A pure inductor of 25.0 mH is connected to a source of 220 V . Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.

Explanation

Magnitude of inductive reactance is,
`X_(L) = omega L `
`= 2 pi f L`
`= ( 2) ( 3.14 ) ( 50) ( 25 xx 10^(-3))`
` = 7.85 Omega`
Here, `V_(rms) = I_(rms) X_(L)`
`:. I_(rms) = ( V_(rms)/ X_(L))`
`= ( 220)/( 7.85)`
`= 28.025 A`