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Unformatted text preview: brown (twb493) – 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. 1 H 14 Ω 1283 Ω 10 . 7 V S b a What is the current in the circuit a long time after the switch has been in position “ a ”? Correct answer: 0 . 764286 A. Explanation: L R 2 R 1 E S b a Let : R 2 = 14 Ω and E = 10 . 7 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 = 10 . 7 V 14 Ω = . 764286 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: 991 . 279 V. Explanation: Let : R 1 = 1283 Ω and I = 0 . 764286 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 . 764286 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 = (0 . 764286 A) (1283 Ω) + (0 . 764286 A) (14 Ω) = 991 . 279 V . 003 (part 3 of 3) 10.0 points How much time elapses before the voltage across the inductor drops to 13 V? Correct answer: 3 . 34159 ms. Explanation: Let : L = 1 H and V L = 13 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 . brown (twb493) – Ch23h3 – chiu – (56565) 2 The time constant is τ = L R t = L R 1 + R 2 = 1 H 1283 Ω + 14 Ω = 0 . 00077101 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 . 00077101 s) × ln bracketleftbigg (13 V) (0 . 00077101 s) (1 H) (0 . 764286 A) bracketrightbigg = 0 . 00334159 s = 3 . 34159 ms . 004 10.0 points A(n) 66 . 1 cm length of wire when used as a re sistor has a resistance of 0 . 00537 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 . 0698 T / s....
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This note was uploaded on 12/01/2011 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Turner
 Physics

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