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A 15.0mu F capacitor is connected to a 220 V , 50 z source. Find the capacitive reactance and the current ( rms and peak ) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current ? |
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Answer» Solution :`X_(C ) = (1)/( omega C )` `= (1)/( 2 pi f C )` `= (1)/( 2 xx 3.14 xx 50 xx 10^(-6)) = 212 .3 Omega` `I_(rms) = (V_(rms))/(X_(C ))` `=(220)/( 212.3)` `= 1.0363A` We know, `I_(rms) = ( I_(m))/( sqrt(2))` `:. I_(m) = sqrt( 2) I_(rms)` `= ( 1.414) ( 1.063)` = 1.465 A Here, CURRENT through the capacitor oscillates between `+1` 1.465A and - 1.465A, keeping itself ahead of voltage by amount `( pi )/(2)` rad. After the frequency is MADE double , `X_(C ) = ( 1)/( 2pi f C )` , would become half and so current`I_(rms)` would become doubled . |
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