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The current in an L-R circuit builds upt...

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)/(In2)sec`

B

`(2)/( In 2)sec`

C

`(3)/(In2)sec`

D

`(4)/(In2)sec`

Text Solution

Verified by Experts

The correct Answer is:
B
`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`
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