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A circuit containing a 80 mH inductor and a 60muFcapacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. a. Obtain the current amplitude and rms values.b. Obtain the rms values of potential drops across each element. c. What is the average power transferred to the inductor? d. What is the average power transferred to the capacitor? e. What is the total average power absorbed by the circuit? ['Average' implies "averaged over one cycle'.] |
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Answer» Solution :`I_(max) = (V_(max) )/(sqrt(R^2 + (X_L - X_C)^2) ) "" R = 0, omega= 2pi xx 50 = 100 pi rad//s` `V_0 = 230 sqrt2 ,L = 80 xx 10^3 H, C = 60 xx 10^(-6) F "" therefore X_L = omega L = 25.14 Omega` `X_C = (1)/(omega C) = 53.03 Omega"" therefore I_(max) = 11.6AthereforeI_(max) = (I_(max) )/(sqrt2) = (11.6)/(sqrt2) = 8.24 A` b. `V_L = I_(rms) xx omega L = 207 V` `V_C = I_(rms) xx(1)/(omega C) = 437 V` c. `phi = pi/2` . so current lags VOLTAGE by `pi/2` `P_L= 0` d. `phi = pi/2` . so current leads voltage by `pi/2` `P_C = 0` e. Total average power absorbed = 0 |
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