Question 6
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]
Reference: Past Exam Paper – June 2010 Paper 23 Q1
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.
For 6aii) can i take the full value 0.0205 for calculation so i will get 2.0705 that rounds off to 2.08? instead of considering the ± 1 digit???
ReplyDeleteno, it should be given to the nearest digit.
Deletealso, 2.0705 does not round off to 2.08, it is 2.07