question_answer

In given circuit capacitor initially uncharged. Now at t=0 switch S is closed then current given by source at any time t is

A) \[\frac{2}{R}\left( 1-{{e}^{\frac{-2t}{CR}}} \right)\]

B) \[\frac{\varepsilon }{2R}\left( 1+{{e}^{\frac{-2t}{CR}}} \right)\]

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

D) \[\frac{2\varepsilon }{R}\left( 1-{{e}^{\frac{-2t}{CR}}} \right)\]

Correct Answer: B

Solution :

[b] \[Q=\frac{EC}{2}(1-{{e}^{-t/\tau }})\] \[{{I}_{2}}=\frac{\varepsilon C}{2r}{{e}^{-t/\tau }}=\frac{\varepsilon }{R}{{e}^{-t/\tau }}\]     \[{{I}_{1}}=\frac{Vc}{R}=\frac{\varepsilon /2}{R}(1-{{e}^{-t/\tau }})\] \[{{I}_{1}}=\frac{\varepsilon }{R}-{{e}^{-t/\tau }}+\frac{\varepsilon }{2R}-\frac{\varepsilon }{2R}{{\varepsilon }^{-t/\tau }}\] \[\frac{\varepsilon }{2R}(1+{{e}^{-t/\tau }})=\frac{\varepsilon }{2R}\left( 1+{{e}^{\frac{-2t}{CR}}} \right)\]