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] \[Q=\frac{EC}{2}(1-{{e}^{-t/\tau }})\] \[{{I}_{2}}=\frac{\varepsilon C}{2r}{{e}^{-t/\tau }}=\frac{\varepsilon }{R}{{e}^{-t/\tau }}\] \[{{I}_{1}}=\frac{Vc}{R}=\frac{\varepsilon /2}{R}(1-{{e}^{-t/\tau }})\] \[{{I}_{1}}=\frac{\varepsilon }{R}-{{e}^{-t/\tau }}+\frac{\varepsilon }{2R}-\frac{\varepsilon }{2R}{{\varepsilon }^{-t/\tau }}\] \[\frac{\varepsilon }{2R}(1+{{e}^{-t/\tau }})=\frac{\varepsilon }{2R}\left( 1+{{e}^{\frac{-2t}{CR}}} \right)\]You need to login to perform this action.
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