# 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:

- \( P ~=~ U \, I \)
- \( P ~=~ \frac{U^2}{R} \)
- \( P ~=~ I^2 \, R \)

**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
\[ f_{\text P} ~=~ f_{\text I} ~+~ f_{\text I} ~+~ f_{\text R} ~=~ 11.1 \% ~+~ 11.1 \% ~+~ (+2\%) ~=~ 24.2 \% \]
`