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}}} ) $



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