A 15.0 ` mu 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 ?

Explanation

`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 .