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A circuit containing a 80 mH inductor and a 60 uF capacitor 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' average over one cycle] |
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Answer» Solution :Here, L = 80 mH `=80 xx 10^(-3) H = 0.08 H, C = 60 muF = 60 xx 10^(-6) F = 6 xx 10^(-5) F, V_(rms) = 230 V` and v = 50 Hz or `omega = 2pi v = 2 xx 3.14 xx 50 = 314 s^(-1), R=0` (a) `therefore` Impedance `Z = X_(L) - X_(C) = L. omega -1/(C omega) = 0.08 xx 3.14 - 1/(6 xx 10^(-5) xx 314) = 25.1 - 53.1 = 28 Omega` (capacitance) `therefore I_(rms) = 8.24 A` and `I_(m) = sqrt(2) xx I_(rms) = sqrt(2) xx 8.23 = 11.6 A` (B) `therefore V_(L) = I_(rms) = sqrt(2)xx 8.23 = 11.6` A and `V_( C) = I_(rms) xx X_(C ) = 8.24 xx 53.1 = 437 V` `rArr V_( C) - V_(L) = 437 - 207 = 230 V` ( c) As `V_(L)` leads the current by `pi/2` hence, averae power transfered to the inductor `=V_(L). I_(rms) cos phi = V_(L).I_(rms) cos pi/2 =0` (d) As Vc LAGS behind the current by `pi/2` , hence average power transferred to the capacitor `=V_(C) I_(rms) cos phi = V_( C) I_(rms). cos pi/2=0` (e) Total average power absorbed by the circuit = 0 |
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