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Name: ________________________ Class: ___________________ Date: __________
ID: A
Chapter 21 Multi-format Test
Modified True/False
Indicate whether the statement is true or false. If false, change the identified word or phrase to make the statement true.
____
1. A series circuit contains branching points and multiple paths for current to flow through.
_________________________
____
2. Each branch of a parallel circuit has the same voltage. _________________________
____
3. Kirchhoff’s current law states that all the current entering a branch point must exit the point.
_________________________
____
4. A kilowatt-hour (kWh) is a unit of power. _________________________
____
5. The current from a battery is always direct current (DC). _________________________
Completion
Complete each statement.
Select the correct term to complete each sentence. There are extra terms in the list.
voltage
parallel
zero
increases
current
series
alternating
decreases
resistance
breaker
direct
stays the same
6. When adding resistance in a circuit, the resistance of wires and batteries are usually given a value of
____________________.
7. Short circuits cause the circuit to draw a large amount of ____________________.
8. The value of current is the same at all points in a ____________________ circuit.
9. Each separate resistor in a series circuit creates a(n) ____________________ drop.
10. The total circuit resistance ____________________ as more resistors are added in parallel.
11. A circuit in which current has more than one path is called a ____________________ circuit.
12. A device that protects circuits from high current is a circuit ____________________.
13. In the U.S., the electrical system runs on current that reverses 60 times per second which is referred to as
____________________ current.
Matching
Match each statement as belonging to either a “Series” or “Parallel” circuit.
a. Series
b. Parallel
____ 14. There are multiple paths for current to travel.
____ 15. There is only one path for current to travel.
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____ 16. Current is the same at every point in the circuit.
____ 17. Total battery voltage is equal to the sum of the voltages across each resistor.
____ 18. The voltage across each branch is the same and is equal to the battery voltage.
Short Answer
19. The circuit shown below has 3 identical bulbs connected in series to a battery. If 1 bulb is unscrewed, what
will happen to the 2 remaining bulbs?
20. Give an example of a circuit that uses alternating current and an example of a circuit that uses direct current.
Problem
Three resistors, a 2-ohm, a 4-ohm, and a 6-ohm are connected in series to a 6-volt source. Use this
information to answer the following questions:
21. What is the total resistance of the resistors in the circuit?
22. Three resistors, each with a resistance of 2 ohms, are connected in series to a 6-volt source. What is the
voltage drop across each resistor?
23. A series circuit contains a 9 volt battery and three resistors of 1 ohm, 3 ohms, and 5 ohms. What is the current
in the circuit?
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Three resistors of 1 ohm, 3 ohms and 5 ohms are connected in parallel to a 3-volt source. Use this information
to answer the following questions:
24. What is the current through the 3-ohm resistor?
25. Measured in kilowatt-hours, how much energy does a 4,500-watt water heater element use in 2 hours?
A radio’s label identifies its power rating as 4.0 watts. Assume that the radio is plugged into a 120-volt outlet
as you answer the following questions:
26. The power company charges 10 cents per kilowatt-hour of energy. How much would it cost to run this radio
for 100 hours?
27. In many homes, 75-watt light bulbs are common. If the bulb is connected to a 120-volt source, what is the
resistance to the flow of charge in the bulb?
28. In many homes, 75-watt incandescent light bulbs are common. In one home there are 10 lights, and each light
is on 90 hours per month. If the power company charges 15 cents per kilowatt-hour of energy, how much
money will be saved each month if all the incandescent bulbs are replaced with 25-watt fluorescent bulbs?
Essay
29. In a household parallel circuit, two 100-watt light bulbs operate at 120 volts. If the same bulbs were placed in
a series circuit, would the bulbs appear brighter, dimmer, or remain unchanged? In your answer, describe the
current through the household circuit and the series circuit.
Other
The circuit pictured contains 3 identical light bulbs. They are connected to a voltage source which causes 2
amperes of current to flow through each of the bulbs.
Figure 21-1
30. What is the total resistance of the circuit shown in Figure 21-1?
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Chapter 21 Multi-format Test
Answer Section
MODIFIED TRUE/FALSE
1. ANS: F, parallel circuit
DIF: basic
2. ANS: T
3. ANS: T
4. ANS: F, energy
REF: section 21.1 | section 21.2
DIF: basic
DIF: basic
DIF: intermediate
5. ANS: T
REF: section 21.3
DIF: intermediate
REF: section 21.2
REF: section 21.2
REF: section 21.3
COMPLETION
6. ANS:
zero
0
DIF: basic
7. ANS: current
REF: section 21.1
DIF: basic
8. ANS: series
REF: section 21.1
DIF: basic
9. ANS: voltage
REF: section 21.1
DIF: basic
10. ANS: decreases
REF: section 21.1
DIF: basic
11. ANS: parallel
REF: section 21.2
DIF: basic
12. ANS: breaker
REF: section 21.2
DIF: basic
13. ANS: alternating
REF: section 21.2
DIF:
basic
REF: section 21.3
MATCHING
14. ANS: B
DIF: basic
REF: section 21.1 | section 21.2
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15.
16.
17.
18.
ANS:
ANS:
ANS:
ANS:
A
A
A
B
DIF:
DIF:
DIF:
DIF:
basic
basic
basic
basic
REF:
REF:
REF:
REF:
section 21.1 | section 21.2
section 21.1 | section 21.2
section 21.1 | section 21.2
section 21.1 | section 21.2
SHORT ANSWER
19. ANS:
The other two bulbs will go out when one bulb is unscrewed.
DIF: basic
REF: section 21.1
20. ANS:
Answers may vary. Correct answers include:
Alternating current: household circuits
Direct current: battery circuits
DIF:
intermediate
REF: section 21.3
PROBLEM
21. ANS:
resistancetotal = resistance1 + resistance2 + resistance3
resistancetotal = 2 ohms + 4 ohms + 6 ohms
resistancetotal = 12 ohms
DIF: intermediate REF: section 21.1
22. ANS:
The voltage drop across each resistor in series is part of the total drop. Since each resistor has the same value,
the drop across each will be an equal part of the total;
voltagetotal ÷ 3 resistors = 6 volts ÷ 3 = 2 volts per resistor.
DIF: intermediate REF: section 21.1
23. ANS:
total resistance = 1Ω + 3Ω + 5Ω = 9 Ω
current = voltage ÷ resistance
current = 9 V ÷ 9Ω = 1 amp
DIF: intermediate REF: section 21.1
24. ANS:
current = voltage ÷ resistance (or) I = V ÷ R;
current = 3 volts ÷ 3 ohms
current = 1 amp
DIF:
intermediate
REF: section 21.2
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25. ANS:
energy = power × time
kilowatt-hours = watts ×
1 kilowatt
× hours
1,000 watts
ÊÁ 1 kilowatt ˆ˜
˜˜˜ × 2 hours
energy = ÊÁË 4,500 watts ˜ˆ¯ × ÁÁÁÁ
˜
Ë 1,000 watts ¯
energy = 9 kWh
DIF: intermediate REF: section 21.3
26. ANS:
1 kilowatt
12 watts ×
= 0.012 kilowatts
1000 watts
energy = power × time
energy = 0.012 kW × 100 hrs
energy = 1.2 kWh
cost = energy × rate
cost = 1.2 kWh × $0.10/kWh
cost = $0.12
DIF: advanced
REF: section 21.3
27. ANS:
power = current × voltage (or) P = I × V
current = power ÷ voltage
current = 75 watts ÷ 120 volts
current = 0.625 amperes
resistance = voltage ÷ current (or) R = V ÷ I
resistance = 120 volts ÷ 0.625 amperes
resistance = 192 ohms
DIF:
advanced
REF: section 21.3
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28. ANS:
cost per month = cents/kWh × kWh/month; kWh = (watts ÷ 1000) × hours
For each incandescent light:
kWh/month = (75 W ÷ 1000) × 90 hr/month = 6.75 kWh/month
cost per month = 15 cents/kWh × 6.75 kWh = 101.25 cents, or $1.01 per bulb.
total cost per month = cost per bulb × number of bulbs
total cost per month = $1.01/bulb × 10 bulbs = $10.10.
For each fluorescent light:
kWh/month = (25 W ÷ 1000) × 90 hr/month = 2.25 kWh/month
cost per month = 15 cents/kWh × 2.25 kWh = 33.75 cents, or $0.34 per bulb.
total cost per month = cost per bulb × number of bulbs
total cost per month = $0.34 /bulb × 10 bulbs = $3.40.
cost savings per month = cost for incandescent bulbs - cost for fluorescent bulbs
cost savings per month = $10.10 - $3.40 = $6.70
DIF:
advanced
REF: section 21.3
ESSAY
29. ANS:
The bulbs would appear dimmer. The household circuit is a parallel circuit, and the current flows through one
bulb in each branch. In series circuits, the current must flow through two bulbs instead of one. It would draw
less current which would be split between two bulbs, resulting in much dimmer light from each bulb.
DIF:
advanced
REF: section 21.1 | section 21.2
OTHER
30. ANS:
Resistance 1 + 2 =
R 1 × R 2 1.5 ohms × 1.5 ohms
=
= 0.75 ohm
R 1 + R 2 1.5 ohms + 1.5 ohms
Resistance 1 + 2 + 3 =
DIF:
advanced
R 1 + 2 × R 3 0.75 ohm × 1.5 ohm
=
= 0.5 ohm
R 1 + 2 + R 3 0.75 ohm + 1.5 ohm
REF: section 21.2
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Chapter 21 Multi-format Test [Answer Strip]
A 16.
_____
A 17.
_____
B 18.
_____
F
_____
1.
T
_____
2.
T
_____
3.
F
_____
4.
T
_____
5.
B 14.
_____
A 15.
_____
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