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Unformatted text preview: johnson (rj6247) – hw 13 – Opyrchal – (121014) 1 This printout should have 16 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An inductor has a 66 . 3 Ω reactance at 91 . 2 Hz. What will be the peak current if this in ductor is connected to a 47 . 9 Hz source that produces a 144 V rms voltage? Correct answer: 5 . 84821 A. Explanation: Let : X L = 66 . 3 Ω , f 1 = 91 . 2 Hz , f 2 = 47 . 9 Hz , and V rms = 144 V . The inductance is L = X L 2 π f 1 = 66 . 3 Ω 2 π (91 . 2 Hz) = 0 . 115701 H . Thus, the rms current is I rms = V rms X L = V rms 2 π f 2 L , so the maximum current is I max = √ 2 I rms = √ 2 V rms 2 π f 2 L = √ 2 (144 V) 2 π (47 . 9 Hz) (0 . 115701 H) = 5 . 84821 A . 002 (part 1 of 2) 10.0 points In a purely inductive AC circuit, as in the figure, the maximum voltage is 150 V. L S E If the maximum current is 6 . 18 A at 50 Hz, find the inductance. Correct answer: 77 . 2597 mH. Explanation: Let : Δ V max = 150 V , I 1 = 6 . 18 A , and f 1 = 50 Hz . The inductive reactance is X L = V I 1 = ω L = 2 π f 1 ?L , so the inductance is L = V 2 π f 1 I 1 = 150 V 2 π (50 Hz)(6 . 18 A) · 10 3 mH H = 77 . 2597 mH . 003 (part 2 of 2) 10.0 points At what angular frequency is the maximum current 3 . 95 A? Correct answer: 491 . 52 rad / s. Explanation: Let : I max = 3 . 95 A The inductive reactance is X L = V I 2 = ω 2 L, so the angular frequency is ω 2 = X L L = V I 2 L = 150 V (3 . 95 A)(77 . 2597 mH) · 10 3 mH H = 491 . 52 rad / s ....
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 Spring '10
 Turner
 Alternating Current, Correct Answer, LC circuit

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