Step 1: Understanding the Concept:
At steady state (DC condition), a capacitor acts as an open circuit.
No current flows through the branch containing the capacitor.
The charge on the capacitor depends on the potential difference across the nodes it is connected to. Step 2: Key Formula or Approach:
Charge on a capacitor is $q = C \times \Delta V$.
Ohm's law $V = IR$ is used to find node potentials. Step 3: Detailed Explanation:
In the steady state, the branch containing the $6\ \mu\text{F}$ capacitor carries no current ($i_C = 0$).
We must find the equivalent resistance of the remaining resistive circuit to find the main current.
Based on the provided solution diagram, the circuit is a balanced/symmetric bridge network.
Equivalent resistance $Req$ is calculated to be $10\ \Omega$.
Total current from the battery $I = \frac{V}{R_{eq}} = \frac{12}{10} = 1.2 \text{ A}$.
Due to symmetry, the current splits equally in the parallel branches:
\[ I_1 = I_2 = \frac{I}{2} = 0.6 \text{ A} \]
Within the inner bridge section, current splits again equally:
\[ I_3 = I_4 = \frac{I_2}{2} = 0.3 \text{ A} \]
The potential difference across the nodes connecting the capacitor (say A and B) is the voltage drop across the $10\ \Omega$ resistor in that segment:
\[ \Delta V_C = I_4 \times 10 = 0.3 \text{ A} \times 10\ \Omega = 3 \text{ V} \]
Now, calculate the charge on the capacitor:
\[ q = C \times \Delta V_C = 6\ \mu\text{F} \times 3 \text{ V} = 18\ \mu\text{C} \] Step 4: Final Answer:
The charge on the capacitor is $18\ \mu\text{C}$.
