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Unformatted text preview: terry (ect328) oldhomework 27 Turner (59130) 1 This print-out should have 14 questions. Multiple-choice 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 time-varying high-voltage from a low-voltage source, as shown in the figure. 2 H 13 1272 13 . 1 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 . 00769 A. Explanation: L R 2 R 1 E S b a Let : R 2 = 13 and E = 13 . 1 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 Ohms Law I = E R 2 = 13 . 1 V 13 = 1 . 00769 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: 1294 . 88 V. Explanation: Let : R 1 = 1272 and I = 1 . 00769 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 . 00769 A . From Kirchhoffs 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 . 00769 A) (1272 ) + (1 . 00769 A) (13 ) = 1294 . 88 V . 003 (part 3 of 3) 10.0 points How much time elapses before the voltage across the inductor drops to 11 V? Correct answer: 7 . 42145 ms. Explanation: Let : L = 2 H and V L = 11 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 / . terry (ect328) oldhomework 27 Turner (59130) 2 The time constant is = L R t = L R 1 + R 2 = 2 H 1272 + 13 = 0 . 00155642 s ....
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- Fall '09