Time constant of a C-R circuit is `(2)/(ln (2))s`. Capacitance is discharged at time `t=0`. The ratio of charge on the capacitor at time `t=2s` and `t=6s` is

Options

A. `3:1`
B. `8:1`
C. `4:1
D. `2:1`

Answer

C

Explanation

`q=q_(0) e^(-t//c)`
at `t=2sec rArr q_(1)=q_(0) e^((2)/(2//ln^(2))=(q_(0))/(2))`
at t=6 sec `rArr q_(1) q_(0) e^(2/(2//ln^(2)))e^(6/(2//ln^(2)))=(q_(0))/8rArr q_(1):q_(2):: 4:1`