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 :

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: