Download Chapter 22-23 Assignment Solutions

Chapter 22-23 Assignment Solutions
Table of Contents
Page 610 #39, 44-45, 61-75 .......................................................................................................................... 1
Page 610 #38, 53-57, 82-85, 89-92 ............................................................................................................... 5
Page 636 #36-47, 57-63, 77-81 ..................................................................................................................... 8
Chapter 22 Review ...................................................................................................................................... 12
Chapter 23 Review ...................................................................................................................................... 12
Combined Circuit Practice........................................................................................................................... 12
Page 610 #39, 44-45, 61-75
39) Define the unit of electric current in terms of fundamental MKS units. (22.1)
1A=
1C
1s
44) Describe the energy conversions that occur in each of the following devices. (22.1)
a) An incandescent lightbulb
Electric energy to heat and light
b) A clothes dryer
Electric energy to heat and kinetic energy
c) A digital clock radio
Electric energy to light and sound
45) Which wire conducts electricity with the least resistance: one with a large cross-sectional diameter
or one with a small cross-sectional diameter? (22.1)
A larger-diameter wire has a smaller resistance because there are more electrons to carry the charge.
61) A motor is connected to a 12-V battery, as shown in Figure 22-13.
a) How much power is delivered to the motor?
𝑃 = 𝐼𝑉 = (1.5 A)(12 V) = 18 W
b) How much energy is converted if the motor runs for 15 min?
60 s
𝑈𝐸 = 𝑃𝑡 = (18 W)(15 min) (
) = 1.6 × 104 J
1 min
62) Refer to Figure 22-14 to answer the following questions.
a) What should the ammeter reading be?
𝐼=
𝑉 27 V
=
= 1.5 A
𝑅 18 Ω
b) What should the voltmeter reading be?
27 V
c) How much power is delivered to the resistor?
𝑃 = 𝐼𝑉 = (1.5 A)(27 V) = 41 W
d) How much energy is delivered to the resistor per hour?
60 min 60 s
𝑈𝐸 = 𝑃𝑡 = (41 W)(1 hour) (
)(
) = 1.5 × 105 J
1 hour 1 min
63) Refer to Figure 22-15 to answer the following questions
a) What should the ammeter reading be?
𝐼=
𝑉 27 V
=
= 3.0 A
𝑅 9.0 Ω
b) What should the voltmeter reading be?
27 V
c) How much power is delivered to the resistor?
𝑃 = 𝐼𝑉 = (3.0 A)(27 V) = 81 W
d) How much energy is delivered to the resistor per hour?
60 min 60 s
𝑈𝐸 = 𝑃𝑡 = (81 W)(1 hour) (
)(
) = 2.9 × 105 J
1 hour 1 min
64) Refer to Figure 22-16 to answer the following questions
a) What should the ammeter reading be?
𝐼=
𝑉 9.0 V
=
= 0.50 A
𝑅 18 Ω
b) What should the voltmeter reading be?
9.0 V
c) How much power is delivered to the resistor?
𝑃 = 𝐼𝑉 = (0.50 A)(9.0 V) = 4.5 W
d) How much energy is delivered to the resistor per hour?
60 min 60 s
𝑈𝐸 = 𝑃𝑡 = (4.5 W)(1 hour) (
)(
) = 1.6 × 104 J
1 hour 1 min
65) Toasters: The current through a toaster that is connected to a 120-V source is 8.0 A. What power is
dissipated by the toaster?
𝑃 = 𝐼𝑉 = (8.0 A)(120 V) = 9.6 × 102 W
66) Lightbulbs: A current of 1.2 A is measured through a lightbulb when it is connected across a 120-V
source. What power is dissipated by the bulb?
𝑃 = 𝐼𝑉 = (1.2 A)(120 V) = 1.4 × 102 W
67) A lamp draws 0.50 A from a 120-V generator.
a) How much power is delivered?
𝑃 = 𝐼𝑉 = (0.50 A)(120 V) = 6.0 × 101 W
b) How much energy is converted in 5.0 min?
60 s
𝑈𝐸 = 𝑃𝑡 = (60. W)(5.0 min) (
) = 1.8 × 104 J
1 min
68) A 12-V automobile battery is connected to an electric starter motor. The current through the motor
is 210 A.
a) How many joules of energy does the battery deliver to the motor each second?
𝑈𝐸 = 𝑃𝑡 = 𝐼𝑉𝑡 = (210 A)(12 V)(1 s) = 2500 J
b) What power, in watts, does the motor use?
𝑃 = 2500 W
69) Dryers: A 4200-W clothes dryer is connected to a 220-V circuit. How much current does the dryer
draw?
𝑃 = 𝐼𝑉
𝐼=
𝑃 4200 W
=
= 19 A
𝑉
220 V
70) Flashlights: A flashlight bulb is connected across a 3.0-V potential difference. The current through
the bulb is 1.5 A.
a) What is the power rating of the bulb?
𝑃 = 𝐼𝑉 = (1.5 A)(3.0 V) = 4.5 W
b) How much electric energy does the bulb convert in 11 min?
60 s
𝑈𝐸 = 𝑃𝑡 = (4.5 W)(11 min) (
) = 3.0 × 103 J
1 min
71) Batteries: A resistor of 60.0 Ω has a current of 0.40 A through it when it is connected to the
terminals of a battery. What is the voltage of the battery?
𝑉 = 𝐼𝑅 = (0.40 A)(60.0 Ω) = 24 V
72) What voltage is applied to a 4.0-Ω resistor if the current is 1.5 A?
𝑉 = 𝐼𝑅 = (1.5 A)(4.0 Ω) = 6.0 V
73) What voltage is placed across a motor with a 15-Ω operating resistance if there is 8.0 A of current?
𝑉 = 𝐼𝑅 = (8.0 A)(15 Ω) = 120 V
74) A voltage of 75 V is placed across a 15-Ω resistor. What is the current through the resistor?
𝑉 = 𝐼𝑅
𝐼=
𝑉 75 V
=
= 5.0 A
𝑅 15 Ω
75) Some students connected a length of nichrome wire to a variable power supply to produce between
0.00 V and 10.00 V across the wire. They then measured the current through the wire for several
voltages. The students recorded the data for the voltages used and the currents measured, as shown in
Table 22-2.
a) For each measurement, calculate the resistance.
Voltage, 𝑉 (volts)
Current, 𝐼
Resistance, 𝑅 = 𝑉/𝐼 (ohms)
(amps)
2.00
0.0140
143 Ω
4.00
0.0270
148 Ω
6.00
0.0400
150 Ω
8.00
0.0520
154 Ω
10.00
0.0630
159 Ω
-2.00
-0.0140
143 Ω
-4.00
-0.0280
143 Ω
-6.00
-0.0390
154 Ω
-8.00
-0.0510
157 Ω
-10.00
-0.0620
161 Ω
b) Graph 𝐼 versus 𝑉.
c) Does the nichrome wire obey Ohm’s law? If not for all the voltages, specify the voltage range
for which Ohm’s law holds.
Ohm’s law is obeyed when the resistance of a device is constant, independent of 𝑉. The wire
does not obey Ohm’s law for all 𝑉, but does from about −4 V to about 2 V.
Page 610 #38, 53-57, 82-85, 89-92
38) Complete the concept map using the following terms: watt, current, resistance.
53) Power Lines: Why can birds perch on high-voltage lines without being injured?
The birds do not provide a conduit for current to flow anywhere but back into the wire. Since the birds’
bodies have much higher resistance than the wire, the current takes the path of least resistance –
through the wire – rather than through the bird.
54) Describe two ways to increase the current in a circuit.


Increase the electric potential difference of the power source.
Reduce the resistance of the circuit.
o Add more paths in parallel.
o Use a material with lower resistivity.
o Use wire with a larger cross-section.
o Remove resistors/lamps/etc. from the circuit.
55) Lightbulbs: Two lightbulbs work on a 120-V circuit. One is 50 W and the other is 100 W. Which
bulb has a higher resistance? Explain.
The 50-W bulb.
𝑃=
𝑉2
𝑉2
→ 𝑅=
𝑅
𝑃
Therefore, the lower 𝑃 is caused by the higher 𝑅.
56) If the voltage across a circuit is kept constant and the resistance is doubled, what effect does this
have on the circuit’s current?
𝑉 = 𝐼𝑅 → 𝐼 =
𝑉
𝑅
If the resistance is doubled, the current is cut in half.
57) What is the effect on the current in a circuit if both the voltage and the resistance are doubled?
Explain.
𝑉
No effect. 𝑉 = 𝐼𝑅, so 𝐼 = 𝑅, and if the electric potential difference and the resistance are both doubled,
2𝑉
𝑉
𝐼2 = 2𝑅 = 𝑅 = 𝐼, the current will not change.
82) Batteries: A 9.0-V battery costs $3.00 and will deliver 0.0250 A for 26.0 h before it must be
replaced. Calculate the cost per kWh.
𝑈𝐸 = 𝑃𝑡 = 𝐼𝑉𝑡 = (0.0250 A)(9.0 V)(26.0 h) = 5.9 Wh = 5.9 × 10−3 kWh
𝑅𝑎𝑡𝑒 =
𝐶𝑜𝑠𝑡
$3.00
=
= $510/kWh
𝑈𝐸
5.9 × 10−3 kWh
83) What is the maximum current allowed in a 5.0-W, 220-Ω resistor?
𝑃 = 𝐼2 𝑅
𝑃
5.0 W
𝐼=√ =√
= 0.15 A
𝑅
220 Ω
84) A 110-V electric iron draws 3.0 A of current. How much thermal energy is developed in an hour?
60 min 60 s
𝑄 = 𝑈𝐸 = 𝑃𝑡 = 𝐼𝑉𝑡 = (3.0 A)(110 V)(1 hour) (
)(
) = 1.2 × 106 J
1 hour 1 min
85) For the circuit shown in Figure 22-18, the maximum safe power is 5.0 × 101 W. Use the figure to
find the following:
a) the maximum safe current
𝑃 = 𝐼2 𝑅
𝑃
5.0 × 101 W
𝐼=√ =√
= 1.1 A
𝑅
40.0 Ω
b) the maximum safe voltage
𝑉2
𝑃=
𝑅
1
𝑉 = √𝑃𝑅 = √(5.0 × 10 W)(40.0 Ω) = 45 V
89) If a person has $5, how long could he or she play a 200 W stereo if electricity costs $0.15 per kWh?
𝑈𝐸 = 𝑃𝑡 =
𝑡=
𝐶𝑜𝑠𝑡
𝑅𝑎𝑡𝑒
𝐶𝑜𝑠𝑡
$5
=
(𝑅𝑎𝑡𝑒)(𝑃) ($0.15/kWh)(200
1 kW
W) (1000 W)
= 200 hours
90) A current of 1.2 A is measured through a 50.0-Ω resistor for 5.0 min. How much heat is generated
by the resistor?
𝑄 = 𝑈𝐸 = 𝑃𝑡 = 𝐼 2 𝑅𝑡 = (1.2 A)2 (50.0 Ω)(5.0 min) (
60 s
) = 2.2 × 104 J
1 min
91) A 6.0-Ω resistor is connected to a 15-V battery.
a) What is the current in the circuit?
𝑉 = 𝐼𝑅
𝑉 15 V
𝐼= =
= 2.5 A
𝑅 6.0 Ω
b) How much thermal energy is produced in 10.0 min?
𝑄 = 𝑈𝐸 = 𝑃𝑡 = 𝐼 2 𝑅𝑡 = (2.5 A)2 (6.0 Ω)(10.0 min) (
60 s
) = 2.3 × 104 J
1 min
92) Lightbulbs: An incandescent lightbulb with a resistance of 10.0 Ω when it is not lit and a resistance
of 40.0 Ω when it is lit has 120 V placed across it.
a) What is the current draw when the bulb is lit?
𝑉 = 𝐼𝑅
𝑉 120 V
𝐼= =
= 3.0 A
𝑅 40.0 Ω
b) What is the current draw at the instant the bulb is turned on?
At the instant the bulb is turned on, its resistance is still 10.0 Ω. It takes time (a very small
amount of time, but still some time) for its resistance to increase to 40.0 Ω.
𝑉 = 𝐼𝑅
𝑉 120 V
𝐼= =
= 12 A
𝑅 10.0 Ω
c) When does the lightbulb use the most power?
The instant the bulb is turned on
Page 636 #36-47, 57-63, 77-81
36) Complete the concept map using the following terms: series circuit, 𝑅 = 𝑅1 + 𝑅2 + 𝑅3 , constant
current, parallel circuit, constant potential.
37) Why is it frustrating when one bulb burns out on a string of holiday tree lights connected in series?
(23.1)
When one bulb burns out, the circuit is open and all the bulbs go out.
38) Why does the equivalent resistance decrease as more resistors are added to a parallel circuit?
(23.1)
Each new resistor provides an additional path for the current.
39) Several resistors with different values are connected in parallel. How do the values of the individual
resistors compare with the equivalent resistance? (23.1)
The equivalent resistance will be less than the resistance of each of the resistors.
40) Why is household wiring constructed in parallel instead of in series? (23.1)
Appliances/lights/etc. connected in parallel can be run independently of one another. If they were
connected in series, they would have to all be on or all be off at any given time.
41) Why is there a difference in equivalent resistance between three 60 Ω resistors connected in series
and three 60 Ω resistors connected in parallel? (23.1)
In a series circuit, the current is opposed by each resistor in turn. The total resistance is the sum of the
individual resistances (180 Ω). In a parallel circuit, each resistor provides an additional path for current.
The result is a decrease in total resistance, to 20 Ω.
42) Compare the amount of current entering a junction in a parallel circuit with that leaving the
junction. (A junction is a point where three or more conductors are joined.) (23.1)
The amount of current entering a junction is equal to the amount of current leaving the junction. The is
known as Kirchoff’s junction rule.
43) Explain how a fuse functions to protect an electric circuit. (23.2)
The purpose of a fuse is to prevent conductors from being overloaded with current, causing fires due to
overheating. A fuse is simply a short length of wire that will melt from the heating effect if the current
exceeds a certain maximum. This opens the circuit and stops the current from flowing.
44) What is a short circuit? Why is a short circuit dangerous? (23.2)
A short circuit is a circuit that has extremely low resistance. A short circuit is dangerous because any
potential difference will produce a large current. The heating effect of the current can cause a fire.
45) Why is an ammeter designed to have a very low resistance? (23.2)
An ammeter must have low resistance because it is placed in series in the circuit. If its resistance were
high, it would significantly change the total resistance of the circuit and thus serve to reduce the current
in the circuit, thereby changing the current it is meant to measure.
46) Why is a voltmeter designed to have a very high resistance? (23.2)
A voltmeter is placed in parallel with the portion of the circuit whose electric potential difference is to
be measured. A voltmeter must have very high resistance for the same reason that an ammeter must
have very low resistance. If the voltmeter had low resistance, it would lower the resistance of the
portion of the circuit it is connected across and increase the current in the circuit. This would produce a
larger decrease in electric potential across the part of the circuit where the voltmeter is connected,
changing the electric potential difference it is measuring.
47) How does the way in which an ammeter is connected in a circuit differ from the way in which a
voltmeter is connected? (23.2)
An ammeter is connected in series; a voltmeter is connected in parallel.
57) Ammeter 1 in Figure 23-14 reads 0.20 A.
a) What should ammeter 2 indicate?
0.20 A, because current is constant in a series circuit
b) What should ammeter 3 indicate?
0.20 A, because current is constant in a series circuit
58) Calculate the equivalent resistance of these series-connected resistors: 680 Ω, 1.1 kΩ, and 10 kΩ.
𝑅 = 0.68 kΩ + 1.1 kΩ + 10 kΩ = 12 kΩ
59) ) Calculate the equivalent resistance of these parallel-connected resistors: 680 Ω, 1.1 kΩ, and
10.2 kΩ.
1
1
1
1
=
+
+
𝑅 𝑅1 𝑅2 𝑅3
𝑅=
1
1
=
= 0.40 kΩ
1
1
1
1
1
1
+
+
+
+
𝑅1 𝑅2 𝑅3 0.68 kΩ 1.1 kΩ 10.2 kΩ
60) A series circuit has two voltage drops: 5.50 V and 6.90 V. What is the supply voltage?
According to Kirchoff’s loop rule, the supplied Δ𝑉 must equal the drops in Δ𝑉.
𝑉𝑠𝑜𝑢𝑟𝑐𝑒 = 𝑉1 + 𝑉2 = 5.50 V + 6.90 V = 12.40 V
61) A parallel circuit has two branch currents: 3.45 A and 1.00 A. What is the current in the energy
source.
𝐼𝑠𝑜𝑢𝑟𝑐𝑒 = 𝐼1 + 𝐼2 = 3.45 A + 1.00 A = 4.45 A
62) Ammeter 1 in Figure 23-14 reads 0.20 A.
a) What is the total resistance in the circuit?
𝑅 = 𝑅1 + 𝑅2 = 15 Ω + 22 Ω = 37 Ω
b) What is the battery voltage?
𝑉 = 𝐼𝑅 = (0.20 A)(37 Ω) = 7.4 V
c) How much power is delivered to the 22 Ω resistor?
𝑃 = 𝐼 2 𝑅 = (0.20 A)2 (22 Ω) = 0.88 W
d) How much power is supplied by the batter?
𝑃 = 𝐼𝑉 = (0.20 A)(7.4 V) = 1.5 W
63) Ammeter 2 in Figure 23-14 reads 0.50 A.
a) Find the voltage across the 22-Ω resistor.
𝑉 = 𝐼𝑅 = (0.50 A)(22 Ω) = 11 V
b) Find the voltage across the 15-Ω resistor.
𝑉 = 𝐼𝑅 = (0.50 A)(15 Ω) = 7.5 V
c) What is the battery voltage?
𝑉 = 𝑉1 + 𝑉2 = 11 V + 7.5 V = 19 V
77) Refer to Figure 23-19 and assume that all the resistors are 30.0 Ω. Find the equivalent resistance.
1
1
1
=
+
→ 𝑅𝑝 = 15.0 Ω
𝑅𝑝 30.0 Ω 30.0 Ω
𝑅 = 30.0 Ω + 15.0 Ω = 45.0 Ω
78) Refer to Figure 23-19 and assume that each resistor dissipates 120 mW. Find the total dissipation.
𝑃 = 3(120 mW) = 360 mW
79) Refer to Figure 23-19 and assume that 𝐼𝐴 = 13 mA and 𝐼𝐵 = 1.7 mA. Find 𝐼𝐶 .
𝐼𝐴 = 𝐼𝐵 + 𝐼𝐶
𝐼𝐶 = 𝐼𝐴 − 𝐼𝐵 = 13 mA − 1.7 mA = 11 mA
80) Refer to Figure 23-19 and assume that 𝐼𝐵 = 13 mA and 𝐼𝐶 = 1.7 mA. Find 𝐼𝐴 .
𝐼𝐴 = 𝐼𝐵 + 𝐼𝐶 = 13 mA + 1.7 mA = 15 mA
81) Refer to Figure 23-20 to answer the following questions.
a) Determine the total resistance.
The 30.0-Ω and 20.0-Ω resistors are in series.
𝑅1 = 30.0 Ω + 20.0 Ω = 50.0 Ω
The 30.0-Ω and 20.0-Ω resistors are in series.
𝑅2 = 10.0 Ω + 40.0 Ω = 50.0 Ω
𝑅1 and 𝑅2 are in parallel.
1
1
1
=
+
𝑅 𝑅1 𝑅2
1
1
𝑅=
=
= 25.0 Ω
1
1
1
1
+
+
𝑅1 𝑅2 50.0 Ω 50.0 Ω
𝑅 is in series with the top 25.0-Ω resistor.
𝑅𝑇𝑜𝑡𝑎𝑙 = 25.0 Ω + 25.0 Ω = 50.0 Ω
b) Determine the current through the 25-Ω resistor.
All of the current must flow through the 25-Ω resistor. We can use Ohm’s Law and the total
resistance to calculate the current.
𝑉 = 𝐼𝑅𝑇𝑜𝑡𝑎𝑙
𝑉
25 𝑉
𝐼=
=
= 0.50 A
𝑅𝑇𝑜𝑡𝑎𝑙 50.0 Ω
c) Which resistor is the hottest? Coolest?
The temperature of the resistor is directly proportional to the power consumed by the resistor.
𝑃 = 𝐼2 𝑅
𝑃25Ω = (0.50 A)2 (25.0 Ω) = 6.25 W
The 0.50 A is split evenly between the two parallel branches because the sum of resistances in
each branch are equal.
𝑃30Ω = (0.25 A)2 (30.0 Ω) = 1.9 W
𝑃20Ω = (0.25 A)2 (20.0 Ω) = 1.2 W
𝑃10Ω = (0.25 A)2 (10.0 Ω) = 0.62 W
𝑃40Ω = (0.25 A)2 (40.0 Ω) = 2.5 W
The 25.0-Ω resistor is the hottest. The 10.0-Ω resistor is the coolest.
Chapter 22 Review
(coming soon)
Chapter 23 Review
(coming soon)
Combined Circuit Practice
(coming soon)