Electrical and electronics measurements miscellaneous Difficult Questions and Answers | Page - 17

Electrical and electronics measurements miscellaneous

The RTDs use the principle of change of resistance with temperature. The properties of a conductor material to be used as an element of an RTD should possess which of the following properties?

1. The change in resistance per unit change in temperature should be as small as possible
1. The resistance of the materials should not have a continuous and stable relationship with temperature
1. The change of resistance with temperature should not be a linear function
1. All of these

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NA

Correct Option: D

NA

An unfinished moving iron voltmeter is used to measure the voltage in an a.c. circuit. If a stray d.c. magnetic field having a component along the axis of the meter coil appears, the meter reading would be

1. unaffected
1. decreased
1. increased
1. either decreased or increased depending on the direction of the d.c. field.

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The meter reading will either decrease or increase depending upon the direction of flux.

As shown in figure let FM be the flux due to meter coil and F, is due to stray magnetic field, then resultant will be F. Similarly when F is in opposite direction. Thus the reading will either increase or decrease.

Correct Option: D

The meter reading will either decrease or increase depending upon the direction of flux.

As shown in figure let FM be the flux due to meter coil and F, is due to stray magnetic field, then resultant will be F. Similarly when F is in opposite direction. Thus the reading will either increase or decrease.

A 230 V, 10 A, single-phase energymeter makes 90 revolutions in 3 minutes at half load rated voltage and unity pf. If the meter constant is 1800 revolutions /kWh, then its error at half load will be

1. 13.04% slow
1. 13.04% fast
1. 15% slow
1. 15% fast

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Energy consumed as indicate by energy meter
= 90 = 0.05 kWh
1800

Actual energy consumed
= 200 × 10 × 1 × 3 × 10^-3 = 0.0575 kWh
2 60

∴ Error = 0.0500 – 0.0575 = – 0.0075 kWh

Hence meter is slow by = 0.0075 × 100 = 15%
0.0500

Correct Option: C

Energy consumed as indicate by energy meter
= 90 = 0.05 kWh
1800

Actual energy consumed
= 200 × 10 × 1 × 3 × 10^-3 = 0.0575 kWh
2 60

∴ Error = 0.0500 – 0.0575 = – 0.0075 kWh

Hence meter is slow by = 0.0075 × 100 = 15%
0.0500

In the case of power measur ement by t wo wattmeter method in a balanced 3-phase system with a pure inductive load

1. both the wattmeters will indicates the same value but with opposite signs
1. both the wattmeter will indicate zero
1. both the wattmeter will indicate the same value with same sign
1. one wattmeter will indicate zero and the other will indicate some non-zero value

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Two wattmeter readings are,
W₁ = V_L I_L cos (30 – φ)
= V_L I_L cos (30 – 90°) = V_L I_L cos 60°
W₂ = V_L I_L cos (30 + φ)
= V_L I_L cos 120° = – V_L I_L cos 60°

Correct Option: A

Two wattmeter readings are,
W₁ = V_L I_L cos (30 – φ)
= V_L I_L cos (30 – 90°) = V_L I_L cos 60°
W₂ = V_L I_L cos (30 + φ)
= V_L I_L cos 120° = – V_L I_L cos 60°

For the voltmeter circuit shown in the given figure, the basic D’Arsonval meter used has fullscale current of 1 mA and meter resistance (R_m) of 100 ohms. The values of the series resistance R₁ and R₂ required for 10 V range and 50 V range will be respectively

1. 9.9 kΩ and 40 kΩ
1. 10 kΩ and 50 kΩ
1. 20 kΩ and 30 kΩ
1. 200 kΩ and 250 kΩ

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Full scale meter current = 1 mA
For 10 V range, 10^–3 = 10
(100 + R₁)

or R₁ = 9.9 × 10³ Ω
For 10 V range, 10^–3 = 50
(100 + R₁ + R₂)

or R₁ + R₂ = 49.9 × 10³
∴ R₂ = 40 kΩ

Correct Option: A

Full scale meter current = 1 mA
For 10 V range, 10^–3 = 10
(100 + R₁)

or R₁ = 9.9 × 10³ Ω
For 10 V range, 10^–3 = 50
(100 + R₁ + R₂)

or R₁ + R₂ = 49.9 × 10³
∴ R₂ = 40 kΩ