This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: nguyen (jmn727) – 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 110 V and oscillates at a frequency of 83 Hz. The circuit contains a capacitor whose capacitance can be varied, a 1200 Ω resistor, and a 2 . 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 37 ◦ , with V L leading V max . Correct answer: 0 . 617598 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 = 37 ◦ , f = 83 Hz , L = 2 . 9 H , R = 1200 Ω . From the figure, the phase angle is φ = 90 ◦- α = 90 ◦- 37 ◦ = 53 ◦ , 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 (83 Hz) = 521 . 504 Hz . We know that X L = ω L = (521 . 504 Hz) (2 . 9 H) = 1512 . 36 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 = 1 521 . 504 Hz × 1 1512 . 36 Ω + (1200 Ω) tan 53 ◦ × 10 6 μ F 1 F = . 617598 μ F . 002 10.0 points Consider a series RLC circuit. The applied voltage has a maximum value of 150 V and oscillates at a frequency of 88 Hz. The circuit contains an inductor whose inductance can be varied, a 1000 Ω resistor, and a 1...
View Full Document
This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
- Spring '08