Answer
(a) the same as
(b) I
Work Step by Step
(a) We know that the capacitive reactance $ X_C=\frac{1}{\omega_C}$ and inductive reactance $ X_L=\omega L $. As the circuits are connected to a battery that is a dc source, so the frequency is zero $\implies \omega=0$
Now $ X_L=0(L)=0\Omega $ and $ X_C=\frac{1}{(0)\times C}=\infty $
Thus, the dc inductor behaves like an ideal wire and the capacitor breaks the circuit; in both cases the effective resistance is the same, i.e. resistance $=R $. Hence, the current in the two circuits is the same.
(b) We know that the best explanation is option (I) -- that is, the circuits have the same current because the capacitor acts like an open circuit and the inductor acts like a short circuit.
