A 100 W bulb $B_1$ and two 60 W bulbs $B_2$ and $B_3$ are connected to a  $250V$ source as shown. Now $W_1$ , $W_2$ and $W_3$ are the output powers of the bulbs $B_1,B_2$ and $B_3$ 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 ,   $W_3=\frac{{250}^2}{R_3}={\left(\frac{250}{V}\right)}^2.60$

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

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

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

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

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

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