An inductor (L) - resistance (R) battery circ | vedclass

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Question

(A) The current in an $LR$ circuit at time $t$ is given by $I(t) = \frac{E}{R}(1 - e^{-t/T})$,where $T = L/R$ is the time constant.The charge $q$ pass

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$\frac{ET}{eR}$

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(A) The current in an $LR$ circuit at time $t$ is given by $I(t) = \frac{E}{R}(1 - e^{-t/T})$,where $T = L/R$ is the time constant.The charge $q$ passing through the battery in time $T$ is the integral of current with respect to time:$q = \int_{0}^{T} I(t) dt = \int_{0}^{T} \frac{E}{R}(1 - e^{-t/T}) dt$Integrating the expression:$q = \frac{E}{R} \left[ t - (-T)e^{-t/T} \right]_{0}^{T}$$q = \frac{E}{R} \left[ t + Te^{-t/T} \right]_{0}^{T}$Substituting the limits:$q = \frac{E}{R} \left[ (T + Te^{-T/T}) - (0 + Te^{0}) \right]$$q = \frac{E}{R} \left[ T + \frac{T}{e} - T \right]$$q = \frac{ET}{eR}$

An inductor $(L)$ - resistance $(R)$ battery circuit is switched on at $t = 0$. If the $e.m.f.$ of the battery is $E$,what is the charge passing through the battery in time constant $T$?

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