Test: Electrical & Electronic Measurements- 6
10 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Test: Electrical & Electronic Measurements- 6
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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.
Vavg = 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.
fclk = 400kHz
t = 20m sec
⇒ Clock pulses, n = t fclk = 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 dual-slope 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 auto-zero phase time into consideration) in millisecond is______.
Given that, Vref = 100 mV
T1 = 100 msec
Vin = 100 mV
We know that,
Vin T1 = Vref T2
⇒ (100) (100) = (100) (T2)
⇒ T2 = 100 msec
Total conversion time
T = T1 + T2
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,
Vavg = 2 V
For triangular wave shape, peak value of voltage Vm = 2 Vavg = 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 D2?
In the absence of D2, leakage current would have flowed in D1. To bypass this reverse leakage current of D1 during the negative half cycle of the input. The absence of D2 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. R1 = 3 kΩ, R2 = 2 kΩ and the internal resistance of the voltmeter is 40 kΩ.
The actual voltage of R2 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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22 docs|274 tests
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22 docs|274 tests
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