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A capcacitor of capacitance 10muF is connected to an AC source and an AC ammeter . If the source voltage varies as V=50sqrt2 sin 100t, the reading of the ammeter is |
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Answer» `50mA` Comparing with `V=V_(0)sinomegat` `V_(0)=50sqrt""2V,omega=100` `therefore V_(rms)=(V_(0))/(SQRT(2))=(50sqrt(2))/(sqrt(2))=50V` `X_(C)=(1)/(omegaC)=(1)/(100xx10xx10^(-6))=10^(3)=1000OMEGA` `therefore I_(rms)=(V_(rms))/(X_(C))=(50)/(1000)=50mA` = Ammeter reading |
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