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An a.c. source of 220 V 50 Hz is connected in series with a 50 Omega resistance 150 mu F capacitors and 0.5 H inductor in series. Calculate the current through the combination. |
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Answer» Solution :`Z = SQRT(R^(2) + (X_(L) - X_(C ))^(2))` `X_(L) = 2pifL = 157 Omega` `X_(C ) = (1)/(2pi fC) = 21.23 Omega` To find `Z = 144.68 Omega` `I = (V)/(Z) = (220)/(144.68) = 1.52 A` |
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