Download PE PE PE work PE PE work 48 J 12 J 60 J ∆ = - = = + = +

Physics 100 Energy in Today’s World Homework Ch. 10
1.
Prof. Menningen
p. 1 of 4
The electric potential energy of a charged object at point A is known to be 48 J. If it requires
12 J of work to move the object from A to B, what is the potential energy at point B?
PE  PE B  PE A  work
PE B  PE A  work  48 J  12 J  60 J
2.
How much work does a 8.0-Volt battery do in pushing 2.9 milliCoulombs of charge
through a circuit containing one light bulb?
V
work
 work  charge  V   0.0029 C 8.0 V   0.0232 J
charge
3.
When you stack three flashlight batteries in the same direction, you get a voltage of 3 × 1.5
volts = 4.5 volts. What voltage do you get if one of the batteries is turned around end-forend?
a. 10.0 V
V  1.5  1.5  1.5 V  1.5 V
b. 3.0 V
c. 1.5 V
d. −1.5 V
4.
What are the analogs to the following terms in the water model of electricity?
a. wire
pipe
____________
friction
b. resistance ____________
c. voltage
pressure or GPE
____________
d. charge
water
____________
water wheel
e. light bulb ____________
f. current
flow rate
____________
pump
____________
h. switch
valve
____________
g. battery
5.
If the voltage in a circuit is doubled, what happens to the current?
a. It increases by a factor of four.
b. It decreases by a factor of four.
c. It is doubled.
d. It is cut in half.
6.
Label as "series" or "parallel" the following circuit characteristics:
series
a. Has a large total resistance: ____________
parallel
b. Used in household circuits: ____________
7.
c. Same voltage throughout:
parallel
____________
d. Same current throughout:
series
____________
In what unit is a household's consumption of electricity measured?
a. watt/hour
b. kilowatt/hour
c. watt·sec
d. kilowatt·hour
e. kilowatt·sec
Physics 100 Energy in Today’s World Homework Ch. 10
8.
Prof. Menningen
p. 2 of 4
Why do power companies use high voltages for their long-distance transmission lines?
Write a brief but complete explanation.
Using a high voltage reduces the current required to produce the same
amount of power. Reducing the current greatly reduces the power loss
Ploss = I2R, where R is the resistance of the wires.
9.
A 750-MW power plant sends its power out on a 530,000 V high-voltage line. If the total
resistance of the line is 0.70 Ω, what percentage of the power is lost due to resistive
heating?
P 750  106 W
P  VI  I  
 1415 A
V
530,000 V
Ploss  I 2 R  1415 A   0.70    1.40  106 W  1.40 MW
2
% loss 
Ploss 1.40 MW

 100  0.00187  0.187%
P
750 MW
10. For the following questions assume household electricity is provided at 120 V.
(a) What is the resistance of a light bulb that draws 0.570 A of current?
V  IR  R 
V
120 V

 211 
I 0.570 A
(b) What is the power used by a toaster that draws a current of 9.95 A?
P  VI  120 V  9.95 A   1194 W  1.19 kW
(c) A 1250-W heater for a sauna requires 65 minutes to heat the sauna to 120 °F.
What does this cost if electricity sells for 9¢/kWh?
P
E
1 hr 

 E  Pt  1.250 kW   65 min 
  1.354 kWh
t
60 min 

cost  1.354 kWh  9¢/kWh  12.2¢
Physics 100 Energy in Today’s World Homework Ch. 10
Prof. Menningen
p. 3 of 4
11. (a) Three resistors with values 1.0 Ω, 4.0 Ω, and 7.0 Ω are connected in series with a
9.00 V battery. How much current is drawn from the battery?
Rtotal  1.0  4.0  7.0   12.0 
I
V
9.00 V

 0.750 A
Rtotal 12.0 
(b) If the same three resistors are each connected in parallel to the same battery, how
much current is drawn from the battery?
I1 
V 9.00 V
9.00 V
9.00 V

 9.0 A, I 2 
 2.25 A, I 3 
 1.29 A
R1 1.0 
4.0 
7.0 
I total  9.00  2.25  1.29 A  12.5 A
12. Two V = 1.65-V batteries are connected in parallel to a 7.00-Ω resistor. How much current
does each battery supply?
V 1.65 V

 0.236 A
R 7.00 
That's the total current through the resistor. Each battery
V  IR  I 
supplies half the current:
1
2
 0.236 A  
0.118 A
13. The common exit sign in a public building uses two 15 W incandescent bulbs. These could
be replaced with light emitting diodes (LEDs) at 1.9 W per sign. If there are 45 million
signs in use in the United States, what savings in electricity use for an entire year (8,760
hours) could be secured with this changeover? Use $0.10/kWh as the cost of electricity.
 P   2  15 W  1.9 W   28.1 W  0.0281 kW per sign
 E   Pt   0.0281 kW  8760 h   246 kWh per sign
savings   45  106 signs   246 kWh/sign   $0.10 kWh 1 
 $1.11  109  1.11 billion dollars
Physics 100 Energy in Today’s World Homework Ch. 10
Prof. Menningen
p. 4 of 4
14. (a) Suppose you have an electric water heater with efficiency of 86%, and you take a
shower using 54 kg of water (~15 gallons). At a price of $0.10 per kWh, how much did you
pay for the shower? Assume it heated the water from 16 °C (~61 °F) to 42 °C (~108 °F).
Q  mcT   54 kg  4186 J/kg/ C  42  16C   5.877  106 J
That's the heat added to the water. The energy input to the heater is:
Q 5.877  106 J
E 
 6.834  106 J
e
0.86
1 kWh
$0.10
E  6.834  106 J 

 $0.190
6
3.6  10 J kWh
(b) Consider the above scenario. If the power plant is 34% efficient and the transmission
lines are 84% efficient, how many pounds of coal were burned for your shower? Each
pound of coal releases about 1.2×107 J of thermal energy.
The energy sent into the transmission lines by the power plant:
W
E
etransmission
6.834  106 J

 8.136  106 J
0.84
The fuel energy supplied to the power plant is:
QH 
W
epower plant
8.136  106 J

 2.393  107 J
0.34
QH  2.393  107 J 
1 lb coal
 1.99 lb of coal
1.2  107 J
(c) How many pounds of coal will be burned each year to provide showers for a family of
four, assuming one shower per day for each family member?
1.99 lb 4 shower

 7.98 lb/day  365 day/year
shower
day
 2911 lb of coal/year  1.46 ton of coal/year
Note that this calculation does not include the cost of keeping the water heater hot during
the rest of the day!