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Unformatted text preview: nguyen (jmn727) – homework 37 – Turner – (59070) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 1 . 4 mF capacitor is connected to a standard outlet ( rms voltage 57 . 3 V, frequency 81 Hz ). Determine the magnitude of the current in the capacitor at t = 0 . 00551 s, assuming that at t = 0, the energy stored in the capacitor is zero. Correct answer: 54 . 4838 A. Explanation: Let : V rms = 57 . 3 V , C = 1 . 4 mF = 0 . 0014 F , f = 81 Hz , and t = 0 . 00551 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 (81 Hz) (0 . 0014 F) (57 . 3 V) × sin [2 π (81 Hz)(0 . 00551 s) + 90 ◦ ] = − 54 . 4838 A , so  I  = 54 . 4838 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. both very low frequencies or very high frequencies correct 2. the frequency ω = √ LC 3. the frequency ω = 1 √ LC 4. very high frequencies 5. very low frequencies 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....
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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