This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: terry (ect328) homework 37 Turner (59130) 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 . 5 mF capacitor is connected to a standard outlet ( rms voltage 97 . 9 V, frequency 44 Hz ). Determine the magnitude of the current in the capacitor at t = 0 . 00898 s, assuming that at t = 0, the energy stored in the capacitor is zero. Correct answer: 45 . 3929 A. Explanation: Let : V rms = 97 . 9 V , C = 1 . 5 mF = 0 . 0015 F , f = 44 Hz , and t = 0 . 00898 s . The capacitive reactance is X C = 1 C = 1 2 f C . The maximum current is I max = V max X C = 2 V rms X C = 2 (2 f C ) V rms . Because the current leads the voltage across a capacitor by 90 , at time t , I = I max sin ( t + 90 ) = 2 2 f C V rms sin (2 f t + 90 ) = 2 2 (44 Hz) (0 . 0015 F) (97 . 9 V) sin [2 (44 Hz)(0 . 00898 s) + 90 ] = 45 . 3929 A , so  I  = 45 . 3929 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. the frequency = LC 2. very low frequencies 3. both very low frequencies or very high frequencies correct 4. very high frequencies 5. the frequency = 1 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....
View
Full
Document
 Fall '09
 Turner

Click to edit the document details