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Unformatted text preview: hussain (wh4892) – hw 13 – Opyrchal – (121104) 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 34 . 9 Ω reactance at 90 . 2 Hz. What will be the peak current if this in ductor is connected to a 51 . 1 Hz source that produces a 47 . 8 V rms voltage? Correct answer: 3 . 41903 A. Explanation: Let : X L = 34 . 9 Ω , f 1 = 90 . 2 Hz , f 2 = 51 . 1 Hz , and V rms = 47 . 8 V . The inductance is L = X L 2 π f 1 = 34 . 9 Ω 2 π (90 . 2 Hz) = 0 . 0615799 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 (47 . 8 V) 2 π (51 . 1 Hz) (0 . 0615799 H) = 3 . 41903 A . 002 (part 1 of 2) 10.0 points In a purely inductive AC circuit, as in the figure, the maximum voltage is 120 V. L S E If the maximum current is 6 . 43 A at 50 Hz, find the inductance. Correct answer: 59 . 4046 mH. Explanation: Let : Δ V max = 120 V , I 1 = 6 . 43 A , and f 1 = 50 Hz . The inductive reactance is X L = V I 1 = ω L = 2 π f 1 suppressL , so the inductance is L = V 2 π f 1 I 1 = 120 V 2 π (50 Hz)(6 . 43 A) · 10 3 mH H = 59 . 4046 mH . 003 (part 2 of 2) 10.0 points At what angular frequency is the maximum current 1 . 97 A? Correct answer: 1025 . 4 rad / s. Explanation: Let : I max = 1 . 97 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 = 120 V (1 . 97 A)(59 . 4046 mH) · 10 3 mH H = 1025 . 4 rad / s ....
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 Spring '08
 moro
 Alternating Current, Correct Answer, LC circuit

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