power at voltage variation
Power, voltage and current are related by the formula $$ P = VI \tag 1$$.

For a resistive load we can make some substitutions from Ohm's Law, \$ V = IR \$. This results in $$ P = \frac {V^2}{R} \tag 2$$ and $$ P = I^2 R \tag 3$$.

From (2) we can see that if R remains constant and we halve the voltage then the power will be 1/4. So if your 220 V, 5 W load example is a resistor then on 110 V it will dissipate 5/4 = 1.25 W.



Figure 1. A bulb running on reduced voltage.

Your bulb example is a resistor load but the resistance is temperature dependent. At high temperatures the resistance increases. This means that at 110 V a 220 V, 5 W bulb might consume somewhat more than 1/4 - maybe 1/3 of its rated power, 5/3 = 1.6 W. The bulb will glow a nice warm shade of orange.