Two electric bulb 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?

(C)
current for bulb 1
\(\begin{array}{l}I_1=\frac{P_1}{V_1} \\ I_1=\frac{25}{220}\end{array}\)
\(I_1=0.114 A \text { (current capacity) }\)
and
\(\begin{array}{l}R_1=\frac{V_1^2}{P_1} \\ R_1=\frac{220 \times 220}{25} \\ R_1=1936 \Omega\end{array}\)
current for bulb 2
\(\begin{array}{l}I_2=\frac{P_2}{V_2} \\ I_2=\frac{100}{220}\end{array}\)
\(I_2=0.454 A\)
(current capacity)
and
\(\begin{array}{l}R_2=\frac{V_2^2}{P_2} \\ R_2=\frac{220 \times 220}{100} \\ R_2=484 \Omega\end{array}\)
Total resistance of circuit
\(\begin{array}{l}R=R_1+R_2 \\ R=1936+484 \\ R=2420 \Omega\end{array}\)
current flowing through circuit
\(\begin{array}{l}I=\frac{V}{R} \\ I=\frac{440}{2420} \\ I=0.181 A\end{array}\)
Now \(I > I _1\) and \(I < I _2\) so bulb 1 will fuse.