Consider the LR circuit shown in the figure. | vedclass

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Question

(B) The current in an $LR$ circuit as a function of time $t$ after closing the switch is given by
$i(t) = \frac{E}{R}\left(1 - e^{-\frac{Rt}{L}}\righ

Vedclass · Accepted Answer

$\frac{EL}{2.7R^2}$

Vedclass · Answer

(B) The current in an $LR$ circuit as a function of time $t$ after closing the switch is given by
$i(t) = \frac{E}{R}\left(1 - e^{-\frac{Rt}{L}}\right)$.
The charge $q$ that passes through the battery in time interval $t = 0$ to $t = \frac{L}{R}$ is given by the integral of current with respect to time:
$q = \int_{0}^{\frac{L}{R}} i(t)\, dt = \int_{0}^{\frac{L}{R}} \frac{E}{R}\left(1 - e^{-\frac{Rt}{L}}\right) dt$
$q = \frac{E}{R} \int_{0}^{\frac{L}{R}} \left(1 - e^{-\frac{Rt}{L}}\right) dt$
$q = \frac{E}{R} \left[ t + \frac{L}{R} e^{-\frac{Rt}{L}} \right]_{0}^{\frac{L}{R}}$
$q = \frac{E}{R} \left[ \left(\frac{L}{R} + \frac{L}{R} e^{-1}\right) - \left(0 + \frac{L}{R} e^{0}\right) \right]$
$q = \frac{E}{R} \left[ \frac{L}{R} + \frac{L}{Re} - \frac{L}{R} \right]$
$q = \frac{E}{R} \cdot \frac{L}{Re} = \frac{EL}{eR^2}$
Since $e \approx 2.718$, we have:
$q = \frac{EL}{2.7R^2}$

Consider the $LR$ circuit shown in the figure. If the switch $S$ is closed at $t = 0$, then the amount of charge that passes through the battery between $t = 0$ and $t = \frac{L}{R}$ is

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