A 100 W bulb B₁ and two 60 W bulbs B₂ and B₃ are connected to a 250 V source as shown. Now W₁, W₂ and W₃ are the output powers of the bulbs B₁ , B₂ and B₃ respectively, then :

$\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\text{ so, }\mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}}$

${\mathrm{R}}_1=\frac{{\mathrm{V}}^2}{100} ; {\mathrm{R}}_2=\frac{{\mathrm{V}}^2}{60} ; {\mathrm{R}}_3=\frac{{\mathrm{V}}^2}{60}$

Potential drop across bulbs are

${\mathrm{V}}_1=\frac{{\mathrm{R}}_1}{{\mathrm{R}}_1+{\mathrm{R}}_2}\mathrm{V}=\frac{3}{8}\mathrm{V}$

${\mathrm{V}}_2=\frac{5}{8}\mathrm{V}$

${\mathrm{V}}_3=\mathrm{V}$

Output power from bulbs,

${\mathrm{W}}_1=\frac{{{\mathrm{V}}_1}^2}{{\mathrm{R}}_1}=14.06\mathrm{W}$

${\mathrm{W}}_2=\frac{{{\mathrm{V}}_2}^2}{{\mathrm{R}}_2}=23.43\mathrm{W}$

${\mathrm{W}}_3=\frac{{{\mathrm{V}}_3}^2}{{\mathrm{R}}_3}=60\mathrm{W}$

Thus, ${\mathrm{W}}_1<{\mathrm{W}}_2<{\mathrm{W}}_3$