Physics 2B
Quiz 3
Summer Session I 2009
Name:
PID:
Version 0
Quiz ID:
READ FIRST: Make sure to write/fill in your scantron form your name,
quiz ID and the test version.
e ≈ 1.6 × 10−19 C , k ≈ 9.0 × 109 N m2 /C2 , 0 ≈ 8.9 × 10−12 C2 /N m2 ,
Unless otherwise specified, assume potential is zero at infinity.
1. A wire has a cross-sectional area of 0.1 mm2 . If there are 5.0 × 1028 atoms per cubic
meter and each contributes, on average, 2 free electrons, what is the drift speed of the
electrons when the current in the wire is 4.0 A?
Solution: The current density is simply the current per area. So J = I/A =
4.0 A/0.1 mm2 = 4.0 × 107 A/m2 . The drift speed is related to the current density
by:
4.0 × 107 A/m2
J
=
= 0.0025m/s
vd =
ne
2 · 5.0 × 1028 m−3 · 1.6 × 10−19 C
A. 2.5 mm/s
B. 5.0 mm/s
C. 7.5 mm/s
D. 1.0 mm/s
2. Mercury has resistivity 9.84 × 10−7 Ω·m. A tube of mercury has a 35.0-V/m electric
field inside. How much current is flowing, if the diameter of the tube is 2.0 mm?
Solution: Relating conductivity to resistivity, the current density is:
J = σE =
35.0 V/m
E
=
≈ 3.56 × 107 A/m2
ρ
9.84 × 10−7 Ω · m
The current is then just the product of current density and the cross-sectional area:
I = JA = 3.56 × 107 A/m2 · π(2.0 × 10−3 m)2 /4 ≈ 111 A
A. 111 A
B. 1110 A
C. 555 A
D. 1700 A
3. How much current will be flowing through a 51.0-m length of copper wire with radius
5.1 mm if it is connected to a source supplying 36.0 V? (The resistivity of copper is
1.68 × 10−8 Ω·m.)
Physics 2B
Quiz 3
Summer Session I 2009
Version 0
Solution: We first find the resistance of this particular cut of copper wire:
R=
ρl
1.68 × 10−8 Ω · m × 51.0 m
=
≈ 0.01Ω
A
π (5.1 × 10−3 m)2
Then use Ohm’s Law to find the current:
A. 3600 A
B. 2.2 GA
I=
V
= 3600A
R
C. 130 nA
D. 900 A
4. A component with a 14-Ω resistance is rated for user at power levels not exceeding 11
W. How much current can safely flow through the component?
Solution: The most convenience relation to use in this case is:
r
r
P
11 W
=
≈ 0.886A
I=
R
14 Ω
A. 0.89 A
B. 1.27 A
C. 154 A
D. 0.24 A
5. What is the equivalent resistance of the following network of four identical resistors, each
with resistance R?
Solution: Going from right to left:
1. First two resistors are in parallel: R1−2 =
R·R
R+R
=
R
.
2
2. The third one is in series with the first two: R1−3 = R1−2 + R = 32 R.
3. The fourth resistor is in parallel with the first three: R1−4 =
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R1−3 ·R
R1−3 +R
= 35 R.
Physics 2B
Quiz 3
A. 3R/5
Summer Session I 2009
B. 4 R
C. R/4
Version 0
D. 5R/3
6. A rechargeable battery that is completely drained of electric energy can be completely
charged by applying 4.0 mA of current for 1.0 h. If the battery acquires an emf of 2.0 V
in the process of being charged, how much energy is the battery capable of storing.
Solution: The energy store is the rate at which it is charged, i.e. the power, times
the time:
E = P δt = IV δt = 4.0 mA × 2.0 V × 3600 s = 28.8 J
A. 29 J
B. 8.0 mJ
C. 0.50 J
D. 440 mJ
7. In the following circuit. If an ideal voltmeter is inserted between B and C. What would
be its reading?
Solution: The total resistance of the three resistance is 9 Ω. The current in the
circuit is I = V /R = 1 A. The potential difference between B and C is the sum of
potential difference across the 3 Ω and 4 Ω resistors. Since the current is same for
both of them, the potential difference is 3 V and 4 V, respectively. The total is 7 V.
A. 7 V
B. 2 V
C. 5 V
D. 6 V
8. A 7.0-µF capacitor is charged and has a 600.0-V potential difference across its plates.
If the capacitor is discharged through a 300.0-kΩ resistor, how long does it take for the
potential difference across the plates of the capacitor to drop to 0.000001 of its initial
voltage.
Solution: The RC constant is τ = R C = 7.0 µF · 300 kΩ = 2.1 s. The voltage across
a discharging capacitor as function of time is: V = V0 e−t/τ . The time for the required
voltage is:
V
= −(2.1 ln 0.000001)s ≈ 29.0s
t = −τ ln
V0
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Physics 2B
Quiz 3
A. 29.0 s
Summer Session I 2009
B. 29000 s
C. 2.1 ms
Version 0
D. 2.1 µs
9. Consider the following circuit, what is the current through the 6-Ω resistor?
Solution: One terminal of the 6-Ω is held at 12 V, and the other at 9 V. The
potential difference across the resistor is 3 V. The current through it is 0.5 A.
A. 0.5 A
B. 4.0 A
C. 1.5 A
D. 4.5 A
10. In the following circuit, what is the current through the 3 Ω resistor immediately after
the switch is closed.
Solution: Immediately after the circuit is close when the capacitor is still uncharged,
the voltage across the capacitor is zero. The capacitor acts as a short. Therefore,
the total resistance of the circuit is 3 Ω. The current is 3 A.
A. 3A
B. 1A
C. 0A
D. 4.5A
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