The reason an integrating digital voltmeter is described as taking a mean rather than a sample is that its converter
Answer: D
The integrator's output at the end of the charging phase is proportional to the time integral of the input, which divided by the fixed period is the mean. Nothing is sampled at all, and that is why periodic interference cancels so completely.
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A dual slope digital voltmeter has a gate time of 400 ms reflecting the measurement cycle. The frequency bandwidth of this instrument is :
MH SET - 2018, SET Exam - 2018,
Answer: B
The bandwidth is 1.25 Hz.
A dual slope voltmeter integrates the input over a fixed time and so behaves like an averaging filter. For an integration (measurement) time T, the effective bandwidth of such an averaging process is taken as \(1/2T\):
\[ \begin{aligned} B &= \frac{1}{2T} = \frac{1}{2 \times 0.4\ \text{s}} \\ &= 1.25\ \text{Hz} \end{aligned} \]
The option 2.5 Hz is \(1/T\), the measurement rate, not the bandwidth. The values 0.8 Hz and 0.6 Hz do not follow from the gate time.
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Match List-I with List-II :
| List-I : Number on Display | List-II : ADC |
|---|---|
| (a) 1072 | (i) 8-bit |
| (b) 893 | (ii) 10-bit |
| (c) 132 | (iii) 12-bit |
| (d) 59705 | (iv) 16-bit |
MH SET - 2019, SET Exam - 2019,
Answer: B
The correct matching is a-iii, b-ii, c-i, d-iv.
An n-bit ADC can show the codes 0 to 2n - 1, so a display reading needs the smallest ADC whose top code is not exceeded:
\[ \text{8-bit: }0\text{ to }255 \]
\[ \text{10-bit: }0\text{ to }1023 \]
\[ \text{12-bit: }0\text{ to }4095 \]
\[ \text{16-bit: }0\text{ to }65535 \]
132 is at most 255, so an 8-bit ADC is enough: (c) with (i). 893 is above 255 and at most 1023, so 10-bit: (b) with (ii). 1072 is above 1023 and at most 4095, so 12-bit: (a) with (iii). 59705 is above 4095 and at most 65535, so 16-bit: (d) with (iv).
The other codes break at least one of these ranges, for example putting 893 on an 8-bit ADC, which cannot show a number above 255.
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Which of the following would need a 4½ digit digital multimeter ?
MH SET - 2023, SET Exam - 2023,
Answer: D
The reading that needs a 4½ digit meter is 1.4856 mV.
A 4½ digit display has four full digits (0-9) and a leading half digit that can only be 0 or 1, so its maximum count is 19999. The reading 1.4856 has five digits with a leading 1, which fits exactly.
1.999 mA and -1.999 µV have four digits with a leading 1, which a 3½ digit meter (maximum count 1999) can already show, and so can 1000 V. Only 1.4856 mV requires the extra digit.
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The resolution of a 3½ digit DMM on 1 V and 10 V is :
MH SET - 2023, SET Exam - 2023,
Answer: D
The resolutions are 0.001 V and 0.01 V.
A 3½ digit meter has a maximum count of 1999, so its resolution is one count in the last digit. On the 1 V range the display reads up to 1.999 V, so the smallest step is 0.001 V. On the 10 V range it reads up to 19.99 V, so the smallest step is 0.01 V.
\[ \begin{aligned} \frac{1\ \text{V}}{1000} &= 0.001\ \text{V} \\ \frac{10\ \text{V}}{1000} &= 0.01\ \text{V} \end{aligned} \]
The other options are ten or a hundred times too coarse on one or both ranges.
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The most suitable instrument for the measurement of voltage is
UGC NET - 2012,
Answer: C
The key takes the CRO as the most suitable instrument for measuring voltage (option C). It shows the actual waveform, so it measures the peak, peak-to-peak and instantaneous values of any waveform, including non-sinusoidal and fast signals, and it has a high input impedance.
An analog voltmeter (B) and a DMM (D) are calibrated for steady or sinusoidal values and respond slowly to a changing signal, and a DVM (A) gives a number without showing the waveform. The question has no stated condition, so the answer rests on the key.
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In a DVM, the accuracy of the measurement depends on
I. No. of A/D conversions in a second.
II. The resolution of A/D converter.
III. A statistical processing of samples and a huge memory.
IV. Speed of the processor should be high.
UGC NET - 2012,
Answer: B
The key takes statements II, III and IV as the conditions to be fulfilled, option (B). The accuracy of a digital voltmeter depends on the resolution of the A/D converter, statement II, because the smallest step sets the quantisation error. It also depends on statistical processing of the samples with enough memory, statement III, which averages out noise, and on a fast processor, statement IV, so that the processing keeps up. Statement I, the number of conversions in a second, is not counted by the key as a determinant of accuracy, which it treats as a matter of speed. Options (A), (C) and (D) include I. This item follows the key.
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