solutions 16 - Patel Kinal Homework 16 Due Apr 1 2008 7:00...

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Patel, Kinal – Homework 16 – Due: Apr 1 2008, 7:00 pm – Inst: Weathers 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A lightbulb is connected to a 60 Hz power source having a maximum voltage of 158 V. What is the resistance of the light bulb that uses an average power of 96 . 5 W? Correct answer: 129 . 347 Ω. Explanation: Let : V max = 158 V and P av = 96 . 5 W . The rms voltage is V rms = V max 2 = 158 V 2 = 111 . 723 V . The average power is P av = V 2 rms R R = V 2 rms P av = V 2 max 2 P av = (158 V) 2 2 (96 . 5 W) = 129 . 347 Ω . keywords: 002 (part 1 of 1) 10 points An inductor has a 51 . 6 Ω reactance at 60 Hz. What will be the maximum current if this inductor is connected to a 50 Hz source that produces a 195 . 3 V rms voltage? Correct answer: 6 . 42316 A. Explanation: Let : X L = 51 . 6 Ω , f = 60 Hz , f 0 = 50 Hz , and V rms = 195 . 3 V . The maximum voltage of the circuit is V max = V rms 2 , and the inductive reactance is X L = ω L , so X 0 L = ω 0 ω X L = f 0 f X L , and the maximum current is I max = V max X 0 L = 2 V rms f f 0 X L = 2 (195 . 3 V) (60 Hz) (50 Hz) (51 . 6 Ω) = 6 . 42316 A . keywords: 003 (part 1 of 3) 10 points An 890 Ω resistor, a 1 . 52 μ F capacitor, and a 3 H inductor are connected in series across a 220 Hz AC source for which the maximum voltage is 50 V. Calculate the impedance of the circuit.
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