Question

An inductor of inductance 200 mH is connected to an A.C. source of peak e.m.f. 210 V and frequency 50 Hz. Calculate the peak current and instantaneous voltage of the source when the current is at its peak value.

Answer 1

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

e_{0 =} 210 V

f = 50 Hz

To find: Peak current = ?

Instantaneous voltage = ?

Peak current `I_0 = (e_0)/(X_L) = (e_0)/(2pifL)`

= `(210)/(2 xx 3.142 xx 50 xx 0.2)`

= `(210)/(62.84)`

= 3.342 A

Since, in an inductive AC circuit, current lags behind the emf by `pi/2`, so the voltage is zero when the current is at its peak value.