Question

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200V induced, calculate the self-inductance of the circuit.

Answer 1

Initial current, $I_1=5.0A$

Final current, $I_2=0A$

Change in current, $dl=I_1-I_2=5A$

Time taken for the charge, $t=0.1s$

Average emf, $e=200V$

For self inductance(L) of the coil, we have the relation for average emf as:

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

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

$=\frac{200}{\frac{5}{0.1}=4H}$

Hence the self inductance of the coil is $4H$.