Homework 11-solutions

Homework 11-solutions - im(gi768 Homework 11 markert(55325...

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im (gi768) – Homework 11 – markert – (55325) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of3)10.0points Consider the LC circuit shown below. Switch S is initially open, and the capaci- tor has a charge Q m on its plates. At t = 0 the switch is closed. L Q m C S What will be the energy U C stored in the capacitor as a function of time? 1. U C = parenleftbigg Q 2 m 2 C parenrightbigg cos parenleftBig t L C parenrightBig 2. U C = parenleftbigg Q 2 m 2 C parenrightbigg cos 2 parenleftbigg t L C parenrightbigg correct 3. U C = parenleftbigg Q 2 m 2 C parenrightbigg cos parenleftbigg t L C parenrightbigg 4. U C = parenleftbigg Q 2 m C parenrightbigg sin 2 parenleftbigg t L C parenrightbigg 5. U C = parenleftbigg Q 2 m 2 C parenrightbiggparenleftbigg 1 - exp bracketleftbigg - t L C bracketrightbiggparenrightbigg 6. U C = parenleftbigg Q 2 m C parenrightbigg cos 2 parenleftbigg t L C parenrightbigg 7. U C = parenleftbigg Q 2 m 2 C parenrightbigg sin 2 parenleftBigg t radicalbigg L C parenrightBigg 8. U C = parenleftbigg Q 2 m 2 C parenrightbigg exp parenleftbigg - t L C parenrightbigg 9. U C = parenleftbigg Q 2 m 2 C parenrightbigg sin 2 parenleftbigg t L C parenrightbigg 10. U C = Q 2 m 2 C Explanation: The charge on the capacitor in the LC circuit satisfies d 2 Q dt 2 = - 1 L C Q . The solution is Q = Q m cos parenleftbigg t L C parenrightbigg , where Q m is the initial charge on the capaci- tor. Thus the energy is given by U c = Q 2 2 C = Q 2 m 2 C cos 2 parenleftbigg t L C parenrightbigg . 002(part2of3)10.0points What is the energy U L stored in the inductor as a function of time? 1. U L = parenleftbigg Q 2 m 2 C parenrightbigg sin parenleftbigg t L C parenrightbigg 2. U L = parenleftbigg Q 2 m C parenrightbigg cos 2 parenleftBigg t radicalbigg L C parenrightBigg 3. U L = Q 2 m 2 C 4. U L = parenleftbigg Q 2 m 2 C parenrightbigg sin 2 parenleftbigg t L C parenrightbigg correct 5. U L = parenleftbigg Q 2 m 2 C parenrightbigg cos 2 parenleftBig t L C parenrightBig 6. U L = parenleftbigg Q 2 m 2 C parenrightbigg exp parenleftbigg - t L C parenrightbigg 7. U L = parenleftbigg Q 2 m 2 C parenrightbigg sin 2 parenleftBigg t radicalbigg L C parenrightBigg 8. U L = parenleftbigg Q 2 m C parenrightbigg sin 2 parenleftbigg t L C parenrightbigg 9. U L = parenleftbigg Q 2 m 2 C parenrightbiggparenleftbigg 1 - exp bracketleftbigg - t L C bracketrightbiggparenrightbigg 10. U L = parenleftbigg Q 2 m 2 C parenrightbigg cos parenleftbigg t L C parenrightbigg Explanation: Given the charge in the capacitor, we can find the current by I = d Q dt = - Q L C sin parenleftbigg t L C parenrightbigg .
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