Question:
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Solution:
Initial current, $I_{1}=5.0 \mathrm{~A}$
Final current, $l_{2}=0.0 \mathrm{~A}$
Change in current, $d I=I_{1}-I_{2}=5 \mathrm{~A}$
Time taken for the change, $t=0.1 \mathrm{~s}$
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as:
$e=L \frac{d i}{d t}$
$L=\frac{e}{\left(\frac{d i}{d t}\right)}$
$=\frac{200}{\frac{5}{0.1}}=4 \mathrm{H}$
Hence, the self induction of the coil is 4 H.