Capacitor

At $\mathbf{t}<\mathbf{t}_0$,




At $\mathbf{t} < \mathbf{t}_0$

$ \begin{aligned} i & =\frac{\left(\frac{45}{3+R}\right)-5}{1=\frac{3 R}{3+R}} \\\\ 1 & =\frac{45-15-5 R}{3+4 R} \\\\ \Rightarrow R & =3 \Omega \end{aligned} $

(A) is correct.

$ I_1=\frac{\frac{15}{2}}{3+\frac{3}{4}} \Rightarrow I_1=2 \mathrm{~A} $

(B) is correct.

At $\mathbf{t} > \mathbf{t}_0$



$\begin{aligned} \frac{\varepsilon}{r} & =\frac{15}{3}+\frac{5}{1}+\frac{0}{3} \\\\ \frac{1}{r} & =\frac{1}{3}+\frac{1}{1}+\frac{1}{3}=\frac{5}{3} \\\\ \varepsilon & =10 \times \frac{3}{5} \Rightarrow 6 \mathrm{~V} \\\\ \text { At } t \rightarrow \infty & \\\\ q & =\mathrm{C} \varepsilon \\\\ & =2 \times 10^{-6} \times 6=12 \mu \mathrm{C}\end{aligned}$

$\begin{aligned} & \text { Time constant } \tau=R C \\\\ & =(r+3) 2 \mu \mathrm{C} \\\\ & =\left(\frac{3}{5}+3\right) \times 12=7.2 \mu \mathrm{s} \\\\ & i=i . e^{-t / \tau} \\\\ & i_0=\frac{\varepsilon}{r+3}=\frac{6}{\left(3+\frac{3}{5}\right)}\end{aligned}$

$ \begin{aligned} \Rightarrow \quad i & =\frac{6}{\frac{18}{5}} \times e^{-7.2 / 1.2} \\\\ & =\frac{5}{3} \times 0.36=0.6 \mathrm{~A} \end{aligned} $

All options are correct.

Let A be area of each plate and d is the distance between the plates.

The given capacitor is equivalent to two capacitors in parallel with capacitances

${C_1} = {{K{\varepsilon _0}(A/3)} \over d} = {{K{\varepsilon _0}A} \over {3d}}$

${C_2} = {{{\varepsilon _0}(2A/3)} \over d} = {{2{\varepsilon _0}A} \over {3d}}$

$\therefore$ $C = {C_1} + {C_2}$

$ = {{K{\varepsilon _0}A} \over {3d}} + {{2{\varepsilon _0}A} \over {3d}} = {{{\varepsilon _0}A} \over {3d}}(K + 2)$

$\therefore$ ${C \over {{C_1}}} = {{K + 2} \over K}$

Hence, option (d) is correct.

Let V be potential difference between the plates. Then

${E_1} = {V \over d}$ and ${E_2} = {V \over d}$

$\therefore$ ${{{E_1}} \over {{E_2}}} = 1$

Hence, option (a) is correct and option (b) is incorrect.

${Q_1} = {C_1}V = {{K{\varepsilon _0}A} \over {3d}}V$ and ${Q_2} = {C_2}V = {{2{\varepsilon _0}A} \over {3d}}V$

$\therefore$ ${{{Q_1}} \over {{Q_2}}} = {K \over 2}$

Hence, option (c) is incorrect.

$ \mathrm{C}_1=\mathrm{C}_2=\mathrm{C} $

* When $\mathrm{S}_1$ is closed. charge on $\mathrm{C}_1=+2 \mathrm{VC}$ on upper plate and $-2 \mathrm{Vc}$ on lower plate.

* When $S_1$ is released and $S_2$ is closed. charge on $C_1$ and $C_2=\mathrm{VC}$

* When $\mathrm{S}_2$ is released and $\mathrm{C}_3$ is closed. charge on $\mathrm{C}_2=\mathrm{CV}$

* When $\mathrm{S}_2$ is closed ( $\mathrm{S}_1$ open)

As $\mathrm{C}_1=\mathrm{C}_2$ the charge gets distributed equal. The upper plates of $C_1$ and $C_2$ now take charge to $\mathrm{CV}_0$ each and lower plate $-\mathrm{CV}_0$ each.