31.11. Model: Assume ideal connecting wires and an ideal battery for which ∆Vbat = E.
Visualize: Please refer to Figure Ex31.11. We will choose a clockwise direction for I. Note that the choice of the
current’s direction is arbitrary because, with two batteries, we may not be sure of the actual current direction. The
3 V battery will be labeled 1 and the 6 V battery will be labeled 2.
Solve: (a) Kirchhoff’s loop law, going clockwise from the negative terminal of the 3-V battery is
∆Vclosed loop = ∑ ( ∆V )i = ∆Vbat 1 + ∆VR + ∆Vbat 2 = 0
i
⇒ +3 V – (18 Ω) I + 6 V = 0 ⇒ I =
9V
= 0.50 A
18 Ω
Thus, the current through the 18 Ω resistor is 0.50 A. Because I is positive, the current is left to right (i.e., clockwise).
(b)
Assess: The graph shows a 3 V gain in battery 1, a −9 V loss in the resistor, and a gain of 6 V in battery 2. The
final potential is the same as the initial potential, as required.
31.15. Model: The 100 W rating is for operating at 120 V.
Solve: A standard bulb uses ∆V = 120 V. We can use the power dissipation to find the resistance of the filament:
∆V 2
∆V 2 (120 V )
=
= 144 Ω
⇒ R=
R
P
100 W
2
P=
But the resistance is related to the filament’s geometry:
R=
ρL
A
=
ρL
ρL
⇒r=
=
π r2
πR
The filament’s diameter is d = 2r = 23.6 µm.
( 9.0 × 10
−7
)
Ω m ( 0.070 m )
π (144 Ω )
= 1.18 × 10 −5 m = 11.8 µ m
31.31. Model: The connecting wires are ideal with zero resistance.
Solve:
For the first step, the resistors 30 Ω and 45 Ω are in parallel. Their equivalent resistance is
1
1
1
=
+
⇒ Req 1 = 18 Ω
Req 1 30 Ω 45 Ω
For the second step, resistors 42 Ω and Req 1 = 18 Ω are in series. Therefore,
Req 2 = Req 1 + 42 Ω = 18 Ω + 42 Ω = 60 Ω
For the third step, the resistors 40 Ω and Req 2 = 60 Ω are in parallel. So,
1
1
1
=
+
⇒ Req 3 = 24 Ω
Req 3 60 Ω 40 Ω
The equivalent resistance of the circuit is 24 Ω.
31.37. Model: Assume ideal wires as the capacitors discharge through the two 1 kΩ resistors.
Visualize: The circuit in Figure Ex37.37 has an equivalent circuit with resistance Req and capacitance Ceq.
Solve: The equivalent capacitance is
1
1
1
⇒ Ceq = 1 µF
=
+
Ceq 2 µ F 2 µ F
and the equivalent resistance is Req = 1 kΩ + 1 kΩ = 2 kΩ. Thus, the time constant for the discharge of the
capacitors is
τ = ReqCeq = (2 kΩ)(1 µF) = 2 × 10−3 s = 2 ms
Assess: The capacitors will be almost entirely discharged 5τ = 5 × 2 ms = 10 ms after the switch is closed.
31.38. Model: Assume ideal wires as the capacitors discharge through the two 1 kΩ resistors.
Visualize: The circuit in Figure Ex37.38 has an equivalent circuit with resistance Req and capacitance Ceq.
Solve: The equivalent capacitance is Ceq = 2 µF + 2 µF = 4 µF, and the equivalent resistance is
1
1
1
=
+
⇒ Req = 500 Ω
Req 1 kΩ 1 kΩ
Thus, the time constant for the discharge of the capacitors is
τ = ReqCeq = (500 Ω)(4 µF) = 2 × 10−3 s = 2 ms
31.39. Model: The capacitor discharges through a resistor. Assume that the wires are ideal.
Solve:
The decay of the capacitor charge is given by the Equation 31.38: Q = Q0 e−t/τ. The time constant is
τ = RC = (1.0 × 103 Ω)(10 × 10−6 F) = 0.010 s
The initial charge on the capacitor is Q0 = 20 µC and it decays to 10 µC in time t. That is,
 10 µ C 
t
10 µC = (20 µC)e−t/0.010 s ⇒ ln 
⇒ t = (0.010 s) ln 2 = 6.93 ms
=−
0.010 s
 20 µ C 
31.57. Model: The batteries are ideal, the connecting wires are ideal, and the ammeter has a negligibly small
resistance.
Visualize: Please refer to Figure P31.57.
Solve: Kirchhoff’s junction law tells us that the current flowing through the 2.0 Ω resistance in the middle branch
is I1 + I2 = 3.0 A. We can therefore determine I1 by applying Kirchhoff’s loop law to the left loop. Starting
clockwise from the lower left corner,
+9.0 V – I1(3.0 Ω) – (3.0 A)(2.0 Ω) = 0 V ⇒ I1 = 1.0 A ⇒ I2 = (3.0 A – I1) = (3.0 A – 1.0 A) = 2.0 A
Finally, to determine the emf Ε, we apply Kirchhoff’s loop law to the right loop and start counterclockwise from
the lower right corner of the loop:
+Ε − I2(4.5 Ω) − (3.0 A)(2.0 Ω) = 0 V ⇒ Ε – (2.0 A)(4.5 Ω) – 6.0 V = 0 V ⇒ Ε = 15.0 V
31.71. Model: The battery and the connecting wires are ideal.
Visualize:
The figure shows how to simplify the circuit in Figure P31.71 using the laws of series and parallel resistances.
Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit
to find the current and potential difference of each resistor.
Solve: (a) From the last circuit in the figure and from Kirchhoff’s law, I = 100 V 10 Ω = 10 A . Thus, the current
through the battery is 10 A. Now as we rebuild the circuit, we note that series resistors must have the same current I
and that parallel resistors must have the same potential difference.
In Step 1 of the above diagram, we return the 10 Ω resistor to the 4 Ω, 4 Ω, and 2 Ω resistors in series. These
resistors must have the same 10 A current as the 10 Ω resistance. That is, the current through the 2 Ω and the 4 Ω
resistors is 10 A. The potential differences are
∆V2 = (10 A)(2 Ω) = 20 V
∆V4 (left) = (10 A)(4 Ω) = 40 V
∆V4 (left) = (10 A)(4 Ω) = 40 V
In Step 2, we return the left 4 Ω resistor to the 20 Ω and 5 Ω resistors in parallel. The two resistors must have the
same potential difference ∆V = 40 V. From Ohm’s law,
I5 =
40 V
= 8A
5Ω
I 20 =
40 V
=2A
20 Ω
The currents through the various resistors are I2 = I4 = 10 A, I5 = 8 A, and I20 = 2 A.
2
(b) The power dissipated by the 20 Ω resistor is I 20
( 20 Ω ) = ( 2 A ) ( 20 Ω ) = 80 W .
2
(c) Starting with zero potential at the grounded point, we travel along the outside path to point a and add/subtract
the potential differences on the way:
0 V + (20 Ω)I20 + (2 Ω)I2 = (20 Ω)(2 A) + (2 Ω)(10 A) = 60 V = Va
31.75. Model: The connecting wires are ideal. The capacitors discharge through the resistors.
Visualize:
The figure shows how to simplify the circuit in Figure P31.75 using the laws of series and parallel resistors and the
laws of series and parallel capacitors.
Solve: The 30 Ω and 20 Ω resistors are in parallel and are equivalent to a 12 Ω resistor. This 12 Ω resistor is in
series with the 8 Ω resistor so the equivalent resistance of the circuit Req = 20 Ω. The two 60 µF capacitors are in
series producing an equivalent capacitance of 30 µF. This 30 µF capacitor is in parallel with the 20 µF capacitor so
the equivalent capacitance Ceq of the circuit is 50 µF. The time constant of this circuit is
τ = ReqCeq = (20 Ω)(50 µF) = 1.0 ms
The current due to the three capacitors through the 20 Ω equivalent resistor is the same as through the 8 Ω resistor.
So, the voltage across the 8 Ω resistor follows the decay equation V = V0e−t/τ. For V = V0/2, we get
t
1
V0/2 = V0e−t/1.0 ms ⇒ ln   = −
⇒ t = 0.69 ms
1.0 ms
2