Electrical and electronics measurements miscellaneous Difficult Questions and Answers | Page - 29

Electrical and electronics measurements miscellaneous

A simple d.c. potentiometer is to be standardised by keeping the sidewire setting at 1. 0813 V. If by mistake, the setting is at 1.0138 V and the standardisation is made to obtain a source voltage of 1.0318 V, then the reading of the potentiometer will be

1. 1.0138 V
1. 1.0183 V
1. (1.0138)²/ 1.0183 V
1. (1.0138)² V

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The setting of the potentiometer is at 1.0138 instead of 1.0813. Therefore, working current in the slide wise is more than what it would have been if the setting was at 1.0813,
i.e. correct value for applied voltage of 1.0813V,
I = 1.0813, I_c = 1.0138
It the setting is at 1.01138 and the applied voltage is also 1.0138 and the applied voltage is also 1.0138, the standardisation is then correct. (Statement of the problem appears ambiguous).

Correct Option: C

The setting of the potentiometer is at 1.0138 instead of 1.0813. Therefore, working current in the slide wise is more than what it would have been if the setting was at 1.0813,
i.e. correct value for applied voltage of 1.0813V,
I = 1.0813, I_c = 1.0138
It the setting is at 1.01138 and the applied voltage is also 1.0138 and the applied voltage is also 1.0138, the standardisation is then correct. (Statement of the problem appears ambiguous).

The power in a resistor R is estimated by measuring the voltage and current using the voltmeterammeter method. Two different arrangements can be used as shown in circuits I and II. Less erroneous results are obtained by adopting

1. circuit I for low values of R
1. circuit II for values of R
1. circuit I for values of R
1. circuit II for low and high values of R

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For low resistance measurements, if circuit II is used, the ammeter resistance which may be comparable with the low resistance being measured gets added to the unknown and thereby causing larger error.

Correct Option: A

For low resistance measurements, if circuit II is used, the ammeter resistance which may be comparable with the low resistance being measured gets added to the unknown and thereby causing larger error.

A shunt type ohmmeter is shown in the figure below. With R_x disconnected, the meter reads full scale. S represents the meter current as a fraction of full scale current with R_x connected such that
S = R_x
R_xR_p

The value of R_p is given by

1. R_m
1. R₁ + R_m
1. R₁R_x
  (R₁R_m)
1. R₁R_m
  (R₁R_m)

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I_ƒs = E
R₁ + R_m

With R_x connected,
I_m = R_x E
R₁ + R_x R₁ + (R_mR_x)/(R_m + R_x)

= R_{2 + R₃} = ER_x
(R₁ + R_x) R_x + (R₁R_m)/(R₁ + R_m)

= I_ƒs R_x
R_m + R_p

where, R_p = R₁R_m
(R₁ + R_m)

Correct Option: D

I_ƒs = E
R₁ + R_m

With R_x connected,
I_m = R_x E
R₁ + R_x R₁ + (R_mR_x)/(R_m + R_x)

= R_{2 + R₃} = ER_x
(R₁ + R_x) R_x + (R₁R_m)/(R₁ + R_m)

= I_ƒs R_x
R_m + R_p

where, R_p = R₁R_m
(R₁ + R_m)

Consider the following statements about LVDT as a transducer:
1. The relationship between input displacement and output voltage is almost linear.
2. The range of displacement that can be measured is wide.
3. It does not form a loading on the mechanical system. Of these statements

1. 1, 2 and 3 are correct
1. 2 and 3 are correct
1. 1 and 2 are correct
1. 1 and 3 are correct

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NA

Correct Option: D

NA

An LVDT is used to measure displacement. The output of the LVDT is connected to a voltmeter of range 0 to 5 V through an amplifier having a gain of 250. For a displacement of 0.5 mm, the output of the LVDT is 1/2 mV. The sensit ivit y of the instrument would be

1. 0.1 V/ mm
1. 0.5 V/ mm
1. 1 V/ mm
1. 5 V/ mm

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Sensitivity of LVDT = output voltage
displacement

= 1/x × 10^-3 = 1V/mm
0.5

Correct Option: C

Sensitivity of LVDT = output voltage
displacement

= 1/x × 10^-3 = 1V/mm
0.5