Question

Initially, the switch is open for a long time. Now the switch is closed at t=0. Find the charge on the rightmost capacitor as a function of time given that it was initially unchanged.

Answer 1

$i=i_1+i_2+i_3$

$\frac{d{q}_1}{dt}=i,\frac{d{q}_2}{dt}=i_2$

And

$\frac{q_1}{C}=\frac{q_2}{C}=0$ ___(I)

$\Rightarrow q_1=q_2$ ___(II) $\Rightarrow \frac{d{q}_1}{dt}=\frac{d{q}_2}{dt}=i_1=i_2$

$\frac{q_2}{C}=i_{3}R\Rightarrow q_2=i_{3}Rc\Rightarrow i_3=\frac{q_2}{Rc}$ __(III)

$V-iR-\frac{q_1}{C}=0$

(Kirchoff's law in Bigger loop)

$\frac{V}{R}=i+\frac{q_1}{RC}=i_1+i_2+i_3+\frac{q_1}{RC}$

$\Rightarrow \frac{V}{2R}=\frac{2d{q}_1}{dt}+\frac{2q_1}{C}$ (From (II) and (III))

$\Rightarrow \frac{V}{2R}{\int }_0^t{e}^{t/RC}={\int }_{0,0}^{q,t}d.(q,e^{t/RC})$

$\Rightarrow \frac{VC}{2}(e^{t/RC}-1)=q_1.e^{t/RC}$

$\Rightarrow q_1=\frac{VC}{2}(1-e^{-t/RC})$