Page 8 - 6413

160 V
R   200 Оhm
0,8 А

As measured voltage and current values are not accurate, the
value of the resistance found will not be accurate. To calculate the
error of resistance, output the error formula, namely:
First, the working formula is logarithmic:

ln R = ln U – ln I, (1.5)

and secondly, the resulting expression is differentiated

dR dU dI
  , (1.6)
R U I

thirdly, the icons of the differential "d", replaced by "" icons, and
taking into account the rule of adding errors, receive the formula of
relative error,

 R  U I 
  , (1.7)
R U I

where ΔR is the absolute resistance error, U is the absolute error of

measurement of the voltage, I is the absolute error of the measured
current strength. The absolute errors U and I can be easily
determined through the accuracy class of the relevant electrical
measuring instruments. So, if an ammeter was used to measure current

strength, which was calculated for current 1A and had a class of
accuracy of 1, then the absolute error of current strength would be
0.01A. If, however, the voltmeter at 200V with the accuracy class of

1.5 is used to measure the voltage, then the absolute voltage error is
equal to

U = (200  0,015) V = 3 V.

Substituting these values of absolute errors in formula 1.7, we

obtain:

 R 3 0.01
   0.03 (3%)
R 160 0.8

then R = 200 Оhm 0,03 = 6 Оhm.

So, R = (200  6) Оhm.

7