The charge stored by the capacitor C in the given circuit in the steady state is ______ $$\mu C$$.

In second branch capacitor acts as an open circuit when in steady state.

The upper diode (in the 3 Ω branch) points left. Its cathode is at a higher
potential than its anode, meaning it is reverse-biased and acts as an
open circuit

The lower diode (in the 4 Ω branch) points right. Its anode is at a higher
potential, meaning it is forward-biased and acts as a short circuit (assuming an ideal diode)

$$R_{\text{eq}} = 1\,\Omega + 4\,\Omega = 5\,\Omega$$

$$I = \frac{V}{R_{\text{eq}}} = \frac{2.5}{5} = 0.5\text{ A}$$

The voltage across the capacitor, $$V_{C}$$ , is exactly equal to the potential difference between the left and
right common nodes.

$$V_C = I \times 4\,\Omega$$
$$V_C = 0.5 \times 4 = 2\text{ V}$$

$$Q = C \times V_C$$

$$Q = 5\,\mu\text{F} \times 2\text{ V}$$
$$Q = 10\,\mu\text{C}$$