The current in a `LR` circuit builds up to `(3)/(4th)` of its steady state value ion `4s`. The time constant of this circuit is

Options

A. `(1)/("in"2)s`
B. `(2)/("in"2)s`
C. `(3)/("in"2)s`
D. `(4)/(I"n" 2)s`

Answer

B

Explanation

We know that `i=i_(0)[1-e^((-Rt)/(L))]or (3)/(4)i_(0)=i_(0)[1-e^(-t//tau]`
(where `tau=(L)/(R )=`time constant)
`(3)/(4)=1-e^(-t//tau)ore^(-t//tau)=1-(3)/(4)=(1)/(4)`
`e^(t//tau)=4or(t)/(tau)=In4`
`impliestau=(t)/(In 4)=(4)/(2In2)impliestau=(2)/(In2)sec`.