Discussion
A digital instrument specified as 0.05 per cent of reading plus 2 counts, reading 1.0000 V on a 10.000 V range, has an uncertainty of about
Answer: A
Correct answer: 0.0025 V
\[ 0.05\%\ \mathrm{of}\ 1.0000 = 0.0005\ \mathrm{V}, \qquad 2\ \mathrm{counts} = 0.0020\ \mathrm{V}, \] giving 0.0025 V in total. The count term dominates because the reading is at a tenth of the range, which is the digital equivalent of the analogue class problem: both instruments do their best work near full scale.
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