What will happen when a 40 watt-220 volt lamp and a 100 watt-220 volt lamp are connected in series across 40 volt supply?

Solution:

Bulb (I): Rated current ${\mathrm{I}}_1=\frac{\mathrm{P}}{\mathrm{V}}=\frac{40}{220}=\frac{2}{11}\mathrm{A}$

Resistance, ${\mathrm{R}}_1=\frac{{\mathrm{V}}^2}{\mathrm{P}}=\frac{{\left(220\right)}^2}{40}=1210\mathrm{\Omega }$

Bulb (II) : Rated current, ${\mathrm{I}}_2=\frac{100}{220}=\frac{5}{11}\mathrm{A}$

Resistance, ${\mathrm{R}}_2=\frac{{\left(220\right)}^2}{100}=484\mathrm{\Omega }$

When both are connected in series across 40 V supply

Total current through supply

${\mathrm{R}}_1=\frac{40}{{\mathrm{R}}_1+{\mathrm{R}}_2}=\frac{40}{1210+484}=\frac{40}{1694}=0.023\mathrm{A}$

This current is less than the rated current of each bulb. So neither bulb will fuse.