# Thickness of bulb filament affect brightness?

1. Jan 21, 2008

### kapellmeister

Hi,

does the thickness of filament affect the brightness of a bulb? So will a thinner filament be brighter than a thicker filament?

2. Jan 22, 2008

### mgb_phys

For a fixed voltage (ie. in a domestic light bulb) the thicker the filament the lower it's resistance and so less energy is converted into heat - the filament is dimmer.
Brighter filaments are thinner, have higher resistance and so heat up more -> brighter.
This is also why low wattage bulbs in cellars and cupboards last so long.

3. Jan 22, 2008

### pixel01

If the two filaments are at the same temperature and of the same length, the thicker one will be brighter because is has larger surface. In turn, the temperature depends on the length, thichkness, and material, voltage... so the question is not clear.

4. Jan 22, 2008

### kapellmeister

thanks for the replies!

ya, assuming the length, material of the filaments are the same and that they are connected to the same electrical source, then the one with a thinner filament will glow brighter as it offers more resistance. Am I right in saying so?

5. Jan 22, 2008

### vanesch

Staff Emeritus
Uh

If the resistance is LOWER, and the voltage is fixed, then the dissipated power is higher:

P = V^2/R (or if you like, I = V/R and P = V I).

6. Jan 22, 2008

### vanesch

Staff Emeritus
No, it is the other way around... the lower the resistance, the higher the dissipated power.

However, you have to take into account also the equilibrium temperature. The dissipated power (P = V.I = V^2/R) has to be irradiated away through the surface of the filament, and the power irradiated per unit of surface goes to the 4th power of the temperature. So the temperature of the filament will rise until there is as much power irradiated away as there is dissipated electrically inside.
For a cilindrical filament, if you double the radius, you do the section times 4 (pi radius^2), so the resistance lowers by a factor of 4, and hence the power dissipated increases by a factor of 4 (at constant voltage).

Its surface, however, grows only by a factor of 2 (2 pi radius). So per unit of filament surface, twice as much energy has to be irradiated away, and hence the temperature will have to be the fourth root of 2 larger (1.18 times the temperature).

So the thicker filament irradiates 4 times more power away, and does this at a slightly higher temperature.

7. Jan 22, 2008

### mgb_phys

It was early, I'm still jet-lagged and I haven't had enough coffee!!!!!

8. Jan 22, 2008

### Emanresu

I am interested to know why that is ...

E.

9. Jan 22, 2008

### kapellmeister

so a bulb with a thicker filament should glow brighter as more power can be dissipated due to a lower resistance (P= V^2/R).

Am I right in saying that the reason tungsten filament glow is due to its high resistance to electron flow, thereby creating heat. So wouldn't a thinner filament (offering higher resistance) create even more heat, thereby giving a brighter glow?

My question arose because I saw a video showing a bulb filament, moments before it melted. It showed the thinner region of the filament starting to glow brighter and subsequently melted. This led me to think whether filament thickness affects brightness of bulb.

Last edited: Jan 22, 2008
10. Jan 23, 2008

### vanesch

Staff Emeritus
11. Jan 23, 2008

### vanesch

Staff Emeritus
This is correct, but that is because you now have a series of "small resistors". Consider this: you have a small region of resistance r, and the rest, resistance R. The total resistance (series) is r+R.

Now, the current in your overall system is I = V/(r+R). The power, dissipated in the small region, is I^2 r.

So the power dissipated locally, is given by V^2 r/(r+R)^2.

The overall power dissipated, is V x I = V^2 / (r+R)

So we see that if the local region increases its resistance (r increases), with R constant, that:
The overall power diminishes slightly: V^2 / (r + R) decreases a bit if r increases a bit.
The local power dissipation INCREASES strongly:
V^2 r/(r+R)^2 increases.

Take V = 1V, R = 10 ohm, and r goes from 1 to 2 ohm:

overall power goes from 1/11 to 1/12, so it diminishes a bit.

local power goes from 1/121 to 2/144: it almost doubles.

On top of that, the AREA of the local piece has decreased, so locally, we dissipate more power, and we have a smaller area --> temperature increase.