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).

Pav Small R, large Q Δω Large R, small Q Ao- fo Δω Δ/ 31-60 E2 Ro? rms

Fantastic news! We've Found the answer you've been seeking!

Step by Step Answer:

Related Book For  book-img-for-question
Question Posted: