A `60muF` capacitor is connected to a 110V, 60 Hz AC supply determine the rms value of the curent in the circuit.

Explanation

Given, capacitance, C = 60 `mu`F = `60 xx 10^(-6)` F
rms value of voltage , `E_(rms) = 110` V,
frequency,f = 60 Hz rms value of current in the circuit is
`I_(rms) = (E_(rms))/(X_(C))`
`therefore " " I_(rms) = omega C E_(rms)`
= ` 2 pi` fC `e_(rms)`
= ` 2 xx 3.42 xx 60 xx 60 xx 10^(-6) xx 110`
`rArr " " I_(rms) ` = 2.49 A