From the corresponding equivalent circuit, we have

\begin{aligned} V-\frac{I}{3} Z_{ L } & -\frac{I}{6} Z_{ L }-\frac{I}{3} Z_{ L }=0 \\ V & =I \cdot \frac{5}{6} \cdot Z_{ L } \\ \frac{V}{I} & =\frac{5}{6} \cdot Z_{ L } \end{aligned}

Note: If capacitors are used in the circuit, we have

Z_{\text {eq }}=\frac{6}{5} Z_{ C }

Note: For DC input, the inductor acts as short circuit. The capacitor will act as an open circuit in steady state.