Old Midterm 4

# Old Midterm 4 - benavides(jjb2356 – oldmidterm 04 –...

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Unformatted text preview: benavides (jjb2356) – 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 24 Ω, the inductance is 15 mH, and the capac- itance is 20 μ F. The maximum potential is 162 V, and the angular frequency is 100 rad / s. Calculate the maximum current in the cir- cuit. Correct answer: 0 . 324599 A. Explanation: Let : R = 24 Ω , L = 15 mH = 0 . 015 H , C = 20 μ F = 2 × 10 − 5 F , V max = 162 V , and ω = 100 rad / s . The capacitive reactance is X C = 1 ω C = 1 (100 rad / s) (2 × 10 − 5 F) = 500 Ω and the inductive reactance is X L = ω L = (100 rad / s) (0 . 015 H) = 1 . 5 Ω , so the maximum current is I max = V max Z = V max radicalbig R 2 + ( X L − X C ) 2 = 162 V radicalbig (24 Ω) 2 + (1 . 5 Ω − 500 Ω) 2 = . 324599 A . 002 10.0 points A certain capacitor in a circuit has a capaci- tive reactance of 42 . 7 Ω when the frequency is 150 Hz. What capacitive reactance does the capac- itor have at a frequency of 10900 Hz? Correct answer: 0 . 587615 Ω. Explanation: Let : X C = 42 . 7 Ω , f low = 150 Hz , and f high = 10900 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 = (42 . 7 Ω) parenleftbigg 150 Hz 10900 Hz parenrightbigg = . 587615 Ω . 003 10.0 points A 93 mH inductor is connected to a outlet where the rms voltage is 78 . 9 V and the fre- quency is 79 Hz. Determine the energy stored in the inductor at t = 7 . 8 ms, assuming that this energy is zero at t = 0. Correct answer: 0 . 120849 J. Explanation: Let : t = 7 . 8 ms = 0 . 0078 s , f = 79 Hz , L = 93 mH = 0 . 093 H , and V rms = 78 . 9 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 benavides (jjb2356) – 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(78 . 9 V) 2 π (79 Hz) (0 . 093 H) × sin[(2 π (79 Hz) (0 . 0078 s)] = − 1 . 61211 A , so the potential energy stored in the inductor is U = 1 2 LI 2 = 1 2 (0 . 093 H) ( − 1 . 61211 A) 2 = . 120849 J . 004 10.0 points A ideal transformer shown in the figure below having a primary with 40 turns and secondary with 16 turns. The load resistor is 66 Ω. The source voltage is 140 V rms . 140V rms 40turns 16turns 66Ω What is the rms electric potential across the 66 Ω load resistor?...
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## This note was uploaded on 03/26/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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Old Midterm 4 - benavides(jjb2356 – oldmidterm 04 –...

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