Resistance of a tungsten filament bulb at 20 C

[ephphatha16]

I was looking for any credible sources who could reveal the resistance of a normal tungsten filament bulb at room temp 20 C? I heard its cold resistance is about 9.5ohms. Is that the resistance at rtp? Thanks a lot in advance!

 

[Charles Link]

$P=V^2/R$, for the operating temperature, which is about $T=2500$ K, so that a 60 watt bulb, with $V \approx 100$ Volts, makes $R=160 \, \Omega$, or thereabouts. The resistance at room temperature is considerably lower, perhaps by a factor of 5 or more. In order to give very precise numbers, the wattage of the bulb would need to be specified. e.g. at operating temperature, a 100 watt bulb would have $R \approx 100 \, \Omega$. In any case, a number of 10 ohms is in the right ballpark for the room temperature resistance. I have previously measured the resistance of standard light bulbs at room temperature with an ohm meter, but that was quite a few years ago.

 

[anorlunda]

Well, it depends on the wattage of course. The power of a light bulb $P=\frac{V^2}{R}$ where P is in watts, V is RMS voltage, and R is in ohms. If you know P and V, you can calculate R.

For example, P=40 watts V=120V, then R=360 ohms.

But that number is for a hot filament. For a cold filament, I found this answer from an earlier PF post.

Exactly correct on both accounts! A common tungsten light bulb filament runs at 2800K, for a resistance ratio of about 15:1 according to
http://hypertextbook.com/facts/2004/DeannaStewart.shtml
This is close to the ratio you observe.

So 360/15=24 ohms for a 40W bulb coldp.s. I don't know what rtp means.

 

[Charles Link]

Here is some additional data: http://physics.usask.ca/~bzulkosk/L...l-radiation-lab_Tungsten-Temp-Resistivity.pdf

 

[AlexCaledin]

Here, I am holding a 500W 220V bulb, size of a pineapple, its cold resistance is just 7 Ohm.

 

[Charles Link]

Here, I am holding a 500W 220V bulb, size of a pineapple, its cold resistance is just 7 Ohm.

That would match the calculations of @anorlunda with a ratio of 15:1 quite closely.

 

[ephphatha16]

Here is some additional data: http://physics.usask.ca/~bzulkosk/L...l-radiation-lab_Tungsten-Temp-Resistivity.pdf

Thanks a lot for the data! However these are values for resistivity I couldn't use them for resistance, could I?

 

[ephphatha16]

Well, it depends on the wattage of course. The power of a light bulb $P=\frac{V^2}{R}$ where P is in watts, V is RMS voltage, and R is in ohms. If you know P and V, you can calculate R.

For example, P=40 watts V=120V, then R=360 ohms.

But that number is for a hot filament. For a cold filament, I found this answer from an earlier PF post.
So 360/15=24 ohms for a 40W bulb coldp.s. I don't know what rtp means.

Thanks a lot for the values!
rtp: room temperature pressure

 

[Charles Link]

Thanks a lot for the data! However these are values for resistivity I couldn't use them for resistance, could I?

The resistivity is proportional to the resistance for a given resistor. You can write the equation $\frac{\rho_{T=2800K}}{\rho_{T=300K}}=\frac{R_{T=2800K}}{R_{T=300K}}$ where $\rho=$ resistivity, and $R=$ resistance. Thereby, the 15:1 ratio with resistivity for these two temperatures also applies to resistance.

 

[ephphatha16]

The resistivity is proportional to the resistance for a given resistor. You can write the equation $\frac{\rho_{T=2800K}}{\rho_{T=300K}}=\frac{R_{T=2800K}}{R_{T=300K}}$ where $\rho=$ resistivity, and $R=$ resistance. Thereby, the 15:1 ratio with resistivity for these two temperatures also applies to resistance.

Ok... so then if I have a 2.5V bulb of 0.25W at 300K how could i find the resistance?

 

[anorlunda]

Ok... so then if I have a 2.5V bulb of 0.25W at 300K how could i find the resistance?

You should be able to calculate that yourself using the methods in post #2. Show us your calculation.

 

[AlexCaledin]

- measured a "2.5V 0.15A" bulb, the cold resistance is 2 Ohm approximately . . .

. . . awww sorry - forgot that my ohm-meter makes 0.05A current so it's not that cold.

 

[ephphatha16]

You should be able to calculate that yourself using the methods in post #2. Show us your calculation.

P =( V^2)/R
R = (2.5^2)/0.25
= 25ohms
but i don't know whether it is the resistance at 20C / 300K or room temperature...

 

[anorlunda]

P =( V^2)/R
R = (2.5^2)/0.25
= 25ohms
but i don't know whether it is the resistance at 20C / 300K or room temperature...

You missed the second part

But that number is for a hot filament. For a cold filament, I found this answer from an earlier PF post.

So 360/15=24 ohms for a 40W bulb cold

So divide your answer by 15. 25/15=1.67 ohms cold.

 

[Charles Link]

P =( V^2)/R
R = (2.5^2)/0.25
= 25ohms
but i don't know whether it is the resistance at 20C / 300K or room temperature...

This looks like a bulb (P=.25 watts) that might be strung together with about 40 or 50 other bulbs in series as a small Christmas tree type light. You could assume that it also runs at $R=2800 \, K$ approximately. The room temperature resistance would then be approximately $25/15=1.7$ ohms, but unless you have a very good ohmmeter, it is not going to measure a small resistance like this very accurately. Also, the ohmmeter will supply a voltage, and might supply enough voltage and current to raise the temperature of a bulb like this to near the operating temperature of 2800 K. (an ohmmeter might measure 25 ohms, and cause the bulb to light up, or it might dimly light the bulb and read 10 or 15 ohms).

 

[ephphatha16]

You missed the second partSo divide your answer by 15. 25/15=1.67 ohms cold.

No I didnt forget... I wasnt sure that if cold resistance meant at 300 K (20C)(room temperature) Is it?Really grateful for your help!

 

[anorlunda]

Yes.

Of course the numbers are approximate because the filament may have other metals alloyed with the tungsten.

 

[ephphatha16]

This looks like a bulb (P=.25 watts) that might be strung together with about 40 or 50 other bulbs in series as a small Christmas tree type light. You could assume that it also runs at $R=2800 \, K$ approximately. The room temperature resistance would then be approximately $25/15=1.7$ ohms, but unless you have a very good ohmmeter, it is not going to measure a small resistance like this very accurately. Also, the ohmmeter will supply a voltage, and might supply enough voltage and current to raise the temperature of a bulb like this to near the operating temperature of 2800 K. (an ohmmeter might measure 25 ohms, and cause the bulb to light up, or it might dimly light the bulb and read 10 or 15 ohms).

Thanks! The reason why I wanted to obtain this value is because I was doing an experiment for I-V characteristics of filament bulb. The maximum resistance of the filament during experiment was 9.46ohms. Now, if the actual value was 1.7 ohms at rtp, then there would be a huge percentage uncertainty... the bulb used in the experiment approx. is of 0.25W with two cells(1.5V)

 

[Charles Link]

Thanks! The reason why I wanted to obtain this value is because I was doing an experiment for I-V characteristics of filament bulb. The maximum resistance of the filament during experiment was 9.46ohms. Now, if the actual value was 1.7 ohms at rtp, then there would be a huge percentage uncertainty... the bulb used in the experiment approx. is of 0.25W with two cells(1.5V)

This is not really an uncertainty. (You might have measured it quite accurately). It is likely you found that the resistance increases with current or voltage, so that the I-V curve (with current on on the y-axis), will curve over to the right, (since the ratio I/V=1/R), instead of being a straight line, with slope $m=1/R$, like it would be if $R$ were constant. $\\$ I'm not sure how complete of an I-V curve you were able to generate though, if you are limited to just having a couple of batteries for voltage sources, i.e. V=0 volts, V=1.5 Volts, and V=3.0 volts... Your current meter could also have a finite internal resistance so that you might have had slightly less than 3.0 volts across the bulb itself.

 

[ephphatha16]

Thank y'all for your wonderful support and help! I will try to change my values...i suppose...

 

[Charles Link]

Thank y'all for your wonderful support and help! I will try to change my values...i suppose...

I would recommend you simply report what you measured. The resistance of $R=9.46$ ohms that you measured/computed could have been relatively accurate, but it really isn't accurate to two decimal places. (If the $R=9.46$ is a measured value using the ohmmeter, there is very limited information in what the temperature of the bulb was, (caused by the ohmmeter), as the ohmmeter made the reading. Considering the other calculations, just as a ballpark estimate, it looks like the ohmmeter could have caused the temperature of the bulb to go to about 1000 K when it made this measurement, if R=1.7 ohms for room temperature, and R=25 ohms for T=2800 K are good estimates). It is also possible the bulb is not exactly $P=.25$ watts, and there are quite a number of factors that could lead to experimental inaccuracy. $\\$ The $I-V$ curve of a tungsten filament simply is not a straight line with a constant slope $m=1/R$, (because the filament temperature and resistance changes tremendously with the applied voltage and corresponding current), and in that sense, this exercise looks to be quite educational. :)

 

[davenn]

Ok... so then if I have a 2.5V bulb of 0.25W at 300K how could i find the resistance?

you can do all the theory as indicated by previous posters, but that doesn't always play out in reality

so for reality ...
measure it with a multimeter
put meter into Ohms auto range
put meter probes across the globe terminals and read the value
if room temperature isn't 20C, do what you need to do to get to the required temp

 

[sophiecentaur]

For an alternative method of measurement you could make up an AC Bridge type circuit, using a cheap mains wall wart low voltage AC adaptor You could then use another transformer (step up) to scale up the filament resistance. The impedance transformation is proportional to the turns ratio squared so a simple mains to 12V transformer (ratio = 1:20 but you could actually measure the ratio from the voltage ratio and calibrate the system) would give you a multiplication factor of 400 so you would then be measuring a resistance of a few hundred Ohms. A simple Wheatstone Bridge or even potential divider measurement would then do the job and the null on the bridge could be found with a high impedance, high sensitivity meter or 'scope. With 1% tolerance fixed resistors with value of a few hundred Ohms, your resistance measurement would give a similar accuracy and your filament current could be arbitrarily low.

 

[ephphatha16]

Thanks a lot for all the beneficial input regarding the topic!

 

[sophiecentaur]

Thanks a lot for all the beneficial input regarding the topic!

Bridges are a bit of a thing of the past but they are capable of much more than the old school Wheatstone Bridge with the metre of resistance wire. You could spend an entertaining few minutes / hours with google, using terms like " wheatstone bridge applications" and "RF component bridges. State of the art modern equipment is very expensive, if you want the sort of accuracy that some of the old bridges were capable of. The DMM is ok as far as it goes but . . . .

 

[Baluncore]

Here, I am holding a 500W 220V bulb, size of a pineapple, its cold resistance is just 7 Ohm.

I have used filaments removed from light globes as temperature sensors. For 240VAC filament lamps I always used Rcold = 3600 / watts.
3600 / 500W = 7.2 ohms cold.

At 293K the filament will have one tenth of the resistance that it has operating bright at 2930K.
The tungsten filament is a sufficiently pure metal to have a linear temperature to resistance relationship.
You can use that linearity to interpolate or extrapolate at the cold end of operation.

If you have a problem with the filament self warming while measuring the resistance you can always wrap the globe in cloth, carefully break the glass, then immerse the filament directly in a water bath at 20°C while measuring the resistance.

Filament globes increase resistance as they age.