Circuit condition

Initially both switches are open. Then $S_1$ is closed and $S_2$ remains open, so the circuit reduces to a simple RC charging circuit with battery $V$, resistor $R$, and capacitor $C$.

Charging equation of capacitor

For a charging capacitor, $q(t)=CV\left(1-e^{-t/RC}\right)$

Given the time constant $\tau = RC$ so we can write $q(t)=CV\left(1-e^{-t/\tau}\right)$

Check each option

Option B

At $t=\tau$, $q = CV\left(1-e^{-1}\right)$ But the option says $q=\frac{CV}{2}$ Since $1-e^{-1}\approx 0.632 \neq 0.5$, this is false.

Option C

At $t=2\tau$, $q = CV\left(1-e^{-2}\right)$ This matches exactly. So C is correct.

Option D

At $t=2\tau$, $q = CV\left(1-e^{-2}\right)$ not $CV\left(1-e^{-1}\right)$ Hence D is false.

Option A

In charging a capacitor from a battery $V$:

• Work done by battery: $W_{\text{battery}} = CV^2$
• Energy stored in capacitor: $U = \frac{1}{2}CV^2$
• Energy dissipated in resistor: $W_R = W_{\text{battery}} - U = \frac{1}{2}CV^2$

Therefore, $W_{\text{battery}} = CV^2 = 2W_R$ So the battery does twice the work dissipated in the resistor, not half. Hence A is false.

Final answer

The only correct statement is: $\boxed{C}$