Four identical bulbs are connected to a battery as shown in the figure. After the switch S is closed,

initially current through each bulb,${\mathrm{i}}_1=\frac{\mathrm{V}}{2\mathrm{R}}$

After S is closed, equivalent resistance of the circuit, ${\mathrm{R}}_{\mathrm{C}}=\frac{\mathrm{R}\times \mathrm{R}}{\mathrm{R}+\mathrm{R}}+\frac{\mathrm{R}\times \mathrm{R}}{\mathrm{R}+\mathrm{R}}=\mathrm{R}$

So, current through each bulb,  ${\text{i}}_2=\frac{1}{2}\left(\frac{\mathrm{V}}{\mathrm{R}}\right)=\frac{\mathrm{V}}{2\mathrm{R}}={\mathrm{i}}_1$

Since brightness $\propto {\mathrm{i}}^2 \mathrm{and} {\mathrm{i}}_2={\mathrm{i}}_1.$ brightness of the bulbs remains unchanged.