Figure shows a plot of average power P av versus generator frequency for an RLC circuit
Question:
Figure shows a plot of average power Pav versus generator frequency ω for an RLC circuit with a generator. The average power Pav is given by Equation 31-58. The "full width at half-maximum" ∆ω is the width of the resonance curve between the two points where Pav is one-half its maximum value. Show that, for a sharply peaked resonance, ∆ω ≈ R/L and, hence, that Q ≈ ω0/∆ω in this case (Equation 31-60).
Transcribed Image Text:
Pav Small R, large Q Δω Large R, small Q Ao- fo Δω Δ/ 31-60 E2 Ro? rms" Pay 31-58 2 (m? – 03)? +o² R? av
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Fundamentals of Ethics for Scientists and Engineers
ISBN: 978-0195134889
1st Edition
Authors: Edmund G. Seebauer, Robert L. Barry
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