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# solution_pdf30 - tovar(jdt436 – homework 30 – Turner...

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Unformatted text preview: tovar (jdt436) – homework 30 – Turner – (59070) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider a series RLC circuit. The applied voltage has a maximum value of 210 V and oscillates at a frequency of 86 Hz. The circuit contains a capacitor whose capacitance can be varied, a 1000 Ω resistor, and a 3 . 9 H inductor. Determine the value of the capacitor such that the voltage across the capacitor is out of phase with the applied voltage by 50 ◦ , with V L leading V max . Correct answer: 0 . 628085 μ F. Explanation: The phase relationships for the voltage drops across the elements in the circuit are shown in the figure. V R V L V C V m a x φ α Let : α = 50 ◦ , f = 86 Hz , L = 3 . 9 H , R = 1000 Ω . From the figure, the phase angle is φ = 90 ◦- α = 90 ◦- 50 ◦ = 40 ◦ , since the phasors representing I max and V R are in the same direction (in phase). The angular frequency is ω = 2 π f = 2 π (86 Hz) = 540 . 354 Hz . We know that X L = ω L = (540 . 354 Hz) (3 . 9 H) = 2107 . 38 Ω and X C = 1 ω C . From the equation tan φ = X L- X C R we obtain 1 ω C = ω L + R tan φ, C = 1 ω bracketleftbigg 1 ω L + R tan φ bracketrightbigg = 1 ω bracketleftbigg 1 X L + R tan φ bracketrightbigg = 1 540 . 354 Hz × 1 2107 . 38 Ω + (1000 Ω) tan 40 ◦ × 10 6 μ F 1 F = . 628085 μ F . 002 10.0 points Consider a series RLC circuit. The applied voltage has a maximum value of 120 V and oscillates at a frequency of 53 Hz. The circuit contains an inductor whose inductance can be varied, a 700 Ω resistor, and a 1...
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## This note was uploaded on 12/11/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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solution_pdf30 - tovar(jdt436 – homework 30 – Turner...

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