Measurement Errors: What Types There Are (+ Examples)

You want to measure the power \( P \) across the resistor \( R ~=~ 5 \Omega ~\pm~ 2\% \). For this, you have measured the voltage \( x_{\text U} ~=~ 60 \, \text{V} \) with a voltmeter and the current \( x_{\text I} ~=~ 10 \, \text{A} \) with an ammeter.

Determine the relative error \( f_x \) for the following three approaches:

1. \( P ~=~ U \, I \)
2. \( P ~=~ \frac{U^2}{R} \)
3. \( P ~=~ I^2 \, R \)

Data of the Ammeter: End of range value \( x_{\text E} ~=~ 100 \, \text{A} \). Error class: \( K_{\text E} ~=~ 1 \).

Data of the Voltmeter: End of range value \( x_{\text E} ~=~ 100 \, \text{V} \). Error class: \( K_{\text E} ~=~ 1 \).

Tip: Error class is calculated as follows: 1 $$ K_{\text E} ~=~ 100 \cdot \frac{G}{x_{\text E}} $$ where \( G \) is the magnitude of the maximum possible error.

Solution to Exercise #1.1

Rearrange 1 to find \( G \): 2 \[ G ~=~ \frac{ K_{\text E} }{ 100 } \, x_{\text E} \]

Hence, the maximum possible error of the voltmeter is: 3 \[ G ~=~ \frac{ 1 }{ 100 } \, 100 \, \text{V} ~=~ 1 \, \text{V} \]

And similarly for the ammeter: 4 \[ G ~=~ \frac{ 1 }{ 100 } \, 100 \, \text{A} ~=~ 1 \, \text{A} \]

The displayed voltage value, as well as the current value, can deviate by a maximum of \( F_{\text U} ~=~ \pm 1 \text{V} \) and \( F_{\text I} ~=~ \pm 1 \text{A} \) respectively.

The relative error of the voltage value is calculated as: 5 \[ f_{\text U} ~=~ \frac{ F_{\text U} }{ x_{\text U} ~-~ F_{\text U}} ~\cdot~ 100\% \]

The maximum relative error occurs when you set \( F_{\text U} ~=~ +1 \text{V} \): 6 \[ f_{\text U} ~=~ \frac{ 1 \text{V} }{ 60 \text{V} ~-~ 1 \text{V}} ~\cdot~ 100\% ~\approx~ 1.7 \% \]

Similarly for the relative error of the measured current (with \( F_{\text I} ~=~ + 1 \text{A} \)): 7 \[ f_{\text I} ~=~ \frac{ 1 \text{A} }{ 10 \text{A} ~-~ 1 \text{A}} ~\cdot~ 100\% ~\approx~ 11.1 \% \]

The relative error of the power value (calculated using the first power formula) is thus: 8 \[ f_{\text P} ~=~ f_{\text U} ~+~ f_{\text I} ~=~ 1.7 \% ~+~ 11.1 \% ~=~ 12.8 \% \]

Solution to Exercise #1.2

In Exercise #1.1, the maximum possible error of the voltmeter (\(f_{\text U} ~=~ 1.7 \%\), see Eq. 6) was calculated. Furthermore, the relative error of the measured resistance according to the exercise is \( f_{\text R} ~=~ \pm 2 \% \). Therefore, the maximum possible error of the power value, calculated with second power formula, is: 9 \[ f_{\text P} ~=~ f_{\text U} ~+~ f_{\text U} ~-~ f_{\text R} ~=~ 1.7 \% ~+~ 1.7 \% ~-~ (-2\%) ~=~ 5.4 \% \]

Solution to Exercise #1.3

In Exercise #1.1, the maximum possible error of the ammeter (\(f_{\text I} ~=~ 11.1 \%\), see Eq. 7) was calculated. Furthermore, the relative error of the measured resistance is \( f_{\text R} ~=~ \pm 2 \% \). Therefore, the maximum possible error of the power value, calculated with third power formula, is: 10 \[ f_{\text P} ~=~ f_{\text I} ~+~ f_{\text I} ~+~ f_{\text R} ~=~ 11.1 \% ~+~ 11.1 \% ~+~ (+2\%) ~=~ 24.2 \% \]