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A 3 ½ digit dual slope integrated DVM, 2 V full scale is used to measure a time varying voltage V(t) = 1.0 + 2 sin 50 πt. The voltmeter indicates
V(t) = 1 + 2 sin 50 πt
Dual slope integrated DVM measures the average value of input.
V_{avg} = 1 V
Hence the meter reading will be 1.000 V
A digital voltmeter uses a 20 MHZ clock and has a voltage controlled generation switch provides a width of 20 μ sec per volt of unit signal. 25 V of input signal would correspond to pulse count of –
In a digital voltmeter, the oscillator frequency is 400 kHz and the ramp voltage falls from 8V to 0V in 20m sec. The number of pulse counted by the counter is
Frequency = 400 kHz.
f_{clk} = 400kHz
t = 20m sec
⇒ Clock pulses, n = t f_{clk }= 20m sec × 400 kHz = 8000
How would a reading 0.6973 V be displayed on digit digital voltmeter on its 10 V scale.
Resolution on 10 V scale
N = no. of full bits
Reading of 0.6973 = 0.697 volts
A 200 mV full scale dualslope 3 ½ digit DMM has a reference voltage of 100 mV and a first integration time of 100 ms. For an input of [100 + 10 Cos (100πt)] mV, the conversion time (without taking the autozero phase time into consideration) in millisecond is______.
Given that, V_{ref} = 100 mV
T_{1} = 100 msec
V_{in} = 100 mV
We know that,
V_{in} T_{1} = V_{ref} T_{2}
⇒ (100) (100) = (100) (T_{2})
⇒ T_{2} = 100 msec
Total conversion time
T = T_{1} + T_{2}
T = 100 + 100 = 200 ms
A rectifier type instrument uses a bridge rectifier and has its scale calibrated in terms of rms values of a sine wave. It indicates a voltage of 2.22 V. When measuring a voltage of triangular wave shape. Estimate the error –
The meter uses a full wave rectifier circuit it indicates value of 2.22 V the form factor for full wave rectified sinusoidal waveform is 1.11
Average value of voltage,
V_{avg} = 2 V
For triangular wave shape, peak value of voltage V_{m} = 2 V_{avg} = 4 V
A dual slope integrating type A/D converter has 5 μF capacitor and 25 kΩ resistor connected to it. If the reference voltage is taken as 10 V and output voltage of integrator cannot exceed 10 V then time for which reference voltage can be integrated will be – (in msec)
The output voltage of dual slope integrating type AD converter,
= 80 t
Given that, output voltage cannot exceed 10 V
⇒ 80 t = 10 ⇒ 125 msec
In a multimeter circuit, for a.c. voltage measurement, what is the function of diode D_{2}?
In the absence of D_{2}, leakage current would have flowed in D_{1}. To bypass this reverse leakage current of D_{1} during the negative half cycle of the input. The absence of D_{2} reduces the average value for complete cycle than it would have been in half wave rectification.
Determine the error percentage of the voltage reading due to the insertion of a voltmeter. R_{1} = 3 kΩ, R_{2} = 2 kΩ and the internal resistance of the voltmeter is 40 kΩ.
The actual voltage of R_{2} is:
The voltage after inserting the voltmeter will then be
Thus, error percentage
Which meter has a greater sensitivity meter A having a range of 0  15 V and a multiplier resistance of 22 kΩ, or meter B with a range of 0  300 V and multiplier resistance of 324 kΩ? Both meter moments have a resistance of 2 kΩ.
Meter A:
Total resistance = 2 + 22 = 24 kΩ
Meter B:
Total resistance = 2 + 324 = 326 kΩ
Meter A is more sensitive
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