Two electric bulbs marked 25 W - 220 V and 100 W 220 V are connected in series to a 440 V supply. Which of the bulbs will fuse?

Here we will use the formula of power

$\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\text{. }$ So resistance $\mathrm{R}=\frac{{\mathrm{V}}^2}{\mathrm{P}}\text{. }$

Resistance of 25 W bulb

${\mathrm{R}}_1=\frac{(220{)}^2}{25}=1936 \mathrm{ohm}$ and resistance of 100 W bulb

${\mathrm{R}}_2=\frac{(220{)}^2}{100}=484 \mathrm{ohm}$. When they are connected in series with a supply of 440 V, the resultant current will be

$\mathrm{I}=\frac{440}{(1936+484)}=0.1819\mathrm{A}$

Now the voltage drop across 25 W bulb will be

${\mathrm{V}}_1=\mathrm{IR}=0.1819\times 1936=352\mathrm{V}$ and the voltage drop across 100 W bulb will be ${\mathrm{V}}_2=0.1819\times 484=88\mathrm{V}$. Here because of the excessive voltage (much higher than rated value) across 25 W bulb, it will fuse.