A solenoid has an inductance of 60 , H and a | vedclass

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(B) The current in an $LR$ circuit at time $t$ is given by the formula $I(t) = I_0(1 - e^{-t/	au})$,where $I_0 = V/R$ is the maximum steady-state cur

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$2 \, s$

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(B) The current in an $LR$ circuit at time $t$ is given by the formula $I(t) = I_0(1 - e^{-t/	au})$,where $I_0 = V/R$ is the maximum steady-state current and $	au = L/R$ is the time constant of the circuit.We are asked to find the time $t$ when the current reaches $\frac{e-1}{e} \approx 63.2\%$ of its final value $I_0$.This condition corresponds to $I(t) = I_0(1 - e^{-1}) = I_0(1 - 1/e)$.Comparing this with the general formula $I(t) = I_0(1 - e^{-t/	au})$,we see that $t = 	au$.The time constant $	au$ is calculated as $	au = \frac{L}{R} = \frac{60 \, H}{30 \, \Omega} = 2 \, s$.Therefore,it will take $2 \, s$ for the current to reach approximately $63.2\%$ of its final value.

$A$ solenoid has an inductance of $60 \, H$ and a resistance of $30 \, \Omega$. If it is connected to a $100 \, V$ battery,how long will it take for the current to reach $\frac{e-1}{e} \approx 63.2\%$ of its final value?

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