HW26Sol - Platt David – Homework 26 – Due Nov 9 2005...

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Unformatted text preview: Platt, David – Homework 26 – Due: Nov 9 2005, 4:00 am – Inst: Ken Shih 1 This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points An effective AC voltage of 145 V at 60 Hz is applied to an RLC circuit with 27 . 4 mH inductor, 0 . 52 μ F capacitor, and 45 Ω resistor in series. 27 . 4 mH 45 Ω 145 V S . 52 μ F I What is the effective current? Correct answer: 0 . 0284817 A. Explanation: Let : V = 145 V , f = 60 Hz , C = 0 . 52 μ F = 5 . 2 × 10- 7 F , L = 27 . 4 mH = 0 . 0274 H , and R = 45 Ω . For an RLC series circuit at f = 60 Hz, X L = 2 π f L = 2 π (60 Hz) (0 . 0274 H) = 10 . 3296 Ω , and X C = 1 2 π f C = 1 2 π (60 Hz) (5 . 2 × 10- 7 F) = 5101 . 12 Ω , so the total impedance is Z = R 2 + ( X L- X C ) 2 = (45 Ω) 2 + (10 . 3296 Ω)- (5101 . 12 Ω) 2 1 / 2 = 5090 . 99 Ω ....
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HW26Sol - Platt David – Homework 26 – Due Nov 9 2005...

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