(III) Determine a formula for the average power

Question

\overline{P}

dissipated in an LRC circuit in terms of L, R, C, ω and V₀. Find an approximate formula for the width of the resonance peak in average power, Δω, which is the difference in the two (angular) frequencies where

\overline{P}

has half its maximum value. Assume a sharp peak.

Accepted Answer

Welcome back everyone. In this problem, an LR C circuit in a radio receiver consists of an inductor resistor and capacitor in series driven by an ac suicide varying frequencies at resonance max power is dissipated. Find an approximate formula for the width of the resonance peak delta F which is the difference between the two frequencies where the power is half of its max value. A hint is that we can use the relationship at half max power that says it's equal to V knot squared R divided by two multiplied by R squared plus omega L minus one divided by omega C squared. For a sharp peak assume that RCR squared C squared is much smaller than four LC A says that the width of the resonance peak is R divided by two pi lb says it's R divided by four pi LCR divided by two L and Dr divided by pi L. Now, if we're going to figure out the width of the resonance peak, let's make sure we understand exactly what's happening at resonance. OK. No, for starters, our problem tells us that at resonance max power is dissipated. But do we know anything else that happens at resonance. We recall that at resonance, OK, the resonance frequency multiplied by the inductance is going to be equal to one divided by the resonance frequency multiplied by the capacitance. In other words, the react, the inductive reactance is equal to the capacity reactant at resonance. OK. So that tells us then that the resonance angular frequency is going to be equal to one divided by the square root of LC. We also know that because the impedance is minimized. If we take our idea here, that means our maximum power OK. At resonance is going to be equal to or peak voltage squared divided by two multiplied by R or resistance. Now, how is that going to help us to figure out the width of the resonance peak? Well, let's set the power, OK, let's set the power like our problem tells us to half of the maximum value. OK. You know here is where the power is half of its maximum value. And let's try to use that to solve for our angular frequency. And thus help us to find the width of the resonance peak delta F. So solving for angular frequency at P equals a half of P max. Well, we can use what we already know. In other words, we have our formula for half P max. OK, which is gonna be or peak voltage squared divided by sorry, multiplied by R divided by two multiplied by R squared plus omega L minus one divided by Omega C squared. OK. And all of that is gonna be equal to a half of the formula that we found for our max power. So that's a half of V not squared divided by two R which is our peak voltage squared divided by four R. Now we can cancel some common terms here. OK? Can cancel two from both sides. And now if we do some cross multiplication, OK, then we should find that we get two R squared to be equal to R squared plus O mega L minus one divided by Omega C squared. The formula that we had in the denominator. Now, if we solve this formula for R squared, we get R squared to be equal to omega L minus one divided by Omega C all squared. No, if we square root both sides, then we get plus R minus R to be equal to Omega L di sorry minus one divided by Omega C. So we're almost able to solve for angular frequency. Let's try to get omega out of the denominator by multiplying the entire expression by Omega C. OK. And now when we do that, OK, and we take our formula and we set it to zero, we get Omega squared LC, OK plus R minus RC. Omega minus one equal to zero. No, this is good because notice here we have a squared term for Omega, a linear term and a constant. And now if you notice our coefficients here, our formula is a quadratic formula. Thus we can solve for omega quadratic. OK. So solving quadratic OK, using the quadratic formula, then we should get omega to be equal to negative RC plus R minus the square root of RC squared plus four LC. OK. Divided by two LC. OK. Using what we know about the quadratic formula. Now remember that earlier in our problem, we also said that for a sharp peak assume R squared C squared is much less than four LC, much smaller than four LC. And in our problem RC, all squared is going to give us R squared C squared. So when we apply the assumption for a sharp peak only considering positive angular frequencies where the angular width will be the difference between the two positive frequencies. OK, then we get omega to be equal to RC plus R minus the two multiplied by the square root of LC. Because now remember essentially this term has disappeared and the square root of four LC equals two multiplied by the square root of LC. And all of this is being divided by two LC. Now, if we take it a step further and simplify trying to write separate fractions, then we get it to be equal to R divided by two L plus R minus one divided by the square root of LC. Now to find the angular width by finding the difference between the two positive frequencies. As we said, then this means that the difference in angular width will be equal to one divided by the square root of LC plus R divided by two LM, one divided by the square root of LCMR divided by two L. Now, here we notice one divided by root LC minus one divided by root LC equals zero and R divided by two L minus negative R divided by two L equals two R divided by two L which we can simplify as R divided by L. So I change in angular frequency equals are divided by L. All we need to do now to figure out the width of the resonance peak is to convert our frequency. Now we know that F OK is equal to omega divided by two pi. So delta F equals delta omega divided by two pi which we know is going to be R divided by L. OK. Multiplied by one, divided by two pi or R divided by two pi L. So what does this mean? Well, no, this tells us then that for the width of the resonance peak where the difference, sorry for the width of the resonance peak, its approximate value is R divided by two pi L. If we look back at our answer choices that tells us a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Average Power in LRC Circuits

Resonance in LRC Circuits

Width of the Resonance Peak (Δω)

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