Physics 9
Spring 2012
TA:
NAME:
Quiz 6 - Solutions
1. You have three 12 Ω resistors. Draw diagrams showing how you could arrange all three
so that their equivalent resistance is
(a) 4.0 Ω
(b) 8.0 Ω
(c) 18 Ω, and
(d) 36 Ω.
2. The figure shows six identical bulbs connected to
an ideal battery. All the bulbs are glowing. Rank
in order, from brightest to dimmest, the brightness of bulbs A to F. Explain.
————————————————————————————————————
Solution
1. The four diagrams are seen in the figure below.
1
(a) In the first diagram, all three resistors are in parallel. So, the equivalent resistance
is
1
1
1
1
1
1
1
1
=
+
+
=
+
+
= ,
Reqv
R1 R2 R3
12 12 12
4
which gives Reqv = 4 Ω.
(b) In this case the top two resistors are in series and are both in parallel with the
bottom resistor. The top resistors add to give R1 + R2 = 12 + 12 = 24. This
equivalent resistance combines with the resistor on the bottom to give
1
1
1
3
1
=
+
=
= ,
Reqv
24 12
24
8
and so Reqv = 8 Ω.
(c) Next, we have a two resistors in parallel in series with a third resistor. The two in
1
1
1
= 12
+ 12
, giving an equivalent resistance of 6 Ω for
parallel combine to give Reqv
that branch of the circuit. Adding this in series with the other resistance gives
Reqv = 6 + 12 = 18 Ω.
(d) Finally, we just have three resistors in series, which all add together to give
Reqv = R1 + R2 + R3 = 12 + 12 + 12 = 36 Ω.
2. The brightness of the bulbs depends on the power dissipated in the bulb, which depends
on the current. So, whichever bulb has the highest current has the highest power
dissipated, and so is the brightest. Now, because bulb A is in parallel with the rest of
the bulbs, the current from the battery is split between bulb A, and all the rest of the
bulbs! So, A is brightest.
The current next passes through bulb D before splitting up to the rest of the bulbs,
and so D is next brightest. Next, the current passes through the next set of bulbs.
Bulbs C and E are in parallel with each other, so they are equally bright. The current
through F is the sum of the currents through C, E, and B, and so F is brighter than
C and E. Now, B is in parallel with C, E, and F. But, since the CEF combination has
a higher resistance than B, more current passes through B than the others since they
are all at the same potential. So, we have for brightness, A > D > B > F > C = E.
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