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Practice Homework 19 2010

# Practice Homework 19 2010 - choe(dec865 Practice HW 19...

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choe (dec865) – 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 – find all choices before answering. 001 10.0 points Calculate the average power delivered to the series RLC circuit with resistance of 367 Ω, inductance of 0 . 591 H, capacitor of 2 . 75 μ F, and frequency of 219 s 1 , if the maximum value of the applied voltage equal to 165 V. Correct answer: 2 . 0155 W. Explanation: Let : ω = 219 s 1 , L = 0 . 591 H , C = 2 . 75 μ F = 2 . 75 × 10 6 F , R = 367 Ω , and V max = 165 V . The reactances are X L = ω L = (219 s 1 ) (0 . 591 H) = 129 . 429 Ω and X C = 1 ω C = 1 (219 s 1 ) (2 . 75 × 10 6 F) = 1660 . 44 Ω , so the impedance is Z = radicalBig R 2 + ( X L - X C ) 2 = radicalBig (367 Ω) 2 + (129 . 429 Ω - 1660 . 44 Ω) 2 = 1574 . 38 Ω . The rms voltage and current are V rms = V max 2 = 165 V 2 = 116 . 673 V and I rms = V rms Z = 116 . 673 V 1574 . 38 Ω = 0 . 0741069 A . The phase angle is φ = arctan X L - X C R = arctan 129 . 429 Ω - 1660 . 44 Ω 367 Ω = - 76 . 52 , so the average power is P av = I rms V rms cos φ = (0 . 0741069 A) (116 . 673 V) × cos( - 76 . 52 ) = 2 . 0155 W . 002 (part 1 of 3) 10.0 points In a certain RLC circuit, the rms current is 5 . 1 A, the rms voltage is 294 V, and the current leads the voltage by 59 . 2 .

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