question_answer

As shown in the figure, a battery of emf E 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: [JEE MAIN Held on 08-01-2020 Evening]

A)             \[\frac{EL}{{{R}^{2}}}\]        

B) \[\frac{ER}{e{{L}^{2}}}\]

C)             \[\frac{EL}{{{R}^{2}}}\left( 1-\frac{1}{e} \right)\]

D) \[\frac{EL}{e{{R}^{2}}}\]

Correct Answer: D

Solution :

[d] \[i=\frac{E}{R}\left( 1-{{e}^{-\,{t}/{{{t}_{C}}}\;}} \right)\]

\[\because \,\,\,\,\,\,\,\,\,\,\,\,{{t}_{C}}=\frac{L}{R}\]

\[\Rightarrow \,\,\,\,\,\,\,\,\int{dq}=\int{\frac{E}{R}}\left( 1-{{e}^{-{t}/{{{t}_{C}}}\;}} \right)dt\]

\[\Rightarrow \,\,\,\,\,\,q=\frac{E}{R}\left[ t+{{t}_{c}}{{e}^{-{t}/{{{t}_{C}}}\;}} \right]_{0}^{{{t}_{C}}}\]

\[\Rightarrow \,\,\,\,\,\,q=\frac{E}{R}\left[ {{t}_{C}}+\frac{{{t}_{C}}}{e}-{{t}_{C}} \right]\]

\[\Rightarrow \,\,\,\,\,\,q=\frac{E}{R}\frac{L}{\operatorname{Re}}\]

\[\therefore \,\,\,\,\,\,q=\frac{E\,L}{{{R}^{2}}e}\]