# Lightbulb physics problem?

The light in a lightbulb is the result of an increase in temperature. This increase is created by both a highter elektrical resistance in the wire and a highter elektric current. But a higher resistance results in a lower elektric current (I) (I=V/R), so it is both increasing and decreasing the brightness?? Is this correct?

The power flowing into the light bulb can be described by ## P=VI ##
Inserting for ## I=\frac {V}{R} ## , since V is constant, using Ohm s law gives
## P= \frac {V^2} {R} ##
The output power is equal to the output power and approximated by a black body radiator so it is ## P= \alpha (\Delta T)^4 A##

As you can see for a constant voltage source like your connection to the grid, you will draw less power if you increase the resistance.
See ## P= \frac {V^2} {R} ##
So the temperature and brightness of the wire decreases.
Having a low resistive object hooked up will draw lots of power.
For example you could kill your fuses in the house by short circuting a wall socket.
DONT ATTEMPT THAT BTW.  berkeman
Mentor
The light in a lightbulb is the result of an increase in temperature. This increase is created by both a highter elektrical resistance in the wire and a highter elektric current. But a higher resistance results in a lower elektric current (I) (I=V/R), so it is both increasing and decreasing the brightness?? Is this correct?
It's simple negative feedback. The filament initially has a low resistance, so a high current flows briefly to heat it up. As the filament heats up, its resistance increases, and this ends up limiting the current at the stable operating point of the filament.

Khashishi
The resistance of the filament needs to be designed at a sweet spot such that the resistance is considerably higher than the resistance of the wires in your lamp and house, but low enough to draw a lot of current and glow white hot. Say, around 100-1000 ohms. Light bulbs come in various watt ratings which vary in the resistance in the filament, I think.

Merlin3189
Homework Helper
Gold Member
The light in a lightbulb is the result of an increase in temperature. This increase is created by both a highter elektrical resistance in the wire and a highter elektric current. But a higher resistance results in a lower elektric current (I) (I=V/R), so it is both increasing and decreasing the brightness?? Is this correct?
The problem I see with your comment, is the use of one-sided comparisons. Eg. increase from what? higher than what? lower than what?
If you get rid of these (which tell us nothing) it makes more sense.
"The light in a lightbulb is the result of <temperature>. This <temperature> is created by both <elektrical resistance in the wire> and <elektric current>." So far so good. Correct and no problems.
"But a higher resistance results in a lower elektric current (I) (I=V/R), so it is both increasing and decreasing the brightness??"
So is the question: "if we increase the resistance of a bulb, that will decrease the current, so you have two changes which both increase and decrease brightness."
Then your premise is correct. In fact the net effect is that the bulb is less bright.
Similarly, if you decrease the resistance of a bulb, the current increases and the net effect is that the brightness increases.

Both of those presume (as Tazerfish said) a constant voltage source (or near enough) like mains or a car battery, because for constant voltage, power is inversely proportional to resistance. Otherwise the result can be either way.

Both of these results are actually important in the design of light bulbs as Khashishi and Berkeman say. Most lightbulbs have filaments of metal whose resistance increases with temperature (brightness.) If the bulb temperature (brightness) varies, its resistance changes in such a way as to counteract that change. ie. they are stable.
Bulbs have been made with other filaments, such as carbon, whose resistance decreases with temperature. They are unstable: if they get hotter the resistance decreases, so they get even hotter, so the resistance decreases more, so they get even hotter, etc. Such bulbs are operated in series with a metallic ballast resistance, or with a more complex current controlled power supply.

So (correct me if I'm wrong)... if we increase the resistance with x... the power(brightness) will increase with x but the power(brightness) will decrease with x^2 according to P=I^2*R...

Merlin3189
Homework Helper
Gold Member
If you like to use P=I2R and I=V/R, then I2=V2/R2
So P = RV2/R2
sort of making P proportional to R but inversely proportional to R2
but really R cancels and its just inversely proportional to R for a constant voltage supply.

But if it is not a constant voltage supply (equivalent to a very low internal series resistance) then the same relationship does not apply.
In general the bulb (or any resistive load) gives maximum power when it is the same resistance as the internal series resistance. Then any change in resistance reduces the power.

Source Resistance--- Description --------- ---- Bulb Resistance R----Effect of Increasing R ---- Effect of Decreasing R
Very low --------------Voltage source-----------Higher than source----Less power ---------------More power
Intermediate----------Real----------------------- Higher than source----Less power-------------More power until R = source
Intermediate ---------Real -----------------------Equal to source -------Less power ---------------Less power
Intermediate----------Real------------------------Lower than source---More power until R=source--Less power
Very high -------------Current source-----------Lower than source----More power----------------Less power

An ideal voltage source has zero internal resistance, but a real "constant" voltage source will have some.
An ideal current source has infinite internal resistance, but a real "constant current" source will have finite resistance.

But can I please emphasise that the way you are thinking about it is NOT helpful. Power is the product of current and voltage and both can change when the resistance changes. There is no general rule that relates power and resistance in isolation. Just work out the voltage and current and take it from there.