Electro Magnetic Induction And Alternating Currents Question 161
Question: In given circuit capacitor initially uncharged. Now at t=0 switch S is closed then current given by source at any time t is
Options:
A) $ \frac{2}{R}( 1-{{e}^{\frac{-2t}{CR}}} ) $
B) $ \frac{\varepsilon }{2R}( 1+{{e}^{\frac{-2t}{CR}}} ) $
C) $ \frac{\varepsilon }{2R}( 1-{{e}^{\frac{-2t}{CR}}} ) $
D) $ \frac{2\varepsilon }{R}( 1-{{e}^{\frac{-2t}{CR}}} ) $
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Answer:
Correct Answer: B
Solution:
-
$ 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}( 1+{{e}^{\frac{-2t}{CR}}} ) $
