If two bulbs of wattage 25 and 100 respectively each rated at 220 volt are connected in series with the supply of 440 volt, which bulb will fuse ?

Resistance of 25 watt bulb
${\mathrm{R}}_1=\frac{{\mathrm{V}}^2}{\mathrm{P}}=\frac{(220{)}^2}{25}=1936\mathrm{\Omega }$
Resistance of 100 watt bulb
${\mathrm{R}}_2=\frac{{\mathrm{V}}^2}{\mathrm{P}}=\frac{(220{)}^2}{100}=484\mathrm{\Omega }$
Series resistance, $\mathrm{R}={\mathrm{R}}_1+{\mathrm{R}}_2=2420\mathrm{\Omega }$
Current through series combination
$\mathrm{i}=\frac{\left(440\right)}{2420}=\frac{2}{11}$
Potential difference across 25 watt bulb
${\mathrm{V}}_1={\mathrm{R}}_1\times \mathrm{i}=1936\times (2/11)=352 \mathrm{volt}$
Potential difference across 100 watt bulb
${\mathrm{V}}_2={\mathrm{R}}_2\times \mathrm{i}=484\times (2/11)=88 \mathrm{volt}$
Thus the bulb of 25 watt will be fused because it can tolerate only 220 V while the voltage across it is 352 volt