As shown in the figure,a battery of emf ; var | vedclass

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(B) The current in an $LR$ circuit at time $t$ is given by $i(t) = \frac{\varepsilon}{R} (1 - e^{-Rt/L})$.Let $T_c = \frac{L}{R}$ be the time constant

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$\frac{\varepsilon L}{e R^{2}}$

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(B) The current in an $LR$ circuit at time $t$ is given by $i(t) = \frac{\varepsilon}{R} (1 - e^{-Rt/L})$.Let $T_c = \frac{L}{R}$ be the time constant. Then $i(t) = \frac{\varepsilon}{R} (1 - e^{-t/T_c})$.The total charge $q$ flowing from $t=0$ to $t=T_c$ is given by the integral of current with respect to time:$q = \int_{0}^{T_c} i(t) dt = \int_{0}^{T_c} \frac{\varepsilon}{R} (1 - e^{-t/T_c}) dt$$q = \frac{\varepsilon}{R} \left[ t - \frac{e^{-t/T_c}}{-1/T_c} \right]_{0}^{T_c} = \frac{\varepsilon}{R} \left[ t + T_c e^{-t/T_c} \right]_{0}^{T_c}$$q = \frac{\varepsilon}{R} \left[ (T_c + T_c e^{-1}) - (0 + T_c e^{0}) \right]$$q = \frac{\varepsilon}{R} [T_c + T_c e^{-1} - T_c] = \frac{\varepsilon}{R} T_c e^{-1}$Substituting $T_c = \frac{L}{R}$:$q = \frac{\varepsilon}{R} \cdot \frac{L}{R} \cdot \frac{1}{e} = \frac{\varepsilon L}{e R^2}$.

As shown in the figure,a battery of $emf \; \varepsilon$ is connected to an inductor $L$ and resistance $R$ in series. The switch $S$ is closed at $t=0$. The total charge that flows from the battery between $t=0$ and $t=t_{c}$ ($t_{c}$ is the time constant of the circuit) is

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