The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $220 \ V$ and $100 \ W$ is connected to $(220\times 0.8) \ V$ source, then the power would be :

Since Power $P = \dfrac{V^2}{R}$

For the first case, when $V=220 \ V$

$P_{1}=\dfrac{(220)^2}{R_1}$

For the second case, when $V=220 \times 0.8 \ V$

$P_{2}=\dfrac{(220\times 0.8)^{2}}{R_{2}}$

Now, $\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )^{2}}{(220)^{2}}\times \dfrac{R_{1}}{R_{2}}$

$\Rightarrow \dfrac{P_{2}}{P_{1}}=(0.8)^{2}\times \dfrac{R_{1}}{R_{2}}$

Since $V_2<V_1$, voltage has been decreased and is directly proportional to the resistance, from Ohm's law.

Hence, $R_{2}< R_{1}$

$\Rightarrow \dfrac{R_1}{R_2} > 1$

$\Rightarrow \dfrac{P_2}{P_1} > (0.8)^2$

$\Rightarrow P_2 > 100 \times (0.8)^2 \ W$

Also, Since Power $P = Vi$

Hence$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )i_2}{220i_1},$

Since $V_2<V_1$, voltage has been decreased and is directly proportional to the current, from Ohm's law.

Hence, $i_{2}< i_{1}$

$\Rightarrow \dfrac{i_2}{i_1} <1$

So $\dfrac{P_{2}}{P_{1}}< 0.8\Rightarrow P_{2}< (100\times 0.8) \ W$

Hence the actual power would be between $100 \times (0.8)^{2} \ W$ and $100\times (0.8) \ W$