In the circuit shown in figure, if both the bulbs B₁ and B₂ are identical, then

Let I₁ and I₂ be currents through B₁ and B₂, then

${\mathrm{I}}_1\times \sqrt{{\mathrm{R}}^2\times {\mathrm{X}}_{\mathrm{L}}^2}=220$

${\mathrm{I}}_2\times \sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{C}}^2}=220$

$\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}=\frac{\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{C}}^2}}{\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2}}=\frac{\sqrt{{\mathrm{R}}^2+{\left(\frac{1}{\mathrm{C\omega }}\right)}^2}}{\sqrt{{\mathrm{R}}^2+{\mathrm{L}}^2{\mathrm{\omega }}^2}}$

$=\frac{\sqrt{{\mathrm{R}}^2+{\left(\frac{1}{500\times {10}^{-6}\times 2\mathrm{\pi }\times 50}\right)}^2}}{\sqrt{{\mathrm{R}}^2+{\left(10\times {10}^{-3}\times 2\mathrm{\pi }\times 50\right)}^2}}$

$=\sqrt{{\mathrm{R}}^2+40}/\sqrt{{\mathrm{R}}^2+9.87},{\mathrm{I}}_2>{\mathrm{I}}_1$

Bulb B₂ will be brighter. As frequency decreases, X_Ci increases and X_L decreases. I₂ becomes more and I₁ becomes less.

So, brightness of B₁ will decrease and that of B₂ will increase.