Consider a current in a circuit falls from 6.0 A to 1.0 A in 0.2 s. If an average emf of 150 V is induced by the circuit, then the self inductance of the circuit is

Options

A. 2H
B. 6H
C. 4H
D. 8H

Answer

B

Explanation

Given, `I_(1)=6A, I_(2)=1A, Delta =0.2s` and
`e=150` V
Average emf induced in a inductor,
`e=L((I_(1)-I_(2)))/(Deltat)`
Self inductance, `L=(e * Deltat)/(I_(1)-I_(2))=(150xx0.2)/(6-1)=(15xx2)/(5)`
`rArr L=6H`
Hence, the correct option is (b).