How does the current through a capacitor behave in a DC circuit when the circuit reaches a steady state?

Correct option is C

​$$\text{In a DC circuit, when the circuit reaches steady state:} \\[8pt]\bullet \ \text{The capacitor becomes fully charged.} \\[4pt]\bullet \ \text{The voltage across the capacitor becomes constant.} \\[6pt]\text{Since capacitor current is given by} \\[6pt]i_C = C \frac{dV}{dt} \\[8pt]\text{and } \frac{dV}{dt} = 0 \text{ at steady state,} \\[8pt]i_C = 0 \\[10pt]\text{Thus, a capacitor behaves like an open circuit for DC in steady state.}$$​​