The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 volt and 100 watt is connected to a (220  x .8) volt source, then the actual power would be

${\mathrm{P}}_1=\frac{{\left(220\right)}^2}{{\mathrm{R}}_1}\mathrm{and}{\mathrm{P}}_1=\frac{{\left(220\times 0.8\right)}^2}{{\mathrm{R}}_2}$

$\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{{\left(220\times 0.8\right)}^2}{{\left(220\right)}^2}\times \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\Rightarrow \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}={\left(0.8\right)}^2\times \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}$

Here R₂ < R₁

(because voltage decreases from 220 V $\to$ 220 x 0.8 V. It means heat produced $\to$ decreases)

$\mathrm{So}\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}>1\Rightarrow {\mathrm{P}}_2{\left(0.8\right)}^2{\mathrm{P}}_1\Rightarrow {\mathrm{P}}_2>{\left(0.8\right)}^2\times 100\mathrm{W}$

$\mathrm{Also}\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\frac{\left(220\times 0.8\right){\mathrm{i}}_2}{220{\mathrm{i}}_1},\mathrm{Since}{\mathrm{i}}_2<{\mathrm{i}}_1\left(\mathrm{we}\text{\hspace{0.17em}\hspace{0.17em}expect}\right)$

$\mathrm{So}\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}<0.8\Rightarrow {\mathrm{P}}_2<\left(100\times 0.8\right)$

Hence, the actual power would be between 100 x (0.8)² W and (100 x 0.8) W