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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 = Page 2 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 Page 3 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 Page 4