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Unformatted text preview: Niu (qn269) – Homework #8 – antoniewicz – (58855) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 4.0 points An AC power supply with a maximum po tential of 88 V is across a series RLC circuit, where the inductance is 19 mH, the capaci tance is 99 nF, the resistance is 25 Ω. Find the resonant frequency. Correct answer: 3 . 66966 kHz. Explanation: Let : L = 19 mH = 0 . 019 H , and C = 99 nF = 9 . 9 × 10 − 8 F . The resonance frequency for a series RLC circuit is f = 1 2 π radicalbigg 1 LC = 1 2 π radicalBigg 1 (0 . 019 H) (9 . 9 × 10 − 8 F) · 1 kHz 10 3 Hz = 3 . 66966 kHz . 002 (part 2 of 3) 3.0 points Find the amplitude of the current at the res onant frequency. Correct answer: 3 . 52 A. Explanation: Let : R = 25 Ω , and V max = 88 V . At resonance, I max = V max R = 88 V 25 Ω = 3 . 52 A . 003 (part 3 of 3) 3.0 points Find the amplitude of the voltage across the inductor at resonance. Correct answer: 1542 . 06 V. Explanation: The amplitude of the voltage across the inductor at resonance is V L = X L I max = ω LI max = 2 π f LI max = 2 π (3669 . 66 Hz) (0 . 019 H) (3 . 52 A) = 1542 . 06 V . 004 10.0 points Consider the circuit shown in the figure, with resistance 637 Ω and and capacitance . 127 μ F. V in R C V out Calculate the ratio V out /V in for angular ve locity 436 rad / s. Correct answer: 0 . 03525. Explanation: Let : ω = 436 rad / s , C = 0 . 127 μ F = 1 . 27 × 10 − 7 F , and R = 637 Ω . The capacitive reactance is X C = 1 ω C = 1 (436 rad / s) (1 . 27 × 10 − 7 F) = 18059 . 7 Ω . Thus the gain is V out V in = RI Z I = R Z Niu (qn269) – Homework #8 – antoniewicz – (58855) 2 = R radicalBig R 2 + X 2 C = 637 Ω radicalbig (637 Ω) 2 + (18059 . 7 Ω) 2 = . 03525 . 005 (part 1 of 4) 3.0 points The primary coil of a certain transformer has an inductance of 2 . 73 H and a resistance of 68 . 4 Ω. If the primary coil is connected to an ac source with a frequency of 29 . 3 Hz and a voltage of 170 V rms , what is the rms current through the primary? Correct answer: 0 . 335161 A. Explanation: Let : f = 29 . 3 Hz , L = 2 . 73 H , R = 68 . 4 Ω , and V rms = 170 V . X L = 2 π f L = 2 π (29 . 3 Hz)(2 . 73 H) = 502 . 586 Ω , and Z = radicalBig X 2 L + R 2 = radicalBig (502 . 586 Ω) 2 + (68 . 4 Ω) 2 = 507 . 219 Ω . So the rms current is I rms = V rms Z = 170 V 507 . 219 Ω = . 335161 A . 006 (part 2 of 4) 3.0 points If the primary is connected to 93 . 6 V dc, what is the current through the primary? Disregard initial effects. Correct answer: 1 . 36842 A. Explanation: I = V R = 93 . 6 V 68 . 4 Ω = 1 . 36842 A ....
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 Spring '09
 Antoniewicz
 Power, Work

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