Homework 18 Solutions

# Homework 18 - lim(kl9356 – Homework 18 – Weathers –(22202 1 This print-out should have 11 questions Multiple-choice questions may continue on

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Unformatted text preview: lim (kl9356) – Homework 18 – Weathers – (22202) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points An LC circuit is shown in the figure below. The 31 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 16 mH inductor. 16 mH 31 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 . 000308119 A. Explanation: Let : C = 31 pF = 3 . 1 × 10 − 11 F , L = 16 mH = 0 . 016 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 = radicalbigg C L V = radicalbigg 3 . 1 × 10 − 11 F . 016 H (7 V) = . 000308119 A . 002 (part 2 of 3) 10.0 points What is the natural angular frequency of the circuit? Correct answer: 1 . 4199 × 10 6 rad / s. Explanation: ω = 1 √ LC = 1 radicalbig (0 . 016 H) (3 . 1 × 10 − 11 F) = 1 . 4199 × 10 6 rad / s . 003 (part 3 of 3) 10.0 points What is the maximum energy stored in the magnetic field of the inductor? Correct answer: 7 . 595 × 10 − 10 J. Explanation: U max,ind = U max,cap = Q 2 max 2 C = 1 2 C V 2 . Therefore U max,ind = 1 2 C V 2 = 1 2 (3 . 1 × 10 − 11 F) (7 V) 2 = 7 . 595 × 10 − 10 J ....
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## This note was uploaded on 09/07/2010 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Homework 18 - lim(kl9356 – Homework 18 – Weathers –(22202 1 This print-out should have 11 questions Multiple-choice questions may continue on

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