General Physics II (2012)
HW6 solution
1. A hollow cylindrical resistor with inner radius r1 and outer radius
r2, and length l, is made of a material whose resistivity is ρ (Fig.
25-36). (a) Show that the resistance is given by
ρ
r
R =
ln 2
2π l
r1
for current that flows radially outward. [Hint: Divide the resistor
into concentric cylindrical shells and integrate.] (b) Evaluate the
resistance R for such a resistor made of carbon whose inner and
outer radii are 1.0 mm and 1.8 mm and whose length is 2.4 cm.
(Choose ρ=15·10-5 Ω·m.) (c) What is the resistance in part (b)
for current flowing parallel to the axis ?
Fig. 25-36
Sol:
(a)
dr
dR = ρ
2πrl
l
R=ρ
A
r2
R = ∫ dR = ∫ ρ
r1
(b)
(c)
dr
r
ρ
=
ln 2
2πrl 2πl r1
r2 15 ×10 −5 Ω ⋅ m  1.8mm 
ρ
−4
R=
=
×
Ω
ln =
ln
5
.
8
10

2πl r1
2π (0.024m )  1.0mm 
(
)
) (
l
15 ×10 −5 Ω ⋅ m (0.024m )
ρl
R=ρ =
=
= 0.51Ω
2
2
2
2
−
−
3
3
A π r2 − r1
π [ 1.8 × 10 m − 1.0 ×10 m ]
(
)
(
)
2. (a) Determine the current I1, I2, and I3 in the Fig 26-53.
Assume the internal resistance of each battery is r = 1.0 Ω.
(b) What is the terminal votage of the 6.0-V battery.
Fig 26-53
Sol:
(a) Use Kirchhoff’s junction rule
I1 = I 2 + I 3
(1)
Use Kirchhoff’s loop rule
for top loop:
12.0V − I 2 (11 + 1)Ω + 12.0V − I1 (12 + 1 + 22)Ω = 0
⇒ 35 I1 + 12 I 2 = 24
(2)
Use Kirchhoff’s loop rule
for bottom loop:
12.0V − I 2 (11 + 1)Ω + 6.0V − I (18 + 1 + 15)Ω = 0
⇒ −34 I1 + 12 I 2 = 6
(3)
from equation (1), (2), and (3)
I1 = 0.511A
(b)
I 2 = 0.508 A
(
I 3 = 2.97mA
)
6.0V − I 3 r = 6.0V − 2.97 × 10 −3 A (1.0Ω ) = 5.997V ≈ 6.0V
3. A R-C circuit is shown in Fig.1. R1 = 1 kΩ, R2 = 2 kΩ,
R3 = 2 kΩ, C = 5μF, and E = 3 volt. The capacitor is initially
uncharged. The switch S is closed at t = 0.
(a) Immediately after the switch is closed, what is the current
through each resistor? (b) What is the final charge on the
capacitor? (c) Find the charge on the capacitor Q(t) and the
time constant for charging the capacitor in the circuit.
Sol:
(a) Immediately after the switch is
closed, the capacitor behaves like
a conducting wire. Then the
circuit becomes the one shown
R1
below.
(b) As t ∞, capacitor behaves
like a open circuit.
R1
i1
i3
i1
R2
E
i2
R3
R2
E
i2
Vc
R3
C
S
S
1
1
1
1
1
1
=
+
=
+
=
Reff R2 R3 2 ⋅103 2 ⋅103 103
Rtotal = R1 + Reff = 2 ⋅103 Ω
′ = R1 + R2 = 3 ⋅103 Ω
Rtotal
i1 = i2 = E
′
Rtotal
= 1 mA
Veff = Reff ⋅ I total = 1.5 V
Vc = R2 ⋅ I 2 = 2 V
I total = i1 = E
Q = C ⋅Vc = 10 mCoulombs
i2 = i3 =
Rtotal
Veff
R2
= 1.5 mA
= 0.75 mA
R1, i1
(c) Determine the time constant for charging
the capacitor in the circuit.
(1)
⇒ 3 − 10 i1 − 2 ⋅10 i2 = 0
3
dQ3
dt
First, i1 is replaced by i2,
3 − 3 ⋅103 i2 − 103 i3 = 0
2 ⋅103 i2 − 2 ⋅103 i3 − 2 ⋅105 Q3 = 0
and then eliminate i2,
6 − 8 ⋅103 i3 − 6 ⋅105 Q3 = 0
i.e.
(2)
Q3
+ i2 R2 = 0
C
⇒ − 2 ⋅103 i3 − 2 ⋅105 Q3 + 2 ⋅103 i2 = 0
i3 =
i2
R3
2
+ + Q3
- -C
dQ3
3 ⋅ 102 dt
=
−5
10 − Q3
4
−i3 R3 −
and
R2
S
E − i1 R1 − i2 R2 = 0
3
Loop 2:
E
1
i1 = i2 + i3
Loop 1:
i3
dQ3 3 ⋅10−3 − 3 ⋅102 Q3
i3 =
=
dt
4
dQ3
3 ⋅ 102 dt
⇒ ∫ −5
=∫
10
−
Q
4
3
0
0
Q
(3)
t
 10−5 − Q  3 ⋅ 102 t
ln 
=
−5
10
4


−t

−5
Q = 10 ⋅  1 − e


 4 −2 
 ⋅10 
3

4 −2
τ = ⋅10 sec.
3


