Question 92: Electromagnetic Induction

Question

A shown in the figure, a battery of emf $\varepsilon $ is
connected to an inductor L and resistance R in
series. The switch 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 :
![](https://imagex.cdn.examgoal.net/7k9k2k5hh0k9k/d152bcde-5a7e-4566-af3e-f4f0dfd2f1a4/c71a2180-38c6-11ea-92dd-4b9c3519996f/file-7k9k2k5hh0k9l-302w.jpg)

Accepted Answer

$i = {i_0}\left( {1 - {e^{ - t/{t_c}}}} \right)$
q = $\int\limits_0^{{t_c}} {idt} $
= $\int\limits_0^{{t_c}} {{ \in  \over R}\left( {1 - {e^{ - t/{t_c}}}} \right)dt} $
= ${ \in  \over R}\left[ {t - {{{e^{ - t/{t_c}}}} \over {\left( { - {1 \over {{t_c}}}} \right)}}} \right]_0^{{t_c}}$
= ${ \in  \over R}\left[ {\left( {{t_c} + {t_c}{e^{ - 1}}} \right) - \left( {0 + {t_c}} \right)} \right]$
= ${ \in  \over R}{t_c}.{1 \over e}$
= ${ \in  \over R}.{L \over R}.{1 \over e}$
= ${{ \in L} \over {e{R^2}}}$

Answer

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

Answer

${{\varepsilon L} \over {{R^2}}}$

Answer

${{\varepsilon L} \over {{R^2}}}\left( {1 - {1 \over e}} \right)$

Answer

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

Electromagnetic Induction

A shown in the figure, a battery of emf \varepsilon is connected to an inductor L and resistance R in series. The switch 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 :

Choose the correct answer: