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

Options

A. `(1)/(In2)sec`
B. `(2)/( In 2)sec`
C. `(3)/(In2)sec`
D. `(4)/(In2)sec`

Answer

2

Explanation

`I=I_(0)(1-e^(-t/r)` where `taurightarrow` timeconstant
`therefore(3)/(4)I_(0)=I_(0)(1-e^(t)/(tau))Rightarrow(3)/(4)=1-e(-t/tau)`
`Rightarrowe(-t)/(tau)=(1)/(4))Rightarrow(-t)/(tau)Ine=In(1)/(4)`
`Rightarrow(-4)/(tau)=-2In2Rightarrowtau=(2)/In2`