A 100W bulb, ${\mathrm{B}}_1$ and two 60W bulbs   ${\mathrm{B}}_2\mathrm{and}{\mathrm{B}}_3$  are connected to a 250 V source, as shown in the figure. Now ${\mathrm{W}}_1,{\mathrm{W}}_2,\mathrm{and}{\mathrm{W}}_3$  are the output power of the bulbs ${\mathrm{B}}_1,{\mathrm{B}}_2,\mathrm{and}{\mathrm{B}}_3$ respectively. Then

$R_1=\frac{V^2}{100}$

$R_2=\frac{V^2}{60}=R_3$

$W_1={\left(\frac{250}{R_1+R_2}\right)}^2{R}_1$

Now ,   $W_3=\frac{{250}^2}{R_3}={\left(\frac{250}{V}\right)}^2.60$

$(\text{As power}=i^2\mathrm{R})$

${{\mathrm{W}}_1=\left[\frac{250}{\frac{{\mathrm{V}}^2}{100}+\frac{{\mathrm{V}}^2}{60}}\right]}^2\frac{{\mathrm{V}}^2}{100}={\left(\frac{250}{\mathrm{V}}\right)}^2.{\left[\frac{100\times 60}{100+60}\right]}^2\times \frac{1}{100}$

$\Rightarrow {\mathrm{W}}_1={\left(\frac{250}{\mathrm{V}}\right)}^2.(14.0625)$

$W_2={\left(\frac{250}{R_1+R_2}\right)}^2{R}_2={\left[\frac{250}{\frac{V^2}{100}+\frac{V^2}{60}}\right]}^2\left(\frac{V^2}{60}\right)$

$={\left(\frac{250}{V}\right)}^2\times \left(23.4325\right)$

$\Rightarrow {\mathrm{W}}_1<{\mathrm{W}}_2<{\mathrm{W}}_3$