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(B) In a steady state, a capacitor acts as an open circuit because it becomes fully charged and blocks the flow of direct current $(DC)$.In the given

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$0, \frac{\varepsilon}{R_1 + R_3}$

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(B) In a steady state, a capacitor acts as an open circuit because it becomes fully charged and blocks the flow of direct current $(DC)$.In the given circuit, the branch containing $C_1$ and $R_2$ is in series with $C_1$. At steady state, no current flows through this branch, so $I_1 = 0$.The branch containing $C_2$ is also in series with $C_2$, so no current flows through the rightmost branch.However, the branch containing $I_2$ consists only of the resistor $R_3$. This branch is in parallel with the capacitor $C_2$ and in series with $R_1$.At steady state, the current flows from the battery through $R_1$ and $R_3$ only, as the branches with capacitors are open circuits.Using Ohm's law for the loop containing $\varepsilon$, $R_1$, and $R_3$:$I_2 = \frac{\varepsilon}{R_1 + R_3}$.Thus, $I_1 = 0$ and $I_2 = \frac{\varepsilon}{R_1 + R_3}$.

Find the steady-state currents $I_1$ and $I_2$ in the given circuit.

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