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Unformatted text preview: pan (zp695) – Ch23h3 – chiu – (56565) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points One application of an RL circuit is the gen eration of timevarying highvoltage from a lowvoltage 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 = E R 2 = 11 . 2 V 10 Ω = 1 . 12 A . 002 (part 2 of 3) 10.0 points 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 = 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 R 1 + I R 2 = (1 . 12 A) (1238 Ω) + (1 . 12 A) (10 Ω) = 1397 . 76 V . 003 (part 3 of 3) 10.0 points 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 that decays as I = I e t/ τ . pan (zp695) – Ch23h3 – 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 e t/ τ = L I τ e t/τ . Solving for t , we obtain t = τ ln bracketleftbigg V L τ L I bracketrightbigg = (0 . 00240385 s) × ln bracketleftbigg (12 V) (0 . 00240385 s) (3 H) (1 . 12 A) bracketrightbigg = 0 . 0114368 s = 11 . 4368 ms . 004 10.0 points 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....
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This note was uploaded on 02/21/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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