27 - Shie, Gary Homework 27 Due: Nov 8 2004, 4:00 am Inst:...

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Shie, Gary – Homework 27 – Due: Nov 8 2004, 4:00 am – Inst: Turner 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Consider a series RLC circuit. The applied voltage has a maximum value oF 150 V and oscillates at a Frequency oF 82 Hz. The circuit contains a capacitor whose capacitance can be varied, a 1100 Ω resistor, and a 3 . 5 H inductor. Determine the value oF the capacitor such that the voltage across the capacitor is out oF phase with the applied voltage by 33 , with V L leading V max . Correct answer: 0 . 555002 μ ±. Explanation: The phase relationships For the voltage drops across the elements in the circuit are shown in the fgure. V R V L V C V max φ α ±rom the fgure, we see that the phase angle is φ = α - 90 = (90 ) - (33 ) = 57 = 0 . 994838 rad . This is because the phasors representing I max and V R are in the same direction (they are in phase). ±rom the equation tan φ = X L - X C R (1) and ω = 2 π f = 2 π (82 Hz) = 515 . 221 Hz . Substituting the expressions For X L and X C , X L = ω L = (515 . 221 Hz) (3 . 5 H) = 1803 . 27 Ω . and X C = 1 ω C , into the above relation, we obtain 1 2 π f C = 2 π f L + R tan φ, (2) we obtain C = 1 2 π f 1 2 π f L + R tan φ = 1 ω X L + R tan φ - 1 = 1 (515 . 221 Hz) (1803 . 27 Ω) + (1100 Ω) tan(57 ) - 1 = 0 . 555002 μ ± . 002 (part 1 oF 1) 10 points The emF E drives the circuit shown below at an angular velocity ω. R E C L The light bulb (with resistance R ) glows most brightly at 1. both higher Frequecies, 1 LC < ω < , or lower Frequencies, 0 < ω < 1 LC . 2. lower Frequencies, 0 < ω < 1 LC . 3. steady DC voltage, ω = 0 . correct
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Shie, Gary – Homework 27 – Due: Nov 8 2004, 4:00 am – Inst: Turner 2 4. the resonant frequency, ω = 1 LC . 5. higher frequencies, 1 LC < ω < . Explanation:
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27 - Shie, Gary Homework 27 Due: Nov 8 2004, 4:00 am Inst:...

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