In a coil, current falls from 5 A to 0 A in 0.2 s. If an average emf of 150 V is induced, then the self inductance of the coil is

Options

A. 4H
B. 2 H
C. 3 H
D. 6 H

Answer

D

Explanation

Here `(dI)/(dt)=(0-5)/(0.2)=-25A s^(-1),epsi=150V`
`As |epsi|=|(dI)/(dt)|`
`:. L=(|epsi|)/(|dI//dt|)=(250)/(25)=6H`