Ch23-h3-solutions - pan(zp695 Ch23-h3 chiu(56565 This...

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pan (zp695) – Ch23-h3 – chiu – (56565) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of3)10.0points One application of an RL circuit is the gen- eration of time-varying high-voltage from a low-voltage source, as shown in the figure. 3 H 10 Ω 1238 Ω 11 . 2 V S b a What is the current in the circuit a long time after the switch has been in position a ”? Correct answer: 1 . 12 A. Explanation: L R 2 R 1 E S b a Let : R 2 = 10 Ω and E = 11 . 2 V . When the switch is at “ a ”, the circuit com- prises the battery, the inductor L , and the resistor R 2 . A long time after the switch has been in position “ a ”, the current is steady. This means that the inductor has no response to the current. In other words, the circuit can be considered as consisting of E and R 2 only, with the inductor reduced to a wire. In this case, the current is simply found by Ohm’s Law I 0 = E R 2 = 11 . 2 V 10 Ω = 1 . 12 A . 002(part2of3)10.0points Now the switch is thrown quickly from “ a ” to b ”. Compute the initial voltage across the in- ductor. Correct answer: 1397 . 76 V. Explanation: Let : R 1 = 1238 Ω and I 0 = 1 . 12 A . When the switch is thrown from “ a ” to b ”, the current in the circuit is the current passing through R 2 , which was found in Part 1 to be 1 . 12 A . From Kirchhoff’s Loop Law, the initial voltage across the inductor is equal to the initial voltage across R 1 and R 2 . So, we have V L = V R 1 + V R 2 = I 0 R 1 + I 0 R 2 = (1 . 12 A) (1238 Ω) + (1 . 12 A) (10 Ω) = 1397 . 76 V . 003(part3of3)10.0points How much time elapses before the voltage across the inductor drops to 12 V? Correct answer: 11 . 4368 ms. Explanation: Let : L = 3 H and V L = 12 V . The voltage across an inductor is V L = - L dI dt . When the switch is at “ b ”, we are dealing with an RL circuit with an initial current I 0 that decays as I = I 0 e - t / τ .
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pan (zp695) – Ch23-h3 – chiu – (56565) 2 The time constant is τ = L R t = L R 1 + R 2 = 3 H 1238 Ω + 10 Ω = 0 . 00240385 s . Therefore, the voltage across the inductor is V L = - L d dt I 0 e - t / τ = L I 0 τ e - t / τ . Solving for t , we obtain t = - τ ln bracketleftbigg V L τ L I 0 bracketrightbigg = - (0 . 00240385 s) × ln bracketleftbigg (12 V) (0 . 00240385 s) (3 H) (1 . 12 A) bracketrightbigg = 0 . 0114368 s = 11 . 4368 ms . 004 10.0points A(n) 53 . 1 cm length of wire when used as a re- sistor has a resistance of 0 . 00526 Ω. The ends of the wire are connected to form a circular loop, and the plane of the loop is positioned at right angles to a uniform magnetic field that is increasing at the rate of 0 . 0683 T / s. At what rate is thermal energy generated in the wire? Correct answer: 446 . 492 μ W. Explanation: The changing magnetic field generates a current in the wire. The induced potential is V = A d B d t . The radius is found from the circumference, ( C = ), to be r = C 2 π = 53 . 1 cm 2 π . Thus the area is given by A = πr 2 = π parenleftbigg C 2 π parenrightbigg 2 = 1 π parenleftbigg C 2 parenrightbigg 2 and the induced potential is V = 1 π parenleftbigg C 2 parenrightbigg 2 d B dt = 0 . 0015325 V The power dissipated is P = V 2 R = (0 . 0015325 V) 2 0 . 00526 Ω · 10 6 μ Ω Ω = 446 . 492 μ W 005(part1of2)10.0points The magnetic field inside a superconducting
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