A coil has an inductance of 2.5,H and a resis | vedclass

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(D) The current in an $LR$ circuit at time $t$ is given by $I = I_0(1 - e^{-Rt/L})$,where $I_0 = V/R$ is the final steady-state current.When the curre

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$5$

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(D) The current in an $LR$ circuit at time $t$ is given by $I = I_0(1 - e^{-Rt/L})$,where $I_0 = V/R$ is the final steady-state current.When the current rises to $0.63$ of its final value,$I = 0.63 I_0$.Substituting this into the equation: $0.63 I_0 = I_0(1 - e^{-Rt/L})$.$0.63 = 1 - e^{-Rt/L}$,which implies $e^{-Rt/L} = 1 - 0.63 = 0.37$.Since $e^{-1} \approx 0.368 \approx 0.37$,we have $Rt/L = 1$.Therefore,the time $t$ is equal to the time constant $	au = L/R$.Given $L = 2.5\,H$ and $R = 0.5\,\Omega$,we get $t = 2.5 / 0.5 = 5\,s$.

$A$ coil has an inductance of $2.5\,H$ and a resistance of $0.5\,\Omega$. If the coil is suddenly connected across a $6.0\,V$ battery,then the time required for the current to rise to $0.63$ of its final value is.....$s$.

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