Step 1: Identify the steady state.
In the steady state, the capacitor behaves as an open circuit because no current flows through it once it is fully charged. Therefore, the equivalent circuit simplifies to the resistances without considering the capacitor in terms of current flow.
Step 2: Analyze the resistances.
In the given circuit, the resistances are arranged in a combination of series and parallel. First, we need to calculate the equivalent resistance of the resistors.
- The \( 2 \, \Omega \) and \( 3 \, \Omega \) resistors are in parallel. The equivalent resistance \( R_1 \) of these two resistors is:
\[
R_1 = \frac{2 \times 3}{2 + 3} = \frac{6}{5} = 1.2 \, \Omega
\]
- Now, the \( 1 \, \Omega \), \( 1.2 \, \Omega \), and \( 4 \, \Omega \) resistors are in series. So, the total equivalent resistance \( R_{\text{eq}} \) is:
\[
R_{\text{eq}} = 1 + 1.2 + 4 = 6.2 \, \Omega
\]
Step 3: Apply Ohm’s law.
Now, we can use Ohm’s law to find the total current in the circuit. The total voltage is \( 2.5 \, \text{V} \), and the total resistance is \( 6.2 \, \Omega \), so the total current \( I \) is:
\[
I = \frac{V}{R_{\text{eq}}} = \frac{2.5}{6.2} = 0.403 \, \text{A}
\]
Step 4: Calculate the charge on the capacitor.
The charge on the capacitor is given by:
\[
Q = C \times V_{\text{capacitor}}
\]
where \( C = 5 \, \mu \text{F} \) and \( V_{\text{capacitor}} \) is the voltage across the capacitor. In the steady state, the voltage across the capacitor is the same as the voltage across the \( 5 \, \mu \text{F} \) capacitor branch. Using the current through the branch, we can find this voltage by multiplying the current by the total resistance in series with the capacitor:
\[
V_{\text{capacitor}} = I \times R_{\text{capacitor branch}} = 0.403 \times 5 = 2.015 \, \text{V}
\]
Now, calculating the charge:
\[
Q = 5 \times 10^{-6} \times 2.015 = 10.075 \times 10^{-6} = \frac{75}{8} \, \mu \text{C}
\]
Thus, the charge on the capacitor is \( \frac{75}{8} \, \mu \text{C} \).
