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An alternating voltage `E = 200sqrt(2) sin (100 t)V` is applied to a `2 mu F` capacitor through an A.C. ammeter. The reading of the ammeter isA. 4 mAB. 40 mAC. 2 mAD. 3 mA |
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Answer» Correct Answer - B `I_(0)=(e_(0))/(X_(C))` `I_(0)=(e_(0))/((1)/(WC))=wce_(0)` `I=100xx2xx10^(-6)xx200sqrt(2)` `I_(rms)=(I_(0))/(sqrt(2))=(100xx2xx10^(-6)xx200sqrt(2))/(sqrt(2))=40mA`. |
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