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A circuit containing a 80 mH inductor and a 60 mu F capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible. Obtain the current amplitude and rms values. |
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Answer» Solution :(a) For `V=V_(0)sin omega t` `I=(V_(0))/(|omegaL-(1)/(OMEGAC)|) sin (omega t +(pi)/(2)), " if "R=0` where - sign appears if `omegaL gt 1//omegaC`, and + sign appears if `omegaL lt 1//omegaC.` `I_(0)=11.6A, I_("rms")=8.24A` (b) `V_("Lrms")=207V, V_("Crms")=437V` (c) Whatever be the current I in L, actual voltage leads current by `pi//2`. Therefore, AVERAGE POWER CONSUMED by L is zero. (d) For C, voltage lags by `pi//2`. Again, average power consumed by C is zero. (E) Total average power absorbed is zero. |
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