B Resistance of a bulb 100 W, 230 V

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.

 

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.

 

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

 

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.

 

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?

 

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.