Solution
Let $i_{2R}$ be the current flowing through the resistor $2R$ and $I$ be the total current flowing through the resistor $R$.
The power dissipated in the resistor $2R$ is given as $P_{2R} = 2\,\text{W}$. Using the formula $P = i^2 R$:
$$ i_{2R}^2 (2R) = 2 \implies i_{2R}^2 R = 1 \implies i_{2R} = \frac{1}{\sqrt{R}} $$The power dissipated in the resistor $R$ is given as $P_R = 9\,\text{W}$.
$$ I^2 R = 9 \implies I = \frac{3}{\sqrt{R}} $$Applying Kirchhoff’s Current Law (KCL) at the junction, the total current $I$ is the sum of the current through the capacitor branch ($i_c$) and the resistor branch ($i_{2R}$):
$$ I = i_c + i_{2R} $$ $$ \frac{3}{\sqrt{R}} = i_c + \frac{1}{\sqrt{R}} \implies i_c = \frac{2}{\sqrt{R}} $$The capacitor $C$ and the resistor $2R$ are connected in parallel, so the potential difference across them is the same:
$$ V_c = V_{2R} = i_{2R} (2R) = \left(\frac{1}{\sqrt{R}}\right)(2R) = 2\sqrt{R} $$The rate of increase in the energy stored in the capacitor is given by the power supplied to it:
$$ P_{stored} = \frac{dU}{dt} = V_c \cdot i_c $$Substituting the values derived above:
$$ P_{stored} = (2\sqrt{R}) \cdot \left(\frac{2}{\sqrt{R}}\right) = 4\,\text{W} $$Answer: 4 W