If we consider a L-C circuit analogous to a harmonically oscillating springblock system. The electrostatic energy $\frac{1}{2} C V^2$ is analogous to potential energy and energy associated with moving charges (current) that is magnetic energy $\left(\frac{1}{2} L I^2\right)$ is analogous to kinetic energy.

We know that inductive reactance $X_L=2 \pi f L$

and capacitive reactance $X_C=\frac{1}{2 \pi f C}$

For very high frequencies $(f \rightarrow \infty), X_L \rightarrow \infty$ and $X_C \rightarrow 0$

When reactance of a circuit is infinite it will be considered as open circuit. When reactance of a circuit is zero it will be considered as short circuited.

So, $C_1, C_2 \rightarrow$ shorted and $L_1, L_2 \rightarrow$ opened.

So, effective impedance $=R_{\text {eq }}=R_1+R_3$

Let, $$\begin{aligned} & \left(I_{\mathrm{ms}}\right) a=\mathrm{rms} \text { current in circuit (a) } \\ & \left(I_{\mathrm{rms}}\right) b=\mathrm{rms} \text { current in circuit (b) } \\ & \left(I_{\mathrm{rms}}\right) a=\frac{V_{\mathrm{rms}}}{R}=\frac{V}{R} \\ & \left(I_{\mathrm{rms}}\right) b=\frac{V_{\mathrm{rms}}}{Z}=\frac{V}{\sqrt{R^2+\left(X_L-X_C\right)^2}} \end{aligned}$$

(a) When

$$\begin{aligned} \left(I_{\mathrm{rms}}\right) a & =\left(I_{\mathrm{rms}}\right) b \\ R & =\sqrt{R^2+\left(X_L-X_C\right)^2} \end{aligned}$$

$\Rightarrow \quad X_L=X_C$, resonance condition

(b) As $Z \geq R$

$$\begin{aligned} \Rightarrow \quad \frac{\left(I_{\mathrm{rms}}\right) a}{\left(I_{\mathrm{rms}}\right) b} & =\frac{\sqrt{R^2+\left(X_L-X_{\mathrm{C}}\right)^2}}{R} \\ & =\frac{Z}{R} \geq 1 \\ \Rightarrow \quad\left(I_{\mathrm{rms}}\right) a & \geq\left(I_{\mathrm{rms}}\right) b \end{aligned}$$

No, the rms current in circuit (b), cannot be larger than that in (a).