In which way a resistance is to be connected with a bulb labelled 120W-60V when it is connected with 220V DC source so that maximum illumination with proper security is maintained? What is the necessary amount of resistance needed? If I connect it in parallel connection then proper security is maintained but not maximum illumination.On the contrary, series connection cause damage to the bulb though high illumination is obtained. Is there a way to verify and analyse mathematically? Am I right? If yes then how will I calculate the resistance?
A 120W 60V normal incandescent bulb behave like a resistor. So it requires 2A to operate (120W = 60V * 2A). To make this work from 220V you need to drop 160V (220V-60V) across a series resistor. To get this voltage drop you need a resistor of 80 Ohms (160V/2A) - BUT this resistor will need to be capable of dissipating 320W (160V * 2A)
A possible way to do this is to wire 4 of these bulbs in series (each would be slightly dimmer than optimal).
I have used some capacitive power sources for low voltage applications and why one couldn't be used for this one? Since an ideal capacitor is a reactive element, it wouldn't dissipate any active power (more economical and cooler (temperature wise)) You could try connecting a capacitor with a value of about 40 uF (as found in fluorescent light bulb starters) with a 500k to 1M bleeder resistor in series with the bulb.
These caps can get quite expensive, but net more expensive as 300 Watt resistor :D