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oldmidterm 04 - nguyen(jmn727 oldmidterm 04 Turner(59070...

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nguyen (jmn727) – oldmidterm 04 – Turner – (59070) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In a series RLC AC circuit, the resistance is 10 Ω, the inductance is 13 mH, and the capac- itance is 14 μ F. The maximum potential is 70 V, and the angular frequency is 100 rad / s. Calculate the maximum current in the cir- cuit. Correct answer: 0 . 098169 A. Explanation: Let : R = 10 Ω , L = 13 mH = 0 . 013 H , C = 14 μ F = 1 . 4 × 10 5 F , V max = 70 V , and ω = 100 rad / s . The capacitive reactance is X C = 1 ω C = 1 (100 rad / s) (1 . 4 × 10 5 F) = 714 . 286 Ω and the inductive reactance is X L = ω L = (100 rad / s) (0 . 013 H) = 1 . 3 Ω , so the maximum current is I max = V max Z = V max radicalbig R 2 + ( X L X C ) 2 = 70 V radicalbig (10 Ω) 2 + (1 . 3 Ω 714 . 286 Ω) 2 = 0 . 098169 A . 002 10.0 points A certain capacitor in a circuit has a capaci- tive reactance of 36 . 3 Ω when the frequency is 110 Hz. What capacitive reactance does the capac- itor have at a frequency of 5810 Hz? Correct answer: 0 . 687263 Ω. Explanation: Let : X C = 36 . 3 Ω , f low = 110 Hz , and f high = 5810 Hz . The capacitive reactance is X C = C 2 π f , so X C ( high ) X C ( low ) = 2 π f low C 2 π f high C = f low f high . Thus X C ( high ) = X C ( low ) f low f high = (36 . 3 Ω) parenleftbigg 110 Hz 5810 Hz parenrightbigg = 0 . 687263 Ω . 003 10.0 points A 71 mH inductor is connected to a outlet where the rms voltage is 95 . 8 V and the fre- quency is 15 Hz. Determine the energy stored in the inductor at t = 7 . 6 ms, assuming that this energy is zero at t = 0. Correct answer: 6 . 27355 J. Explanation: Let : t = 7 . 6 ms = 0 . 0076 s , f = 15 Hz , L = 71 mH = 0 . 071 H , and V rms = 95 . 8 V . The inductive reactance is X L = ω L = 2 π f L and the maximum current is I max = 2 V rms X L = 2 V rms 2 π f L
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nguyen (jmn727) – oldmidterm 04 – Turner – (59070) 2 The current at time t is I = I max sin( ω t ) = 2 V rms 2 π f L sin(2 π f t ) = 2 (95 . 8 V) 2 π (15 Hz) (0 . 071 H) × sin[(2 π (15 Hz) (0 . 0076 s)] = 13 . 2936 A , so the potential energy stored in the inductor is U = 1 2 L I 2 = 1 2 (0 . 071 H) (13 . 2936 A) 2 = 6 . 27355 J . 004 10.0 points A ideal transformer shown in the figure below having a primary with 45 turns and secondary with 36 turns. The load resistor is 74 Ω. The source voltage is 75 V rms . 75 V rms 45 turns 36 turns 74 Ω What is the rms electric potential across the 74 Ω load resistor? Correct answer: 60 V rms . Explanation: Let : N 1 = 45 turns , N 2 = 36 turns , and V 1 = 75 V rms . The rms voltage across the transformer’s secondary is V 2 = N 2 N 1 V 1 = 36 turns 45 turns (75 V rms ) = 60 V rms , which is the same as the electric potential across the load resistor.
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