# What does a lightbulb prefer? Volts or Amps?

• Ocata
In summary, the light bulb is rated in watts. Watts = energy/time = volts*current. Watts 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.

#### Ocata

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?

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.

Ocata
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.

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Ocata said:
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.

Bystander and Ocata
Ocata said:
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.

Ocata
Thanks guys

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.

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Ocata, Tom.G and Charles Link
Also, incandescent bulbs last much longer with DC compared to AC. If you never turn it off, a DC bulb can last a lifetime.

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

Ocata and Charles Link
Ocata said:
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.

bmhiggs and Ocata
Ocata said:
What does a lightbulb prefer?
Never anthropomorphise lightbulbs. @Ocata -- Why?

Ocata
They are non-binary

Ocata
anorlunda said:
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.c...ps 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.

Ocata and Charles Link
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

Ocata, berkeman and Charles Link

## What is a lightbulb's preferred voltage?

A lightbulb's preferred voltage depends on its wattage rating. Generally, a 40-watt bulb prefers a voltage of around 120 volts, while a 100-watt bulb prefers a higher voltage of around 240 volts. However, it is always best to check the manufacturer's specifications for the specific bulb.

## Does a lightbulb prefer high or low voltage?

A lightbulb's preference for high or low voltage also depends on its wattage rating. Higher wattage bulbs typically prefer higher voltages, while lower wattage bulbs prefer lower voltages. This is because higher voltages allow for more energy to flow through the bulb, producing a brighter light.

## Can a lightbulb function without amps?

No, a lightbulb cannot function without amps. Amps, or amperes, are a unit of electrical current and are necessary for the flow of electricity through the bulb. Without amps, the bulb would not light up.

## What happens if a lightbulb receives too much voltage?

If a lightbulb receives too much voltage, it can cause the bulb to burn out or even explode. This is because the high voltage creates too much heat, which can damage the filament inside the bulb. It is important to use the correct voltage for a bulb to ensure its longevity and safety.

## What happens if a lightbulb receives too many amps?

If a lightbulb receives too many amps, it can cause the bulb to overheat and burn out. This is because the excess amperage creates too much energy for the bulb to handle, causing it to fail. It is important to use the correct amperage for a bulb to ensure its proper functioning and safety.