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# solution13_pdf - taherisefat(mt23852 hw 13 Opyrchal(121012...

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taherisefat (mt23852) – hw 13 – Opyrchal – (121012) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. 001 10.0 points An inductor has a 70 . 1 Ω reactance at 87 . 7 Hz. What will be the peak current if this in- ductor is connected to a 68 Hz source that produces a 104 V rms voltage? Correct answer: 2 . 70596 A. Explanation: Let : X L = 70 . 1 Ω , f 1 = 87 . 7 Hz , f 2 = 68 Hz , and V rms = 104 V . The inductance is L = X L 2 π f 1 = 70 . 1 Ω 2 π (87 . 7 Hz) = 0 . 127215 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 (104 V) 2 π (68 Hz) (0 . 127215 H) = 2 . 70596 A . 002 (part 1 of 2) 10.0 points In a purely inductive AC circuit, as in the Fgure, the maximum voltage is 140 V. L S E If the maximum current is 8 . 35 A at 50 Hz, Fnd the inductance. Correct answer: 53 . 3693 mH. Explanation: Let : Δ V max = 140 V , I 1 = 8 . 35 A , and f 1 = 50 Hz . The inductive reactance is X L = V I 1 = ω L = 2 π f 1 SL , so the inductance is L = V 2 π f 1 I 1 = 140 V 2 π (50 Hz)(8 . 35 A) · 10 3 mH H = 53 . 3693 mH . 003 (part 2 of 2) 10.0 points At what angular frequency is the maximum current 2 . 65 A? Correct answer: 989 . 898 rad / s. Explanation: Let : I max = 2 . 65 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 = 140 V (2 . 65 A)(53 . 3693 mH) · 10 3 mH H = 989 . 898 rad / s . 004 (part 1 of 3) 10.0 points A 4 μ ± capacitor is charged to 24 V and is then connected across a 15 mH inductor.

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solution13_pdf - taherisefat(mt23852 hw 13 Opyrchal(121012...

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