A series LCR circuit with `L= 0.12H, C=480 nF,` and `R=23 Omega` is connected to a `230V` variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum? Find this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum? Obtain the value of maximum power.
(c ) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency?
(d) What is the Q-factor of the circuit?

Explanation

C = 2 `mu F = 2 xx 10^(-6)`F
R = 100 `Omega`
L = 8 H
(i) Current in a circuit is maximum at resonance.
this is called resonant frequency. `omega = (1)/(sqrt(LC))`
(ii) In case of resonance impedance is equal to resistance.
Z = 100 `Omega , epsilon_(0) = 200` V
`I_(0) = (epsilon_(0))/(Z) = (200)/(100) ` = 2 A
(iv) Sharpness of a resonant circuit is described with the help of quality factor.
Quality factor (Q) = `(1)/(R) sqrt((L)/(C))`
(iii)

(iv ) If R increases, Q-factor decreases and sharpness also decreases. Higher the value of Q. sharper is the resonance.
When the resonance is very sharp, the circuit become more selective.