By how much the brightness of incandescent light bulb will change, if voltage is dropped by 5%?

Answer 1

Kirchwey is right. The question asks for the "brightness" which means that you should integrate the radiated spectrum only over the part that the eye can see.

But the ratio of visible to invisible spectral power is not going to change much with a 5% voltage drop.

Let's start with the previous answer that the electrical power will drop by 6.3%.

It is also true that a standard incandescent bulb is less than 3% efficient in producing visible light. Probably less than 20% efficient at radiating at all wavelengths. So lets assume that most of the heat energy is conducted away through the metal filiment mounts.

This then is a Newton's cooling law situation. If the power declines by 6.3%, then the temperature will decline by about 6.3%. (I suppose we could worry about cooling relative to room temperature and radiation calculated in Kelvin relative to absolute zero, but that would be nit picking.)

So the total radiated power should decline according to a T^4 power for black body radiation which gives a 77.1% radiation level or a decline of 22.9%.

If the ratio of visible to total radiated power is about the same for both voltage levels, then we should have a 22.9% decline in "brightness."

Answer 2

The problem is more complicated than the first two answers assume. (The third answer addresses only power, not brightness.) Brightness in the visible spectrum is only proportional to power if the filament remains at the same temperature. This can only happen when you compare, say, a 60-watt bulb and a 100-watt bulb, both running at their nominal voltage. Reduce the voltage to a given bulb and you reduce the power and also the proportion of that reduced power that is radiated in the visible spectrum, as the ref. makes clear. To solve this you'd first have to compute the equilibrium temperatures at the nominal and reduced voltages, by iterating on resistance, resulting current & power, and resulting temperature. (Since power dissipation is partly conductive, i.e., linear, and partly radiative, i.e., 4th-power, converting power to temperature could be tricky). Then you'd have to go to the blackbody radiation formula and integrate over the visible spectrum to get the nominal and reduced brightnesses. Too complicated for me.

Answer 3

We'll say that it's a 100 watt light bulb and 120 volts.

W = IV

W / V = I

100/120 = 0.8333amps (original hot current)

Resistance is 144 ohms at normal temperature (R = V/I, or V^2/W, or W/I^2) <== all equal 144 ohms.

If you energize it at 114 volts, then current will drop slightly and filament resistance will drop as well due to a decrease in temperature.

W = IV (now voltage went down to 95% of it's initial value and current went down as well, so wattage will go down by a factor greater than 5%)

Final wattage will be < 95 watts.

The filament is made of tungsten. We'll say there's a 100C difference between the 120V and 114V.

Electrical Resistivity @ 1727 ºC (microhm-cm) 55.7
Electrical Resistivity @ 2727 ºC (microhm-cm) 90.4

so it goes down by a factor of ~86.5/90

Original hot resistance is 144 ohms so it goes down to ~138.4 ohms

I = V/R = 114/~138.4 = ~0.8237 amps

W = IV = ~0.8237 x 114 = ~93.9 watts

That's about a 6.1% drop in power.

Answer 4

Light bulbs are not linear devices. Their resistance changes with temperature.

However, if you approximate the bulb as a linear circuit element, the brightness is proportional to the power dissipated in the bulb:

P = V*I = (V^2)/R

So, if the voltage goes to 0.95 of its former value, the power goes to (0.95)^2 = 0.9025 of its former value, a 9.65% drop in brightness.

Answer 5

p = k v2/R ; k(a constant) times V squared upon R
now v' =0.95v
=> p' = k (0.95v)2 /R
=> p' = k 0.9025 v2/R
=> p' = 0.9025 p
brightness is proportional to power
drop in power = 9.75%