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A series LCR circuit with L = 0.12 H, C = 480 nF, R = 232 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 ? |
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Answer» <P> Solution :(a) Here L = 0.12 H, C = 480 nF `=480 xx 10^(-9) F, R = 23 Omega` and `V_(rms) = 230 V`Resonant frequency when the current is maximum `v_(r) =1/(2pisqrt(LC)) = 1/(2 xx 3.14 xx sqrt(0.12xx 480 xx 10^(-19))) = 663` Hz and maximum current `I_(m) = V_(m)/R = (sqrt(2) xx 230)/23 = 14.1 A` (b) Power absorbed is maximum at resonance frequency `v_(r)` The maximum power `= I_(rms) V_(rms) = (14.1)/sqrt(2) xx 230 = 2300 V` ( c) For `Deltaomega = +- R/(2L)`, the power transferred is half as that at resonating frequency i.e. `omega_(2) = omega_(r) + R/(2L)` or `v_(2) =v_(r) + R/(4pi L) = 663+ (23)/(4 xx 3.14 xx 0.12) = 678 Hz` At these frequencies, the power transferred to the CIRCUIT is half the power at resonant frequency `P = 1/2 P_("max") = 1/2 I^(2)R = 1/2 xx 1/2 I_(m)^(2)R rArr I = I_(m)/sqrt(2) = (14.1)/sqrt(2) = 10 A` (d) The Q-factor `=(omega_(r)L)/R = (2PI v_(r)L)/R = (2 xx 3.14 xx 663 xx 0.12)/23 = 21.7` |
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