Measuring the resistance of a hot resistor

Try doing the measurements, in circuit? By this, I mean measuring both the voltage across it, in parallel, when in-circuit and measuring the current through it (meter set to measure at least 7A, which is what 100W suggests) using a series arrangement. The two will tell you what you need to know: R=V/I.

Jon

The measurement I am taking requires a value for the resistor itself, independant of reactive elements in the circuit. As such, I need to measure the resistance out of circuit.

Jon

I suspect your circuit is powered up when you're making the hot measurement and not when you're making the cold measurement.

Graham

The resistance will not vary much with temperature. For accurate measurements under load, measure the voltage and current at the same time.

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Jonathan Kirwan wrote in news: snipped-for-privacy@4ax.com:

(meter set to measure at least 7A, which is what 100W suggests)

or 1 mA or 100 Amps or.....

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"Jon Danniken" wrote in news: snipped-for-privacy@mid.individual.net:

Use a DC circuit to indirectly measure the reistance as described (roughly) above.

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"Jon Danniken"

** It will have the same resistance cold or hot, +/- 5%.

You have some silly bad contact issue with your probes.

....... Phil

circuit

resistance,

with

this

Like a thermocouple, it's generating a few DC mV when hot. This upsets the ohmmeter readings. john

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"john jardine"

** The resistance wire in the power resistor cannot do that - but maybe there is a thermocouple created by the terminals or when the DMM's probes are used to make connection - as little as 0.5mV would explain the OP's dilemma.

DMMs are poor at reading low ohms.

If the OP simply connects a resistor of known value in series with the power resistor and applies say 12 volts DC - the voltage will split in the same ratio as the two values.

Should put his false anxiety to rest.

...... Phil

Aha, thanks, John. I measured the resistor again while it was hot and you're right, it develops 0.4mV across itself when it is hot.

Jon

Excellent, thanks for that, Me. I'll cook up a little DC circuit to toss it into for the measurement.

Thanks,

Jon

"Phil Allison"

Hey Dr. Phil, maybe you should go piss off to a psych group and let the rest of us discuss electronics.

Jon

"Jon Danniken CUNTHEAD "

** Huh - what the f*ck is your problem ??

Why so keen to prove to the whole world what a pig ignorant & asinine CUNTHEAD you are.

When it was obvious all along anyhow.

...... Phil

toss it

You could also do a direct measurement with the resistor heated by a hot plate or a hot air gun...

On Sun, 25 Feb 2007 15:21:31 -0800, "Jon Danniken" Gave us:

Put a good ammeter in series with it, and place a volt meter across it. Read both at run time, extrapolate, and you'll get very accurate results.

On Sun, 25 Feb 2007 15:42:55 -0800, "Jon Danniken" Gave us:

No. With the ammeter right next to it, and the voltage reading has ZERO effect, you WILL get the exact resistance at the moment you take the readings. Your imagined parasitics won't exist nor matter.

The value extrapolated WILL be exact.

On Sun, 25 Feb 2007 19:24:21 -0800, "Jon Danniken" Gave us:

Yes. A very precise solution. Rent a thermal imager, and get some real repeatable results!

On Mon, 26 Feb 2007 04:40:17 GMT, Ross Herbert Gave us:

Do you not think that a known voltage and a good current reading would not be more accurate than an unknown calibration level ohm meter?

to

I am fully aware of the method of performing a DC current and voltage measurement and applying Ohm's Law. I was simply suggesting an alternative which would avoid the need to lash up a circuit which requires both an ammeter and a voltmeter simultaneously in circuit in order to achieve best accuracy. Two measurements with meters each having accuracy and tolerance inaccuracy will also magnify any calculation error.

A single meter measuring resistance directly might be more accurate depending on the meter used.