Physics - Current Electricity Question with Solution | TestHub

Resistances of both the bulbs are
${\mathrm{R}}_1=\frac{{\mathrm{V}}^2}{{\mathrm{P}}_1}=\frac{220^2}{25}$${\mathrm{R}}_2=\frac{{\mathrm{V}}^2}{{\mathrm{P}}_2}=\frac{220^2}{100}$
Hence $R_1>R_2$
When connected in series, the voltages divide in them in the ratio of their resistances. The voltage of $440\mathrm{V}$ devides in such a way that voltage across $25\mathrm{w}$ bulb will be more than $220\mathrm{V}$.
Hence $25\mathrm{w}$ bulb will fuse.