7uHU41DbjLt4_1210016792_jwk572

7uHU41DbjLt4_1210016792_jwk572 - . 58947. Explanation: Let...

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inclasstest 08 – KIM, JI – Due: Mar 31 2008, 1:00 pm 1 Question 1, chap -1, sect -1. part 1 of 1 1 points This is a question for showing your atten- dance. Please choose the correct answer be- low, or you will lose 25% of this weekly test. 1. This is the correct answer. correct 2. This is the wrong answer. Explanation: It is correct because the choice does not lie. Question 2, chap 32, sect 4. part 1 of 1 3 points A 40 μ F capacitor is connected in series with a 10 mH inductance and a switch. The capacitor is ±rst charged to a voltage of 120 V. The charging battery is then removed. As soon as the switch is closed, the current be- gins to oscillate back and forth between one direction and the reversed direction. What is the maximum current in the cir- cuit? Correct answer: 7
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Unformatted text preview: . 58947. Explanation: Let : C = 40 μ F = 4 × 10-5 F , V = 120 V , and L = 10 mH = 0 . 01 H . The initial charge is Q max = C V = (4 × 10-5 F) (120 V) = 0 . 0048 C . Therefore, I max = ω Q max = Q max √ L C = . 0048 C r (0 . 01 H)(4 × 10-5 F) = 7 . 58947 A . Derivation of I max = ω Q max : 1. Initially, there is an electric ±eld energy stored in the capacitor, U C = 1 2 Q 2 max C . When the current achieves its maximum, all the energy in the system is stored as magnetic energy in the inductor, U L = 1 2 L I 2 max . From the conservation of energy, U L = U C , we get L I 2 max = Q 2 max C . Therefore I max = Q max √ L C . 2. Q = Q max cos ω t . I = dI dt =-ω Q max sin ω t . Therefore I max = ω Q max ....
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This note was uploaded on 05/05/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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