Two 95 watt light bulbs walk into a bar ....

[Galloleo]

TL;DR

I have a question about this classic problem based on information I read in another post.

Two 95 W (120V) light bulbs are wired in series, then the combination is connected to a 120 V supply.
How much power is dissipated by each bulb?
Answer 24W

I followed the explanations as such:
r1 = 120^2/(95) = 151.8
R = 303
r2 = r1
W = 190
I = P/V
I = 190/120 = 1.58
v1 = 190/1.58 = 60
p1 = 60^2/(151.8) = 23.7w
p2 = p1

I = E/R as well.
My question is about this:
I = 120/303 = 0.396 which is about 1/4th the amps originally calculated.
What have I failed to consider? There is something about this relationship I don't understand.

 

[DaveE]

First, let's pretend you said "resistor" instead of "light bulb"; more on that later.

I didn't follow every step in your thought process, but it started right and you ended up with the right answer. When you are very familiar with this sort of problem, it will take fewer steps. I'm a little unclear on what you are confused about since your math is right, except your comment about "1/4th the amps originally calculated". I guess you are comparing two lights in series compared to two lights in parallel? 1/2 of the current reduction is because you only have one of the two parallel paths, the other half is because each light bulb only has 1/2 of the voltage across it.

Since you got all of the calculations correct, I think your confusion lies in not having a clear idea about what two things you are comparing. I would draw two schematics for the two cases you are comparing and annotate them with the parameters you determine, like V, I, P for each component.

OK, now my annoying pedantic point. Incandescent light bulbs don't act like simple resistors. It turns out that their resistance is a strong function of the temperature of the element inside, which is normally literally "white hot". When you drop the voltage in half, the power decreases by 4x and the element gets a lot cooler. This means that it's resistance also drops increases drops some. Don't worry about the details or how much, you can investigate that later if you care. I think that's not the point of this problem. Just remember that things like incandescent lights and "red hot" heater elements don't have constant resistance values when you change how much power you put into them. Normal resistors, like you find on a circuit board do have a constant value (until they break, LOL).

edit: 3rd try, back to the original. See post #9. Honestly, I mean well, LOL. Eventually I'll converge on a sensible post, with your help.

edit #4: "When you drop the voltage in half, the power decreases by 4x" - nope, that's me being sloppy again. The power drops by a lot, but not 4x since the resistance changed as the temperature dropped.

 

[Galloleo]

First, let's pretend you said "resistor" instead of "light bulb"; more on that later.

I didn't follow every step in your thought process, but it started right and you ended up with the right answer. When you are very familiar with this sort of problem, it will take fewer steps. I'm a little unclear on what you are confused about since your math is right, except your comment about "1/4th the amps originally calculated". I guess you are comparing two lights in series compared to two lights in parallel? 1/2 of the current reduction is because you only have one of the two parallel paths, the other half is because each light bulb only has 1/2 of the voltage across it.

Since you got all of the calculations correct, I think your confusion lies in not having a clear idea about what two things you are comparing. I would draw two schematics for the two cases you are comparing and annotate them with the parameters you determine, like V, I, P for each component.

OK, now my annoying pedantic point. Incandescent light bulbs don't act like simple resistors. It turns out that their resistance is a strong function of the temperature of the element inside, which is normally literally "white hot". When you drop the voltage in half, the power decreases by 4x and the element gets a lot cooler. This means that it's resistance also drops increases some. Don't worry about the details or how much, you can investigate that later if you care. I think that's not the point of this problem. Just remember that things like incandescent lights and "red hot" heater elements don't have constant resistance values when you change how much power you put into them. Normal resistors, like you find on a circuit board do have a constant value (until they break, LOL).

Thank you for your response. Yes, I have little experience with these types of problems and it seems there is a lot more to them than what the formulas on the Ohms/Watts Law wheel imply. I was testing them out and I expected to get equivalent results with I = P/E and I = E/R. Well ... not really expected. 190/120 and 120/303 ... we can sort of tell without a calculator. I immediately blamed my lack of understanding.

Thank you for explaining that there is a relationship between temperature and resistance. I am studying for an electrical exam and I am sure it will come up.

 

[alan123hk]

I followed the explanations as such:
r1 = 120^2/(95) = 151.8
R = 303
r2 = r1
W = 190
I = P/V
I = 190/120 = 1.58
v1 = 190/1.58 = 60
p1 = 60^2/(151.8) = 23.7w
p2 = p1

As mentioned in the #2, please note that the resistance of the filament bulb has a nonlinear relationship with the voltage change.

Nevertheless, let's assume that the resistance of the light bulb is constant, so how the value of W = 190 derived?

 

[Averagesupernova]

...how the value of W = 190 derived?

I was wondering the same thing.

 

[DaveE]

how the value of W = 190 derived?

I thought it was just the 2 95W bulbs running in parallel. There's no other mention of a higher (or different) supply voltage.

 

[Averagesupernova]

I thought it was 95 watts * 2 also. But that doesn't apply in this scenario with them in series. So until this is ironed out, I'm going to say there is a basic problem with the math.

 

[DaveE]

I thought it was 95 watts * 2 also. But that doesn't apply in this scenario with them in series. So until this is ironed out, I'm going to say there is a basic problem with the math.

Yes, you're right. The middle bit where he finds that the voltage is 1/2 of the source for 2 in series isn't right. Right answer, wrong reason.

 

[Tom.G]

...the power decreases by 4x and the element gets a lot cooler. This means that it's resistance also drops increases some.

Uhmm... you had it right the first time, Tungsten has a positive temperature coefficient of resistance. The inrush current of a Tungsten incandescent lamp is roughly 10 to 15 times that of the 'operating' current.

For more info than anyone cares to read, try:
https://www.google.com/search?&q=lamp+inrush+current

 

[DaveE]

Uhmm... you had it right the first time, Tungsten has a positive temperature coefficient of resistance. The inrush current of a Tungsten incandescent lamp is roughly 10 to 15 times that of the 'operating' current.

For more info than anyone cares to read, try:
https://www.google.com/search?&q=lamp+inrush+current

Yes, duh! 100% right. Maybe I should edit the edits? LOL.

 

[DaveE]

relationship between temperature and resistance. I am studying for an electrical exam and I am sure it will come up.

Probably not. That's the sort of detail that they don't often teach in school. They think you'll learn it when you have that sort of problem. I actually suspect whoever wrote that problem doesn't know either, or they would picked a more realistic version.

 

[Windadct]

"OK, now my annoying pedantic point. Incandescent light bulbs don't act like simple resistors."..agreed..

 

[Shane Kennedy]

"OK, now my annoying pedantic point. Incandescent light bulbs don't act like simple resistors."..agreed..

Resistors don't act like "simple" resistors. Incandescent light-bulbs act like the wirewound resistors that they are, just that they are designed to run much hotter.

 

[Averagesupernova]

Resistors don't act like "simple" resistors. Incandescent light-bulbs act like the wirewound resistors that they are, just that they are designed to run much hotter.

In all my years this is the first I've heard that resistors don't act like resistors.
-
An incandescent light bulb acts like a PTC thermistor.
-
A wirewound resistor is expected to maintain it's ohm spec throughout it's entire range in watts. Can't say the same of an incandescent bulb.

 

[DaveE]

A wirewound resistor is expected to maintain it's ohm spec throughout it's entire range in watts.

The key word here is "resistor". That's not just any piece of wire. It is an alloy chosen to have a low TC of resistance, also higher resistivity so they don't have use thin or long pieces.

There is a tremendous amount of specific technical knowledge applied to what most of us only think of a "simple" parts. First you learn in school about resistors as an idealized concept, then when you get out in the real world you (may) have to learn the differences between real world parts; like resistors made from carbon, metal film, ceramic, wirewound,...; capacitors made from... oh never mind, we'll be here all day at this rate.

 

[DaveE]

We should also give credit to Mr. Edison and his minions for figuring out how to make those lights. That filament isn't just any piece of wire either.

 

[Shane Kennedy]

In all my years this is the first I've heard that resistors don't act like resistors.
-
An incandescent light bulb acts like a PTC thermistor.
-
A wirewound resistor is expected to maintain it's ohm spec throughout it's entire range in watts. Can't say the same of an incandescent bulb.

I said "they don't act like SIMPLE resistors. They ALL have a temperature constant, they ALL have inductance and capacitance

 

[Kenwstr]

The initial error was assuming each bulb draws the same power as it would if it was the only load.
In series the, R increases, therefore current decreases reducing power.

My solution (knowing the 2 bulbs are the same power and impedance, assuming 120V is AC):

R1 = 120^2/(95) = 151.58
R2 = R1
R = R1 + R2 = 303
I = V/R = 120 / 303 = 0.396
P = V * I =120 * 0.396 = 47.52
P = P1 + P1 & P1 = P2
P1 = P / 2 = 47.52 / 2 = 23.76W
P2 = P / 2 = 47.52 / 2 = 23.76W

 

[Averagesupernova]

I said "they don't act like SIMPLE resistors. They ALL have a temperature constant, they ALL have inductance and capacitance

Yes they do. And I think 'IDEAL' would be a better choice of a word than 'SIMPLE'.

 

[Rive]

I think 'IDEAL' would be a better choice of a word than 'SIMPLE'.

For these kind of problems only a few years left to 'live' anyway. You will get very different answer (or, at least a weird look) once LED will become the default expectation for a 'bulb'

 

[Frodo]

I have always been meaning to do this - thank you for providing the stimulus for me to do it!

I have just measured a 60W bulb for a 240V supply. At room temperature it was 80 ohms.

Now, if it consumes 60W when on, then 60W = 240 * 240 / R. The resistance when on is therefore ~960 ohms or 12 times greater..

See Resistivity of Tungsten which shows that at 2,800K (~2,500C) tungsten's resistivity is 14.99x higher than at room temperature.

Facts almost agreeing with theory - it must be my lucky day

Or maybe it runs closer to ~2,100C for 12.46x resistivity.

Look at the inrush current. When the bulb is cold it is putting 80 ohms across 240V and is drawing 720W ...

 

[sophiecentaur]

Look at the inrush current. When the bulb is cold it is putting 80 ohms across 240V and is drawing 720W ..

Which accounts for a lot of blown fuses in 5A lighting circuits in the past.

 

[alan123hk]

I have always been meaning to do this - thank you for providing the stimulus for me to do it!

If you use a lux meter to measure the illuminance, you can get the change in illuminance with temperature.