A \(4 \frac 1 2\) digit DMM has the error specification as 0.2% of reading +10 counts. If a d.c. voltage of 100 V is read on its 200 V full-scale, the maximum error that can be expected in the reading is
± 0.3%
This question requires calculating the accuracy of a Digital Multimeter (DMM). The DMM is identified as a \(4 \frac{1}{2}\) digit model. Its accuracy specification is given as 0.2% of reading + 10 counts. The task is to determine the maximum expected error when measuring a d.c. voltage of 100 V on the 200 V full-scale range.
The total error in a DMM's measurement comes from different sources. According to the provided specification, there are two primary error components:
The first part of the error specification is 0.2% of the reading. The measured voltage is 100 V.
Calculation for this component:
Error$_{reading}$ = $0.2\% \times \text{Reading}$
Error$_{reading}$ = $\frac{0.2}{100} \times 100 \, \text{V}$
Error$_{reading}$ = $0.2 \, \text{V}$
The second error component is stated as 10 counts. To find its value in volts, we need to know the voltage represented by one count on the given range.
The DMM is a \(4 \frac{1}{2}\) digit meter, meaning it can display up to 19999. The full-scale range (FS) is 200 V. For calculation, the full scale is considered divided into 20000 parts.
The voltage value for one count is calculated as:
Value per count = $\frac{\text{Full Scale Range}}{\text{Total Counts on Scale}}$
Value per count = $\frac{200 \, \text{V}}{20000}$
Value per count = $0.01 \, \text{V}$ (or 10 mV)
Now, calculate the error from 10 counts:
Error$_{counts}$ = $10 \times \text{Value per count}$
Error$_{counts}$ = $10 \times 0.01 \, \text{V}$
Error$_{counts}$ = $0.1 \, \text{V}$
The total maximum error is the sum of the absolute values of the two error components. This represents the worst-case error.
Total Error = $| \text{Error}_{reading} | + | \text{Error}_{counts} |$
Total Error = $0.2 \, \text{V} + 0.1 \, \text{V}$
Total Error = $0.3 \, \text{V}$
The answer choices are percentages. We convert the total calculated error (0.3 V) into a percentage relative to the measured reading (100 V).
Percentage Error = $\frac{\text{Total Error}}{\text{Measured Reading}} \times 100\%$
Percentage Error = $\frac{0.3 \, \text{V}}{100 \, \text{V}} \times 100\%$
Percentage Error = $0.3\%$
Therefore, the maximum error that can be expected in the reading is $\pm 0.3\%$.
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