# What does a lightbulb prefer? Volts or Amps?

## Main Question or Discussion Point

Lightbulbs are rated in watts. Watts = energy/time = volts*current

Does it matter if you have a very low voltage source with a very high current or a very high voltage with a very low current?

If the voltage and current multiplied meets the power rating, what should it matter what proportion each multiple/property is?

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Homework Helper
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The light bulb is designed to operate at a certain power $P$, but it also has a certain resistance $R$. To achieve that power, you can apply a voltage $V$ so that $P=\frac{V^2}{R}$, or you can use a current source $I$ with $P=I^2 R$. Because of the specific resistance , the voltage and current are not independent, but are related by $V=IR$. $\\$ One additional complicating factor is the resistance increases with temperature, so the resistance of a light bulb will increase with current, but when operating at the designed power level, it does have a specific resistance.

Okay so if I have a 100W lightbulb and measure its resistance to be 100ohms.
Can I say P*R=V^2
100W*100ohms = V^2
10000=V^2
P*W=100V

So for a lightbulb with a 100W power rating, I would need a 100V source if the resistance of the lightbulb is 100ohms?

And the current would be 100v÷100ohms= 1amp.

So does the question become:
If you have a lightbulb if 100ohms resistance and a power rating of 100W, you must find out what two numbers will A) divide to give you 100 (ohms) and also multiply to give you 100 (Watts), which the only two numbers is 100 and 1. Where 100 must be volts and 1 must be current.

Last edited:
russ_watters
Mentor
Does it matter if you have a very low voltage source with a very high current or a very high voltage with a very low current?

If the voltage and current multiplied meets the power rating, what should it matter what proportion each multiple/property is?
Power sources provide voltage and loads draw the current based on whatever in the load regulates it (in this case, the resistance). So if you increase the supply voltage, the load will draw a new, higher current or one of them will die trying.

russ_watters
Mentor
Okay so if I have a 100W lightbulb and measure its resistance to be 100ohms.
Can I say P*R=V^2
100W*100ohms = V^2
10000=V^2
P*W=100V

So for a lightbulb with a 100W power rating, I would need a 100V source if the resistance of the lightbulb is 100ohms?
Correct.

Thanks guys

CWatters
Homework Helper
Gold Member
For the same power a bulb [designed] for a lower voltage had a lower resistance. That means a filament with a larger crossectional area. That potentially makes the filament less fragile..if that matters.

Last edited:
anorlunda
Mentor
Also, incandescent bulbs last much longer with DC compared to AC. If you never turn it off, a DC bulb can last a lifetime.

Also note - you can not (easily) measure the resistance of an incandescent light bulb. The filament's resistance increases significantly when hot ( on ).

sophiecentaur
Gold Member
Okay so if I have a 100W lightbulb and measure its resistance to be 100ohms.
The worst component to cut your teeth on about Resistance and Power dissipation is the humble filament light bulb

As @Windadct mentioned, when you measure the resistance of your light bulb with a multimeter, it puts a very low current through it and measures the Volts across it. As soon as you connect that bulb to a supply and it glows, its resistance can go to a value around ten times what you measured. So the current you calculated that it should take or the Power it should dissipate will be horribly wrong. Forget Ohm's Law because the tungsten filament in the glass envelope doesn't get the chance to take part in the Ohm's Law thing.
The sort of question that the OP should ask, for a well behaved answer, should be about a heater element, used to heat top a water bath. Then the subsequent posts would start to apply.
The more advanced situation involving a light bulb can be answered, of course, but the basics should be dealt with first.

berkeman
Mentor
What does a lightbulb prefer?
Never anthropomorphise lightbulbs. @Ocata -- Why?

They are non-binary

Tom.G
Also, incandescent bulbs last much longer with DC compared to AC. If you never turn it off, a DC bulb can last a lifetime.
It turns out that electromigration can be a problem with DC-driven lamps. Electromigration also limits the minimum conductor size in Integrated Circuit manufacturing/lifetime.

This from a light bulb manufacturer:
http://www.extra.research.philips.com/hera/people/aarts/_Philips Bound Archive/PRRep/PRRep-30-1975A-218.pdf

Below 2700 K, d.c. operation appears to lead to a considerably shorter life than a.c.

Can be current driven or voltage driven. Pros & cons as follows.
Current source drive:
Better on start up, a cold filament has low resistance. A current source results in low voltage & low power on start up. Filament is not highly stressed. Current drive is great for lamps which get cycled a lot.
Not as good spectral consistancy over the life of the bulb. As a filament ages, it loses mass. The thickness gets reduced, resulting in higher resistance. With current source, higher R means higher power & higher temp. Higher temp shifts spectral emission towards blue end. Reduced mass raises temp more.

Voltage source: hard on filament during start up. Low R when cold means start current & power are high. Filament gets stressed hard, notched. Can be mitigated with R-C turn on, gradually increasing current. Spectral emission more consistent over lamp life. Aging filament gets thinner & higher R. Current reduces due to high R. But lower current & power means lower temp except that reduced mass compensates. Lower power would lower the temp but lower mass would raise the temp. These don't cancel exactly, but spectrum is more constant over lamp life.
Current drive increases temp as filament ages accelerating lamp burnout.
I designed lamp drives in the 90s. Did much studying on lamp issues. Either constant current or voltage can be used, voltage source works better.

Claude Abraham
PhD student
EE 39 years