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# OMT 04 - terry(ect328 oldmidterm 04 Turner(59130 This...

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terry (ect328) – oldmidterm 04 – Turner – (59130) 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 23 Ω, the inductance is 37 mH, and the capac- itance is 22 μ F. The maximum potential is 119 V, and the angular frequency is 100 rad / s. Calculate the maximum current in the cir- cuit. Correct answer: 0 . 263606 A. Explanation: Let : R = 23 Ω , L = 37 mH = 0 . 037 H , C = 22 μ F = 2 . 2 × 10 5 F , V max = 119 V , and ω = 100 rad / s . The capacitive reactance is X C = 1 ω C = 1 (100 rad / s) (2 . 2 × 10 5 F) = 454 . 545 Ω and the inductive reactance is X L = ω L = (100 rad / s) (0 . 037 H) = 3 . 7 Ω , so the maximum current is I max = V max Z = V max radicalbig R 2 + ( X L X C ) 2 = 119 V radicalbig (23 Ω) 2 + (3 . 7 Ω 454 . 545 Ω) 2 = 0 . 263606 A . 002 10.0 points A certain capacitor in a circuit has a capaci- tive reactance of 32 . 8 Ω when the frequency is 103 Hz. What capacitive reactance does the capac- itor have at a frequency of 4880 Hz? Correct answer: 0 . 692295 Ω. Explanation: Let : X C = 32 . 8 Ω , f low = 103 Hz , and f high = 4880 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 = (32 . 8 Ω) parenleftbigg 103 Hz 4880 Hz parenrightbigg = 0 . 692295 Ω . 003 10.0 points A 90 mH inductor is connected to a outlet where the rms voltage is 138 V and the fre- quency is 43 Hz. Determine the energy stored in the inductor at t = 5 . 4 ms, assuming that this energy is zero at t = 0. Correct answer: 2 . 8627 J. Explanation: Let : t = 5 . 4 ms = 0 . 0054 s , f = 43 Hz , L = 90 mH = 0 . 09 H , and V rms = 138 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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terry (ect328) – oldmidterm 04 – Turner – (59130) 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 (138 V) 2 π (43 Hz) (0 . 09 H) × sin[(2 π (43 Hz) (0 . 0054 s)] = 7 . 97593 A , so the potential energy stored in the inductor is U = 1 2 L I 2 = 1 2 (0 . 09 H) (7 . 97593 A) 2 = 2 . 8627 J . 004 10.0 points A ideal transformer shown in the figure below having a primary with 15 turns and secondary with 12 turns. The load resistor is 53 Ω. The source voltage is 110 V rms . 110 V rms 15 turns 12 turns 53 Ω What is the rms electric potential across the 53 Ω load resistor? Correct answer: 88 V rms . Explanation: Let : N 1 = 15 turns , N 2 = 12 turns , and V 1 = 110 V rms . The rms voltage across the transformer’s secondary is V 2 = N 2 N 1 V 1 = 12 turns 15 turns (110 V rms ) = 88 V rms , which is the same as the electric potential across the load resistor.
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