Question

Current in a coil falls from 2.5 A to 0.0 A in 0.1 second inducing an emf of 200V. Calculate the value of self inductance.

Answer 1

Rate of change of current $\frac{di}{dt}=\frac{0.0-2.5}{0.1}$

$⟹$ $\frac{di}{dt}=-25$ $A/s$

Induced emf $E=200V$

Using $E=-L\frac{di}{dt}$

$\therefore$ $200=-L\times (-25)$

$⟹$ $L=8H$