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HW26Sol - Platt David Homework 26 Due Nov 9 2005 4:00 am...

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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 – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 3) 10 points An e±ective AC voltage oF 145 V at 60 Hz is applied to an RLC circuit with 27 . 4 mH inductor, 0 . 52 μ ² capacitor, and 45 Ω resistor in series. 27 . 4 mH 45 Ω 145 V S 0 . 52 μ ² I What is the e±ective current? Correct answer: 0 . 0284817 A. Explanation: Let : V = 145 V , f = 60 Hz , C = 0 . 52 μ ² = 5 . 2 × 10 - 7 ² , L = 27 . 4 mH = 0 . 0274 H , and R = 45 Ω . ²or 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 ²) = 5101 . 12 Ω , so the total impedance is Z = q R 2 + ( X L - X C ) 2 = (45 Ω) 2 + h (10 . 3296 Ω) - (5101 . 12 Ω) i 2 ¾ 1 / 2 = 5090 . 99 Ω . Thus the e±ective current is I = V Z = 145 V 5090 . 99 Ω = 0 . 0284817 A . 002 (part 2 oF 3) 10 points What is the power Factor oF the circuit? Correct answer: 0 . 00883915 .
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HW26Sol - Platt David Homework 26 Due Nov 9 2005 4:00 am...

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