RC Trouble

Let after n cycles the charge on the capacitor is $q_{n}$ .

Now when the key connected to point $1$ , the capacitor will gain some charge and let that charge be $q_{k}$ .

To calculate that we can apply Kirchoff's law. $V-IR-\dfrac{q}{C}=0$ Put $I=\dfrac{dq}{dt}$ and rearrange the terms we get. $-\dfrac{dt}{RC}=\dfrac{dq}{q-CV}$ Integrating and applying limits $\Big[-\dfrac{t}{RC}\Big]^{RC}_{0}=\Big[ln(q-CV)\Big]_{q_{n}}^{q_{k}}$ Solving we get $q_{k}=CV\Big(1-\dfrac{1}{e}\big) +\dfrac{q_{n}}{e}$ Now the key will be connected to the point 2.

Again we will apply Kirchoff's law to find the charge on capacitor on after time $RC$ .

$\dfrac{q}{C}=IR$ Put $I=-\dfrac{dq}{dt}$ , and rearrange we will get $-\dfrac{dt}{RC}=\dfrac{dq}{q}$ Integrate and apply limits. Note that the final charge will be $q_{n+1}$ . $\Big[-\dfrac{t}{RC}\Big]_{0}^{RC}=\Big[ln(q)\Big]_{q_{k}}^{q_{n+1}}$ Solving it we get $q_{n+1} = \dfrac{q_{k}}{e}$ Put the previously calculated $q_{k}$ in the above equation we get $eq_{n+1} = CV\Big(1-\dfrac{1}{e}\Big) +\dfrac{q_{n}}{e}$ At infinite the charge on the capacitor will be almost same after every turn. So we can conclude that $\lim_{n\to\infty} q_{n+1} = \lim_{n\to\infty} q_{n}=l(say).$ So in our previous equation we can write that $e^{2}l=CV(e-1) +l$ Hence we get $l=\dfrac{CV}{e+1}$ So from the given equation $x=\dfrac{1}{1+e}$ $\boxed{x=0.2689}$

In first complete cycle, final charge will be: ${VC{(e-1)}{(e^{2}+1)}}/e^{3}$

In second cycle, final charge will be : ${VC{(e-1)}{(e^{6}+e^{4}+e^{2}+1)}}/e^{7}$

Hence, series will be form... And answer will be $0.2627VC$

We can directly write the charge equation as $Q\left( t \right) ={ Q }_{ f }+({ Q }_{ i }-{ Q }_{ f }){ e }^{ -\frac { t }{ \tau } }$