Question

An inductor of inductance 200 mH is connected to an A.C. source of peak e.m.f. 220 V and frequency 50 Hz. Calculate the peak current in the circuit.

Answer 1

Given: L = 200 mH = 200 × 10⁻³ H = 0.2 H

e₀ = 220 V

f = 50 Hz

To find: Peak current = ?

Peak current (I₀) = `(e_0)/(X_L)`

= `(e_0)/(2 pi f L)`

= `(220)/(2 xx 3.142 xx 50 xx 0.2)`

= `(220)/(62.84)`

= 3.50 A