Practice HW 18 solution

Practice HW 18 solution - song (shs546) Practice HW 18...

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Unformatted text preview: song (shs546) Practice HW 18 Solutions weathers (22202) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 0.0 points Initially, the capacitor in a series LC circuit is charged. A switch is closed, allowing the capacitor to discharge, and 1 . 3 s later the energy stored in the capacitor is one-fourth its initial value. Find the inductance if the capacitance is 6 pF. Correct answer: 0 . 256849 H. Explanation: Let : t = 1 . 3 s = 1 . 3 10 6 s and C = 6 pF = 6 10 12 F . The capacitor charge in an oscillating LC circuit is Q ( t ) = Q max cos( t + ) , where = 1 L C . To determine the phase angle in the above equation, we use the initial condition that, at t = 0, the charge is maximum. Thus Q max = Q max cos 1 = cos = 0 rad . The energy stored in the capacitor at any given time is U ( t ) = 1 2 Q ( t ) 2 C . When the energy is one-fourth its initial value, we find that U ( t ) = 1 4 U (0) 1 2 Q ( t ) 2 C = 1 4 parenleftbigg Q 2 max 2 C parenrightbigg Q ( t ) = 1 2 Q max 1 2 = cos parenleftbigg t L C parenrightbigg t L C = cos 1 parenleftbigg 1 2 parenrightbigg = 3 rad . Solving for L gives L = 9 t 2 2 C = 9 (1 . 3 10 6 s) 2 2 (6 10 12 F) = . 256849 H . 002 (part 1 of 3) 0.0 points An LC circuit is shown in the figure below. The 11 pF capacitor is initially charged by the 13 V battery when S is at position a . Then S is thrown to position b so that the capacitor is shorted across the 1 . 5 mH inductor. 1 . 5 mH 11 pF 13 V S b a Find the frequency of the oscillations. Correct answer: 1 . 23902 10 6 Hz. Explanation: L C E S b a Let : C = 11 pF = 1 . 1 10 11 F , L = 1 . 5 mH = 0 . 0015 H , and E = 13 V . song (shs546) Practice HW 18 Solutions weathers (22202) 2 Using the equation = 1...
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Practice HW 18 solution - song (shs546) Practice HW 18...

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