The given R-C circuit has two swithes `S_1 and S_2` Swithc `S_2` is clsoed and `S_1` is open till the capacitor is fully charged to `q_0` the `S_2` is opened and `S_1` is closed simultaneously till the charge on capacitor remains `q_0//2` for which it takes time `t_1`. Now `S_1` is again opened, and `S_2` is closed till chrage on capacitor becomes `3q_0/4`. it takes time `t_2` (see fig for referecne). Find the raito `t_1//t_2`

Explanation

`q_(0) CE, q_(0)//2 = q_(0)e^(t1//RC)` or `t_(1) = RC In 2`
Now `int_(q_(0)//2)^(3q_(0)//4) (dq)/(CE - q) = itn_(0)^(t2) (dt)/(3RC)` or `t_(2) = 3RC In 2`
From here, we get `t_(1) //t_(2) =1//3`.