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The value of the driving frequency at which the voltage across the capacitor becomes maximum in a series RLC ac fed circuit

Physics Asked on August 12, 2021

The circuit diagram is as shown.

As per the book voltage across the capacitor is maximum during resonance that is p=1/√(LC)

But what I found out is a bit different .
And it is.
p=√((1/LC)-(R²/2L²))

I got this result by solving the 2nd order degree 1 differential equation in terms of q (charge) in capacitor and then found the voltage across across the capacitor as q/c and took its amplitude and differentiated it to get the result. I am not sure if I am right or wrong . please help me out enter image description here

3 Answers

As per the book voltage across the capacitor is maximum during resonance that is p=1/√(LC)

is an incorrect statement, rather when that condition is satisfied you have current resonance and the current through and the voltage across the resistor is a maximum.
You are asked about charge resonance when the charge stored on the capacitor and the voltage across the capacitor is a maximum.
You have found that it occurs at a different frequency.
This is a related link.

Answered by Farcher on August 12, 2021

"As per the book voltage across the capacitor is maximum during resonance that is p=1/√(LC)"

EDIT: I looked closer after seeing R.W. Bird's answer. Updated below accordingly. You can learn something every day if you pay attention. ;--)

That is approximately correct statement for AC excitation when R is relatively small, but, apparently the OP equation is correct (and the exact solution).

Case 1: I ran a frequency scan for a series RLC circuit with these parameters: R = 1Ω, L = 26.5mH, and C = 10μF. A plot of the capacitor voltage amplitude vs. frequency is shown below. For a series circuit the resonance is a series resonance, |XC| = |XL|, so the current will be maximum at the resonant frequency.

$$ f = frac{1}{2pisqrt{LC}} = frac{1}{2pisqrt{(26.5E-3)(10E-6)}}= 309Hz$$ $$ f = frac{sqrt{frac{1}{LC}-frac{R^2}{2L^2}}}{2pi}= 309Hz$$

enter image description here

Case 2: I change R from 1Ω to 30Ω and re-ran. This time the peak voltage across the capacitor occured at 297Hz as per the OP's formula.

$$ f = frac{1}{2pisqrt{LC}} = frac{1}{2pisqrt{(26.5E-3)(10E-6)}}= 309Hz$$ $$ f = frac{sqrt{frac{1}{LC}-frac{R^2}{2L^2}}}{2pi}= 297Hz$$

enter image description here

Below is the same plot but i've added the voltage across the resistor too (red trace). Interestingly, the peak voltage across the R occurs at 309Hz - which jives with the link that Farcher gives in his answer.

enter image description here

Answered by relayman357 on August 12, 2021

I solved this problem many years ago and scanned the solution into my computer (messy derivative). My result agrees with yours. I also have a spreadsheet which solves RLC circuits (with numbers). I used “solver” to maximize the voltage on the capacitor, and found that the predicted angular frequency matched that suggested by your formula for several different sets of circuit parameters.

Answered by R.W. Bird on August 12, 2021

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