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 (in second)

ln \frac{10}{9}

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2

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\frac{2}{ln \frac{10}{9}}

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2 ln \frac{10}{9}

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Solution

The correct option is D $2 ln \frac{10}{9}$
$i = i_{0} e^{- t / τ}$
Put $i_{0} = 20$ , $i = 18$ and $t = 2 s$ we get time constant $τ = t \times l o g_{e} (i_{0} / i) = 2 \times l o g_{e} (10 / 9)$
Option D is correct as $l o g_{e}$ and $l n$ are same.

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The time constant of an LR circuit is 40 ms. The circuit is connected at $t = 0$ and the steady-state current is found to be 2.0 A. Find the current at (a) $t = 10 m s$ (b) $t = 20 m s,$ (c) $t = 100 m s$ and (d) $t = 1 s .$

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In an LR circuit current at t=0 is 20A. After 2s it reduces to 18A. The time constant of the circuit is (in second)

The correct option is D 2ln109i=i0e−t/τPut i0=20 , i=18 and t=2s we get time constant τ=t×loge(i0/i)=2×loge(10/9)Option D is correct as loge and ln are same.

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 (in second)

The correct option is D $2 ln \frac{10}{9}$
$i = i_{0} e^{- t / τ}$
Put $i_{0} = 20$ , $i = 18$ and $t = 2 s$ we get time constant $τ = t \times l o g_{e} (i_{0} / i) = 2 \times l o g_{e} (10 / 9)$
Option D is correct as $l o g_{e}$ and $l n$ are same.