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*Answer can only contain numeric values

QUESTION: 1

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

Solution:

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

*Answer can only contain numeric values

QUESTION: 2

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 –

Solution:

*Answer can only contain numeric values

QUESTION: 3

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

Solution:

Frequency = 400 kHz.

f_{clk} = 400kHz

t = 20m sec

⇒ Clock pulses, n = t f_{clk }= 20m sec × 400 kHz = 8000

QUESTION: 4

How would a reading 0.6973 V be displayed on digit digital voltmeter on its 10 V scale.

Solution:

Resolution on 10 V scale

N = no. of full bits

Reading of 0.6973 = 0.697 volts

*Answer can only contain numeric values

QUESTION: 5

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______.

Solution:

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

QUESTION: 6

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 –

Solution:

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

*Answer can only contain numeric values

QUESTION: 7

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)

Solution:

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

QUESTION: 8

In a multimeter circuit, for a.c. voltage measurement, what is the function of diode D_{2}?

Solution:

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.

*Answer can only contain numeric values

QUESTION: 9

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.

Solution:

The actual voltage of R_{2} is:

The voltage after inserting the voltmeter will then be

Thus, error percentage

QUESTION: 10

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.

Solution:

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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