JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 11)
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 = tc (tc is the time constant of the
circuit) is :
_8th_January_Evening_Slot_en_11_1.png)
$${{\varepsilon L} \over {e{R^2}}}$$
$${{\varepsilon L} \over {{R^2}}}$$
$${{\varepsilon L} \over {{R^2}}}\left( {1 - {1 \over e}} \right)$$
$${{\varepsilon R} \over {e{L^2}}}$$
Explanation
$$i = {i_0}\left( {1 - {e^{ - t/{t_c}}}} \right)$$
q = $$\int\limits_0^{{t_c}} {idt} $$
= $$\int\limits_0^{{t_c}} {{ \in \over R}\left( {1 - {e^{ - t/{t_c}}}} \right)dt} $$
= $${ \in \over R}\left[ {t - {{{e^{ - t/{t_c}}}} \over {\left( { - {1 \over {{t_c}}}} \right)}}} \right]_0^{{t_c}}$$
= $${ \in \over R}\left[ {\left( {{t_c} + {t_c}{e^{ - 1}}} \right) - \left( {0 + {t_c}} \right)} \right]$$
= $${ \in \over R}{t_c}.{1 \over e}$$
= $${ \in \over R}.{L \over R}.{1 \over e}$$
= $${{ \in L} \over {e{R^2}}}$$
q = $$\int\limits_0^{{t_c}} {idt} $$
= $$\int\limits_0^{{t_c}} {{ \in \over R}\left( {1 - {e^{ - t/{t_c}}}} \right)dt} $$
= $${ \in \over R}\left[ {t - {{{e^{ - t/{t_c}}}} \over {\left( { - {1 \over {{t_c}}}} \right)}}} \right]_0^{{t_c}}$$
= $${ \in \over R}\left[ {\left( {{t_c} + {t_c}{e^{ - 1}}} \right) - \left( {0 + {t_c}} \right)} \right]$$
= $${ \in \over R}{t_c}.{1 \over e}$$
= $${ \in \over R}.{L \over R}.{1 \over e}$$
= $${{ \in L} \over {e{R^2}}}$$
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