The current in an LR circuit builds up to 3/4 | vedclass

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(B) The growth of current in an $LR$ circuit is given by the formula: $I = I_0(1 - e^{-t/	au})$,where $	au$ 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 $LR$ circuit is given by the formula: $I = I_0(1 - e^{-t/	au})$,where $	au$ is the time constant.Given that at $t = 4 \, s$,$I = \frac{3}{4} I_0$.Substituting these values into the equation:$\frac{3}{4} I_0 = I_0(1 - e^{-4/	au})$$\frac{3}{4} = 1 - e^{-4/	au}$$e^{-4/	au} = 1 - \frac{3}{4} = \frac{1}{4}$Taking the natural logarithm on both sides:$-\frac{4}{	au} = \ln(\frac{1}{4})$$-\frac{4}{	au} = -\ln(4) = -\ln(2^2) = -2 \ln 2$$	au = \frac{4}{2 \ln 2} = \frac{2}{\ln 2} \, s$.

The current in an $LR$ circuit builds up to $3/4$ of its steady state value in $4 \, s$. The time constant of this circuit is:

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