Q#6 (Past Exam Paper – June 2010 Paper 23 Q1)
A digital voltmeter with a three-digit display is used to measure the potential difference across a resistor. The manufacturers of the meter state that its accuracy is ±1% and ±1 digit.
The reading on the voltmeter is 2.05 V.
(a) For this reading, calculate, to the nearest digit,
(i) a change of 1% in the voltmeter reading, [1]
(ii) the maximum possible value of the potential difference across the resistor. [1]
(b) The reading on the voltmeter has high precision. State and explain why the reading may not be accurate. [2]
Solution:
(a) (i)
{The accuracy is 1% of the reading (value displayed).
1% of 2.05 = 0.0205
We are asked to give the answer to the nearest digit, that is, a single (non-zero) digit. So,}
1% of ±2.05 is ±0.02
(ii)
{The reading on the voltmeter is 2.05 V.
Accuracy of the meter = 1% of reading = ±0.02 (as calculated above).
Thus, from this information, reading = 2.05 ± 0.02 V
That is, the reading can be from (2.05 – 0.02 =) 2.03 V to (2.05 + 0.02 =) 2.07 V}
{But the voltmeter also has an uncertainty of ±1 digit. That is, the last digit can fluctuate by ±1.
Thus, the reading can be from (2.03 – 0.01 =) 2.02 V [minimum value] to (2.07 + 0.01 =) 2.08 V [maximum value].}
max. value is 2.08 V
(b)
{Precision deals with the smallest digit (or decimal place) that can be obtained using the voltmeter.
Accuracy is how close the reading is to the actual (true) value of the quantity being measured. So, we need to provide reasons that can make the value read far from the actual value.}
There may be a zero error/calibration error/systematic error which makes all readings either higher or lower than true value.
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