A `15.0 mu F` capacitor is connected to a 220 V, 50 Hz source. Find the capactive reactance and the current (rms and peak) in the circuit. If the frequency is doubled. What happens to the capactive reactance and current ?

Explanation

The capacitive reactance is
`X_(C )=(1)/(2pi v C)=(1)/(2pi (50Hz)(15.0xx10^(-6)F))=212Omega`
The rms current is
`I=(V)/(X_(C ))=(220V)/(212Omega)=1.04A`
The peak current is
`i_(m)=sqrt(2)I=(1.41)(1.04A)=1.47A`
This current oscillates between +1.47A and –1.47 A, and is ahead of the voltage by `pi//2`.
If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.