PHYS 202
Exercise Set #11
W 2010, due March 1
1
1. (6 points) (A) Rank the bulbs in the circuit below according to brightness, starting
with the brightest. (B) Suppose the battery voltage is 6.0 V and bulb A drops 4.0 V.
Find the voltage drops across each of the other bulbs.
A
B
+
−
D
C
E
(A) The current splits evenly between A and B. When the current splits for C-(D/E),
more current goes through C than through D-E. Therefore C is brightest. A and B are
next, and finally D and E, which pass less than half the total current.
C>A=B>D=E
(B) B must drop the same as A, since they are in parallel. So VB = 4.0 V
Look at the battery-A-C loop. By Kirchoff’s loop rule, the sum of voltage drops across
A and C must add up to the battery voltage. Since A drops 4.0 V, C must drop 2.0
V, so that these add up to 6.0 V.
The sum of D and E voltage drops must add up to that of C, and they are equal since
they are identical bulbs and are in series. Thus, each drops 1.0 V.
Summary:
B: 4.0 V
C: 2.0 V
D: 1.0 V
E: 1.0 V
NOTE: I made a mistake on this. If these are identical bulbs, and C is brightest, then
of course C must have the greatest voltage drop. Sorry!
PHYS 202
Exercise Set #11
W 2010, due March 1
2
2. (6 points) (A) Determine the total current supplied by the battery in the circuit below.
(B) Determine the voltage drop across the leftmost 10-Ω resistor. (C) Determine the
voltage drop between points B and A in the circuit. (All of these calculations are easy
enough to do in your head.)
5Ω
5Ω
+
8.0 V
10 Ω
−
B
10 Ω
10 Ω
A
(A) The easiest way is to find the equivalent resistance of the network. Start from the
place farthest from the battery. The two 10- Ω resistors in parallel reduce to a single
5- Ω resistor. This is in series with a 5- Ω resistor, giving a 10- Ω resistor. This is in
parallel with the middle 10- Ω resistor, giving an equivalent 5- Ωresistor. Finally, this
is in series with the left 5- Ω resistor, giving a total resistance of 10 Ω. Thus
I=
V
8.0 V
=
= 0.80 A
R
10 Ω
(B) As we showed above, the rightmost 4 resistors reduce to a single 5- Ω resistor. So,
we have two 5- Ωresistors in series. The voltage drop across each of these is then the
same, 4.0 V. This is the voltage from the top node (black dot) to the bottom node.
Now consider the three rightmost resistors. This looks like two 5- Ω resistors in series
(since the two 10- Ω resistors reduced to a 5- Ω resistor). Each of these 5- Ω resistors
must have the same voltage drop, and the sum adds up to 4.0 V. Therefore each has a
drop of 2.0 V. The bottom 5- Ω resistor is between the A-B nodes of the question, so
2.0 V is the answer.
PHYS 202
Exercise Set #11
W 2010, due March 1
3
3. (4 points) Find the equivalent resistance for this network, between points A and B
2Ω
Α
5Ω
10 Ω
5Ω
Β
4 Ω
20 Ω
We work from “inside out.” The 5- Ω and 20- Ω resistors may be combined:
1
1
1
+
=
Req
5 Ω 20 Ω
= 0.25 Ω−1
Req = 4.0 Ω
This 4- Ω resistor is in series with another 4- Ω resistor, giving 8 Ω on the “bottom”
leg. The 2- Ω and 10- Ω resistors add to give a 12- Ω resistor. So now we have a 12- Ω
and8- Ω resistor in parallel. We find this equivalent:
1
1
1
+
=
Req
12 Ω 8 Ω
= 0.2083 Ω−1
Req = 4.8 Ω
Finally, this 4.8- Ω resistor is in series with the 5- Ω resistor, giving a total resistance
of 9.8 Ω .