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# solution_pdf37 - tovar(jdt436 homework 37 Turner(59070 This...

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tovar (jdt436) – homework 37 – Turner – (59070) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 0 . 52 mF capacitor is connected to a stan- dard outlet ( rms voltage 75 V, frequency 22 Hz ). Determine the magnitude of the current in the capacitor at t = 0 . 00952 s, assuming that at t = 0, the energy stored in the capacitor is zero. Correct answer: 1 . 92198 A. Explanation: Let : V rms = 75 V , C = 0 . 52 mF = 0 . 00052 F , f = 22 Hz , and t = 0 . 00952 s . The capacitive reactance is X C = 1 ω C = 1 2 π f C . The maximum current is I max = V max X C = 2 V rms X C = 2 (2 π f C ) V rms . Because the current leads the voltage across a capacitor by 90 , at time t , I = I max sin ( ω t + 90 ) = 2 2 π f C V rms sin (2 π f t + 90 ) = 2 2 π (22 Hz) (0 . 00052 F) (75 V) × sin [2 π (22 Hz)(0 . 00952 s) + 90 ] = 1 . 92198 A , so | I | = 1 . 92198 A . keywords: 002 10.0 points The emf E can drive the circuit below at any given frequency. E C L R A B D At what frequency does the light bulb glow most brightly? 1. very high frequencies 2. very low frequencies 3. both very low frequencies or very high frequencies correct 4. the frequency ω = 1 LC 5. the frequency ω = LC Explanation: Since the brightness of the bulb is propor- tional to the power dissipated in it, P = I 2 rms R, . In order to obtain maximum brightness we should adjust the frequency so that the rms current I rms through the bulb is the largest. The capacitive reactance X C = 1 ω C is in- versely proportional to ω , and the inductive reactance X L = ω L is proportional to ω , so Z = R + 1 radicalBigg parenleftbigg 1 X C 1 X L parenrightbigg 2 = R + 1 radicalBigg parenleftbigg ω C 1 ω L parenrightbigg 2 = R + ω L

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