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IV.
Score________
[20 points total] Assume that all bulbs are identical, and all batteries are ideal and identical.
A. Consider circuit I at right.
i.
[5 pts] Suppose the switch is now closed. Would the brightness of
bulb B be greater than, less than, or equal to what it was when the
switch was open? Explain.
Equal to. The magnitude of the potential difference across bulb B
equals that of battery 1, ΔVbulb B = ΔVbatt 1 . This follows from
Kirchhoff’s loop rule applied to loop L shown at right. The conclusion
is true whether the switch is open or closed, since nothing changes in
loop L. Since the potential difference across bulb B does not change,
neither does its brightness.
L’
2
A
1
B
L
Circuit I
ii. [5 pts] After the switch has been closed, is the brightness of bulb A greater than, less than, or
equal to the brightness of bulb B? If any bulbs are not lit, state so explicitly. Explain.
Less than; bulb A is not lit. Apply Kirchhoff’s loop rule to loop L’, shown above right. Since
batteries 1 and 2 are identical but oppositely oriented in loop L’, there is no potential drop or
gain across these two elements in the loop. Only bulb A remains so it must have zero potential
difference across it. Therefore, it does not light. Alternatively, one could have used the
outermost loop and the fact that the potential drop across bulb B (for counterclockwise current)
equals the potential drop across one battery.
B. A third battery has been added to circuit I, and the switch is closed. This
gives circuit II shown at right.
i.
[5 pts] In Circuit II, is the brightness of bulb A greater than, less than,
or equal to the brightness of bulb B? If any bulbs are not lit, state so
explicitly. Explain.
L’
2
A
1
B
Equal to. Kirchhoff’s loop rule applied to loop L shown at right gives
ΔVbulb B = ΔVbatt 1 . Kirchhoff’s loop rule applied to loop L’, also
L
shown at right, gives ΔVbulb A = −ΔVbatt 1 + ΔVbatt 3 + ΔVbatt 2 . (The
relative minus sign between the potential difference across battery 1
Circuit II
and that of the other two batteries comes from the fact that battery 1 is
oppositely oriented to the other two batteries in loop L’.) Since all batteries are identical, the
previous equation reduces to ΔVbulb A = ΔVbatt 1 . Since both bulbs have the same potential
difference across them, they are equally bright.
ii.
[5 pts] In Circuit II, is the magnitude of the current flowing through battery 3
greater than, less than, or equal to the current flowing through battery 1? If any
currents are zero, state so explicitly. Explain.
3
A
IA
I1
Greater than; there is no current through battery 1. Bulbs A and B are equally
IB
bright so they must have equal nonzero currents flowing through them, IA = IB .
1
B
Note that the current flowing through bulb A is the same current flowing through
battery 3. (The current through bulb A must flow in the direction indicated in the
diagram of the junction at right since the potential gain across batteries 2 and 3 is greater in
magnitude than the potential drop across battery 1. Also, the current through bulb B must flow in
the direction indicated or else Kirchhoff’s loop rule fails along the outermost loop of Circuit II.)
Suppose that the current through battery 1, if it exists, flows in the direction indicated above.
Then, conservation of current requires that IA + I1 = IB . Our two equations are consistent if and
only if I1 = 0 . Thus, no current flows through battery 1.
Physics 122B, Autumn 2009
Exam 3
EM-UWA122B094T-E3