A series LCR circuit with `L = 0.12 H, C = 480 nF, R = 23 Omega` is connected to a 230 V variable frequency supply.
What is the source frequency for which current amplitude is maximum. Obtain this maximum vlaue.

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