Download Solutions to Problems

Solutions to Problems
1.
Use the definition of current, Eq. 18-1.
Q
1.30 C 1 electron
I
 1.30 A 

 8.13  1018 electrons s
19
t
s
1.60  10 C
2.
Use the definition of current, Eq. 18-1.
Q
I
 Q  I t   6.7 A  5.0 h  3600 s h   1.2  105 C
t
3.
Use the definition of current, Eq. 18-1.
19
Q 1200 ions  1.60 10 C ion
I

 5.5 1011 A
6
t
3.5 10 s
4.
Use Eq. 18-2a for resistance.
V 120 V
R 
 29 
I 4.2 A
5.
Use Eq. 18-2b for the voltage.


V  IR   0.25 A  3800    950 V
6.
(a)
(b)
7.
(a)
(b)
8.
Use Eq. 18-2a for resistance.
V 120 V
R 
 16 
I 7.5 A
Use the definition of current, Eq. 18-1.
Q
I
 Q  I t   7.5 A 15 min  60 s min   6.8  103 C
t
Use Eq. 18-2b to find the current.
V 240 V
V  IR  I  
 25 A
R 9.6 
Use the definition of current, Eq. 18-1.
Q
I
 Q  I t   25 A  50 min  60 s min   7.5  10 4 C
t
Find the current from the voltage and resistance, and then find the number of electrons from the
current.
V 9.0 V
V  IR  I  
 5.625 A
R 1.6 
5.625 A  5.625
C

1 electron
19

60s
s 1.60 10 C 1min
 2.1 1021 electrons min
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47
Physics: Principles with Applications, 6th Edition
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9.
Find the potential difference from the resistance and the current.
R  2.5  105  m 4.0  102 m  1.0  10 6 





V  IR   2800 A  1.0  106   2.8  103 V
10. (a) If the voltage drops by 15%, and the resistance stays the same, then by Eq. 18-2b, V  IR , the
current will also drop by 15%.
I final  0.85 I initial  0.85  6.50 A   5.525 A  5.5 A
(b) If the resistance drops by 15% (the same as being multiplied by 0.85), and the voltage stays the
same, then by Eq. 18-2b, the current must be divided by 0.85.
I
6.50 A
I final  initial 
 7.647 A  7.6 A
0.85
0.85
11.
(a)
(b)
Use Eq. 18-2a to find the resistance.
V
12 V
R 
 20 
I 0.60 A
An amount of charge Q loses a potential energy of  Q  V as it passes through the
resistor.
The amount of charge is found from Eq. 18-1.
PE   Q  V   I t  V   0.60 A  60 s 12 V   430 J
12.
Use Eq. 18-3 to find the diameter, with the area as A   r 2   d 2 4 .
R
13.
A

4L
d
2
 d
4L
R

4 1.00 m  5.6  108  m

  0.32  
L
A

4L
d
2

 1.68  108  m

4  3.5 m 
 1.5  10 m 
4
m
3
2
 3.3  102 
Use Eq. 18-3 to calculate the resistances, with the area as A   r 2   d 2 4 , and so
L
4L
R 
.
A
d2
4L
 Al Al2
2
2.65  108  m 10.0 m  2.5 mm 
RAl
 d Al  Al LAl d Cu2



 1.2
2
8
4 LCu Cu LCu d Al2
RCu
1.68

10

m
20.0
m
2.0
mm



 Cu 2
 d Cu


15.
  4.7 10
Use Eq. 18-3 to calculate the resistance, with the area as A   r 2   d 2 4 .
R
14.
L


Use Eq. 18-3 to express the resistances, with the area as A   r 2   d 2 4 , and so
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48
Physics: Principles with Applications, 6th Edition
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R
L
A

4L
d2
.
RW  RCu   W
4L
d
2
W
 Cu
4L

 d Cu2
W
5.6  108  m
  2.5 mm 
 4.6 mm
Cu
1.68  108  m
The diameter of the tungsten should be 4.6 mm.
d W  d Cu
16.
Since the resistance is directly proportional to the length, the length of the long piece must be 4.0
times the length of the short piece.
L  Lshort  Llong  Lshort  4.0 Lshort  5.0 Lshort  Lshort  0.20 L , Llong  0.80 L
Make the cut at 20% of the length of the wire .
Lshort  0.20 L , Llong  0.80 L  Rshort  0.2 R  2.0  , Rlong  0.8R  8.0 
17.
Use Eq. 18-4 multiplied by L A so that it expresses resistance instead of resistivity.
R  R0 1   T  T0   1.15R0  1   T  T0   1.15 
T  T0 
0.15

0.15

 
.0068 C
o
1
 22 Co
So raise the temperature by 22 Co to a final temperature of 42 o C .
18.
Use Eq. 18-4 for the resistivity.
T Cu  0 Cu 1   Cu T  T0   0 W 
T  T0 
19.
1  0 W

 Cu  0 Cu

 1  20o C 

 5.6  108  m 
 1  363.1o C  360o C

o 1 1.68  10 8  m

0.0068  C  
1
Use Eq. 18-4 multiplied by L A so that it expresses resistances instead of resistivity.
R  R0 1   T  T0  
T  T0 
1 R

1
 140  
o
 1   1798o C  1800o C
  1  20 C 
1 
o
  R0 

0.0060  C   12 
20.
Calculate the voltage drop by combining Ohm’s Law (Eq. 18-2b) with the expression for
resistance,
Eq. 18-3.
4 1.68  108  m  26 m 
L
4 L
V  IR  I
I
 12 A 
 2.5 V
2
A
d2
 1.628  103 m




21. In each case calculate the resistance by using Eq. 18-3 for resistance.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
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49
Physics: Principles with Applications, 6th Edition
Giancoli
22.
5
2
 Lx  3.0  10  m 1.0  10 m 

 3.75  104   3.8  104 
2
2
Ayz
 2.0 10 m  4.0 10 m 
(a)
Rx 
(b)
Ry 
(c)
 Lz
Rz 
Axy
 Ly
Axz
 3.0 10  m  2.0 10 m   1.5 10
1.0 10 m  4.0 10 m 
 3.0 10  m  4.0 10 m   6.0  10

1.0 10 m  2.0 10 m 
5

2
2
3

3

2
5
2
2
2
The wires have the same resistance and the same resistivity.
Rlong  Rshort 
 Llong
A1

 Lshort
A2

 4  2Lshort
 dlong
2

4 Lshort
d

2
short
dlong
dshort
 2
The original resistance is R0  V I 0 , and the high temperature resistance is R  V I , where the
23.
two
voltages are the same. The two resistances are related by Eq. 18-4, multiplied by L A so that it
expresses resistance instead of resistivity.


1 R
1V I
1I

R  R0 1   T  T0   T  T0    1   T0  
 1   T0   0  1 
  R0 
  V I0 
 I

 20.0o C 
1
 
0.00429 C
o
1
 0.4212 A 
o
 0.3618 A  1  58.3 C


24.
(a)
Calculate each resistance separately using Eq. 18-3, and then add the resistances together
to find
the total resistance.
RCu 
8
 Cu L 4  Cu L 4 1.68  10  m   5.0 m 


 0.10695 
2
A
d2
 1.0  103 m 
RAl 
8
 Al L 4  Al L 4  2.65  10  m   5.0 m 


 0.16870 
2
A
d2
 1.0  103 m 
Rtotal  RCu  RAl  0.10695   0.16870   0.27565   0.28 
(b)
(c)
The current through the wire is the voltage divided by the total resistance.
V
85  103 V
I

 0.30836 A  0.31A
Rtotal 0.27565 
For each segment of wire, Ohm’s Law is true. Both wires have the current found in (b)
above.
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50
Physics: Principles with Applications, 6th Edition
Giancoli
VCu  IRCu   0.30836 A  0.10695    0.033 V
VAl  IRAl   0.30836 A  0.16870   0.052 V
Notice that the total voltage is 85 mV.
25.
of
The total resistance is to be 4700 ohms  Rtotal  at all temperatures. Write each resistance in terms
Eq. 18-4 (with T0  0 C ), multiplied by L A to express resistance instead of resistivity.
o
Rtotal  R0C 1   CT   R0N 1   NT   R0C  R0C CT  R0N  R0N NT
 R0C  R0N   R0C C  R0N N  T
For the above to be true, the terms with a temperature dependence must cancel, and the terms
without a temperature dependence must add to Rtotal . Thus we have two equations in two unknowns.
R0C C
0   R0C C  R0 N N  T  R0 N  
R0C C
Rtotal  R0C  R0 N  R0C 
N

N
R0C  N   C 
N

0.0004  Co 
N
R0C  Rtotal
  4700  
 2100 
1
1
 N   C 
0.0004  Co   0.0005  Co 
1
R0 N  Rtotal  R0C  2600 
26.
Use Eq. 18-6b to find the resistance from the voltage and the power.
P
27.
V2
 R
R
V2
P

 240 V 2
3300 W
 17 
Use Eq. 18-5 to find the power from the voltage and the current.
P  IV   0.32 A  3.0 V   0.96 W
28.
Use Eq. 18-6b to find the voltage from the power and the resistance.
V2
P
 V  RP   2700   0.25 W   26 V
R
29.
Use Eq. 18-6b to find the resistance, and Eq. 18-5 to find the current.
(a) P 
V2
R
 R
P  IV  I 
(b)
P
V2
R
V2
P
P
V


120 V 2
75 W
75 W
120 V
 R
 192   190 
 0.625 A  0.63 A
V2
P

120 V 2
440 W
 32.73   33 
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51
Physics: Principles with Applications, 6th Edition
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P  IV  I 
30.
(a)
(b)
(a)
V

440 W
 3.667 A  3.7 A
120 V
Use Eq. 18-5 to find the current.
P 110 W
P  IV  I  
 0.96 A
V 115 V
Use Eq. 18-6b to find the resistance.
P
31.
P
Since P 
V2
V2
R
 R
R
 R
V2
V2
P
P

115 V 2
110 W
 120 
says that the resistance is inversely proportional to the power
for a
constant voltage, we predict that the 850 W setting has the higher resistance.
32.
(b)
R
(c)
R
V2
P
V2
P


120 V 2
850 W
120 V 2
1250 W
 17 
 12 
The power (and thus the brightness) of the bulb is proportional to the square of the voltage,
V2
according to Eq. 18-6a, P 
. Since the resistance is assumed to be constant, if the voltage is cut
R
in half from 240 V to 120V, the power will be reduced by a factor of 4. Thus the bulb will appear
only about 1/4 as bright in the United States as in Europe.
33. To find the kWh of energy, multiply the kilowatts of power consumption by the number of hours in
operation.
 1kW 
 1h 
15 min  

  0.1375 kWh  0.14 kWh
 1000 W 
 60 min 
Energy  P  in kW  t  in h    550 W  
To find the cost of the energy used in a month, multiply times 4 days per week of usage, times 4
weeks per month, times the cost per kWh.
kWh   4 d  4 week   9.0 cents 

Cost   0.1375

  20 cents month


d   1week  1month   kWh 

34.
To find the cost of the energy, multiply the kilowatts of power consumption by the number of
hours
in operation times the cost per kWh.
 1kW 
 24 h   $0.095 
Cost   25 W  
 365day  
  $ 21


 1000 W 
 1day   kWh 
35.
(a)
Calculate the resistance from Eq. 18-2a, and the power from Eq. 18-5.
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52
Physics: Principles with Applications, 6th Edition
Giancoli
R
V
I
3.0 V
 6.667   6.7 
P  IV   0.45 A  3.0 V   1.35 W  1.4 W
0.45 A
If four D-cells are used, the voltage will be doubled to 6.0 V. Assuming that the

(b)
resistance of
the bulb stays the same (by ignoring heating effects in the filament), the power that the bulb
V2
would need to dissipate is given by Eq. 18-6b, P 
. A doubling of the voltage means the
R
power is increased by a factor of 4 . This should not be tried because the bulb is probably not
rated for such a high wattage. The filament in the bulb would probably burn out, and the glass
bulb might even explode if the filament burns violently.
36.
The A h rating is the amount of charge that the battery can deliver. The potential energy of the
charge is the charge times the voltage.
 3600s 
PE  QV   85 A h  
12 V   3.7 106 J

 1h 
37.
Each bulb will draw an amount of current found from Eq. 18-5.
P
P  IV  I bulb 
V
The number of bulbs to draw 15 A is the total current divided by the current per bulb.
VI
P
120 V 15 A 
I total  nI bulb  n
 n  total 
 18 bulbs
V
P
100 W
38.
Find the power dissipated in the cord by Eq. 18-6a, using Eq. 18-3 for the resistance.
8  2.7 m 
2L
2
P  I 2R  I 2
 15.0 A  1.68  108  m
 16 W
2
A
 0.129  102 m

39.



Find the current used to deliver the power in each case, and then find the power dissipated in the
resistance at the given current.
P
P2
P  IV  I 
Pdissipated = I 2 R  2 R
V
V
 6.20 10 W   3.0    8008 W

1.2 10 V 
2
5
Pdissipated
12,000 V
4
 6.20 10 W   3.0    461W

 5 10 V 
2
5
Pdissipated
50,000 V
2
4
difference  8008 W  461W  7.5  103 W
2
40. The water temperature rises by absorbing the heat energy that the electromagnet dissipates. Express
both energies in terms of power, which is energy per unit time.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
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53
Physics: Principles with Applications, 6th Edition
Giancoli
Pelectric  Pto heat  IV 
Qheat water
t
water
m
t

IV
cT


mcT
17.5 A  240 V 
 4186 J kg C  7.50 C 
o

t
o
 0.134 kg s  0.13 kg s
This is about 134 ml per second.
41.
(a)
By conservation of energy and the efficiency claim, 60% of the electrical power
dissipated by
the heater must be the rate at which energy is absorbed by the water.
Q
mcT
0.6 Pemitted by  Pabsorbed  0.6  IV   heat water 

t
t
electromagnet
by water
I
(b)
42.
mcT
 0.120 kg  4186 J kg   95o C  25o C 

 10.17 A  10 A
 0.6 12 V  480s 
0.6Vt
Use Ohm’s Law to find the resistance of the heater.
V
12 V
V  IR  R  
 1.2 
I 10.17 A
Use Ohm’s law and the relationship between peak and rms values.
V
220 V
I peak  2 I rms  2 rms  2
 0.14 A
R
2.2  103 
43.
Find the peak current from Ohm’s law, and then find the rms current from the relationship
between
peak and rms values.
I peak 
Vpeak
R

180 V
330 
 0.54545 A
I rms  I peak
0.55 A
2   0.54545 A 
2  0.39 A
44.
(a)
When everything electrical is turned off, no current will be flowing into the house, even
though
a voltage is being supplied. Since for a given voltage, the more resistance, the lower the
current, a zero current corresponds to an infinite resistance.
(b)
Assuming that the voltage is 120 V, use Eq. 18-6a to calculate the resistance.
P
V2
R
 R
V2
P
=
120 V 2
75 W
 192   1.9  102 
45.
The power and current can be used to find the peak voltage, and then the rms voltage can be
found
from the peak voltage.
I peak
2 1500 W 
2P
P  I rmsVrms 
Vrms  Vrms 

 3.9  102 V
I peak
5.4 A
2
46.
Use the average power and rms voltage to calculate the peak voltage and peak current.
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54
Physics: Principles with Applications, 6th Edition
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(a)
Vpeak  2Vrms  2  660 V   933.4 V  9.3 102 V
(b)
P  I rmsVrms 
I peak
2
Vrms  I peak 
2P
Vrms
2 1800 W 

660 V
 3.9 A
47.
(a)
We assume that the 3.0 hp is the average power, so the maximum power is twice that, or
6.0 hp,
as seen in Figure 18-22.
 746 W 
6.0 hp 
 4476 W  4.5  103 W

 1hp 
(b)
Use the average power and the rms voltage to find the peak current.
P  I rmsVrms 
48.
9c.
(a)
I peak
2
Vrms  I peak 
2P
Vrms
2  12  4476 W 

240 V
 13A
The average power used can be found from the resistance and the rms voltage by Eq. 18-
P
2
Vrms
 240 V 2

 1694 W  1.7 103 W
R
34 
(b) The maximum power is twice the average power, and the minimum power is 0.
Pmax  2 P  2 1694 W   3.4 103 W
49.
We follow exactly the derivation in Example 18-14, which results in an expression for the drift
velocity.
I
I
4I m
vd 


2
neA N 1 mole 
N  D e d 2
 D e  12 d 
m 1 mole 



m 1.60  10
4 2.3  10 6 A 63.5  10 3 kg
 6.02 10 8.9 10
23
50.
Pmin  0 W
3
kg
3
19

 
C  0.65  10 m
3

2
 5.1  1010 m s
Use Ohm’s Law to find the resistance.
V 22.0  103 V
V  IR  R  
 0.02933   0.029 
I
0.75 A
Find the resistivity from Eq. 18-3.
(a)
(b)
R
L
 
RA

R r 2
 0.02933    1.0  103 m 

 5.80 m 
A
L
L
Find the number of electrons per unit volume from Eq. 18-10.
(c)
2
 1.6  108  m
I  neAvd 
n
I
eAvd

I
e r vd
2

1.60 10
19
 
 0.75 A 
C  1.0  10 m
3
 1.7 10
2
5
ms

 8.8  1028 e  m 3
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55
Physics: Principles with Applications, 6th Edition
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51. We are given a charge density and a speed (like the drift speed) for both types of ions. From that we
can use Eq. 18-10 to determine the current per unit area. Both currents are in the same direction in
terms of conventional current – positive charge moving north has the same effect as negative charge
moving south – and so they can be added.
I  neAvd 
I
A

 
  nevd He   nevd O   2.8  1012 ions m3 2 1.60  1019 C ion
 2.0 10
6

m s  
 7.0  1011 ions m3 1.60  1019 C ion  7.2  106 m s  
 2.6 A m 2 , North
52.
53.
The magnitude of the electric field is the voltage change per unit meter.
V 70  103 V
E 

 7.0  106 V m
8
x 1.0  10 m
The speed is the change in position per unit time.
x 7.20  102 m  3.40  102 m
v

 35 m s
t
0.0063s  0.0052s
Two measurements are needed because there may be a time delay from the stimulation of the nerve
to the generation of the action potential.
54. The energy required to transmit one pulse is equivalent to the energy stored by charging the axon
capacitance to full voltage. In Example 18-15, the capacitance is estimated at 1.0  10 8 F , and the
potential difference is about 0.1 V.


E  12 CV 2  0.5 108 F  0.1V   5 1011 J
2
The power is the energy per unit time, for 10,000 neutrons transmitting 100 pulses each per
second.
P
E
t
 5 10

11

J neuron 1.0 104 neurons
0.01s
  5 10
5
W
55. The power is the work done per unit time. The work done to move a charge through a potential
difference is the charge times the potential difference. The charge density must be multiplied by the
surface area of the cell (the surface area of an open tube, length times circumference) to find the
actual charge moved.
P
W

QV

Q
V
t
mol 
ions 

19 C 
6
  3  107 2  6.02  1023
 1.6  10
  0.10 m   20  10 m  0.030 V 
m s 
mol 
ion 

t
t


 5.4  10 9 W
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56
Physics: Principles with Applications, 6th Edition
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56.
 1C s   3600 s 
  3600 C

 1A   1 h 
1.00A h  
57.
Use Eq. 18-5 to calculate the current.
P 746 W
P  IV  I  
 6.22 A
V 120 V
58.
The energy supplied by the battery is the energy consumed by the lights.
Esupplied  Econsumed  QV  Pt 
t
QV
P

 95A h  3600 s h 12V 
92 W
 1h 
  12.39 h  12 h
 3600 s 
 44609 s 
59.
Use Eq. 18-6 to express the resistance in terms of the power, and Eq. 18-3 to express the
resistance
in terms of the wire geometry.
P
4
V2
R
L
d
2
 R

V2
P
V2
P
 d
R
4  LP
V
2
L
A


L
r
2
 4

L
d2

4 9.71  108  m  5.4 m 1500 W 
 110 V 
 2.9  104 m
2
60.
From Eq. 18-2a, if R  V I , then G  I V
I 0.73 A
G 
 0.24S
V
3.0 V
61.
To deliver 10 MW of power at 120 V requires a current of I 
P
V

107 W
120 V
 83, 300 A . Calculate
the
power dissipated in the resistors using the current and the resistance.
2 1.0 m 
L
L
L
2
P  I 2 R  I 2   I 2  2  4I 2 
 4  83, 300 A  1.68  10 8  m
2
A
r
d
 5.0  103 m




2
 1.187  107 W
1kW 
$0.10 
1h  
  1000


W
kWh

Cost   Power  time  rate per kWh   1.187  107 W 




 $1187  $1, 200 per hour per meter
62.
(a)
Calculate the total kWh used per day, and then multiply by the number of days and the
cost per
kWh.
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57
Physics: Principles with Applications, 6th Edition
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1.8 kW  3.0 h d   4  0.1kW  6.0 h d    3.0 kW 1.4 h d    2.0 kWh d 
 14.0 kWh d
 $0.105 
  $44.10 per month
 kWh 
Cost  14.0 kWh d  30 d  
(b)
The energy required by the household is 35% of the energy that needs to be supplied by
the
power plant.
Household Energy  0.35  coal mass  coal energy per mass  
coal mass 
Household Energy
 0.35  coal energy per mass 
14.0 kWh d  365 d  
1000 W   3600 s 


 kW   1h 
kcal   4186 J 

 0.35   7000


kg  1kcal 


 1794 kg  1800 kg of coal
63.
the
The length of the new wire is half the length of the original wire, and the cross-sectional area of
new wire is twice that of the original wire. Use Eq. 18-3.
1
L
L
L
L
R0   0
L  12 L0 A  2 A0 R     2 0  14  0  14 R0
A0
A
2 A0
A0
The new resistance is one-fourth of the original resistance.
64.
(a)
Use Eq. 18-6 to relate the power to the voltage for a constant resistance.
P
V2

R
P105
P117
105 V 

2
117 V 
2
105 V 

2
R 117 V 
2
R
 0.805 or a 19.5% decrease
(b) The lower power output means that the resistor is generating less heat, and so the resistor’s
temperature would be lower. The lower temperature results in a lower value of the resistance,
which would increase the power output at the lower voltages. Thus the decrease would be
somewhat smaller than the value given in the first part of the problem.
65.
The
Assume that we have a meter of wire, carrying 35 A of current, and dissipating 1.8 W of heat.
power dissipated is PR  I 2 R , and the resistance is R 
PR  I 2 R  I 2
L
A
 I2
A
.
L
4 L
 I2

2
r
d2
L
d I
 2I
 2  35 A 
PR
PR
2
L
4 L
1.68 10
8

 m 1.0 m 
1.8 W  
 3.8  103 m
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58
Physics: Principles with Applications, 6th Edition
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66.
(a)
(b)
(c)
The angular frequency is   210 rad s .
 210 rad s
f 

 33.4 Hz
2
2
The maximum current is 1.80 A.
I
1.80 A
I rms  max 
 1.27 A
2
2
For a resistor, V  IR .
V  IR  1.80 A  sin 210 t  42.0     75.6sin 210 t  V
67.
(a)
(b)
The power delivered to the interior is 65% of the power drawn from the source.
P
950 W
Pinterior  0.65 Psource  Psource  interior 
 1462 W  1500 W
0.65
0.65
The current drawn is current from the source, and so the source power is used to calculate
the
current.
Psource  IVsource  I 
68.
Psource
Vsource

1462 W
120 V
 12.18 A  12 A
The volume of wire is unchanged by the stretching. The volume is equal to the length of the wire
times its cross-sectional area, and so since the length was increased by a factor of 3, the area was
decreased by a factor of 3. Use Eq. 18-3.
L
3L
L
L
R0   0
L  3L0 A  13 A0 R     1 0  9  0  9 R0  9.00 
A0
A
A0
A0
3
69.
The long, thick conductor is labeled as conductor number 1, and the short, thin conductor is
labeled
V2
as number 2. The power transformed by a resistor is given by Eq. 18-6, P 
,and both have the
R
same voltage applied.
L
L
R1   1
R2   2
L1  2 L2 A1  4 A2  diameter1  2diameter2 
A1
A2
P1
P2
2

V1 R1
2
2
V
R2

R2
R1


L2
A2 L2 A1 1

 4  2
L1
L1 A2 2

A1
P1 : P2  2 :1
70. The heater must heat 124 m3 of air per hour from 5o C to 20 o C , and also replace the heat being lost
at a rate of 850 kcal/h. Use Eq. 14-2 to calculate the energy needed to heat the air. The density of
air is found in Table 10-1.
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59
Physics: Principles with Applications, 6th Edition
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Q  mcT 
Q

t
Power required  408
71.
(a)
m3  
kg  
kcal 
kcal
15Co   408
1.29 3   0.17

o 
h 
m 
kg C 
h

kcal
h
 850
kcal
h
 1258
kcal  4186 J   1h 


  1463 W  1500 W
h  kcal   3600s 
V2
11cents
kWh
 R
V2
mcT
0.75  Poven 


 240 V 2
 26.18   26 
R
P
2200 W
Only 75% of the heat from the oven is used to heat the water.
0.75  Poven  t  Heat absorbed by water  mcT 
t
(c)
t

cT   124
Use Eq. 18-6.
P
(b)
m
 1kg 
o
o
  4186 J kg C  85 C 
1L


 25.88s  26 s
 0.120 L  
0.75  2200 W 
 2.2 kW  25.88s 
1h
3600 s
 0.17 cents
72. (a) The horsepower required is the power dissipated by the frictional force, since we are neglecting
the energy used for acceleration.
 1m s 
 1hp 
P  Fv   240 N  45 km hr  
 3000W 

  4.0 hp
 3.6 km hr 
 746 W 
(b)
The charge available by each battery is Q  95A h  95C s 3600s  3.42 105 C , and so
the
total charge available is 24 times that. The potential energy of that charge is the charge times
the voltage. That energy must be delivered (batteries discharged) in a certain amount of time to
produce the 3000 W necessary. The speed of the car times the discharge time is the range of the
car between recharges.
PE QV
QV d
P

 t


t
t
P
v
d  vt  v
73.
the
QV
P
v
QV
Fv

QV
F



24 3.42  105 C 12 V 
240 N
 410 km
The mass of the wire is the density of copper times the volume of the wire, and the resistance of
wire is given by Eq. 18-3. We represent the mass density by  m and the resistivity by  .
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60
Physics: Principles with Applications, 6th Edition
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R
L
L
mR
A
74.
75.
A
m 
L
R
L
m   m LA   m L
R
L

R
 0.018 kg 12.5  

8.9 10
  d
1
2
3

kg m3 1.68  10 8  m
 d
2
4 L
R


 38.79 m  38.8 m

4 1.68  10 8  m  38.79 m 

 12.5  
 2.58  104 m
Use Eq. 18-6.
(a)
P
(b)
P
V2

R
V2

R
120 V 2
12 
120 V 2
140 
 1200 W
 103 W  100 W
Model the protons as moving in a continuous beam of cross-sectional area A. Then by Eq. 18-10,
N
, where N is the
I  neAvd . The variable n is the number of protons per unit volume, so n 
Al
number of protons in the beam and l is the circumference of the ring. The “drift” velocity in this
case is the speed of light.
N
N
I  neAvd 
eAvd  evd 
Al
l
3
11  10 A  6300 m 
Il
N

 1.4  1012 protons
evd
1.60  10 19 C 3.00  108 m s


76.
 A



The power is given by P  IV .
(a)
P  IV  12 A  220 V   2640W  2600 W
The power dissipated is given by PR  I 2 R , and the resistance is R 
(b)
PR  I R  I
2
2
L
A
.
8
L
2 4 1.68  10  m  15 m 
2 4 L
I
I
 12 A 
 17.433 W
2
A
 r2
d2
 1.628  103 m 
L
2
 17 W
4 L
 12 A 


4 1.68  108  m 15 m 
(c)
PR  I 2
(d)
The savings is due to the power difference.
d2
2
  2.053  10 m 
3
2
 10.962 W  11W
 1kW 
 12 h  $0.12 
 30 d  



 1000 W 
 1d  1kWh 
Savings  17.433 W  10.962 W  
 $0.2795 / month  28 cents per month
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
61
Physics: Principles with Applications, 6th Edition
Giancoli
77. The resistance can be calculated from the power and voltage, and then the diameter of the wire can
be calculated from the resistance.
V2
V2
L
L
V2
L
P
 R
R




2
2
R
P
A   12 d 
P   12 d 
4  LP
d
78.
79.
V


2

4 100  10 8  m  3.8 m  95 W 
 120 V 
2
 1.8  10 4 m
Use Eq. 18-6 for the power in each case, assuming the resistance is constant.
V2 R
P13.8 V
13.82
13.8 V
 2

 1.3225  32% increase
P12.0 V
12.02
V R


(a)
12.0 V
The current can be found from Eq. 18-5.
I PV
(b)


I A  PA VA  40W 120 V  0.33 A
The resistance can be found from Eq. 18-6.
R
(c)
I B  PB VB  40W 12 V  3.3 A
120 V 2
VA2
V2
RA 

 360 
P
PA
40 W
The charge is the current times the time.
RB 
VB2
PB

12 V 2
40 W
 3.6 
QA  I At   0.33 A  3600 s   1200 C
Q  It
QB  I B t   3.3 A  3600 s   12, 000 C
(d)
The energy is the power times the time, and the power is the same for both bulbs.
E A  EB   40 W  3600 s   1.44  105 J
E  Pt
(e) Bulb B requires a larger current, and so should have larger diameter connecting wires to avoid
overheating the connecting wires.
80.
The current in the wire can be found by I  P V .
(a)
PR  I 2 R 

(b)
PR 
P2
V2
R
 2250 W 
2
120 V 
P2 4 L
V d
2
2
2
P2  L

P2  L

P2 4 L
V 2 A V 2  r2 V 2  d 2
4 1.68  10 8  m  25.0 m 
 28.0 W
2
 2.59  103 m



 2250 W 

2
120 V 
2



4 1.68  108  m  25.0 m 
  4.12  103 m 
2
 11.1W
81. Eq. 18-3 can be used. The area to be used is the cross-sectional area of the pipe.
1.68  10 8  m 10.0 m 
L
L
R


 1.34  104 
2
2
2
2

2

2
A  r r
  2.50  10 m  1.50  10 m 

outside
inside




 

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
62
Physics: Principles with Applications, 6th Edition
Giancoli
82. The volume of wire  Volume  length  cross-sectional area  remains constant as the wire is
stretched from the original length of L0 to the final length of 2L0 . Thus the cross-sectional area
1
2
changes from A0 to
R0 
P0 
 L0
A0
V2
A0 . Use Eq. 18-3 for the resistance and 18-6 for the power dissipated.
R
P
L
V2
A


 2 L0
1
2
V2
A0

1
4
4
V2
R0
R 4 R0
R0
The power is reduced by a factor of 4 .
 L0
A0
 4 R0
 14 P0
83. The resistance of the filament when the flashlight is on is R 
V

3.2 V
I 0.20 A
with a combination of Eqs. 18-3 and 18-4 to find the temperature.
R  R0 1   T  T0  
T  T0 
 16  . That can be used
1 R

1
 16  
o
 1  2168o C  2200o C
  1  20 C 


1
o
  R0 
0.0045  C   1.5  
© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the
publisher.
63