31.1. Solve: From Table 30.1, the resistivity of carbon is ρ = 3.5 × 10−5 Ω m. From Equation 31.3, the resistance
of lead from a mechanical pencil is
R=
ρL
A
−5
ρ L ( 3.5 × 10 Ω m ) ( 0.06 m )
=
= 5.5 Ω
2
π r2
π ( 0.35 × 10 −3 m )
=
31.6. Solve: The slope of the I versus ∆V graph, according to Ohm’s law, is the resistance R. From the graph in
Figure Ex31.6, the slope of the material is
∆V 100 V
=
= 50 Ω
I
2A
31.7. Solve:
From the circuit in Figure Ex37.1, we see that 50 Ω and 100 Ω resistors are connected in series across the battery.
Another resistor of 75 Ω is also connected across the battery.
31.14. Model: Assume ideal connecting wires and an ideal battery.
Visualize: Please refer to Figure Ex31.14.
Solve: The power dissipated by each resistor can be calculated from Equation 31.18, PR = I2R, provided we can
find the current through the resistors. Let us choose a clockwise direction for the current and solve for the value of I
by using Kirchhoff’s loop law. Going clockwise from the negative terminal of the battery,
∑ ( ∆V )
i
= ∆Vbat + ∆VR1 + ∆VR2 = 0 ⇒ +9 V – IR1 – IR2 = 0
i
⇒ I=
9V
9V
1
=
= A
R1 + R2 12 Ω + 15 Ω 3
The power dissipated by resistors R1 and R2 is:
PR1 = I 2 R1 = ( 31 A ) (12 Ω ) = 1.33 W
2
PR2 = I 2 R2 = ( 13 A ) (15 Ω ) = 1.67 W
2
31.20. Visualize: Please refer to Figure Ex32.20.
Solve:
The three resistors are in series. The total resistance of this combination is
Req = R + 50 Ω + R = 2R + 50 Ω
Thus, Req > 50 Ω as long as R > 0 Ω.
31.25. Model: Assume ideal connecting wires but not an ideal battery.
Visualize: The circuit for an ideal battery is the same as the circuit in Figure Ex31.25, except that the 1 Ω resistor
is not present.
Solve: In the case of an ideal battery, we have a battery with E = 15 V connected to two series resistors of 10 Ω
and 20 Ω resistance. Because the equivalent resistance is Req = 10 Ω + 20 Ω = 30 Ω and the potential difference
across Req is 15 V, the current in the circuit is
I=
15 V
∆V
E
=
=
= 0.5 A
Req Req 30 Ω
The potential difference across the 20 Ω resistor is
∆V20 = IR = (0.5 A)(20 Ω) = 10 V
In the case of a real battery, we have a battery with E = 15 V connected to three series resistors: 10 Ω, 20 Ω, and an
internal resistance of 1.0 Ω. Now the equivalent resistance is
Req′ = 10 Ω + 20 Ω + 1.0 Ω = 31 Ω
The potential difference across Req is the same as before ( E = 15 V). Thus,
I′ =
∆V ′
E
15 V
=
=
= 0.4839 A
Req′
Req′
31 Ω
Therefore, the potential difference across the 20 Ω resistor is
∆V20′ = I ′R = ( 0.4839 A )( 20 Ω ) = 9.68 V
That is, the potential difference across the 20 Ω resistor is reduced from 10 V to 9.68 V due to the internal
resistance of 1 Ω of the battery. The percentage change in the potential difference is
 10 V − 9.68 V 

 × 100 = 3.23%
10 V


31.28. Visualize: The three resistors in Figure Ex31.28 are equivalent to a resistor of resistance Req.
Solve: Because the three resistors are in parallel,
( 200 Ω ) R =
1
1
1
1 2
1
400 Ω + R
= +
+ = +
=
⇒ Req =
Req R 200 Ω R R 200 Ω ( 200 Ω ) R
( 400 Ω + R )
200 Ω
 400 Ω 
1+ 

 R 
From this equation, we see that (i) Req = 0 Ω if R = 0 Ω and (ii) Req = 200 Ω if R → ∞. Thus, Req < 200 Ω for R < ∞.
31.29. Model: Assume ideal connecting wires.
Visualize: Please refer to Figure Ex31.29.
Solve: The resistance R is given by Ohm’s law, R = ∆VR I R . To determine IR we use Kirchhoff’s junction law.
The input current I splits into the three currents I10, I15, and IR. That is,
2.0 A = I10 + I15 + IR =
8V
8V
+
+ I R ⇒ IR = 2.0 A – 0.8 A – 0.533 A = 0.667 A
10 Ω 15 Ω
Using this value of IR in Ohm’s law,
R=
8V
= 12.0 Ω
0.667 A
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.
31.34. Model: Grounding does not affect a circuit’s behavior.
Visualize: Please refer to Figure Ex31.34.
Solve: Because the earth has Vearth = 0 V, point d has a potential of zero. In going from point d to point a, the
potential increases by 9 V. Thus, point a is at a potential of 9 V. Let us calculate the current I in the circuit before
calculating the potentials at points b and c. Applying Kirchhoff’s loop rule, starting clockwise from point d,
∑ ( ∆V )
i
i
= ∆V9 V bat + ∆V2 Ω + ∆V3 V bat + ∆V4 Ω = 0
⇒ +9 V – I(2 Ω) – 3 V – I(4 Ω) = 0 ⇒ I =
6V
=1A
6Ω
There is a drop in potential from point a to point b by an amount IR = (1 A)(2 Ω) = 2 V. Thus, the potential at point
b is 9 V − 2 V = 7 V. The potential decreases from 7 V at point b to 7 V − 3 V = 4 V at point c. There is a further
decrease in potential across the 4 Ω resistor of IR = (1 A)(4 Ω) = 4 V. That is, the potential of 4 V at c becomes 0 V
at point d, as it must. In summary, the potentials at a, b, c, and d are 9 V, 7 V, 4 V, and 0 V.
31.36. Solve: Noting that the unit of resistance is the ohm (V/A) and the unit of capacitance is the farad (C/V),
the unit of RC is
RC =
V C C
C
× = =
=s
A V A Cs
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.
31.40. Model: The capacitor discharges through a resistor. Assume ideal wires.
Visualize: The switch in the circuit in Figure Ex31.40 is in position a. When the switch is in position b the circuit
consists of a capacitor and a resistor.
Solve: (a) The switch has been in position a for a long time. That means the capacitor is fully charged to a charge
Q0 = C∆V = C E = (2 µF)(9 V) = 18 µC
Immediately after the switch is moved to the b position, the charge on the capacitor is Q0 = 18 µC. The current
through the resistor is
I0 =
∆VR
9V
=
= 0.18 A = 180 mA
R
50 Ω
Note that as soon as the switch is closed, the potential difference across the capacitor ∆VC appears across the 50 Ω
resistor.
(b) The charge Q0 decays as Q = Q0 e−t/τ, where
τ = RC = (50 Ω)(2 µF) = 100 µs
Thus, the charge is
Q = (18 µ C ) e−50 µs 100 µs = (18 µ C ) e−0.5 = 10.9 µ C
The resistor current is
I = I 0 e−t τ = (180 mA ) e−50 µs 100 µs = 109 mA
(c) Likewise, the charge is Q = 2.44 µC and the current is I = 24.4 mA.
31.54. Model: Assume that the connecting wire and the battery are ideal.
Visualize: Please refer to Figure P31.54.
Solve: The middle and right branches are in parallel, so the potential difference across these two branches must
be the same. The currents are known, so these potential differences are
∆Vmiddle = (3.0 A)R = ∆Vright = (2.0 A)(R + 10 Ω)
This is easily solved to give R = 20 Ω. The middle resistor R is connected directly across the battery, thus (for an
ideal battery, with no internal resistance) the potential difference ∆Vmiddle equals the emf of the battery. That is
E = ∆Vmiddle = (3.0 A)(20 Ω) = 60 V
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
31.64. Model: The voltage source/battery and the connecting wires are ideal.
Visualize: Please refer to Figure P31.64.
Solve: Let us first apply Kirchhoff’s loop law starting clockwise from the lower left corner:
+Vin – IR – I (100 Ω) = 0 V ⇒ I =
Vin
R + 100 Ω
The output voltage is
Vout
Vin


100 Ω
Vout = (100 Ω ) I = (100 Ω ) 
=
 ⇒
R
+
100
Ω
V
R
+ 100 Ω


in
For Vout = Vin 10 , the above equation can be simplified to obtain R:
Vin 10
100 Ω
=
⇒ R + 100 Ω = 1000 Ω ⇒ R = 900 Ω
Vin
R + 100 Ω
31.76. Model: The battery and the connecting wires are ideal.
Visualize: Please refer to Figure P31.76.
Solve: (a) A very long time after the switch has closed, the potential difference ∆VC across the capacitor is E.
This is because the capacitor charges until ∆VC = E, while the charging current approaches zero.
(b) The full charge of the capacitor is Qmax = C(∆VC)max = C E.
(c) In this circuit, I = + dQ dt because the capacitor is charging, that is, because the charge on the capacitor is
increasing.
(d) From Equation 31.40, capacitor charge at time t is Q = Qmax(1 − e−t/τ). So,
I=
(
)
dQ
d
1
 1  −t τ E −t τ
= CE
1 − e−t τ = CE   e−t τ = CE 
e = R e
dt
dt
τ
 
 RC 
A graph of I as a function of t is shown below.
31.79. Model: The battery and the connecting wires are ideal.
Visualize: Please refer to Figure 31.38(a).
Solve: After the switch closes at t = 0 s, the capacitor begins to charge. At time t, let the current and the charge in
the circuit be i and q, respectively. Also, assume clockwise direction for the current i. Using Kirchhoff’s loop law
and starting clockwise from the lower left corner of the loop,
+E − iR −
q
dq
q
dq
dt
=0⇒E=
R + ⇒ RC dq = ( EC – q)dt ⇒
=
C
dt
C
EC − q RC
Integrating both sides,
Q
t
Q
dq
dt
t
t
∫0 EC − q = ∫0 RC ⇒ −  ln ( EC − q ) 0 = RC ⇒ − ln ( EC − Q ) + ln ( EC ) = RC
t
EC − Q
 EC − Q 
⇒ ln 
=−
⇒
= e−t RC ⇒ Q = EC 1 − e−t RC

RC
EC
 EC 
(
Letting Qmax = EC and τ = RC, we get Q = Qmax (1 − e−t/τ).
)