In the circuit shown in fig. If both the bulbs L₁ and L₂ are identical:

At current frequency of 500 Hz, X_c < X_L. Therefore, the voltage drop across bulb L₁ will be greater than that of L₂. So, bulb L₁ will glow brighter than bulb L₂.

$$\begin{gathered} {\mathrm{X}}_{\mathrm{C}}=\frac{1}{\mathrm{\omega C}}, {\mathrm{X}}_{\mathrm{L}}=\mathrm{\omega L}\text{ as frequency increases }{\mathrm{X}}_{\mathrm{L}}>{\mathrm{X}}_{\mathrm{C}} \mathrm{and} {\mathrm{i}}_2<{\mathrm{i}}_1. \\ \mathrm{So}, \mathrm{the} \mathrm{brightness} \mathrm{of} \mathrm{bulb} L_1 \mathrm{will} \mathrm{increase} \mathrm{and} \mathrm{that} \mathrm{of} {\mathrm{L}}_2 \mathrm{will} \mathrm{decrease}. \end{gathered}$$