# Measurement Errors: What Types There Are (+ Examples)

## Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

### Exercise #1: Relative Error of Electric Power

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 $f_{\text P} ~=~ f_{\text I} ~+~ f_{\text I} ~+~ f_{\text R} ~=~ 11.1 \% ~+~ 11.1 \% ~+~ (+2\%) ~=~ 24.2 \%$