Question 144: AC fundamentals

Question

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 :
![](https://imagex.cdn.examgoal.net/3rsa0w4kejxsg1zkg/a66a0fd8-c656-418d-a4fc-ff3d7414448f/e9b71500-a06d-11e9-a4ff-cd603c22e5ad/file-3rsa0w4kejxsg1zkh.png)

Accepted Answer

$I = {E \over R}\left( {1 - {e^{ - {t \over 	au }}}} ight)$ where $	au $ = ${L \over R}$
We know $dq = Idt$
$ \Rightarrow $ $\int\limits_0^Q {dq}  = {E  \over R}\int\limits_0^	au  {\left( {1 - {e^{ - {t \over 	au }}}} ight)dt} $
$ \Rightarrow $ Q $ = {E  \over R}\left[ t ight]_0^	au  + {E  \over R} 	imes 	au \left[ {{e^{ - {t \over 	au }}}} ight]_0^	au $
= ${E  \over R}	au  + {E  \over R}	au \left[ {{e^{ - 1}} - 1} ight]$
= ${E  \over R}	au  + {E  \over R}	au {e^{ - 1}} - {E  \over R}	au $
= ${{E	au } \over {eR}}$
= ${{E	au } \over {2.7R}}$ [ as e = 2.7 ]
= ${{EL} \over {2.7{R^2}}}$ [as $	au $ = ${L \over R}$ ]

Answer

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

Answer

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

Answer

${{7.3EL} \over {{R^2}}}$

Answer

${{EL} \over {7.3{R^2}}}$

AC fundamentals

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 :

Choose the correct answer: