For the circuit shown in the related figure, if a capacitor having a capacitance of 4 F is connected between the points A and B , determine the electrostatic energy stored in it during steady state.

Step-by-step solution

Let the potential at point B be V_B = 0 V . In steady state, the capacitor acts as an open circuit, so it does not draw any DC current. The current distribution in the resistor network remains unchanged. Applying Kirchhoff's Current Law (KCL) at node A: V_A - 5 10 + V_A - 5 10 + V_A - 0 20 = 0 Multiplying the equation by 20 , we get: 2(V_A - 5) + 2(V_A - 5) + V_A = 0 4V_A - 20 + V_A = 0 5V_A = 20 V_A = 4 V The potential difference across the capacitor connected between A and B is V = V_A - V_B = 4 V . The electrost