In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced

In an inductor when current changes from 2 A to 18 A in 0.05 sec, the e.m.f. induced is 20 V. The inductance L is

(a) 62.5 mH
(b) 625 mH
(c) 6.25 mH
(d) 0.625 mH

62.5 mH

Induced e.m.f. |e| = L (dI / dt) = L(18 – 2) / 0.05 or,

L = (20 × 0.05) / 16 = 0.0625 H = 62.5 mH

OA2006

admin: