Question

The current in the coil decreases from 10 Amp to 5 Amp in 0.1 second... Calculate the co-efficient of self-inductance if induced e.m.f. is 5 V.

Answer 1

Change in current $I=10-5=5A$

$dt=0.1sec$ $E=5V$

$\therefore$ $e=L\frac{di}{dt}$

$\Rightarrow$ $5=L\left(\frac{5}{0.1}\right)$

$L=\frac{5}{50}=0.1H$