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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?

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  • \$\begingroup\$ someone there?please help me with this thing. I don't have a clear concept on this. Some people say series, some says parallel. Or they give the necessity of using a shunt. \$\endgroup\$ – Tagore Dec 11 '15 at 8:19
  • \$\begingroup\$ Is it a normal 'incandescent' bulb a fluorescent or LED? \$\endgroup\$ – Icy Dec 11 '15 at 8:23
  • \$\begingroup\$ @Icy normal bulb \$\endgroup\$ – Tagore Dec 11 '15 at 8:24
  • \$\begingroup\$ Please define 'security'. Please define what you mean by 'series' and 'parallel', because if they mean the same to you as they mean to me, it shows a deeper level of misunderstand on your part. Perhaps edit your post to include a circuit diagram of a lamp in series or parallel with a resistor, in fact doing this exercise might solve your problem for you by the teddy-bear effect. \$\endgroup\$ – Neil_UK Dec 11 '15 at 14:19
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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)

schematic

simulate this circuit – Schematic created using CircuitLab

A possible way to do this is to wire 4 of these bulbs in series (each would be slightly dimmer than optimal).

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  • \$\begingroup\$ Why won't we connect it in parallel ? What problems would have raised if we connected the resistance in parallel? \$\endgroup\$ – Tagore Dec 11 '15 at 8:38
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    \$\begingroup\$ +1 for the 4 bulb in series but you could also replace it with 4 bulbs of 30W each and save the power that would have been wasted in the load resistor. \$\endgroup\$ – KalleMP Dec 11 '15 at 8:39
  • \$\begingroup\$ @Tagore connection in parallel would just blow 4 60V bulbs across the 220V mains instead of just one. In series each bulb will have 55V to work with and if they are matched bulbs will have a long lifetime \$\endgroup\$ – KalleMP Dec 11 '15 at 8:40
  • \$\begingroup\$ @Tagore You would still have 220V across the bulb - therefore it would draw 220V/30R = 7.3A - or 1613 Watts ! - and have a very short but bright life ! (assuming the 220V source is not current limited in some way) \$\endgroup\$ – Icy Dec 11 '15 at 8:41
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    \$\begingroup\$ Get a bulb rated at 220-240V ! \$\endgroup\$ – Icy Dec 11 '15 at 8:42
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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

schematic

simulate this circuit – Schematic created using CircuitLab

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