The current in an `L-R` circuit builds upto `3//4th` of its steady state value in `4sec`. Then the time constant of this circuit is

Options

A. `(1)/("ln 2")s`
B. `(2)/("ln 2")s`
C. `(3)/("ln 2")s`
D. `(4)/("ln 2")s`

Answer

B

Explanation

In L-R circuit , for growth of current,
`i=i_(0)(1-e^(-Rt//L))or (3)/(4)i_(0)=i_(0)(1-e^(-t//tau_(L)))` (given)
(where, `tau_(L)=(L)/(R)`=time constant)
`therefore" "e^(-t//tau_(L))=1-(4)/(4)=(1)/(4)`
`e^(t//tau_(L))=4 or (t)/(tau_(L))=log_(e)4`
`tau_(L)=(t)/(log_(e)4)=(4)/(2log_(e)2)rArrtau_(L)=(2)/(log_(e)2)s`