InterviewSolution
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InterviewSolution
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In a series `LCR` circuit with an `ac` source of `50 V,R=300 Omega`,frequency `v=50/piHz`.The average electric field energy, stored in the capacitor and average magnetic energy stored in the coil are `25 mJ` and `5 mJ` respectively.The `RMS` current in the circuit is `0.10 A`.Then findA. capacitance `C` of capacitor is `20 mu F`B. inductance `L` of inductor is `2 H`C. peak voltage of source is `50 V`D. the sum of rms voltage across the three elements is `35.4 V` |
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Answer» Correct Answer - A,B,C,D Let `I_(r)` be the rms current through the circuit then `I_(r ) = 2A, (I_(r ))/(omega C) = 20 V, I_(r ) omega C = 20 V` and `I_(r ) R = 10 V` solving we get `R = 50 Omega, C = (1)/(pi) xx 10^(-3) F` and `L = (1)/(10 pi) H` `:. V_(s)` = source voltage = `I_(r ) = sqrt(R^(2) + (omega L - (1)/(omega C))^(2))` `= sqrt((I_(r ) R)^(2) + (I_(r) omega L - (I)/(omega C))^(2))` `= sqrt(10^(2) + (20 - 20)^(2)) = 10` volts After in inductor is shorted `I_(r ) = (V_(s))/(sqrt(R^(2) + (1)/(omega^(2) C^(2)))) = (10)/(sqrt(25 + 10)) = (2)/(sqrt(5)) A` `v_(1) = I_(r ) R = 2 sqrt(5)` volts, `v_(2) = (I_(r ))/(omega C) = 4 sqrt(5)` volts |
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