A `1.5 mu F` capacitor has a capacitative reactance of `12 Omega`. What is the frequency of the source ? If frequency of source is double, what will be the capacitative reactance ?

Explanation

Here, `C = 1.5 mu F = 1.5 xx 10^(-6) F`,
`X_(C) = 12 Omega, V = ?`
As `X_(C) = (1)/(omega C) = (1)/(2 pi v c)`
`:. V = (1)/(2 pi C X_(C)) = (1)/(2 xx 3.14 xx 1.5 xx 10^(-6) xx 12)`
`v = 8846 Hz`
As `X_(C) prop (1)/(v)`, therefore, when v is doubled, `X_(C)`
becomes half, i.e., `(12)/(2) = 6 Omega`