Calculating Error in Resistance Measurements

[khizman]

Rounded Numbers --> error?

Say i am measuring resistance of a wire, i have a voltmetre and an ammetre set up to work out the resistance.

Votmetere says 1.45 and ammetre says 0.45, this wint be enirely aacurate would it? arnt the numbers rounded off on the screen?

So what part does this play in errors? absolute and relative>?

 

[dav2008]

Obviously the voltage is not exactly 1.45 volts and the current isn't exactly .45 amps.

As far as the type of error, think of the way it rounds. If the voltage is read as 1.45 that could mean that the actual voltage is anywhere from 1.445 to 1.455. Notice that 1.45 is exactly in the middle of the range of actual readings that yield 1.45 when rounded. You can express that reading as 1.45 +/- .005. (Someone please correct me if I made a mistake in the decimals. It doesn't really matter when it comes to answering the original question though)

What kind of error does that represent then?

Of course I assumed that the meter actually rounds the number instead of just chopping the extra digits off.

 

[Hootenanny]

As dav said the maximum error is \pm 0.005, however, you have to be carefull when calculating the resultant error on your resistance calculations. Resistance is given by;

R = \frac{V}{I}

Therefore, the maximum error for your resistance would be given by the largest possible voltage divided by the smallest possible resistance. For example, if your intruments displayed values of 1.45V and 0.45A, then your largest possible resistance would be;

R = \frac{1.45 + 0.005}{0.45 - 0.005} = \frac{1.455}{0.445} \approx 3.26966

The smallest your resistance could possible be is;

R = \frac{1.45 - 0.005}{0.45 + 0.005} = \frac{1.445}{0.455} \approx 3.17582

Your 'nominal value' (what you actually measured) would be;

R = \frac{1.45}{0.45} = \frac{29}{9} \approx 3.22222

You would then need to calculate the difference between your nominal and maximum and minimum and you would quote your answer as;

R = 3.22 [+0.047 / - 0.046] \Omega

It is very time consuming to do this by hand and I usually set up a speadsheet to do the calculations for me.

Regards
-Hoot

(Correct me if my calcs are wrong)