Author Topic: Need help (0.03%+2d)/0.005%=?????????  (Read 1492 times)

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Offline 001Topic starter

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Need help (0.03%+2d)/0.005%=?????????
« on: September 05, 2018, 11:42:37 am »
Hi
Sorry
I`m old but havent any grafuate in electronix

If I read voltage with (0.03%+2d) voltmeter around 0.005% resistor what is result current tolerance?

Thanx!
 

Offline mzzj

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Re: Need help (0.03%+2d)/0.005%=?????????
« Reply #1 on: September 05, 2018, 01:02:14 pm »
Hi
Sorry
I`m old but havent any grafuate in electronix

If I read voltage with (0.03%+2d) voltmeter around 0.005% resistor what is result current tolerance?

Thanx!
Assuming that that you are measuring current shunt resistor..
Simple answer: you can pretty much ignore the resistor tolerance, voltmeter accuracy is dominating the accuracy because its over 6 times worse than the current shunt.
Long answer: you don't want to know  >:D ( about 16 pages of explanations, two dozen unknown variables and slight headache)
 

Offline awallin

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Re: Need help (0.03%+2d)/0.005%=?????????
« Reply #2 on: September 05, 2018, 01:08:01 pm »
how about the 5th row in the table here:
https://en.wikipedia.org/wiki/Propagation_of_uncertainty
 
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Offline try

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Re: Need help (0.03%+2d)/0.005%=?????????
« Reply #3 on: September 05, 2018, 01:19:13 pm »
Hi
Sorry
I`m old but havent any grafuate in electronix
Me neither, but I am maybe not so old.
Quote
If I read voltage with (0.03%+2d) voltmeter around 0.005% resistor what is result current tolerance?

Thanx!

Without telling the forum the number of counts, the measurement value and range used nobody can give you an exact figure.

With no correlation effects the uncertainties will add up so that the resulting uncertainty is at least 0,035%.
Low readings in a range will make that figure explode due to the increasing weight of 2d.
=> The value of 0,035% is pretty meaningless for you.

Regards
try


« Last Edit: September 05, 2018, 03:17:04 pm by try »
 


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