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HW #10-solutions

# HW #10-solutions - Brannon(blb675 HW#10 Shih(58250 This...

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Brannon (blb675) – HW #10 – Shih – (58250) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points An LC circuit is shown in the figure below. The 12 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 12 pF 13 V S b a Find the frequency of the oscillations. Correct answer: 1 . 18627 × 10 6 Hz. Explanation: L C E S b a Let : C = 12 pF = 1 . 2 × 10 - 11 F , L = 1 . 5 mH = 0 . 0015 H , and E = 13 V . Using the equation ω = 1 L C , the frequency is f = ω 2 π = 1 2 π L C = 1 2 π (0 . 0015 H) (1 . 2 × 10 - 11 F) = 1 . 18627 × 10 6 Hz . 002 (part 2 of 3) 10.0 points What is the maximum value of charge on the capacitor? Correct answer: 1 . 56 × 10 - 10 C. Explanation: The initial charge on the capacitor equals the maximum charge, so the maximum charge is Q max = C E = (1 . 2 × 10 - 11 F) (13 V) = 1 . 56 × 10 - 10 C . 003 (part 3 of 3) 10.0 points What is the maximum value of current in the circuit? Correct answer: 0 . 00116276 A. Explanation: From I = - ω Q max sin( ω t ) = - I max sin( ω t ) , we see that the maximum current is I max = ω Q max = 2 π f Q max = 0 . 00116276 A . 004 (part 1 of 2) 10.0 points A 0 . 244 H inductor is connected to a capacitor and a 40 . 4 Ω resistor along with a 47 Hz, 35 . 5 V generator. To what value would the capacitor have to be adjusted to produce resonance? Correct answer: 46 . 9953 μ F. Explanation:

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Brannon (blb675) – HW #10 – Shih – (58250) 2 Let : f 0 = 47 Hz , and L = 0 . 244 H . From f 0 = 1 2 π L C , the capacitance is C = 1 4 π 2 L f 0 2 = 1 4 π 2 (0 . 244 H) (47 Hz) 2 · 10 6 μ F 1 F = 46 . 9953 μ F . 005 (part 2 of 2) 10.0 points Find the voltage drop across the resistor at resonance. Correct answer: 35 . 5 V. Explanation: Let : V = 35 . 5 V . At resonance, X L = X C . Therefore, Z across the capacitor-inductor combination is zero, and the voltage drop across the resistor is 35 . 5 V.
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