PHYS 202
Exercise Set #12
W 2010, due March 5
1
1. (6 points) In the circuit shown, the 15-Ω resistor drops a voltage of 4.0 V. (A) Find
the battery voltage. You should not have to calculate any currents.
(B) Determine the power dissipated by the 40-Ω resistor.
(C) How would your answers to (A) and (B) be affected if we removed the 60-Ω resistor?
40 Ω
+
60 Ω
V
−
30 Ω
15 Ω
(A) The parallel 30- Ω and 15- Ω resistor may be combined to an equivalent 10- Ω
resistor. We know that the voltage drop across it is 4.0 V. This resistor is in series
with the 40- Ω resistor, so they both carry the same current. Since V = IR, and I
is the same through both, the voltage drop must be proportional to resistance. The
40- Ω resistor has 4 times the resistance of the 10- Ω resistor, so it must drop 4 times as
much, or 16 V. The total drop across these two is then 20 V. These are hooked directly
to the battery, so the battery voltage must also be 20 V .
(B) We use P = V 2 /R:
P =
16 V
V2
=
= 6.4 W
R
40 Ω
(C) They would not be affected at all.
PHYS 202
Exercise Set #12
W 2010, due March 5
2
2. (5 points) Use Kirchoff’s Rules to find the voltage V (of the middle battery) needed
so that no current flows through the 20-Ω resistor, in the circuit below.
I
A
−
+
−
+
16Ω
4Ω
20 Ω
−
V
9.0 V
−
B
+
−
+
+
6.0 V
I have labeled the top and bottom nodes “A” and “B” in the diagram. If no current
flows through the 20- Ω resistor, then there is no voltage drop across that resistor. That
means that the battery voltage must be equal to the voltage difference between those
two nodes. If we can find the potential of “A” relative to “B”, we have the battery
voltage V.
First we have to find the current I. All the current is around the outside loop. The
total voltage is 15.0 V, and the total resistance is 20 Ω. Therefore the current is I =
(15.0 V)/(20 Ω) = 0.75 A.
Now that we know that, we can find the voltage drop across either of the resistors,
which will allow us to get what we want. The drop across the 16- Ω resistor is V = IR
= (0.75 A)(16 Ω) = 12 volts. I have put in the circuit which side is positive relative to
the other. Now, follow the path from node B along the left-hand loop. First we see a
+9.0 V rise across the battery, follwed by a -12 V drop across the resistor. This adds
up to -3.0 V. Therefore, V = 3.0 V . (The sign is not used when you give a battery
voltage. The direction is indicated by the circuit diagram. You may verify for yourself
that going around the right-hand loop gives the same answer.
PHYS 202
Exercise Set #12
W 2010, due March 5
3
3. (5 points) In the circuit below, the voltage drops across bulbs A and C are given. The
battery voltage is 14 V. Find the voltage drop across each of the other bulbs.
A
+
2V
D
B
C
1V
E
14 V
−
F
Bulbs A and D are in series and are identical bulbs, so they have the same voltage
drop: VD = 2 V. The voltage drops around loop battery-A-D-F must add up to zero.
Start at the bottom. There is a +14-V rise, a -2 V drop across A, a -2 V drop across
D, and some unknown drop across F. The sum must be zero, so VF = 10 V. (The top
side is positive relative to the lower side.)
B and C are in parallel, so they must have the same drop: VB = 1.0 V. The drop across
the bulbs A and D (together) is 4.0 V. This must be the same as the drop along the
path B - E (or along C - E). This is because the A-D combination is in parallel with
the BC-E network. Since we know that B drops 1.0 V, E must drop 3.0 V, so that the
total is 4.0 V.
Summary:
D:
B:
E:
F:
2V
1V
3V
10 V
PHYS 202
Exercise Set #12
W 2010, due March 5
4
4. (5 points) Find the voltages across resistors A, C, E, G, and H in the circuit below.
The voltage drops across resistors B, D, F, and J are given.
A
+
+
B
+
4V
E
12 V
−
C
F
D
+ 2V
G
H
5V
+
J
4V
We use Kirchoff’s loop rule. As an example, look at the loop H-F-J. Start at the node
below H, and go clockwise. First a drop or rise VH , then a drop of -2 V across F, then
a drop of -4 V across J. These must add up to zero. F + J gives a total drop of -6 V,
so we must have a rise of 6 V across H. I indicate the signs in the diagram.
Now look at the loop C-D-F. Starting in the “center” node, we have VC , then a -5 V
drop across D, then a +2 V rise across F. These must add up to zero, so we must have
a 3 V rise across C. We indicate the signs in the diagram.
If VC is 5 V, let’s look at the B-D-E loop. Starting at the node between B and G, we
have a +4 V rise, then a -5 V drop across C, and a rise or drop VE . Clearly VE must
be 1 V, with the left side positive relative to the right side.
Now we can do the loop G-E-H. Start at the node between B and G and go clockwise.
First there is a 1 V rise across E, then a -6 V drop across H, then a rise across G. To
make things add up, the rise across G must be 5 V.
Now, so find VA , we can use the path A-B-G, A-C-H, or A-D-J, plus the battery. They
will all give the same answer. We could have used the path battery-A-D-J right at the
first, before we knew the other voltages. Start at the negative terminal of the battery
and go clockwise. First a 12 V rise, the a change VA , then a -5 V drop, then a -4 V
drop, and we are back. The voltage changes add up to +3 V. Therefore to make the
sum zero, there must be a -3 V change (drop) across A.
Summary:
A:
C:
E:
G:
H:
3
3
1
5
6
V
V
V
V
V
PHYS 202
Exercise Set #12
5
W 2010, due March 5
5. (5 points) In the network shown, some of the resistors have a resistance R, and others
have resistance 2R. Determine the equivalent resistance between points A and B, in
terms of R. (Hint: work from the end opposite A and B, using series/parallel reduction.
You will quickly see a pattern.)
A
B
2R
R
2R
R
2R
2R
Start at the bottom and work up. The bottom 2R parallel pair gives an equivalant R
resistance. This is in series with the left-hand R resistor, giving a sum 2R. This is in
parallel with the next 2R resistor, giving resistance R again. This combination is in
series with the top-left R, adding up to 2R. Finally, this is in parallel with the top 2R
resistor, giving a total equivalent R .
A
B
2R
R
2R
B
2R
2R
ETC.
R
R
R
A
B
A
2R
2R
R
2R
R
2R
2R
PHYS 202
Exercise Set #12
W 2010, due March 5
6
6. (5 points) In the circuit of the previous problem, let R = 10Ω. Suppose we hooked an
80.0-volt battery between points A and B. Find the current through the “lowest” 2R
resistor. (That is, the resistor farthest from the battery.)
The total current will be (80 V)/(10 Ω) = 8.0 A. This current is called I in the diagram,
top left. When it reaches the first node, there are two paths to split into: the 2R resistor
and the rest of the network, which also has equivalent resistance 2R. So the current
will split evenly. The current continuing below the node encounters the next node,
where again there are two paths, both with resistance 2R. It therefore splits in half
again. This process continues until we get to the bottom resistor. As the diagram
indicates, its current is 1/8 of the original 8 amps, or 1.0 A .
V
+
I
2R
I/2 R
I/2
Equivalent resistance 2R
2R
I/4
R
I/4
Equivalent resistance 2R
2R
Equivalent resistance R
I/8
I/8
2R