Discussion
The same instrument, with the same specification, reads 10.0 V. The error as a percentage of the reading is
Answer: D
Correct answer: 2.5 per cent
With \( \pm0.5\% \) of reading and \( \pm2 \) counts on a range whose count is 0.1 V: \[ 0.005\times10.0 + 0.2 = 0.05 + 0.2 = 0.25\ \mathrm{V}, \] which is \( 0.25/10.0 = 2.5\% \) of the reading. The fixed count term dominates at low readings, which is why a measurement should be made on the lowest range that will accommodate it - at 100 V the same specification gave only 0.7 per cent.
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