What makes a low watt incandescent bulb different from a higher watt bulb?

MrChips

Joined Oct 2, 2009
27,686
Use Ohm’s Law and power formula to help you understand this.

Ohm’s Law
I = V / R

Power = I x V = I x I x R = V x V / R

For a given voltage, smaller R increases power.
 

DickCappels

Joined Aug 21, 2008
9,512
@MrChips has given the key formula.

Thicker, possibly longer (makes sense) and maybe the choice of how to coil the filament is different. You would need pretty good microscope to get a good look at the differences.

The idea is to get the tungsten filament to draw the desired power and to get it to the desired temperature. The lower the resistance, the higher the power, but to have that higher power at a given color temperature requires a greater filament length for than the lower power filament.

In addition there may be a difference in preparing the wire in a process called swagging in which the wire is beaten with "hammers" and worked to obtain a specific desired resistance and tolerance to power cycling.

It was not that long ago that I had a custom incandescent lamp designed. The product never made it to market. Know anybody who wants to buy 10,000 8 ma incandescent lamps? Great for stabilizing Wein bridge oscillators or limiting AC current :)
 

WBahn

Joined Mar 31, 2012
27,886
For example, a 30 watt bulb versus a 100 watt bulb. Is the filament thicker? Thinner? Longer?
Since the voltage is the same, to get higher power you need a lower resistance.

There are a few ways to accomplish this. At first glance, the simplest is to decrease the length of the filament or increase it's thickness. However, the first will result in a higher current density in the filament, which will cause it to heat up a lot more. Using this approach, if you increase the wattage by a factor of three, then you need to increase the power dissipation per length of filament by a factor of nine (you want three times the power in one-third the length of the filament). You quickly get limited by the mechanical and thermal limits of the filament. If you increase it's thickness, you keep the same length and increase the power per length by a factor of three, but since the current density is the same, the filament will heat to roughly the same temperature.

Another approach is to put multiple filaments in parallel. If you put three in parallel, you now have three filaments operating, more-or-less, just like they were before with each filament putting out about the same amount of light that it did before, hence about three times the light overall.

Yet another alternative is to use a different filament alloy, but I don't think this is done much in practice because tungsten (or, rather, a particular alloy of tungsten) has overall properties that make it hard to beat.
 

sparky 1

Joined Nov 3, 2018
732
Assuming the bulbs are the older type incandescent. Here are some things that could differ.
Measure the resistance.
Measure the heat given off and the globe size.
Sample the rarefied gas.
Compare the tungsten filament's physical dimensions.

I recall grinding a 1/2-Watt rated carbon resistor and the resistance went up. The cross section (thickness) was reduced
When the resistance goes up the current goes down. So, I think the smaller thickness (cross sectional area) is the higher Watt.
In a circuit sourced with 12Vdc and 1 Amper load you could compare heat given off of a 10 Ohm and a 1000 Ohm.
Unlike the old textbooks depicting a bulb as a resistor, the properties of tungsten in rarefied gas are not as simple as most resistors.
 
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WBahn

Joined Mar 31, 2012
27,886
I recall grinding a 1/2-Watt rated carbon resistor and the resistance went up. The cross section (thickness) was reduced
When the resistance goes up the current goes down. So, I think the smaller thickness (cross sectional area) is the higher Watt.
I don't follow the reasoning here. If the smaller thickness resulted in greater resistance, then for the same voltage applied across it the power dissipated will go down. Now, if you used a constant current source, the power would go up, but that is not the usual case.

In a circuit sourced with 12Vdc and 1 Amper load you could compare heat given off of a 10 Ohm and a 1000 Ohm.
If the circuit is sourced with 12 Vdc and has a 1 ampere load, then the resistance is neither 10 Ω nor 1000 Ω, but rather 12 Ω.

What does it mean to say that the circuit is sourced with both a specific voltage and a specific current and then talk about using different resistances.
 

sparky 1

Joined Nov 3, 2018
732
Further explanation about using resistors to replace a light bulb in showing heat difference.

I should have said 10 Ohm resistor in series with a load can be compared to a 1000 Ohm resistor in series with a load will show that the heat given off is greater for the 10 Ohm resistor. The values 12Vdc at 1 Amp should demonstrate heat but I cannot recommend using 120Vac because of a safety concern. Be careful. I am sure there is a better and safer demonstration where resistors are used instead of a light bulb. It's about showing warmth and not having flames.

When the person walks away from a demonstration, they should have confidence in the math and the logic
regarding the difference in heat between the small and large value resistors. Some of the tools for
answering a question about light bulbs can be deduced after an introduction to Ohms law has been presented.
 

WBahn

Joined Mar 31, 2012
27,886
Further explanation about using resistors to replace a light bulb in showing heat difference.

I should have said 10 Ohm resistor in series with a load can be compared to a 1000 Ohm resistor in series with a load will show that the heat given off is greater for the 10 Ohm resistor. The values 12Vdc at 1 Amp should demonstrate heat but I cannot recommend using 120Vac because of a safety concern. Be careful. I am sure there is a better and safer demonstration where resistors are used instead of a light bulb. It's about showing warmth and not having flames.
I still don't know what the 1 Amp is all about. How do you put a 1000 Ω resistor in series with a load and get 12 Vdc at 1 A?

Whether the 10 Ω gives off more heat than the 1000 Ω depends on the load.

If the load is small enough, the resistors see a voltage source and the smaller resistor will put out more heat. But if the load is large enough, then the resistors will see a current source and the 1000 Ω will give off more heat.

This implies that there is a load that is just right so that the two resistors give off the same amount of heat. That resistance is

Ro = √(R1·R2) = √((10 Ω)(1 kΩ)) = 100 Ω

Let's see if this is true.

For the 10 Ω resistor, the total resistance is 110 Ω and so the total current is

I = 12 V / 110 Ω = 109.1 mA

The power dissipated by the resistor is therefore

P = (109 mA)²·(10 Ω) = 119.0 mW

For the 1000 Ω resistor, the total resistance is 1100 Ω and so the total current is

I = 12 V / 1100 Ω = 10.91 mA

The power dissipated by the resistor is therefore

P = (10.91 mA)²·(1 kΩ) = 119.0 mW
 
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