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Unformatted text preview: Anderson, Matthew – Homework 13 – Due: Apr 29 2008, noon – Inst: Vitaly 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points An LC circuit is shown in the figure below. The 20 pF capacitor is initially charged by the 7 V battery when S is at position a . Then S is thrown to position b so that the capacitor is shorted across the 10 mH inductor. 10 mH 20 pF 7 V S b a What is the maximum value for the oscil lating current assuming no resistance in the circuit? Correct answer: 0 . 00031305 A. Explanation: Let : C = 20 pF = 2 × 10 11 F , L = 10 mH = 0 . 01 H , and E = 7 V . The maximum charge on the capacitor is Q max = C V . The maximum current is I max = ω Q m = ω C V , where ω is the oscillatory frequency given by ω = 1 √ LC . Thus I max = r C L V = r 2 × 10 11 F . 01 H (7 V) = . 00031305 A . 002 (part 2 of 3) 10 points What is the natural angular frequency of the circuit? Correct answer: 2 . 23607 × 10 6 rad / s. Explanation: ω = 1 √ LC = 1 p (0 . 01 H)(2 × 10 11 F) = 2 . 23607 × 10 6 rad / s . 003 (part 3 of 3) 10 points What is the maximum energy stored in the magnetic field of the inductor? Correct answer: 4 . 9 × 10 10 J....
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This note was uploaded on 03/11/2010 for the course PHYS 121 taught by Professor Opyrchal during the Spring '08 term at NJIT.
 Spring '08
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