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In an LR circuit, current at t = 0 is 20...

In an `LR` circuit, current at `t = 0` is `20 A`. After `2 s` it reduces to `18 A`. The time constant of the circuit is :

A

`ln((10)/(9))`

B

`2`

C

`(2)/(ln(10/9))`

D

`2 ln((10)/(9))`

Text Solution

Verified by Experts

`I = I_(0)e^(-t//tau) rArr 18 = 20 e^(-2//tau) rArr (10)/(9) = e^(2//tau)`
`rArr ln (10)/(9) = (2)/(tau) rArr tau = (2)/(ln((10)/(9)))`
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