Practice HW 19 solution

Practice HW 19 solution - song (shs546) Practice HW 19...

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song (shs546) – Practice HW 19 Solutions – weathers – (22202) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 0.0 points Calculate the average power delivered to the series RLC circuit with resistance oF 161 Ω, inductance oF 0 . 597 H, capacitor oF 3 . 08 μ ±, and Frequency oF 330 s 1 , iF the maximum value oF the applied voltage equal to 129 V. Correct answer: 2 . 0767 W. Explanation: Let : ω = 330 s 1 , L = 0 . 597 H , C = 3 . 08 μ ± = 3 . 08 × 10 6 ± , R = 161 Ω , and V max = 129 V . The reactances are X L = ω L = (330 s 1 ) (0 . 597 H) = 197 . 01 Ω and X C = 1 ω C = 1 (330 s 1 ) (3 . 08 × 10 6 ±) = 983 . 865 Ω , so the impedance is Z = r R 2 + ( X L - X C ) 2 = r (161 Ω) 2 + (197 . 01 Ω - 983 . 865 Ω) 2 = 803 . 157 Ω . The rms voltage and current are V rms = V max 2 = 129 V 2 = 91 . 2168 V and I rms = V rms Z = 91 . 2168 V 803 . 157 Ω = 0 . 113573 A . The phase angle is φ = arctan X L - X C R = arctan 197 . 01 Ω - 983 . 865 Ω 161 Ω = - 78 . 4362 , so the average power is P av = I rms V rms cos φ = (0 . 113573 A) (91 . 2168 V) × cos( - 78 . 4362 ) = 2 . 0767 W . 002 (part 1 of 3) 0.0 points
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Practice HW 19 solution - song (shs546) Practice HW 19...

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