Homework Assignment #4

Homework Set 4: Solutions
Due: Wednesday, September 22, 2010
Chapter 19: Questions
(1) 6 points Explain why birds can sit on power lines safely, whereas leaning a metal ladder up against
a power line to fetch a stuck kite is extremely dangerous.
The birds are safe because they are not grounded. Both of their legs are essentially at the same voltage
(the only difference being due to the small resistance of the wire between their feet), and so there is no
current flow through their bodies since the potential difference across their legs is very small. If you lean
a metal ladder against the power line, you are making essentially a short circuit from the high potential
wire to the low potential ground. A large current will flow at least momentarily, and that large current
will be very dangerous to anybody touching the ladder.
(4) 8 points Two lightbulbs of resistance R 1 and R 2 (R 2 > R 1 ) are connected in series. Which is brighter
What if they are connected in parallel Explain.
If the lightbulbs are in series, each will have the same current. The power dissipated by the bulb as heat
and light is given by P = I 2 R. Thus the bulb with the higher resistance (R 2 ) will be brighter. If the
bulbs are in parallel, each will have the same voltage. The power dissipated by the bulb as heat and light
is given by P = ∆V 2 /R. Thus the bulb with the lower resistance (R 1 ) will be brighter.
(7) 6 points If two identical resistors are connected in series to a battery, does the battery have to supply
more power or less power than when only one of the resistors is connected Explain.
The power supplied by the battery is the product of the battery voltage times the total current flowing
from the battery. With the two resistors in series, the current is half that with a single resistor. Thus
the battery has to supply half the power for the two resistors than for the single resistor.
(19) 6 points In an RC circuit, current flows from the battery until the capacitor is completely charged. Is
the total energy supplied by the battery equal to the total energy stored by the capacitor
If not, where does the extra energy go
The total energy supplied by the battery is greater than the total energy stored by the capacitor. The
extra energy was dissipated in the form of heat in the resistor while current was flowing. That energy
will not be “recovered” during the discharging process.
(22) 8 points Explain why an ideal ammeter would have zero resistance and an ideal voltmeter infinite
resistance.
Chapter 19: Problems
An ideal ammeter would have zero resistance so that there was no voltage drop across it, and so it would
not affect a circuit into which it was placed. A zero resistance will not change the resistance of a circuit
if placed in series, and hence would not change the current in the circuit. An ideal voltmeter would have
infinite resistance so that it would draw no current, and thus would not affect a circuit into which it was
placed in parallel. An infinite resistance will not change the resistance of a circuit if placed in parallel,
and hence would not change the current in the circuit.
(2) 8 points (I) Four 1.5-V cells are connected in series to a 12-Ω lightbulb. If the resulting current
is 0.45 A, what is the internal resistance of each cell, assuming they are identical and
neglecting the wires
Using Kirchhoff’s loop rules, we find that 4ε−4Ir−IR = 0 where I = 0.45 A, ε = 1.5 V, R = 12 Ω, and
r is the internal resistance of each cell. Solving for r we find that
r = ε I − R 4 = 1.5 V
0.45 A − 12 Ω = 0.33 Ω.
4

(8) 8 points (I) Given only one 25-Ω and one 35-Ω resistor, list all possible values of resistance that can
be obtained.
The possible resistances are each resistor considered individually (25 Ω and 35 Ω), the series combination
(60 Ω), and the parallel combination (15 Ω).
(26) 10 points For the circuit shown in Fig. 19-46, find the potential difference between points a and b.
Each resistor has R = 75 Ω and each battery is 1.5 V.
Using Kirchhoff’s loop rule, we find that 2ɛ−4IR = 0 where ɛ = 1.5 V and R = 75 Ω. Solving for the
current I in the circuit we find that
I = ε
2R .
Therefore, the potential difference between points a and b is
( ε
)
∆V ab = ε−2IR = ε−2 R = 0.
2R
Thus points a and b are at the same electric potential.
(32) 14 points (III) (a) Determine the currents I 1 , I 2 , and I 3 in Fig. 19-50. Assume the internal resistance
of each battery is r = 1.0 Ω. (b) What is the terminal voltage of the 6.0-V battery
(a) Since there are three currents to determine, there must be three independent equations to determine
those currents. One comes from Kirchhoff’s junction rule
I 1 = I 2 +I 3 .
Another equation comes from Kirchhoff’s loop rule applied to the top loop
which simplifies to
(12.0 V)−I 2 (1.0 Ω)−I 2 (10 Ω)−I 1 (12 Ω)+(12.0 V)−I 1 (1.0 Ω)−I 1 (8.0 Ω) = 0,
24 = 21I 1 +11I 2 .
The final equation comes from Kirchhoff’s loop rule applied to the bottom loop
which simplifies to
(12.0 V)−I 2 (1.0 Ω)−I 2 (10 Ω)+I 3 (18 Ω)+I 3 (1.0 Ω)−(6.0 V)+I 3 (15 Ω) = 0
6 = 11I 2 −34I 3 .
Substituting I 1 = I 2 +I 3 into the top loop equation yields
24 = 32I 2 +21I 3 .
Multiplying both sides of the bottom loop equation by (32/11) and subtracting it from this equation
yields a result for I 3
I 3 = 0.0546A = 0.055 A.
With this result, we also find that
I 1 = 0.77 A and I 2 = 0.71 A.
Therefore, I 1 = 0.77 A, I 2 = 0.71 A and I 3 = 0.055 A.
(b) The terminal voltage of the 6.0-V battery is
∆V = ε−I 3 r = (6.0 V)−(0.0546 A)(1.0 Ω) = 5.9 V.

(38) 10 points (II) If 26.0 V is applied across the whole network of Fig. 19-52, calculate the voltage across
each capacitor.
The full voltage is across the 2.00 µF capacitor, and so ∆V 2.00 = 26.0 V. To find the voltage across the
two capacitors in series, we first note that they have the same charge (conservation of charge). Therefore
Therefore,
and
26.0 V =
Q
3.00×10 −6 F + Q
4.00×10 −6 F
∆V 3.00 = 4.46×10−5 C
3.00×10 −6 F = 14.9 V
∆V 4.00 = 4.46×10−5 C
4.00×10 −6 = 11.1 V.
F
so Q = 4.46×10 −5 C.
(52) (16 points) (III) Two resistors and two uncharged capacitors are arranged as shown in Fig. 19-58.
Then a potential difference of 24 V is applied across the combination as shown. (a) What
is the potential at point a with switch S open (Let V = 0 at the negative terminal of the
source.) (b) What is the potential at point b with the switch open (c) When the switch
is closed, what is the final potential of point b (d) How much charge flows through the
switch S after it is closed
(a) Using Kirchhoff’s loop rule we find that ∆V − IR 1 −IR 2 = 0 where ∆V = 24 V, R 1 = 8.8 Ω and
R 2 = 4.4 Ω. Solving for I we find that
I = ∆V 24 V
= = 1.818 A.
R 1 +R 2 8.8 Ω+4.4 Ω
Therefore, the voltage at point a is the voltage across the 4.4 Ω resistor:
V a = IR 2 = (1.818 A)(4.4 Ω) = 8.0 V.
(b) Using Kirchhoff’s loop rule we find that ∆V − Q C 1
− Q C 2
= 0 where ∆V = 24 V, C 1 = 0.48 µF and
C 2 = 0.24 µF. The charge Q on each capacitor is the same due to conservation of charge. Solving for Q
we find that
Q = C 1C 2 (0.48 µF)(0.24 µF)
∆V = (24 V) = 3.84 µC.
C 1 +C 2 0.48 µF+0.24 µF
The voltage at point b is the voltage across the 0.24 µF capacitor
V b = Q C 2
=
3.84 µC
= 16 V.
0.24 µF
(c) The switch is now closed. After equilibrium has been reached a long time, there is no current flowing
in the capacitors, and so the resistors are again in series, and the voltage of point a must be 8.0 V.
(d) Part (c) implies that the voltage across C 1 is 16 V while the voltage across C 2 is 8.0 V. Therefore,
the charge on each capacitor is
Q 1 = C 1 ∆V 1 = (16 V)(0.48 µF) = 7.68 µC
Q 2 = C 2 ∆V 2 = (8.0 V)(0.24 µF) = 1.92 µC.
When the switch was open, point b had a net charge of 0, because the charge on the negative plate C 1
had the same magnitude as the charge on the positive plate C 2 . With the switch closed, these charges
are not equal. The net charge at point b is the sum of the charge on the negative plate of C 1 and the
charge on the positive plate of C 2 : Q b = Q 2 −Q 1 = −5.76 µC. Thus, 5.76 µC has passed through the
switch.