 320 C 20.0 C

13.7

4.90 10 m

LA T T

R

A

17.31 (a) From R L A , the initial resistance of the mercury is

7

23

9.4 10 m 1.000 0 m1.2

1.00 10 m 4

ii

i

LR

A

118 CHAPTER 17

(b) Since the volume of mercury is constant, f f i iV A L A L gives the final cross-

sectional area as f i i fA A L L . Thus, the final resistance is given by

2

f f

f

f i i

L LR

A A L. The fractional change in the resistance is then

22

1 1 1f i f f i i f

i i i i i

R R R L A L L

R R L A L

2

4100.041 8.0 10

100.00 or a 0.080% increase

17.32 The resistance at 20.0°C is

0 1-3

0

200.0 217

1 1+ 3.92 10 C 0 C 20.0 C

RR

T T

Solving 0 01R R T T for T gives the temperature of the melting potassium as

00 13

0

253.8 217 20.0 C 63.3 C

3.92 10 C 217

R RT T

R

17.33 (a) The power consumed by the device is I VP , so the current must be

3

2

1.00 10 W8.33 A

1.20 10 VI

V

P

(b) From Ohm’ s law, the resistance is 21.20 10 V

14.4 8.33 A

VR

I

17.34 (a) The energy used by a 100-W bulb in 24 h is

100 W 24 h 0.100 kW 24 h 2.4 kWhE tP

and the cost of this energy, at a rate of $0.12 per kilowatt-hour is

2.4 kWh $0.12 kWh $0.29cost E rate

Current and Resistance 119

(b) The energy used by the oven in 5.0 h is

3

1 kW20.0 C s 220 J C 5.0 h 22 kWh

10 J sE t I V tP

and the cost of this energy, at a rate of $0.12 per kilowatt-hour is

22 kWh $0.12 kWh $2.6cost E rate

17.35 The power required is 0.350 A 6.0 V 2.1 WI VP

120 CHAPTER 17

17.36 (a) The power loss in the line is

22 7

loss 1000 A 0.31 km 160 km 5.0 10 W 50 MWI RP

(b) The total power transmitted is

3 8

input 700 10 V 1000 A 7.0 10 W 700 MWV IP

Thus, the fraction of the total transmitted power represented by the line losses is

loss

input

50 MW 0.071

700 MWfraction loss

P

P or 7.1%

17.37 The energy required to bring the water to the boiling point is

50.500 kg 4186 J kg C 100 C 23.0 C 1.61 10 JE mc T

The power input by the heating element is

input 120 V 2.00 A 240 W 240 J sV IP

Therefore, the time required is

5

input

1.61 10 J 1 min672 s 11.2 min

240 J s 60 s

EtP

17.38 (a) 590 W 1 h 90 J s 3 600 s 3.2 10 JE tP

(b) The power consumption of the color set is

120 V 2.50 A 300 WV IP

Therefore, the time required to consume the energy found in (a) is

5

33.2 10 J 1 min1.1 10 s 18 min

300 J s 60 s

EtP

Current and Resistance 121

17.39 The energy input required is

51.50 kg 4186 J kg C 50.0 C 10.0 C 2.51 10 JE mc T

and , if this is to be added in 10.0 min 600 st , the power input needed is

52.51 10 J

419 W600 s

E

tP

The power input to the heater may be expressed as 2

V RP , so the needed

resistance is

2 2120 V

34.4 419 W

VR

P

17.40 (a) At the operating temperature,

120 V 1.53 A 184 WV IP

(b) From 0 01R R T T , the temperature T is given by 00

0

R RT T

R. The

resistances are given by Ohm’ s law as

120 V

1.53 A

VR

I, and 0

0

0

120 V

1.80 A

VR

I

Therefore, the operating temperature is

13

120 1.53 120 1.8020.0 C 461 C

0.400 10 C 120 1.80T

122 CHAPTER 17

17.41 The resistance per unit length of the cable is

2

5

2 2

2.00 W m2.22 10 m

300 A

R I L

L L I

P P

From R L A , the resistance per unit length is also given by R L A . Hence, the

cross-sectional area is 2r AR L

, and the required radius is

8

-5

1.7 10 m0.016 m 1.6 cm

2.22 10 mr

R L

17.42 (a) The rating of the 12-V battery is 55 A hI t . Thus, the stored energy is

Energy stored 12 V 55 A h 660 W h 0.66 kWht V I tP

(b) 0.66 kWh $0.12 kWh $0.079 7.9 centscost

17.43 3 3 5 675 10 V 0.20 10 A 1.5 10 W 15 10 W 15 WV IP

17.44 (a) 440.0 W 14.0 d 24.0 h d 1.34 10 Wh 13.4 kWhE tP

13.4 kWh $0.120 kWh $1.61cost E rate

(b) 20.970 kW 3.00 min 1 h 60 min 4.85 10 kWhE tP

-24.85 10 kWh $0.120 kWh $0.005 83 0.583 cents

cost E rate

(c) 5.20 kW 40.0 min 1 h 60 min 3.47 kWhE tP

3.47 kWh $0.120 kWh $0.416 41.6 centscost E rate

Current and Resistance 123

17.45 The energy saved is

3

high low 40 W 11 W 100 h 2.9 10 Wh 2.9 kWhE tP P

and the monetary savings is

2.9 kWh $0.080 kWh $0.23 23 centssavings E rate

17.46 The power required to warm the water to 100°C in 4.00 min is

2

0.250 kg 4186 J kg °C 100 C 20°C3.5 10 W

4.00 min 60 s 1 min

mc TQ

t tP

The required resistance (at 100°C) of the heating element is then

2 2

2

120 V41

3.5 10 W

VR

P

so the resistance at 20°C would be

0 -3 10

41 40

1 1+ 0.4 10 °C 100°C 20°C

RR

T T

We find the needed d imensions of a Nichrome wire for this heating element from 2 2

0 0 0 04 4R L A L d L d , where L is the length of the wire and d is its

d iameter. This gives

8

2 80

0

4 150 10 m44.8 10 m

40 d L L L

R

Thus, 2 8any combination of length and diameter satisfying the relation 4.8 10 md L will

be suitable. A typical combination might be

3.0 mL and 8 44.8 10 m 3.0 m 3.8 10 m 0.38 mmd

Yes , such heating elements could easily be made from less than 30.5 cm of Nichrome.

The volume of material required for the typical wire given above is

124 CHAPTER 17

242 6 3

7 3 3

3

3.8 10 m 10 cm3.0 m 3.4 10 m 0.34 cm

4 4 1 m

dV AL L

17.47 The energy that must be added to the water is

6

J 1 kWh200 kg 4186 80 C 15 C 15 kWh

kg C 3.60 10 JE mc T

and the cost is 15 kWh $0.080 kWh $1.2cost E rate

17.48 (a) For tungsten, Table 17.1 from the textbook gives the resistivity at

0 20.0°C 293 KT as 8

0 5.6 10 m and the temperature coefficient of

resistivity as 13 3 14.5 10 °C 4.5 10 K . Thus, for a tungsten wire having a

radius of 1.00 mm and a length of 15.0 cm long, the resistance at 0 293 KT is

2

8 3

0 0 0 22 3

15.0 10 m5.6 10 m 2.7 10

1.00 10 m

L LR

A r

(b) From Stefan’ s law, the radiated power is 4AeTP , where A is the area of the

radiating surface. Note that since we are computing the radiated power, not the net

energy gained or lost as a result of rad iation, the ambient temperature is not needed

here. In the case of a wire, this is the cylindrical surface area 2A rL . The

temperature of the wire when it is rad iating a power of 75.0 WP must be

1 41 4

-8 2 4 3

75.0 W

5.669 10 W m K 2 1.00 10 m 0.150 m 0.320T

Ae

P

or 31.45 10 KT

(c) Assuming a linear temperature variation of resistance, the resistance of the wire at

this temperature is

3 3 1 3

0 01 2.7 10 1 4.5 10 K 1.45 10 K 293 KR R T T

giving 21.7 10 R

Current and Resistance 125

(d) The voltage drop across the wire when it is rad iating 75.0 W and has the resistance

found in part (c) above is given by 2

V RP as

21.7 10 75.0 W 1.1 VV R P

(e) Tungsten bulbs release very little of the energy consumed in the form of visible light ,

making them inefficient sources of light.

17.49 The battery is rated to deliver the equivalent of 60.0 amperes of current (i.e., 60.0 C/ s)

for 1 hour. This is

560.0 A 1 h 60.0 C s 3 600 s 2.16 10 CQ I t

17.50 The energy available in the battery is

3Energy stored 12 V 90 A h 1.1 10 W ht V I t V I tP

The two head lights together consume a total power of 2 36 W 72 WP , so the time

required to completely d ischarge the battery is

3Energy stored 1.1 10 W h

15 h72 W

tP

17.51 Assuming a constant resistance, the power consumed by the device is proportional to

the square of the applied voltage, 2

V RP . Thus,

2

22

2

1 1

V R

V R

P

P or

2 2

22 1

1

6.0 V15 W 6.7 W

9.0 V

V

VP P

17.52 The temperature variation of resistance is described by 0 01R R T T , where 0R

is the resistance at 0 20°CT . Thus, if an aluminum wire

133.9 10 °C has

02R R , we have 2 1 20°CT , and its temperature must be

2

13

1 120 C 20 C 2.8 10 °C

3.9 10 °CT

126 CHAPTER 17

17.53 From 2

V RP , the total resistance needed is

2 220 V

8.3 48 W

VR

P

Thus, from R L A , the length of wire required is

6 2

3

8

8.3 4.0 10 m1.1 10 m 1.1 km

3.0 10 m

R AL

17.54 The resistance of the 4.0 cm length of wire between the feet is

8

6

2

1.7 10 m 0.040 m1.79 10

0.011 m

LR

A,

so the potential d ifference is

6 550 A 1.79 10 8.9 10 V 89 VV IR

17.55 Ohm’ s law gives the resistance as R V I . From R L A , the resistivity is given

by R A L . The results of these calculations for each of the three wires are

summarized in the table below.

mL R m

0.540 10.4 61.41 10

1.028 21.1 61.50 10

1.543 31.8 61.50 10

The average value found for the resistivity is

6

av 1.47 10 m3

i

which d iffers from the value of 8 6150 10 m 1.50 10 m given in Table 17.1

by 2.0% .

Current and Resistance 127

17.56 At temperature T, the resistance of the carbon wire is 0 01c c cR R T T , and that of

the nichrome wire is 0 01n n nR R T T . When the wires are connected end to end,

the total resistance is

0 0 0 0 0c n c n c c n nR R R R R R R T T

If this is to have a constant value of 10.0 k as the temperature changes, it is necessary

that

0 0 10.0 kc nR R [1]

and 0 0 0c c n nR R [2]

From Equation [1], 0 010.0 kc nR R , and substituting into Equation [2] gives

1 13 3

0 010.0 k 0.50 10 C 0.40 10 C 0n nR R

Solving this Equation gives 0 5.6 k nichrome wirenR

Then, 0 10.0 k 5.6 k 4.4 k carbon wirecR

17.57 (a) The total power you now use while cooking breakfast is

1200 500 W 1.70 kWP

The cost to use this power for 0.500 h each day for 30.0 days is

h1.70 kW 0.500 30.0 days $0.120 kWh $3.06

daycost t rateP

(b) If you upgraded, the new power requirement would be:

2 400 500 W=2 900 WP

and the required current would be 2 900 W

26.4 A 20 A110 V

IV

P

No , your present circuit breaker cannot handle the upgrade.

128 CHAPTER 17

17.58 (a) The charge passing through the conductor in the interval

0 5.0 st is represented by the area under the vs I t

graph given in Figure P17.58. This area consists of two

rectangles and two triangles. Thus,

rectangle 1 rectangle 2 triangle 1 triangle 2

5.0 s 0 2.0 A 0 4.0 s 3.0 s 6.0 A 2.0 A

1 1 3.0 s 2.0 s 6.0 A 2.0 A 5.0 s 4.0 s 6.0 A 2.0 A

2 2

18 C

Q A A A A

Q

(b) The constant current that would pass the same charge in 5.0 s is

18 C

3.6 A5.0 s

QI

t

17.59 (a) From V IP , the current is 38.00 10 W

667 A12.0 V

IV

P

(b) The time before the stored energy is depleted is

7

storage 3

3

2.00 10 J2.50 10 s

8.00 10 J s

Et

P

Thus, the d istance traveled is

3 420.0 m s 2.50 10 s 5.00 10 m 50.0 kmd v t

Current and Resistance 129

17.60 The volume of aluminum available is

3

5 3

3 3

115 10 kg4.26 10 m

2.70 10 kg m

massV

density

(a) For a cylinder whose height equals the d iameter, the volume is

2 3

4 4

d dV d

and the d iameter is

1 35 31 3 4 4.26 10 m4

0.037 85 mV

d

The resistance between ends is then

8

7

2

4 2.82 10 m49.49 10

0.037 85 m4

L dR

A dd

(b) For a cube, 3V L , so the length of an edge is

1 31 3 54.26 10 m 0.034 9 mL V

The resistance between opposite faces is

8

7

2

2.82 10 m8.07 10

0.034 9 m

L LR

A LL

17.61 The current in the wire is 15.0 V

150 A0.100

VI

R

Then, from dv I nqA , the density of free electrons is

22 4 19 3

150 A

3.17 10 m s 1.60 10 C 5.00 10 md

In

v e r

or 28 33.77 10 mn

130 CHAPTER 17

17.62 Each speaker has a resistance of 4.00 R and can handle 60.0 W of power. From 2I RP , the maximum safe current is

max

60.0 W3.87 A

4.00 I

R

P

Thus, the system is not adequately protected by a 4.00 A fuse.

17.63 The cross-sectional area of the conducting material is 2 2

outer innerA r r

Thus,

5 2

7

2 22 2

3.5 10 m 4.0 10 m3.7 10 37 M

1.2 10 m 0.50 10 m

LR

A

17.64 The volume of the material is

3

6 3

3 6 3

50.0 g 1 m6.36 10 m

7.86 g cm 10 cm

massV

density

Since V A L , the cross-sectional area of the wire is A V L

(a) From 2L L L

RA V L V

, the length of the wire is given by

6 3

-8

1.5 6.36 10 m9.3 m

11 10 m

R VL

(b) The cross-sectional area of the wire is 2

4

d VA

L. Thus, the d iameter is

6 3

44 6.36 10 m4

9.3 10 m 0.93 mm9.3 m

Vd

L

Current and Resistance 131

17.65 The power the beam delivers to the target is

6 3 54.0 10 V 25 10 A 1.0 10 WV IP

The mass of cooling water that must flow through the tube each second if the rise in the

water temperature is not to exceed 50°C is found from m t c TP as

51.0 10 J s

0.48 kg s4186 J kg C 50 C

m

t c T

P

17.66 (a) At temperature T, the resistance is L

RA

, where 0 01 T T ,

0 01L L T T , and 2

0 0 0 01 1 2A A T T A T T

Thus,

0 0 0 0 00 0

0 0 0

1 1 1 1

1 2 1 2

T T T T R T T T TLR

A T T T T

(b)

8

0 00 2

-30

1.70 10 m 2.00 m1.082

0.100 10

LR

A

Then 0 01R R T T gives

31.082 1 3.90 10 C 80.0 C 1.420 R

The more complex formula gives

6

6

1.420 1 17 10 C 80.0 C1.418

1 2 17 10 C 80.0 CR

Note: Some rules for handing significant figures have been deliberately viola ted in

this solution in order to illustrate the very small d ifference in the results obtained

with these two expressions.

132 CHAPTER 17

17.67 Note that all potential d ifferences in this solution have a value of 120 VV .

First, we shall do a symbolic solution for many parts of the problem, and then enter the

specified numeric values for the cases of interest.

From the marked specifications on the cleaner, its internal resistance (assumed constant)

is

2

1

1

where 535 Wi

VR P

P Equation [1]

If each of the two conductors in the extension cord has resistance cR , the total resistance

in the path of the current (outside of the power source) is

2t i cR R R Equation [2]

so the current which will exist is tI V R and the power that is delivered to the cleaner

is

2 2 2 4

2

delivered 2

1 1

i i

t t t

V VV VI R R

R R RP

P P Equation [3]

The resistance of a copper conductor of length L and d iameter d is

CuCu Cu 22

4

4c

LL LR

A dd

Thus, if , maxcR is the maximum allowed value of

cR , the minimum acceptable d iameter

of the conductor is

Cumin

, max

4

c

Ld

R Equation [4]

(a) If 0.900 cR , then from Equations [2] and [1],

2 2

1

120 V2 0.900 1.80 1.80

535 Wt i

VR R

P

and , from Equation [3], the power delivered to the cleaner is

4

delivered 22

120 V470 W

120 V1.80 535 W

535 W

P

Current and Resistance 133

(b) If the minimum acceptable power delivered to the cleaner is minP , then Equations

[2] and [3] give the maximum allowable total resistance as

4 2

, max , max

min 1 min 1

2t i c

V VR R R

P P P P

so

2 2 2 2

, max

1 1min 1 min 1 min 1

1 1 1 1

2 2 2c i

V V V VR R

P PP P P P P P

When min 525 WP , then

2

, max

120 V 1 10.128

2 535 W525 W 535 WcR

and , from Equation [4],

8

min

4 1.7 10 m 15.0 m1.60 mm

0.128 d

When min 532 WP , then

2

, max

120 V 1 10.037 9

2 535 W532 W 535 WcR

and ,

8

min

4 1.7 10 m 15.0 m2.93 mm

0.037 9 d