Name ______________________________________Student ID _________________ Score_______
last
first
III. [20 points] All bulbs in the circuit at right are identical and all
batteries are ideal and identical.
3
1
A. [8 pts] Rank the bulbs according to brightness in order
from brightest to dimmest. Explain your reasoning.
4
5
2
From current conservation, 1=2 and 3=6. (Each pair is in series
6
so all the current through 1 goes through 2. The same is true for 3
and 6.) By symmetry, 4=5. Moreover, 4 and 5 each has half the
current through 3 (or 6) so 3=6 > 4=5. Finally, since the left branch has less resistance than the right
branch, the current through bulb 1 and through bulb 2 is greater than that through bulbs 3 and 6. Thus
1=2 > 3=6 > 4=5.
B. Bulb 5 is replaced by a battery as shown at right.
i.
[6 pts] Does the brightness of bulb 6 increase,
decrease, or remain the same? Explain your
reasoning.
3
1
4
2
Call the potential difference across each identical battery V0.
Using the loop that contains all three batteries, we see that the
6
sum of the potential differences across bulbs 3 and 6 is 3V0.
Since bulbs 3 and 6 are identical and connected in series, they
each have 1.5V0. In the original circuit, the potential difference across bulb 6 was less than V0. Hence the
potential difference across, and therefore the brightness of, bulb 6 has i ncr e as e d.
ii. [6 pts] Rank the bulbs according to brightness in order from brightest to dimmest. Explain
your reasoning.
As shown in (i), V3=V6=1.5V0. V4=V0 because its terminals are connected directly to the terminals of a
battery. The voltage across 1 and 2 together is 2V0 since they are connected across two batteries in
series. Since they are identical, V1=V2=V0. Hence V3=V6>V1=V2=V4, and the brightness ranking is
3=6>1=2=4.
Physics 122, Summer 2005
Exam 3
August 2005
EM-UWA122A053T-E3(CK2,CK3)