JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 15)
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 = $${L \over R}$$
is :
_12th_April_Evening_Slot_en_15_1.png)
$${{2.7EL} \over {{R^2}}}$$
$${{EL} \over {2.7{R^2}}}$$
$${{7.3EL} \over {{R^2}}}$$
$${{EL} \over {7.3{R^2}}}$$
Explanation
$$I = {E \over R}\left( {1 - {e^{ - {t \over \tau }}}} \right)$$ where $$\tau $$ = $${L \over R}$$
We know $$dq = Idt$$
$$ \Rightarrow $$ $$\int\limits_0^Q {dq} = {E \over R}\int\limits_0^\tau {\left( {1 - {e^{ - {t \over \tau }}}} \right)dt} $$
$$ \Rightarrow $$ Q $$ = {E \over R}\left[ t \right]_0^\tau + {E \over R} \times \tau \left[ {{e^{ - {t \over \tau }}}} \right]_0^\tau $$
= $${E \over R}\tau + {E \over R}\tau \left[ {{e^{ - 1}} - 1} \right]$$
= $${E \over R}\tau + {E \over R}\tau {e^{ - 1}} - {E \over R}\tau $$
= $${{E\tau } \over {eR}}$$
= $${{E\tau } \over {2.7R}}$$ [ as e = 2.7 ]
= $${{EL} \over {2.7{R^2}}}$$ [as $$\tau $$ = $${L \over R}$$ ]
We know $$dq = Idt$$
$$ \Rightarrow $$ $$\int\limits_0^Q {dq} = {E \over R}\int\limits_0^\tau {\left( {1 - {e^{ - {t \over \tau }}}} \right)dt} $$
$$ \Rightarrow $$ Q $$ = {E \over R}\left[ t \right]_0^\tau + {E \over R} \times \tau \left[ {{e^{ - {t \over \tau }}}} \right]_0^\tau $$
= $${E \over R}\tau + {E \over R}\tau \left[ {{e^{ - 1}} - 1} \right]$$
= $${E \over R}\tau + {E \over R}\tau {e^{ - 1}} - {E \over R}\tau $$
= $${{E\tau } \over {eR}}$$
= $${{E\tau } \over {2.7R}}$$ [ as e = 2.7 ]
= $${{EL} \over {2.7{R^2}}}$$ [as $$\tau $$ = $${L \over R}$$ ]
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