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# ohw24-solutions - howard(cah3459 ohw24 turner(56705 This...

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howard (cah3459) – ohw24 – turner – (56705) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 7) 10.0 points Consider a parallel circuit with a 99 Ω re- sistor, a 6 . 1 μ ± capacitor, a 120 mH induc- tor, and a 718 rad / s AC supply operating at 120 V rms , as shown in the fgure below. 120 mH 6 . 1 μ ± 99 Ω 120 V rms A typical phasor diagram is shown below. V S I R C L φ ω t The instantaneous voltages and rms volt- ages across the three circuit elements are the same, and each is in phase with the current through the resistor. The currents in C and L lead or lag behind the current in the resistor, as shown above. What is the resonance Frequency oF the cir- cuit? Correct answer: 186 . 022 cycles / s. Explanation: Let : ω = 718 rad / s , L = 120 mH = l u , C =6 . 1 μ ± = 6 . 1 × 10 - 6 ± . I S = ± I 2 R +[ I C - I L ] 2 , = V S Z = ² ³ V S R ´ 2 + ³ V S X C - V S X L ´ 2 1 Z = ² 1 R 2 + ³ 1 X C - 1 X L ´ 2 Z = 1 ² 1 R 2 + ³ 1 X C - 1 X L ´ 2 = 1 ² 1 R 2 + ³ ω C - 1 ω L ´ 2 , ThereFore, I S = V S ² 1 R 2 + ³ - 1 ´ 2 , and I S is minimum when = 1 ; i.e. , at its reso- nance Frequency. f = ω 2 π = 1 2 π LC = 1 2 π µ (0 . 12 H) (6 . 1 × 10 - 6 ±) = 186 . 022 cycles / s . 002 (part 2 of 7) 10.0 points At 718 rad / s, calculate the rms current in the resistor. Correct answer: 1 . 21212 A rms . Explanation: Let : V S = 120 V rms and R = 99 Ω . I R = V S R = 120 V rms 99 Ω = 1 . 21212 A rms . 003 (part 3 of 7) 10.0 points At 718 rad / s, calculate the rms current in the inductor.

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