Question

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?
A. The source frequency `663 Hz`, current amplitude is maximum and this maximum value is `14.1 A`.
B. At the source frequency `663 Hz` average power absorbed by the circuit is maximum and the value of this maximum power is `2300 W`
C. At the frequencies `648 Hz, 678 Hz` of the source, the power transferred to the circuit is half the power at resonant frequency. The current amplitude at these frequencies is `10 A`
D. The `Q`-factor of the given circuit is 21.7

Answer 1

Correct Answer - A,B,C,D
(a) Amplitude of current is maximum at resonance frequency.
`f_(r) = (1)/(2 pi sqrt(LC))`
`= (1)/(2 xx 3.14 sqrt(0.12 xx 480 xx 10^(-9))) = 663 Hz`
`I_("rms") = (V_("rms"))/(R ) = (230)/(23) = 10 A I_(0) = I_("rms") sqrt(2) = 14.14 A`
(b) Resonance frequency, powe absorbed by circuit is maximum.
So `f = 663 Hz`
`P_(av) = I_("rms")^(2) R = (10)^(2) xx 23 = 2300` watt
(c ) `Q` factor `= (omega_(r))/(2 Delta omega) = (omega_(r) L)/(R ) implies Delta omega = (R )/(2 L)`
`rArr Delta omega = (23)/(2 xx 0.12) = (23)/(0.24) = 9.833`
Let `f` be half power frequency
`2 pi (f ~ f_(0)) = 95.833 implies f ~ f_(0) = 15 Hz`
`f = f_(0) +- 15 = 663 +- 15 = 648 Hz` and `678 Hz`
`I^(2) R = (1)/(2) P_("max") = (1)/(2) I_(0)^(2) R implies I = (I_(0))/(sqrt(2)) = 10` Amp
(d) `Q` factor
`(1)/(R ) sqrt((L)/(C)) = (1)/(23) sqrt((0.12)/(480 xx 10^(-9))) = 21.7`