Question

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

Answer 1

`i = i_(0)(i-e^(-(Rt)/(L))), i = (3i_(0))/(4)` when `t = 4 sec`.
`(3 cancel(i_(0)))/(4) = cancel(i_(0))(1-e^(-(4R)/(L))), e^(-4//tau) = (1)/(4)`
`e^(4//tau) = 4`
`(4)/(2)= ln4 , tau = (4)/(2 ln 4) = (2)/(ln 2)`