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AI-তৈরি নমুনা উত্তর

In the steady state, the capacitor is fully charged and draws no direct current, acting as an open circuit.

Let the electric potential at point B be $V_B = 0\text{ V}$ and at point A be $V_A = 12\text{ V}$.

1. Current flowing through the path ACB:
   $I_1 = \frac{V_A - V_B}{3\,\Omega + 9\,\Omega} = \frac{12\text{ V}}{12\,\Omega} = 1\text{ A}$

Potential at node C:
$V_C = V_A - I_1 \times 3\,\Omega = 12 - (1 \times 3) = 9\text{ V}$

2. Current flowing through the path ADB:
   $I_2 = \frac{V_A - V_B}{4\,\Omega + 2\,\Omega} = \frac{12\text{ V}}{6\,\Omega} = 2\text{ A}$

Potential at node D:
$V_D = V_A - I_2 \times 4\,\Omega = 12 - (2 \times 4) = 4\text{ V}$

3. Potential difference across the capacitor:
   $V = |V_C - V_D| = |9\text{ V} - 4\text{ V}| = 5\text{ V}$

4. Amount of energy stored in the capacitor:
   $U = \frac{1}{2} C V^2 = \frac{1}{2} \times (6 \times 10^{-6}\text{ F}) \times (5\text{ V})^2 = 75 \times 10^{-6}\text{ J} = 7.5 \times 10^{-5}\text{ J} = 75\,\mu\text{J}$

Answer: $75\,\mu\text{J}$ (or $7.5 \times 10^{-5}\text{ J}$)