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Topic: Measuring Instrument , Most expected Questions From the selected topic.

Q1. Resolution of an instrument may be defined as:

(a) the maximum quantity it can measure.
(b) the minimum quantity in can measure.
(c) the minimum non-linearity of instrument.
(d) the maximum non-linearity of instrument.

Q2. Torque/weight ratio of an instrument indicates
(a) selectivity
(b) accuracy
(c) fidelity
(d) sensitivity.

Q3. The error introduced by an instrument is categorized in:
(a) Random Errors
(b) Gross Errors
(c) Systematic Errors
(d) Environmental Errors.

Q4. In a moving iron instrument, the hysteresis error may be reduced by using:
(a) High speed steel
(b) stainless steel.
(c) silver coating.
(d) Mumetal or Permalloy.

Q5. The deflecting torque of moving iron instrument is:
(a) I²dL/dθ
(b) 1/2 I²dL/dθ
(c) I dL/dθ
(d) 1/2 I dL/dθ

Q6. AC signal with RMS value 2√2 V, is applied to a PMMC instrument, then the reading of the instrument will be:
(a) 2√2 V
(b) 4 V
(c) Zero
(d) 1/√2 V

Q7. The value of Resistance Rs to be added in series with ammeter whose full-scale defection is of 0.1 mA and internal resistance is of 500Ω, to make it suitable to measure (0-10) V is.
(a) 99.5 K Ω
(b) 0.04 K Ω
(c) 96.5 K Ω
(d) 199.5 K Ω

Q8. The minimum number of watt-meter(s) required to measure 3-phase 3-wire balanced or unbalanced power is:
(a) 3
(b) 1
(c) 2
(d) 4

Q9. If an energy meter disc makes 10 revolutions in 100 seconds when a load of 450 W is connected to it, the meter constant (in rev/kWh) is
(a) 400
(b) 600
(c) 800
(d) 1000

Q10. Megger is an instrument used for measuring:
(a) A very high voltage
(b) A very high Resistances
(c) A very low voltage
(d) A very low Resistances


S1. Ans.(b)
Sol. Resolution can be defined as the minimum quantity in can measure.

S2. Ans.(d)
Sol. sensitivity.

S3. Ans.(c)
Sol. Systematic error

S4. Ans.(d)
Sol. Mumetal or Permalloy.

S5. Ans.(b)
Sol. 1/2 I²dL/dθ

S6. Ans.(c)
Sol. With high frequency signal the PMMC will reflect the average value which will be Zero.

S7. Ans.(a)
Sol. Resistances Required to connect in series:
R series = V/I_m -R_m
= 10/(0.1×10^(-3) )-500
= 99.5 KΩ

S8. Ans.(c)
Sol. Minimum Two watt-meter are required.

S9. Ans.(c)
Sol. Energy consumed in 100 second.
Given load= 450 W
=(P×hours)/1000=450/1000×100/3600=0.0125 kwh.
Meter constant in revolution per kWh
= (Revolution by meter)/(Energy consumed)
= 10/0.0125 rev/kWh
= 800

S10. Ans.(b)
Sol. Megger is used to measure very high resistance.

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