The current changes in a coil from 10 A to 2 A in 0.2 s. The induced emf produced is 50 V. The self inductance of the coil is:

1.25 H

The correct answer is Option (1) → 1.25 H

Induced emf, $E = -L \frac{\Delta I}{\Delta t}$ [By faraday law]

where:

$E$, induced emf = 50 V

L, self-inductance of the coil

ΔI, change in current = $I_2-I_1=-8A$

Δt, time interval = 0.2 sec

$50=L×\frac{8}{0.2}$

$L=\frac{50×0.2}{8}$

$=1.25 H$