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A circuit, containing an 80 mH inductor and a 250 uF capacitor in series, is connected to a 240 V, 100 rad s^(-1) supply. The resistance of the circuit is negligible. (i) Obtain rms value of current. (ii) What is the total average power consumed by the circuit ? |
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Answer» Solution :Here L = 80 mH = 0.08 H, C = 250 `muF = 250 xx 10^(-6) F, R =0, V_(rms) = 240 V` and `omega = 100 rad s^(-1)` (i) `therefore` Impedance of CIRCUIT `Z = sqrt(R^(2) + (X_(L)-X_(C))^(2)) = sqrt(R^(2) + (L omega - 1/(C omega)^(2))` `=sqrt(0 + (0.08 xx 100 -1/(250 xx 10^(-6) xx 100)^(2))) = (8-40) = 32 Omega` `therefore` rms value of current `I_(rms) = V_(rms)/Z = 240/32 A = 7.5 A` (ii) The power factor `cos PHI = R/Z = 0/(32 Omega) =0` Since power factor of circuit is zero, hence the total average power consumed by the circuit is zero. |
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