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InterviewSolution
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A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23Omegais 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» Solution :a. At resonance , `omega= (1)/(sqrtLC) = (1)/(sqrt(0.12 xx 480 xx 10^(-9) ) ) = 4167 rad//sec"" becausev_t = (omega_t)/(2pi) = 66 Hz` `I_(max) = (V_(max) )/(R ) = (230 sqrt2)/(23) = 14.14 A` B. `P_(max) = 1/2 I_(max)^2 R = 1/2 xx (14.14)^2 xx 23= 2300 W` c. ` DELTA omega= (R )/(2L) = (23)/( 2 xx 0.24) = 95.8 rad// sec ` ` Delta v = (Delta omega)/(2pi) = 15.2 Hz ` Power ABSORBED = Half of `v= 663 pm 15 Hz` Current amplitude ` = (I_0)/(sqrt2) = (14.14)/(sqrt2) = 10 A` d. `Q = (X_L)/(R ) = (omega L)/(R ) = (4.67 xx 0.12)/(23) = 21.7 ` |
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