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:
Answer & explanation
Correct answer: option 1
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
$50=L×\frac{8}{0.2}$
$L=\frac{50×0.2}{8}$
$=1.25 H$