Resistance of a bulb 100 W, 230 V

Please view the youtube video below:

This is not a homework problem.

The reason seems rather weird. Though the video seems incomplete, I would like to know the reason why 39.2 Ohms showed up. It seems rather weird. I know that resistance does not depend on the Voltage or the current if I am not wrong. If I have a bulb it has different resistance in different places? If I have a resistor measured lets say 1 Ohm. If I change the voltage or the current the resistance will change? It does'nt really make sense to me.

Though I would like to point out that temperature dependence is what could be affecting the resistance. But by the amount that would V^2/R.

Could someone please assist me in answering the question posed in the video/ explain the reason given by the Professor?

Answers and Replies

berkeman
Mentor
Though I would like to point out that temperature dependence is what could be affecting the resistance.
That's the reason the filimant resistance changes as it heats up. The resistance goes up with temperature, and stabilizes at the light bulb's operating power and temperature. I didn't watch the video, but it seems that is what is causing your confusion.

Yes, I believe that's the reason. But the data is going quite against it. I think I messed it up somwhere.

But if I take the data given below for resistivity (4.5 * 10^-4 K^-1) and temperature inside a light bulb as (3500 K, which is higher than its melting point) And assuming the room temperature is 300 K

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/rstiv.html

http://hypertextbook.com/facts/1999/AlexanderEng.shtml

Tungsten is lit by around 3300K (Hyper textbook result). So if I calculate on that basis it will make that around 100 ohms. It still comes around 429 ohms. The difference between the measured and the actual reading.

cnh1995
Homework Helper
Gold Member
I calculate on that basis it will make that around 100 ohms.
The tables in those links show the relation between resistivity and temperature. How did you calculate the resistance? Resistance of the filament will depend on the physical dimensions.

The physical dimensions will remain the same before and after. I know its not true, but then it should approximately come out to be true. Even if we include all that the difference will stand out to be ~400 Ohms. (Which is still too big of an error)

(I multiplied both sides by A/l where A is the area of the thin tungsten wire and l is the length of the wire.)

I checked out the linear expansion of Tungsten, though. It's 4.3 * 10^-6. Not enough to affect our results greatly.

http://www.engineeringtoolbox.com/linear-expansion-coefficients-d_95.html

If you have a clamp on ammeter you can read the current and figure the resistance. When it is cold it is going to be very low resistance and that resistance goes up as you go up in temperature, it is not like a nice little resistor. Just looking at the numbers, 100 watts, 230 volts, P=I*E so I=P/E I =.43 amp. R=E/I so 230/.43 gives 529 ohms. That is just based on the numbers given, and that is operating resistance. A clamp on ammeter wrapped around one of the power leads would give the actual current flow and a DVM on the bulb would give the actual voltage and would therefore give better results. The calculations for AC into a resistive load is pretty close to DC so the numbers are simple to calculate, no phase angle stuff to worry about. 39 or 40 ohms would represent over 1300 watts, not gonna happen with a 100 watt bulb.

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If you have a clamp on ammeter you can read the current and figure the resistance. When it is cold it is going to be very low resistance and that resistance goes up as you go up in temperature, it is not like a nice little resistor. Just looking at the numbers, 100 watts, 230 volts, P=I*E so I=P/E I =.43 amp. R=E/I so 230/.43 gives 529 ohms. That is just based on the numbers given, and that is operating resistance. A clamp on ammeter wrapped around one of the power leads would give the actual current flow and a DVM on the bulb would give the actual voltage and would therefore give better results. The calculations for AC into a resistive load is pretty close to DC so the numbers are simple to calculate, no phase angle stuff to worry about. 39 or 40 ohms would represent over 1300 watts, not gonna happen with a 100 watt bulb.

I did not understand most part of it. No, it's not because I have not learnt AC. (Basics)

If you have read the previous response. I including the factor of temperature change and since someone mentioned I even included the change of length of the wire due to heating. With that, the error still pops out to be ~ 400 ohms.

The calculation for AC resistive load would be the same as that of DC resistive load to the best of my knowledge.

cnh1995
Homework Helper
Gold Member
A is the area of the thin tungsten wire and l is the length of the wire.
You mean you used the dimensions of the filament in that bulb? You actually measured the area and length?

You mean you used the dimensions of the filament in that bulb? You actually measured the area and length?

Have you seen the video completely? (No off.) I think that is why there is a communication gap.

The resistance reported by the multi-meter is 39.2 ohms which is approximated to 40 ohms. Using that data. I don't have to calculate A/l. But that as per berkeman is the resistance at room temperature (~300K). So I included the change in resistance due to increase in temperature.

berkeman
Mentor
Have you seen the video completely? (No off.) I think that is why there is a communication gap.

The resistance reported by the multi-meter is 39.2 ohms which is approximated to 40 ohms. Using that data. I don't have to calculate A/l. But that as per berkeman is the resistance at room temperature (~300K). So I included the change in resistance due to increase in temperature.
So 40 Ohms at room temperature and 400 Ohms at hot seem like reasonable numbers. I think you may have listed a tempco for the tungsten filament in one of your posts... is that the problem, that the tempco seems off?

Prannoy Mehta and Charles Link
Charles Link
Homework Helper
Gold Member
2020 Award
I agree with Berkeman in post #2. A couple of us once did a similar experiment with a light bulb at room temperature and an ohm meter. The room temperature resistance was a factor of 4 or thereabouts below what the calculations of P=V^2/R say it needs to be at T=2500 K. Linear extrapolations of resistivity as a function of temperature are also not likely to give this result. The behavior of the resistance is likely to be non-linear with increasing temperature. And no, I didn't watch the video either, but this one is very much to be expected. The resistance of the tungsten filament increases dramatically with temperature at higher temperatures.

Prannoy Mehta
I did not understand most part of it. No, it's not because I have not learnt AC. (Basics)

If you have read the previous response. I including the factor of temperature change and since someone mentioned I even included the change of length of the wire due to heating. With that, the error still pops out to be ~ 400 ohms.

The calculation for AC resistive load would be the same as that of DC resistive load to the best of my knowledge.
I watched the video and the problem is the video is incomplete. It stopped before he was able to give the answer. It is about 40 ohms at room temperature but it will go up to about 530 ohms at operating temperature. He used a DVM to show the resistance but that only puts a small current through the load so will not heat up the filament. Putting in 230 volts, there will be what is called "Inrush' current, because at the start the temperature of the filament is room temperature for a short time, 1 second say, then quickly gets hot so the power jumps up for that one second peaking at a thousand watts of power just for the short time it takes to get the filament up to its operating temperature. The equations I used, I should have indicated units. P= power in watts, I=current in amps, E=voltage in volts, R=resistance in ohms.

Sorry about that. So Power can be calculated by E squared divided by R (voltage squared divided by resistance) and another formula is I^2 * R, Current in amps squared times R, resistance in ohms. Also, Power can be calculated by P=I*E current in amps times voltage, for example, one amp and one hundred volts =100 watts. Resistance can be calculated by R= E/I , voltage divided by current in amps. The one answer one guy gave was correct, 529 ohms but ONLY with 230 volts running through the bulb. So R= 230 divide by 0.43 amps, notice if you divide by a number smaller than one your answer will be bigger than the numerator (top number) so that comes out at 529 ohms AT FULL OPERATING TEMPERATURE. Does that help?

Thank you for all your help. I have understood it, it was a misconception in my mind. (litup, berkeman, Charles Link ch995)

1) The temperature vs resistance graph is non linear so the previously derived resistance at 2500K.
2) The numbers could be a off for resistivity, I guess.

I watched the video and the problem is the video is incomplete. It stopped before he was able to give the answer. It is about 40 ohms at room temperature but it will go up to about 530 ohms at operating temperature. He used a DVM to show the resistance but that only puts a small current through the load so will not heat up the filament. Putting in 230 volts, there will be what is called "Inrush' current, because at the start the temperature of the filament is room temperature for a short time, 1 second say, then quickly gets hot so the power jumps up for that one second peaking at a thousand watts of power just for the short time it takes to get the filament up to its operating temperature. The equations I used, I should have indicated units. P= power in watts, I=current in amps, E=voltage in volts, R=resistance in ohms.

Sorry about that. So Power can be calculated by E squared divided by R (voltage squared divided by resistance) and another formula is I^2 * R, Current in amps squared times R, resistance in ohms. Also, Power can be calculated by P=I*E current in amps times voltage, for example, one amp and one hundred volts =100 watts. Resistance can be calculated by R= E/I , voltage divided by current in amps. The one answer one guy gave was correct, 529 ohms but ONLY with 230 volts running through the bulb. So R= 230 divide by 0.43 amps, notice if you divide by a number smaller than one your answer will be bigger than the numerator (top number) so that comes out at 529 ohms AT FULL OPERATING TEMPERATURE. Does that help?

I have a typo over there now.

Yes, that is why I was wondering.

Oh no, thats not required I get what you are trying to say. Thanks again.

berkeman
nasu
Gold Member
When you read the temperature coefficient you made an error of one order of magnitude.
It is 4.5 x 10-3 (and not -4).

The resistivity versus temperature is not perfectly linear but linear relation is pretty good approximation.
At 2500 K the resistivity increase by a factor of about 10, as in this graph
https://encrypted-tbn2.gstatic.com/..._2SHZYBdqo4rx0mkmySkmmvSC2hU2nyV2WiiZtNlQBXLy

The graphs versus voltage are strongly nonlinear but this is another story.
I think your confusion is due just to a small error (one order of magnitude :) ) in the value used for the coefficient.

berkeman and Prannoy Mehta