Digital Voltmeters Questions and Answers

Answer: C

Either an analogue computing circuit or a digital calculation from samples performs those three operations, so the reading is correct whatever the waveform. The specification will state a crest factor limit beyond which even a true rms instrument cannot cope.

Answer: D

A very peaky waveform drives the input circuits into clipping long before the rms value reaches full scale, so a limit of three or five is stated. Exceeding it gives a low reading with no warning, which is a trap when measuring the current of a rectifier load.

Answer: A

A moving pointer conveys direction and rate of change immediately, whereas a flickering set of digits conveys neither. This is why many digital instruments now include a bar graph alongside the numerals, to recover what the pointer gave.

Answer: B

The count runs to 2000, so the range of two hundred volts is divided into two thousand parts of a tenth of a volt. The display would read from 0.0 to 199.9 volts, which is why such a range is described as the 200 V range rather than the 199.9 V range.

Answer: C

On the 200 V range one count is 0.1 V, so two counts is 0.2 V, and half of one per cent of the reading is 0.5 V. The two add to 0.7 V, that is 0.7 per cent of the reading, and the count term would dominate entirely at a much smaller reading.

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.

Answer: A

Correct answer: 0.1 mV

Four and a half digits give 19999 counts, so on the 2 V range each count is \[ \frac{2\ \mathrm{V}}{20{,}000} = 0.1\ \mathrm{mV}. \] Comparing with the 20 V range shows the general rule: dropping a decade of range gains a decade of resolution, which is exactly why the lowest usable range is chosen.

Answer: B

The input is the reference multiplied by the ratio of discharge time to charge time, that is \( 10 \times 40/100 = 4 \) V. Neither the capacitance nor the clock frequency enters, which is the structural advantage of the method.

Answer: C

The whole cycle is fifty milliseconds, so twenty readings a second. An integrating instrument is necessarily slow for this reason, and a successive approximation converter of the same resolution would manage thousands of readings a second.

Answer: D

Eight bits give 256 levels, so the step is \( 5/256 = 0.0195 \) V. The maximum quantisation error is half a step, that is about ten millivolts, which sets a floor on the accuracy however good the rest of the instrument may be.