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Day 16: Review
Name ________________________
A
C
D
E
B
For each question, assume that the batteries, bulbs and wires are ideal.
The students should use analogies, rules, symmetry, conservation and/or experiences in their explanations.
Comparing Brightness
1) In the circuit on the left
A) A is as bright as B
B) A is brighter than B
C) B is brighter than A
A) There is no difference between the bulbs or their situation. The order of the loads does not matter; it is not
like being first in line at a buffet.
2) Comparing the left and middle circuit.
A) A is as bright as C
B) A is brighter than C
C) C is brighter than A
C) The single bulb gets the full voltage drop, whereas it is split for the other bulbs.
3) Comparing the middle and the right circuits
A) C is as bright as D
B) C is brighter than D
C) D is brighter than C
A) All three bulbs get the same voltage drop. The second circuit is putting out
Comparing Current
4) Bulb A’s current is
A) equal to bulb B’s
B) double bulb B’s
C) half bulb B’s
A) No current can be lost, so all that goes through A must also go through B. Have students draw the ammeter
connected properly in the circuit and label the current’s value. I = 0.45 A.
5) Bulb A’s current is
A) equal to bulb C’s
B) double bulb C’s
C) half bulb C’s
C) All current must pass through two bulbs in the left circuit so there is double the resistance and so half the
current. In the center I = 0.9 A.
6) The right battery’s current is
A) equal to the center battery’s
B) double the center battery’s
C) half the center battery’s
B) Each bulb gets the same current and the battery on the right supplies two bulbs. I = 1.8 A. Notice how the
current of the left battery is ¼ that of the right battery. Have the students lengthen the direction arrows they
drew to represent the relative amounts of current.
Comparing Voltage
7) The voltage across bulb A is
A) 4.5 V
B) 9.0 V
C) 2.25 V
B) The voltage up is split into two equal parts for the voltage down. Voltage is like the height at a ski hill. You
come down as much as the lift brings you up. The voltage is twice 4.5 V = 9.0 V
8) The voltage across bulb C is
A) 4.5 V
B) 9.0 V
C) 2.25 V
B) The batteries are identical. The voltage of the bulb is equal and opposite that of the battery.
9) The voltage across bulb D is
A) 4.5 V
B) 9.0 V
C) 2.25 V
B) Each bulb has a voltage equal and opposite that of the battery. This is like a 9.0V ski lift and two identical
routes down.
Calculating Resistance and Power in the circuit. The resistance of 1 bulb is 10 .
10) What is the resistance of the circuit in the middle? on the left? On the right?
A) 5 
B) 10 .
C) 20 
B) middle,
C) left
A) right
11) What is the power produced by the middle circuit? right circuit? left circuit?
A) 4.05 W
B) 8.1 W
C) 16.2 W
B) middle,
C) right,
A) left This answers the question about which is the least bright circuit, because
brightness refers to the visible power output. The left circuit has two bulbs at ¼ the power (brightness) of the
middle bulb. Therefore, the left circuit has half the brightness of the right one.
12) Suppose these were real batteries and do not last forever. The left battery will last
A) ½ as long as the right
B) the same as the right
C) twice as long as the right
None of the answers are right, though C is the best answer. The left is ¼ as bright as the one on the right.
Power is the rate of using energy and the one on the right is using it 4 times faster. The one on the left lasts 4
times as long.
Changing Connections
13) You unscrew bulb A. What happens to bulb B?
A) It goes out
B) it stays the same C) it gets twice as bright
D) it gets half as bright
A) The circuit is broken
14) You unscrew bulb E. What happens to bulb D?
A) It goes out
B) it stays the same C) it gets twice as bright
D) it gets half as bright
B) It stays the same. Note: The real bulb got a bit brighter, but not twice. This is because the battery could not
double the current easily.
15) You connect a wire between the two ends of one bulb in each circuit. What happens?
The bulb goes out because the wire has much lower resistance. The other bulb in the left circuit gets much
brighter. The other two circuits burn up because all the current goes through the wire. It is a short circuit.
Textbook Reinforcement
Pages 588 – 593, Unit E Review : Questions 1-10, 39 -53, 55-62, 65-67, 79-, 80, 83.
Pages 594 – 595, Self-Quiz: Questions 2-24 .The answers for many of these questions are on page 664.
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Go through some of the multiple-choice electrostatic and issues questions to provide balance to the
sheet’s circuit focus.
Provide lots of time in class for individual or quiet pair studying.
Make most of the test from these textbook questions with small alterations.
Take a second day for review, with whole class work and then individual or quiet pair studying.
Email the parents so that they know that no homework is being collected – but there is lots of studying
to be done.
Take up some questions involving diagrams and explanatory paragraphs to go over what is needed in
paragraphs and diagrams using the 4 C’s.