Practice HW 8 solution

# Practice HW 8 solution - song(shs546 – Practice HW 8...

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Unformatted text preview: song (shs546) – Practice HW 8 Solutions – weathers – (22202) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A fully charged capacitor stores 7 . 4 J of en- ergy. How much energy remains when its charge has decreased to half its original value? Correct answer: 1 . 85 J. Explanation: Let : U = 7 . 4 J . Q = C V , so the energy is U = 1 2 C V 2 = 1 2 Q 2 C . If Q is halved, U falls to one-quarter, as shown U f = 1 2 Q 2 2 C = 1 2 Q 2 4 C = 1 4 parenleftbigg 1 2 Q 2 C parenrightbigg = 1 4 U = 1 4 (7 . 4 J) = 1 . 85 J . 002 10.0 points Calculate the energy stored in a(n) 7 . 7 μ F capacitor when it is charged to a potential of 146 V. Correct answer: 82 . 0666 mJ. Explanation: Let : C = 7 . 7 μ F = 7 . 7 × 10 − 6 F and V = 146 V . The energy stored in a capacitor is U = 1 2 C V 2 = 1 2 ( 7 . 7 × 10 − 6 F ) × (146 V) 2 parenleftbigg 10 3 mJ 1 J parenrightbigg = 82 . 0666 mJ . 003 (part 1 of 2) 10.0 points Two capacitors of 21 μ F and 5 . 9 μ F are con- nected parallel and charged with a 52 V power supply. Calculate the total energy stored in the two capacitors. Correct answer: 0 . 0363688 J. Explanation: Let : C 1 = 21 μ F and C 2 = 5 . 9 μ F . U = C V 2 2 C parallel = C 1 + C 2 . Each capacitor has voltage V, so U = 1 2 ( C 1 + C 2 ) V 2 = 1 2 (21 μ F + 5 . 9 μ F) (52 V) 2 = . 0363688 J . 004 (part 2 of 2) 10.0 points What potential difference would be required across the same two capacitors connected in series in order for the combination to store the same energy as in the first part?...
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Practice HW 8 solution - song(shs546 – Practice HW 8...

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