In an L-R circuit,the current reaches frac{3} | vedclass

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(B) The growth of current in an $L-R$ circuit is given by $i = i_0(1 - e^{-t/	au})$,where $	au = \frac{L}{R}$ is the time constant.Given that at $t

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$\frac{2}{\ln 2} \ s$

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(B) The growth of current in an $L-R$ circuit is given by $i = i_0(1 - e^{-t/	au})$,where $	au = \frac{L}{R}$ is the time constant.Given that at $t = 4 \ s$,$i = \frac{3}{4}i_0$.Substituting these values: $\frac{3}{4}i_0 = i_0(1 - e^{-4/	au})$.$\frac{3}{4} = 1 - e^{-4/	au} \implies e^{-4/	au} = 1 - \frac{3}{4} = \frac{1}{4}$.Taking the natural logarithm on both sides: $-\frac{4}{	au} = \ln(\frac{1}{4}) = -\ln(4)$.$\frac{4}{	au} = \ln(2^2) = 2 \ln 2$.$	au = \frac{4}{2 \ln 2} = \frac{2}{\ln 2} \ s$.

In an $L-R$ circuit,the current reaches $\frac{3}{4}$ of its maximum value in $4 \ s$. What is the time constant of the $L-R$ circuit?

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