Electrical and electronics measurements miscellaneous
- In a digital voltmeter, the oscillator frequency is 400 kHz and ramp voltage falls from 8V to 0 V is 20 m sec. The number of pulses counted by the counter is __________
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Assuming that gate opens when the ramp is 8V and closes when the ramp reached zero.
∴ Number of pulses counted by counter = 400 × 10³ × 20 × 10–3 = 8000.Correct Option: A
Assuming that gate opens when the ramp is 8V and closes when the ramp reached zero.
∴ Number of pulses counted by counter = 400 × 10³ × 20 × 10–3 = 8000.
- An average response rectifier type electronic ac voltmeter has dc voltage of 10 V applied to it. The meter reading will be ___________V
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A rectifier type instrument is calibrated to read rms values for sinusoidal waveforms Thus it shows 1.11 time and average value. Therefore, reading should be 10 × 1.11 = 11.1V.
Correct Option: A
A rectifier type instrument is calibrated to read rms values for sinusoidal waveforms Thus it shows 1.11 time and average value. Therefore, reading should be 10 × 1.11 = 11.1V.
- An analog voltage signal whose highest significant is 1 kHz is to be digitally coded with a resolution of 0.01% covering the voltage range of 0 to 10 volts. To avoid loss of information, the MINIMUM number of bits in the digital code should be __________
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NA
Correct Option: A
NA
- In the cirucuit given in the figure, the limiting error in the power dissipation I2R in the resistor R is __________%
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P = I²R
∴ dP = 2dI + dR P 1 R
Therefore, limiting error in the power
= 2 × 0.05 + 0.002 = 0.102 = 10.2%Correct Option: A
P = I²R
∴ dP = 2dI + dR P 1 R
Therefore, limiting error in the power
= 2 × 0.05 + 0.002 = 0.102 = 10.2%
- In a dual stop integrating type digital voltmeter, the first integration is carried out for 10 periods of the supply frequency of 50 Hz. If the reference voltage used is 2 V, the total conversion time for an input of 1V is __________sec.
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In a dual slope integrating type digital voltmeter,
t2 
Vin = Vref t1
where, t1 = first integration time= 10 × 1 = 0.2s 50
But Vin = 1V1, and Vref = 2∴ t2 = Cint1 = 1 × 0.2 = 0.1 sec Vref 2 Correct Option: C
In a dual slope integrating type digital voltmeter,
t2 Vin = Vref t1
where, t1 = first integration time= 10 × 1 = 0.2s 50
But Vin = 1V1, and Vref = 2∴ t2 = Cint1 = 1 × 0.2 = 0.1 sec Vref 2
