Resistor calculation
"how will I calculate the amount of resistance I would need"




1) Calculate how much current needs to flow through the bulb at its nominal power rating. 

$$I_{nom} = \frac{P}{U} = \frac{100W}{120V} = 0.83A$$

2) Calculate the amount of voltage you need to get rid of. That's the voltage drop over the reistor R.
$$U_{drop} = 240V - 120V = 120V$$

3) Calculate the resistor value needed to cause such voltage drop
$$R = \frac{U_{drop}}{I} =  \frac{120}{0.83} = 145Ω$$

4) Then calculate the power rating of the resistor
$$P = I².R =  (0.83)².145 = 100W$$

Such resistor could look like one of these and would get very hot. Because the required voltage drop is exactly 50% in this case, you can use the exact same bulb as well as others have pointed out. 




The efficiency would be: 

$$η = \frac{P_{eff}}{P_{tot}} = \frac{100W}{100W + 100W} = 0.5$$ or 50%, which is obviously not good.

2) More efficient alternative:  Since you use AC you can replace the resistor with a coil or capacitor. These components can also provoke the required voltage drop but have the advantage that they don't use real power but only so-called reactive power. In other words they don't warm up. The apparent resistance they represent is called impedance (X).



1) Calculate the resistance of the bulb
$$R_{bulb} = \frac{U}{I} =  \frac{120}{0.83} = 145Ω$$

2) Calculate the total required impedance 
$$X_{tot} = \frac{U}{I} =  \frac{240}{0.83} = 290Ω$$

3) Calculate the impedance of the capacitor or coil
$$X_c = \sqrt(X_{tot}² - R_{bulb}²) = \sqrt((290)² - (145)²) = 251Ω$$

4) If the line frequency is 60Hz at the place where you live, this would require a capacitor of:
$$C = \frac{1}{2.π.f.X_c} = \frac{1}{2.π.60.251} = 10μF$$

So if you wire a 10μF/300VAC capacitor like the one below it will also do the trick, with an efficiency of 100%, inefficiencies of the bulb not included of course.

The above only works with incandescent bulbs not LED lighting.