Voltmeter relative error question. (Computer Controlled Measurement)

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SUMMARY

The discussion centers on calculating the relative error of a voltmeter measurement of 50V with a maximum input voltage of 300V. The correct relative error is determined to be +/- 0.8%, factoring in both absolute voltage error and percentage of full scale error. The absolute error is 0.1V, and the percentage of full scale error is 0.1%, leading to a total error of +/- 0.4V. The initial calculation of 0.0016% was incorrect due to misunderstanding the error components involved.

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Homework Statement



You measure 50V with a voltmeter, which has 300V as the max. input voltage. What will be the relative error (h%) of the measurement?

Homework Equations





The Attempt at a Solution


hi i divided 50v / 300 V and get = 1,66 and divided it by 100 to get relative error in % and get %0.0016 just wanted to know my answer is correct or not thank you from now.
 
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Hard to say. As in 'impossible' with what you gave us.

Voltmeters usually have an absolute voltage error plus a "% of full scale" error. So if the
reading is 50V, the absolute error is 0.1V and the % of full scale is 0.1% the error would be +/- (0.1V + 300*0.001V) = +/- 0.4V. So the 'relative eror' of your measurement would be +/- 0.4/50 = 0.8%.
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just got an email from the teacher he is mistaken he haven't complated the question and i solved it anyway thanks for your helps
 

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