In an LR circuit,the current increases to one | vedclass

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(C) The growth of current in an $LR$ circuit is given by $I = I_0 (1 - e^{-t/	au})$,where $	au$ is the time constant.Given that at $t = 4 \, sec$,$I

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$\frac{4}{\log_e (4/3)}$

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

In an $LR$ circuit,the current increases to one-fourth of its maximum value in $4 \, sec$. What is the time constant of the circuit?

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