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

Explanation

(a) `omega_(0)=4167 "rad s"^(-1), v_(0)=663Hz`
`I_(0)^("max")=14.1A`
(b) `barP=(1//2)I_(0)^(2)R` which is maximum at the same frequency (663 Hz) for which `I_(0)` is maximum `barP_("max")=(1//2)(I_("max"))^(2)R=2300W`.
(c ) At `omega=omega_(0)pm Deltaw` [Approximation good if `(R//2L) lt lt omega_(0)`].
`Delta omega=R//2L=95.8"rad s"^(-1), Deltav=Delta omega//2pi =15.2Hz.`
Power absorbed is half the peak power at n = 648 Hz and 678 Hz.
At these frequencies, current amplitude is `(1//sqrt(2))` times `I_(0)^("max")`, i.e., current amplitude(at half the peak power points) is 10 A.
(d) Q = 21.7