question_answer

In the circuit shown here, the point 'C' is kept connected to point 'A' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point 'A' and connected to point 'B' at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to :   JEE Main  Solved  Paper-2014

A) −1          

B) \[\frac{1-e}{e}\]

C) \[\frac{e}{1-e}\]                                              

D) 1

Correct Answer: A

Solution :

\[i={{i}_{0}}{{e}^{\frac{-tR}{L}}}\] \[\therefore \]\[{{V}_{R}}=iR={{i}_{0}}{{e}^{\frac{-tR}{L}}}\] \[{{V}_{R}}=R{{i}_{0}}{{e}^{-1}}\]at\[t=\frac{L}{R}\] \[{{V}_{L}}=-L\left( \frac{di}{dt} \right)\]                              \[=L{{i}_{0}}\left( \frac{-R}{L} \right){{e}^{\frac{-tR}{L}}}\] \[=-R{{i}_{0}}{{e}^{-1}}\]at\[t=\frac{L}{R}\]        \[\therefore \]\[\frac{{{V}_{R}}}{{{V}_{L}}}=-1\]