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Unformatted text preview: homework 10 – KANUNGO, ANIL – Due: Sep 27 2007, 4:00 am 1 Question 1, chap 26, sect 4. part 1 of 2 10 points A parallelplate capacitor has a charge of 4 . 9 μ C when charged by a potential difference of 1 . 88 V . a) Find its capacitance. Correct answer: 2 . 60638 × 10 − 6 F (tolerance ± 1 %). Explanation: Let : Q = 4 . 9 μ C and Δ V 1 = 1 . 88 V . The capacitance is given by C = Q Δ V = 4 . 9 × 10 − 6 C 1 . 88 V = 2 . 60638 × 10 − 6 F . Question 2, chap 26, sect 4. part 2 of 2 10 points b) How much electrical potential energy is stored when this capacitor is connected to a 2 . 00 V battery? Correct answer: 5 . 21277 × 10 − 6 J (tolerance ± 1 %). Explanation: Let : Δ V 2 = 2 . 00 V PE electric = 1 2 C (Δ V ) 2 = 1 2 (2 . 60638 × 10 − 6 F) (2 V) 2 = 5 . 21277 × 10 − 6 J Question 3, chap 26, sect 4. part 1 of 2 10 points A 3 μ F capacitor is charged to 100 V. How much energy is stored in the capacitor? Correct answer: 15 mJ (tolerance ± 1 %). Explanation: Let : C = 3 μ F = 3 × 10 − 6 F and V = 100 V . The energy is U i = 1 2 C V 2 = 1 2 (3 × 10 − 6 F) (100 V) 2 × 1000 mJ 1 J = 15 mJ . Question 4, chap 26, sect 4....
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 Fall '08
 Turner
 Capacitance, Charge, Electric Potential, Energy, Work, Correct Answer, 0.00118651 J

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