Two bulbs rated \(25 \mathrm{W}-220 \mathrm{V}\) and \(100 \mathrm{W}-220 \mathrm{V}\) are connected in series to a 440 V supply. The,

Resistance of a bulb
\(R=\frac{V^{2}}{P}\)
\(R_{1}=\frac{(220)^{2}}{25}=1936 \Omega\)
\(R_{2}=\frac{(220)^{2}}{100}=484 \Omega\)
Since, \(R_{1}\) and \(R_{2}\) are in series
\(R_{n e t} =R_{1}+R_{2} \)
\(=1936+484=2420 \Omega \)
\(\text {rent } I =\frac{V}{R}\)
\(\text {Hence, current } I=\frac{V}{R}\)
\(=\frac{440}{2420}=\frac{2}{11} \mathrm{~A}\)
Potential difference across \(25 \mathrm{~W}\) bulb
\(\begin{aligned}V_{1} &=I R_{1} \\&=\frac{2}{11} \times 1936=352 \mathrm{~V}\end{aligned}\)
Potential difference across \(100 \mathrm{~W}\) bulb
\(V_{2}=I R_{2}=\frac{2}{11} \times 484=88 \mathrm{~V}\)
Potential difference across \(25 \mathrm{~W}\) bulb in this combination is \(352 \mathrm{~V}\), but is can tolerate only \(220 \mathrm{~V}\).
Hence, it will fuse.