The given R-C circuit has two switches `S_1 and S_2`. Switch `S_2` is closed and `S_1` is open till the capacitor is fully charged to `q_0` then `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 charge on capacitor becomes `3q_0//4`. it takes time `t_2` (see fig for reference). Find the ratio `t_1//t_2`.

Explanation

When the switch `S_(2)` is closed, the capacitor is charged to a potential V. Now the switch `S_(2)` is opened and the switch `S_(1)` is closed
The instantaneous charge on the capacitor is
`impliesq=q_(0)e^((1)/(RC))`
where`q_(0)=CV`Putting`t=t_(1)`
for`q=q-(0)//2`,`weobaine^(t_(1)/(R_(1)C))=(1)/(2)`
`implies(t_(1)=R_(1)CIn2`
Again, the switch `S_(1)` is opened and `S_(2)`is closed. Therefore, the capacitor starts charging from charge `q-(0)//2 to (3)/(4)q_(0)`
Now, instantaneous charge on the capacitor is
`q=q_(0)[1-e^((t)/(R_(1)+(R_(2))C)]]+q_(0)/(2)e^((t)/(R_(1)+R_(2)C))=q_(0)[1-(1)/(2)e^((t)/(R_(1)+R_(2)c)]]`
`At,t=t_(2),q^(.)=3q_(0)//4implies3q_(0)/(4)=(q_(0)(1)/(2)e^((t_(2))/((R_(1)+R_(2))C)))`
`impliese^(t^(2)/(R_(1)R_(2)C))=(1)/(2)impliest_(2)=(R_(1)+R_(2))CIn2`
`therefore`
The required ratio of the times =`(t_(1)/(t_(2))(R_(1)CIn2)/((R_(1)+R_(2))CIn2)`
or,`(t_(1))/(t_(2))=(R_(1))/((R_(1)+R_(2))`