Download Homework #5: Definition of Current and Resistance

Homework #1: Definition of Current and Resistance.
1. In a particular television picture tube, the measured beam current is 60.0 μA. How many electrons strike
the screen every second?
2. A lightbulb has a resistance of 240 Ω when operating at a voltage of 120 V. What is the current in the
bulb?
3. A 1.00-V potential difference is maintained across a 10.0-Ω resistor for a period of 20.0 s. What total
charge passes through the wire in this time interval?
4. What is the resistance of a toaster if 120 V produces a current of 4.2 A?
5. What voltage will produce 0.25 A of current through a 3800- resistor?
6. A bird stands on a dc electric transmission line carrying 2800 A. The line has 2.5 10 5  resistance
per meter, and the bird’s feet are 4.0 cm apart. What is the potential difference between the bird’s feet?
Homework #2: Resistivity.
7. Calculate the diameter of a 2.0-cm length of
tungsten filament in a small lightbulb if its
resistance is 0.050 Ω.
8. Eighteen-gauge wire has a diameter of
1.024 mm. Calculate the resistance of 15 m
of 18-gauge copper wire at 20°C.
9. A potential difference of 12 V is found to
produce a current of 0.40 A in a 3.2-m length
of wire with a uniform radius of 0.40 cm.
What is (a) the resistance of the wire?
(b) the resistivity of the wire?
10. A wire 50.0 m long and 2.00 mm in
diameter is connected to a source with a
potential difference of 9.11 V, and the
current is found to be 36.0 A. Assume a
temperature of 20°C, and, using Table 17.1,
identify the metal out of which the wire is made.
11. A wire of initial length L0 and radius r0 has
a measured resistance of 1.0 Ω. The wire is
drawn under tensile stress to a new uniform
radius of r = 0.25r0. What is the new resistance
of the wire?
12. Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper. If the wire is to have a
resistance R = 0.500 Ω, and if all of the copper is to be used, what will be (a) the length and (b) the
diameter of the wire? Hint: The density of copper is 8.92 g/cm3, and the volume of the copper does not
change.
Homework #3: Temperature Dependence.
13. A certain lightbulb has a tungsten filament with a resistance of 19 Ω when cold and 140 Ω when hot.
Assume that R  Ro 1   T  To  can be used over the large temperature range involved here, and


find the temperature of the filament when it is hot. Assume an initial temperature of 20°C.
14. At 20°C, the carbon resistor in an electric circuit connected to a 5.0-V battery has a resistance of 200 Ω.
What is the current in the circuit when the temperature of the carbon rises to 80°C?
15. A 100-cm-long copper wire of radius 0.50 cm has a potential difference across it sufficient to produce
a current of 3.0 A at 20°C. (a) What is the potential difference? (b) If the temperature of the wire is
increased to 200°C, what potential difference is now required to produce a current of 3.0 A?
Homework #4: Power.
16. A toaster is rated at 600 W when connected to a 120-V source. What current does the toaster carry, and
what is its resistance?
17. If electrical energy costs 12 cents, or $0.12, per kilowatt-hour, how much does it cost to (a) burn a
100-W lightbulb for 24 h? (b) operate an electric oven for 5.0 h if it carries a current of 20.0 A at 220 V?
18. A high-voltage transmission line with a resistance of 0.31 Ω/km carries a current of 1,000 A. The line
is at a potential of 700 kV at the power station and carries the current to a city located 160 km from the
station. (a) What is the power loss due to resistance in the line?
2
{Hint: Use I R for the power loss to the wire.}
(b) What fraction of the transmitted power does this loss represent?
{Hint: Use V I for the power delivered to the city.}

 
19. The heating element of a coffeemaker operates at 120 V and carries a current of 2.00 A. Assuming that
the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.500 kg of
water from room temperature (23.0°C) to the boiling point.
20. What is the required resistance of an immersion heater that will increase the temperature of 1.50 kg of
water from 10.0°C to 50.0°C in 10.0 min while operating at 120 V?
21. A small motor draws a current of 1.75 A from a 120-V line. The output power of the motor is 0.20 hp.
(a) At a rate of $0.060/kWh, what is the cost of operating the motor for 4.0 h? (b) What is the efficiency of
the motor? 1 horsepower = 746 W
22. It has been estimated that there are 270 million plug-in electric clocks in the United States,
approximately one clock for each person. The clocks convert energy at the average rate of 2.50 W. To
supply this energy, how many metric tons of coal are burned per hour in coal-fired electric generating
plants that are, on average, 25.0% efficient? The heat of combustion for coal is 33.0 MJ/kg.
23. A steam iron draws 6.0 A from a 120-V line. (a) How many joules of internal energy are produced in
20 min? (b) How much does it cost, at $0.080/kWh, to run the steam iron for 20 min?
Solutions:
1. Q  I t and the number of electrons is
6
Q I t  60.0 10 C s  1.00 s
n


 3.75 1014 electrons
-19
e
e
1.60 10 C
V 120 V

 0.500 A  500 m A
R
240 
V 
 Q V
3. The current is I
. Thus, the change that passes is Q  

  t , giving
t
R
 R 
 1.00 V 
Q  
  t   0.100 A  20.0 s  2.00 C
 10.0  
V 120 V
 29 
4. R  
I 4.2 A
2. I
5. V  IR   0.25 A  3800    950 V
6. 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
7. From R 
d
4 L

R
8. R 
9. (a)
L
A

R
(b)
L
A
, we obtain A 
4 5.6 108   m
d
2
4

d 4
2

R
 2.0 10
2
  0.050  
L
L
41.7  108   m
 1.024 10
3
, or
m
  1.7  10
 15 m  
m
2
4
m  0.17 m m
0.31 
V
12 V

 30 
I 0.40 A
From, R 
L
A
,
2
2


R  A  30     0.40 10 m  


 4.7  10 4   m
L
3.2 m
V 9.11 V

 0.253  , so the resistivity of the metal is
10. The resistance is R 
I 36.0 A
2
3
R  A R   d 4  0.253     2.00 10 m



L
L
4 50.0 m 
Thus, the metal is seen to be silver .

2
 1.59 108   m
11. The volume of material, V  AL0   r02  L0 , in the wire is constant. Thus, as the wire
is stretched to decrease its radius, the length increases such that  rf2  Lf   r02  L0 giving
2
2
r
 r 
2
Lf   0  L0   0  L0   4.0 L0  16L0
 rf 
 0.25r0 
 
The new resistance is then
Rf  
Lf
Af

Lf

r
2
f
16L0
  r0 4
2
L 
2
 16 4   02   256R 0  256 1.00    256 
  r0 
12. The volume of the copper is
V
1.00 103 kg
m

density 8.92 103 kg m
3
 1.12 107 m
Since, V  A  L , this gives A  L  1.12 107 m
(a) From R 
L
A
3
3
(1)
, we find that
 1.70 108   m 

8
A   L  
 L   3.40 10 m  L
0.500 
R


Inserting this expression for A into equation (1) gives
 3.40 10
8
From equation (1), A 
(b)
d
m  L2  1.12 107 m 3 , which yields
4 1.12 107 m
L
3

 d2
4

1.12 107 m
L
41.12 107 m
  1.82 m 
3
L  1.82 m
3
, or

 2.80  104 m  0.280 m m
13. Solving R  R0 1   T  T0  for the final temperature gives
T  T0 
14. At 80°C,
R  R0
140   19 
 20C 
 1.4 103 C
1
3
 R0
 4.5 10  C   19  


I
or
V
V
5.0 V


R
R0 1   T  T0    200   1   0.5 103 C 1   80C  20C 


I 2.6 102 A  26 m A
15. (a) At 20°C, the resistance of this copper wire is R0    L A     L  r2  and the
potential difference required to produce a current of I 3.0 A is
1.00 m 
 L 
V0  IR0  I  2    3.0 A  1.7  108   m 
 r 
  0.50 102 m
or

2
V0  6.5 104 V  0.65 m V
(b)
At T  200C , the potential difference required to maintain the same
current in the copper wire is
V  IR  IR0 1   T  T0   V0 1  T  T0 
or
V   0.65 m V  1  3.9 103 °C -1   200C  20C    1.1 m V
16. I
P
600 W

 5.00 A
V 120 V
V
120 V

 24.0 
I 5.00 A
17. (a) The energy used by a 100-W bulb in 24 h is
and
R
E Pt 100 W
 24 h   0.100 kW  24 h  2.4 kW h
and the cost of this energy, at a rate of $0.12 per kilowatt-hour is
cost E  rate  2.4 kW h   $0.12 kW h   $0.29
(b)
The energy used by the oven in 5.0 h is

 1 kW  
E Pt I V   t  20.0 C s 220 J C  
   5.0 h   22 kW h
3

 10 J s 
and the cost of this energy, at a rate of $0.12 per kilowatt-hour is
cost E  rate  22 kW h   $0.12 kW h   $2.6
18. (a) The power loss in the line is
Ploss I2R   1000 A   0.31  km
2
(b)
 160 km   5.0 107 W
 50 M W
The total power transmitted is
Pinput  V  I  700 103 V  1000 A   7.0 108 W  700 M W
Thus, the fraction of the total transmitted power represented by the line losses is
fraction loss
Ploss
50 M W

 0.071 or 7.1%
Pinput 700 M W
19. The energy required to bring the water to the boiling point is
E  m c T    0.500 kg  4186 J kg  C   100C  23.0C   1.61 105 J
The power input by the heating element is
Pinput  V  I  120 V   2.00 A   240 W  240 J s
Therefore, the time required is
t
E
Pinput

1.61 105 J
 1 m in 
 672 s
  11.2 m in
240 J s
 60 s 
20. The energy input required is
E  m c T    1.50 kg  4186 J kg  C   50.0C  10.0C   2.51 105 J
and, if this is to be added in t 10.0 m in  600 s, the power input needed is
P
E 2.51 105 J

 419 W
t
600 s
The power input to the heater may be expressed as P   V  R , so the needed resistance
is
2
R
 V 
P
2

120 V 
419 W
2
 34.4 
21. (a) The power input to the motor is
Pinput  V  I 120 V 1.75 A   210 W  0.210 kW
At a rate of $0.06/kWh, the cost of operating this motor for 4.0 h is


cost  Energy used  rate  Pinputt  rate
  0.210 kW
(b)
 4.0 h   6.0 cents kW h   5.0 cents
The efficiency is
Eff
Poutput
Pinput

 0.20 hp 0.746 kW
hp 
0.210 kW
 0.71 or 71%
22. The total power converted by the clocks is
P  2.50 W
  270 106   6.75 108 W
and the energy used in one hour is
E  Pt  6.75 108 W
  3600 s  2.43 10
12
J
The energy input required from the coal is
Ecoal 
2.43 1012 J
E

 9.72 1012 J
efficiency
0.250
The required mass of coal is thus
m
Ecoal
9.72 1012 J

 2.95 105 kg
6
heatof com bustion 33.0 10 J kg
 1 m etric ton 
m   2.95 105 kg  
  295 m etric tons
3
 10 kg 
23. (a) The power required by the iron is
or
à   V  I 120 V  6.0 A   7.2 102 W
and the energy transformed in 20 minutes is
J 

 60 s  
5
E  Ã  t  7.2 102   20 m in  
   8.6 10 J
s 

 1 m in  
(b)
The cost of operating the iron for 20 minutes is
cost E  rate

 1 kW h  
  8.6 105 J 
   $0.080 kW h   $0.019  1.9 cents
6
 3.60 10 J 
