Forty posts in and you re-ask the original question?

 

Yesterday I wanted to build a rig and test the theory. Unfortunately all the spare bulbs I've kept have two filaments with one burned out. So testing on automotive bulbs for the moment seems to be on hold. I DID find two bulbs, one with the high intensity filament and one with a low intensity filament, not knowing their wattages, and powered them with 12 VDC. To my surprise the higher intensity bulb lit near full intensity while the lower intensity bulb barely glowed. I was going to film it but I can't find another bulb holder. When I get that part sorted out I'll see if I can shoot a video and either prove or disprove that a dual filament bulb predicts that the higher intensity lights, not the lower. I may have to go to the auto parts store and buy a bulb just for chits & giggles.

Hi,

Well you also have to know the lumens per watt, or the lumens. If one has higher than the other, it will be brighter with just the original voltage across it.

 

I just did an experiment with a 40W 120V bulb wired in series with a 4W 120V bulb. With 120V across the pair, the 4W bulb glowed brightly and had 119V across it; the 40W bulb had only 1.0V across it and was dark, as expected.

Hello there,

That is using two bulbs with normal current ratios of 10 to 1. That is too high to be too meaningful.
The bigger bulb just acts as a current limit for the smaller bulb, so the smaller bulb gets almost all the current it normally gets.
For a 10:1 ratio we get approximately 0.99 for the ratio of current to normal current in the smaller bulb, and that means the smaller bulb gets very nearly the normal current while the bigger bulb gets only 1/10 of the current it normally gets. That puts the estimated voltage across the smaller bulb at around 118 volts. So the smaller bulb gets nearly all the power it needs while the bigger one gets almost no power at all.

To make it more interesting, use a 4 to 1 ratio of bulb powers, like say a 40 watt and 10 watt., or even 40 and 20.

A few different current calculations for different bulb power ratios:
10:1, 0.99a, 118v
4:1, 0.95a, 110v
2:1, 0.86a, 92.5v
1.2:1, 0.73a, 69v

The currents above are based on a normal small bulb current of 1 amp. The voltages are based on 120v bulbs.

[corrected typo on 03/21/2019]

 

There is a much easier way to evaluate this.

This problem is nothing more than a voltage divider circuit. Because each bulb represents a resistance (ignoring impedance to keep it simple, including thermal effects etc), each bulb's current needs MUST be satisfied. Because the voltage source is limited to 12 volts, neither bulb will receive 12 volts.

I = P/E
I = 5/12 = 416mA
I = 40/12 = 3.33A

R = E/I
R = 12/416mA = 28.846 Ohms
R = 12/3.33A = 3.603 Ohms

Series Resistance: 32.449 Ohms

5W bulb: 10.667V @ 416mA
40w bulb: 1.332V @ 3.33A

The short answer is- the 5W bulb has enough potential to move some of the current, and the 40W bulb does not have enough potential to move enough current.

 

Forty posts in and you re-ask the original question?

Forty posts in and Tonyr1084 hasn't understood that without knowing the wattage of each filament, the experiment can't give valid results.

 

This problem is nothing more than a voltage divider circuit.

It is, but it's a bit more complex than might seem so on the surface because the resistance of a tungsten filament at operating temperature (≈ 2800 °K) is around 14X its resistance at room temperature, as shown in this plot of resistance vs. voltage for a 100W bulb.

So, with the smaller bulb operating at a high temperature and the larger bulb comparatively cooler, the ratio of their resistances is a lot greater than you would predict by simply calculating based on their resistances at nominal operating voltage.

 

That is using two bulbs with normal current ratios of 10 to 1. That is too high to be too meaningful. (...) To make it more interesting, use a 4 to 1 ratio of bulb powers, like say a 40 watt and 10 watt., or even 40 and 20.

Good point. I repeated the experiment using a 40W bulb and a 25W bulb. With 120V applied to the two in series, the voltages across each were 28V and 92V, respectively, with the 40W bulb glowing a dull red and the 25W bulb showing quite brightly.

 

Due to the nonlinear relation of resistance to voltage in an incandescent bulb, the voltage division between two bulbs in series will be greater than predicted by their hot operating resistances at rated voltage (lower wattage bulb will hog proportionally more of the voltage).

 

The non-linear relation of filament current to luminosity is also a confounding effect. The resistance versus temperature is well known and you could make a reasonable estimate based on internet sources. But have you ever seen how filament brightness depends on the current through it?

For instance if you took a 120W bulb (1A nominal), what would the brightness be as that current is reduced to 0.9, 0.8, 0.7 and so on?

I think you might be able to estimate brightness by estimating temperature, which in turn could be estimated from the resistance-vs. temperature curve. You could calculate the power dissipation at any current level, and that could be used to calculate the estimated temperature of the filament.

 

It is, but it's a bit more complex than might seem so on the surface because the resistance of a tungsten filament at operating temperature (≈ 2800 °K) is around 14X its resistance at room temperature, as shown in this plot of resistance vs. voltage for a 100W bulb.

So, with the smaller bulb operating at a high temperature and the larger bulb comparatively cooler, the ratio of their resistances is a lot greater than you would predict by simply calculating based on their resistances at nominal operating voltage.

Agreed- but what you're referencing is really insignificant in this case. The OP wanted to know which would light and why. Everything beyond it being a voltage divider is insignificant to the fundamental question. The OP did not give us brand, or technical data about each bulb. All we have is the voltage rating, and the wattage.

Therefore, using what we _know_, not what we guess at, we can still arrive at the correct answer to the question with a valid why and be well within the margin of error for correctness .v. wrongness.

 

I tried this experiment years ago and I saw the LIGHT!

 

I tried this experiment years ago and I saw the LIGHT!

I can't believe there are still those that have not after all these posts!

 

There is a much easier way to evaluate this.

This problem is nothing more than a voltage divider circuit. Because each bulb represents a resistance (ignoring impedance to keep it simple, including thermal effects etc), each bulb's current needs MUST be satisfied. Because the voltage source is limited to 12 volts, neither bulb will receive 12 volts.

I = P/E
I = 5/12 = 416mA
I = 40/12 = 3.33A

R = E/I
R = 12/416mA = 28.846 Ohms
R = 12/3.33A = 3.603 Ohms

Series Resistance: 32.449 Ohms

5W bulb: 10.667V @ 416mA
40w bulb: 1.332V @ 3.33A

The short answer is- the 5W bulb has enough potential to move some of the current, and the 40W bulb does not have enough potential to move enough current.

There's a serious flaw in this line of reasoning -- you are assuming that the resistance of each bulb is independent of the power it is dissipating. This is not a case of making a simplifying assumption to make the math a bit easier at the expense of a bit of error in the quantitative results, but more a case of a major error that calls into question even the qualitative results.

The good news is that it is salvageable with only a modicum of handwaving.

The cold resistance of a typical tungsten-filament incandescent bulb is a factor of 5x to 15x lower at room temperature compared to operating temperature.

Now for the handwaving.

At any given temperature, the ratio of the resistances of the lower wattage bulb to the higher wattage bulb is going to be pretty constant (at about 8, in this case). Since they have the same current flowing through them, at the same temperature the lower wattage bulb will be dissipating about 8 times the power of the higher wattage one. But this means that the lower wattage one will heat more (assuming roughly comparable ability to shed heat at a given power level), meaning that its resistance will increase more than the higher wattage bulb's will, so the ratio of power will shift even more in the direction of the lower wattage bulb. This is similar to a thermal runaway situation with paralleled LEDs.

Another factor that supports the 5W bulb putting out more light is that the amount of light given off by a typical incandescent bulb is not linear with power, but the curves I've seen almost look like it's a power law on that. One decent graph I saw looks like it is putting out about 25% of the light at 50% of the power and at 25% of the power it is putting out maybe 5% of the light.

EDIT: Fix typo - left out "of the resistances" in the next to last paragraph. Thanks TheElectrician for pointing it out.

EDIT 2: It was actually a decent graph I saw, not a graph that you see as you are going down (a descent graph).

 

Agreed- but what you're referencing is really insignificant in this case. The OP wanted to know which would light and why. Everything beyond it being a voltage divider is insignificant to the fundamental question. The OP did not give us brand, or technical data about each bulb. All we have is the voltage rating, and the wattage.

Oh, it is VERY significant.

Imagine that the relationship went the opposite way and that the resistance went down as temperature went up (which wouldn't bode well for running bulbs from voltage sources -- and which is why you shouldn't run LEDs from voltage sources).

 

There's a serious flaw in this line of reasoning -- you are assuming that the resistance of each bulb is independent of the power it is dissipating. This is not a case of making a simplifying assumption to make the math a bit easier at the expense of a bit of error in the quantitative results, but more a case of a major error that calls into question even the qualitative results.

The good news is that it is salvageable with only a modicum of handwaving.

The cold resistance of a typical tungsten-filament incandescent bulb is a factor of 5x to 15x lower at room temperature compared to operating temperature.

Now for the handwaving.

At any given temperature, the ratio of the lower wattage bulb to the higher wattage bulb is going to be pretty constant (at about 8, in this case). Since they have the same current flowing through them, at the same temperature the lower wattage bulb will be dissipating about 8 times the power of the higher wattage one. But this means that the lower wattage one will heat more (assuming roughly comparable ability to shed heat at a given power level), meaning that its resistance will increase more than the higher wattage bulb's will, so the ratio of power will shift even more in the direction of the lower wattage bulb. This is similar to a thermal runaway situation with paralleled LEDs.

Another factor that supports the 5W bulb putting out more light is that the amount of light given off by a typical incandescent bulb is not linear with power, but the curves I've seen almost look like it's a power law on that. One descent graph I saw looks like it is putting out about 25% of the light at 50% of the power and at 25% of the power it is putting out maybe 5% of the light.

It seems to me that you haven't used units properly.

What does the phrase "the ratio of the lower wattage bulb to the higher wattage bulb" mean?

Does it mean "the ratio of (the current in) the lower wattage bulb to (the current in) the higher wattage bulb"?

Or does it mean "the ratio of (the voltage across) the lower wattage bulb to (the voltage across) the higher wattage bulb"?

Or does it mean "the ratio of (the resistance of) the lower wattage bulb to (the resistance of) the higher wattage bulb"?

Or does the word "bulb" refer to something that has units of its own?

 

Thanks @WBahn and @MrAl. As it turns out I totally agree but could not express as eloquently as you.

It seems to me that you haven't used units properly.

What does the phrase "the ratio of the lower wattage bulb to the higher wattage bulb" mean?

Does it mean "the ratio of (the current in) the lower wattage bulb to (the current in) the higher wattage bulb"?

Or does it mean "the ratio of (the voltage across) the lower wattage bulb to (the voltage across) the higher wattage bulb"?

Or does it mean "the ratio of (the resistance of) the lower wattage bulb to (the resistance of) the higher wattage bulb"?

Or does the word "bulb" refer to something that has units of its own?

You omitted one. The ratio of the rated watts of the lower wattage bulb to the rated Watts if the higher rated bulb.

 

It seems to me that you haven't used units properly.

What does the phrase "the ratio of the lower wattage bulb to the higher wattage bulb" mean?

Does it mean "the ratio of (the current in) the lower wattage bulb to (the current in) the higher wattage bulb"?

Or does it mean "the ratio of (the voltage across) the lower wattage bulb to (the voltage across) the higher wattage bulb"?

Or does it mean "the ratio of (the resistance of) the lower wattage bulb to (the resistance of) the higher wattage bulb"?

Or does the word "bulb" refer to something that has units of its own?

Simple typo. I'll fix it. The context should, however, make it clear that since I was just talking about resistance of the bulbs as a function of temperature and the sentence in question is predicated with "At any given temperature...", it is referring to the ratio of the resistances. This is further reinforced by the numerical value of 8 provided and the fact that the next sentence uses this statement to conclude that, due to the series connection, the power would be eight times as great in the lower wattage bulb as in the higher wattage one if both are at the same temperature.

 

Thanks @WBahn and @MrAl. As it turns out I totally agree but could not express as eloquently as you.


You omitted one. The ratio of the rated watts of the lower wattage bulb to the rated Watts if the higher rated bulb.

What about the cost? The size? The weight? The carbon footprint? The....?

 

What kind of base does the bulb use?

https://www.bulbs.com/learning/basechart.aspx

 

What kind of base does the bulb use?

https://www.bulbs.com/learning/basechart.aspx

Just curious -- why would this be a factor?