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))`

In an `LR` circuit, current at `t = 0` is `20 A`. After `2 s` it reduc

1.	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))`
Answer» `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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