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An inductor of unknown value, a capacitor of 100 uF and a resistor of 10 Omega are connected in series to a 200 V, 50 Hz a.c. source. It is found that the power factor of the circuit is unity. Calculate the inductance of the inductor and the current amplitude. |
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Answer» SOLUTION :Here `C = 100 muF = 10^(-4) F, R = 10 OMEGA, V_(rms) = 200 V` and v= 50 HZ As the power factor is unity hence, `X_(L) = X_(C)` or `Lomega = 1/(C omega)` `therefore L = 1/(C omega) = 1/(C.4pi^(2) v^(2)) - 1/(10^(-4) XX 4 xx (3.14)^(2) xx (50)^(2)) = 0.1 H` Current amplitude `I_(m) = (sqrt(2)V_(rms))/R = (sqrt(2) xx 200)/10 = 28.2 A` |
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