d
The current upto which bulb of marked \(25\,W - 220\,V,\) will

not fuse \(I_{1}=\frac{W_{1}}{V_{1}}=\frac{25}{220} \,\mathrm{Amp}\)

Similarly, \(\mathrm{I}_{2}=\frac{W_{2}}{V_{2}}=\frac{100}{220} \,\mathrm{Amp}\)

The current flowing through the circuit

\(I=\frac{440}{R_{e f f}}, \quad \mathrm{R}_{\mathrm{eff}}=R_{1}+R_{2}\)

\(R_{1}=\frac{V_{1}^{2}}{P_{1}}=\frac{(220)^{2}}{25} ; \quad R_{2}=\frac{V_{2}^{2}}{P}=\frac{(220)^{2}}{100}\)

\(I=\frac{440}{\frac{(220)^{2}}{25}+\frac{(220)^{2}}{100}}=\frac{440}{(220)^{2}\left[\frac{1}{25}+\frac{1}{100}\right]}\)

\(I=\frac{40}{220} \,\mathrm{Amp}\)

\({I_1}\left( { = \frac{{25}}{{220}}A} \right) < I\left( { = \frac{{40}}{{220}}A} \right) < {I_2}\left( { = \frac{{100}}{{200}}A} \right)\)

Thus the bulb marked \(25\, \mathrm{W}-220\) will fuse