The correct answer is **(B) 2 A**. --- ### Step-by-Step Solution #### 1. Understand the Behavior of a Capacitor in Steady State When a DC circuit containing a capacitor reaches a **steady state**, the capacitor becomes fully charged. Once fully charged, it acts as an **open circuit** (infinite resistance). This means that no current can flow through the branch containing the capacitor. $$I_C = 0\text{ A}$$ #### 2. Simplify the Circuit Diagram Since the capacitor branch is an open circuit, we can conceptually remove it from our analysis. The current from the source must now flow entirely through $R_1$ and then directly through $R_2$ in a single path. Therefore, in a steady state, the resistors $R_1$ and $R_2$ are effectively connected in **series** with the voltage source. #### 3. Calculate Total Equivalent Resistance ($R_{\text{eq}}$) For a series circuit, the total resistance is the sum of the individual resistances: $$R_{\text{eq}} = R_1 + R_2$$ $$R_{\text{eq}} = 2\ \Omega + 4\ \Omega = 6\ \Omega$$ #### 4. Calculate the Steady-State Current Using Ohm's Law ($I = \frac{V}{R}$), we can find the total current flowing through the circuit: $$I = \frac{V}{R_{\text{eq}}}$$ $$I = \frac{12\text{ V}}{6\ \Omega} = 2\text{ A}$$ Since it is a single loop series circuit in steady state, this same current of **2 A** flows through both resistor $R_1$ and resistor $R_2$.