Current in a circuit falls form 0.5 A to 0.0 A in 0.1 s. If an average e.m.f. of 200 V is induced, give an estimate of the self inductance of the circuit ?

Explanation

Initial current, `I_(1)` = 5.0 A
Final current, `I_(2)` = 0.0 A
Change in current, `dI=I_(1)-I_(2)= 5 A`
Time taken for the change, t = 0.1 s
Average emf, e = 200 V
For self-inductance (L) of the coil, we have the relation for average emf as :
e = L`(di)/dt)`
`L=e/(((di)/(dt)))`
=`200/(5/0.1)=4H`
Hence, the self induction of the coil is 4 H.