# OHW 27 - terry(ect328 – oldhomework 27 – Turner...

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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 Ω 1272 Ω 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 Ω 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 Ohm’s Law I = E R 2 = 13 . 1 V 13 Ω = 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 Ω 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 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 . 00769 A) (1272 Ω) + (1 . 00769 A) (13 Ω) = 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 / τ . terry (ect328) – oldhomework 27 – Turner – (59130) 2 The time constant is τ = L R t = L R 1 + R 2 = 2 H 1272 Ω + 13 Ω = 0 . 00155642 s ....
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OHW 27 - terry(ect328 – oldhomework 27 – Turner...

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