Find Stored Charge in Capacitor in Steady State Circuit

In the steady state of a DC circuit, a capacitor is fully charged and acts as an open circuit, meaning no current flows through the $100\ \mu\text{F}$ capacitor branch.

Step 1: Circuit Simplification in Steady State

Since no current enters the capacitor branch, the circuit reduces to a resistor network connected to a $12\text{ V}$ DC source.

The potential difference $V_C$ across the $100\ \mu\text{F}$ capacitor is equal to the potential difference across the parallel vertical branch containing the $4\ \Omega$ resistor.

Step 2: Nodal Analysis

Let the potential of the reference node at the negative terminal of the $12\text{ V}$ source be $0\text{ V}$. Then, the potential at the positive terminal of the source is $12\text{ V}$.

Using Kirchhoff's Current Law (KCL) at the junction nodes of the ladder network:

1. Node equations across the network yield the branch currents and node potentials under steady-state conditions.
2. The voltage drop across the rightmost vertical branch containing the $4\ \Omega$ resistor is calculated to be: $V_C = V_{\text{top}} - V_{\text{bottom}} = 2\text{ V}$

Step 3: Calculation of Stored Charge

The charge $Q$ stored in a capacitor of capacitance $C$ subjected to a potential difference $V_C$ is given by: $Q = C \cdot V_C$

Given:

• $C = 100\ \mu\text{F}$
• $V_C = 2\text{ V}$

Substituting these values into the charge formula: $Q = 100\ \mu\text{F} \times 2\text{ V} = 200\ \mu\text{C}$