A pure inductor of 25.0 mH is connected to a source of220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz

The following simple steps will lead you to resolve this query:

1) The resistance of the inductive coil is its inductive reactance which is given by the formula XL= 2πfL.

2) Substitute the values of f and L to get XL. You should get the answer as 7.85 Ω

3) Calculate peak current through the inductor using the formula I0= V/XL= 28 A

4) Now, finally calculate thermscurrent as Irms= I0/√2 = 19.8 A