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oldhw 27 - nguyen(jmn727 oldhomework 27 Turner(59070 This...

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nguyen (jmn727) – oldhomework 27 – Turner – (59070) 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. 3 H 14 Ω 1154 Ω 10 . 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: 0 . 728571 A. Explanation: L R 2 R 1 E S b a Let : R 2 = 14 Ω and E = 10 . 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 = 10 . 2 V 14 Ω = 0 . 728571 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: 850 . 971 V. Explanation: Let : R 1 = 1154 Ω and I 0 = 0 . 728571 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 0 . 728571 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 = (0 . 728571 A) (1154 Ω) + (0 . 728571 A) (14 Ω) = 850 . 971 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: 11 . 169 ms. Explanation: Let : L = 3 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 0 that decays as I = I 0 e - t / τ .
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nguyen (jmn727) – oldhomework 27 – Turner – (59070) 2 The time constant is τ = L R t = L R 1 + R 2 = 3 H 1154 Ω + 14 Ω = 0 . 00256849 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 .
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