In the circuit shown in figure `C_(1) = 2 C_(2)`. Capacitance `C_(1)` is charged to a potential of V. The current in the circuit just after the switch S is closed is

Options

A. 0
B. `2V//R`
C. `oo`
D. `V//2R`

Answer

D

Explanation

Here, `I_(1) = (V)/(R )e^(-t//RC), I_(2) = (V)/(R )e^(-t//2RC)`
`:. (I_(1))/(I_(2))=e^(-t//RC +t//2RC) = e^(-t//2RC) = (1)/(e^(t//2RC)`
From this expression, it is clear that when t increases, ratio
decrease.