Answer
The maximum current in circuit 1 is greater than the maximum current in circuit 2.
Work Step by Step
The current amplitude $I$ is $\omega Q$
Note that the current amplitude $I$ is the maximum current.
Note that $\omega = \frac{1}{\sqrt{LC}}$
Therefore, the maximum current is $~~I = \frac{Q}{\sqrt{LC}}$
In part (a), we found that the inductance $L$ in circuit 1 is less than the inductance in circuit 2.
Since $Q$ and $C$ are the same in both circuits, the maximum current in circuit 1 is greater than the maximum current in circuit 2.
