Answer
Please see the work below.
Work Step by Step
We know that the capacitve reactance $ X_C=\frac{1}{\omega C}$ and the inductive reactance is given as $ X_L=\omega L $. It is given that the circuits are connected to a battery that is a dc source, which means $\omega=0$.
Now $ X_L=0\times (L)=0\Omega $ and $ X_C=\frac{1}{0\times C}=\infty $. Thus, the inductor behaves as an ideal wire and the capacitor breaks the circuit. The effective resistance of the circuit is $\frac{R}{2}$ and we conclude that the current in the figure (24-30) is greater than the current in the figure (24-31).
