100 Most Important MCQ Of Measurement and Instrumentation with explanation

Ques.31. Ohmmeter is

  1. A meter to record ohm
  2. Used to measure resistance
  3. Combination of Ohm and Meter
  4. An indicatng instrument

Answer.2. Used to measure resistance

Explanation:-

The ohmmeter is an instrument that is generally part of a multimeter (usually including a voltmeter and an ammeter) and is used to measure the resistance of a component. In addition, the ohmmeter may also be used to determine the condition of semiconductor devices such as diodes and transistors.

 

 

Ques.32. A moving coil galvanometer has a resistance of 4 ohms and gives full-scale deflection when carrying 30 milliamperes. The instruments can be used to measure 150 volts by connecting in sereis with the instrument a resistance of

  1. 9996 ohms
  2. 5004 ohms
  3. 5000 ohms
  4. 4996 ohms

Answer.4. 4996 ohms

Explanation

The resistance of the galvanometer RG = 4Ω

Full-scale deflection Current IF = 30 mA = 30 × 10−3A

Voltmeter Range V = 150 V

Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance is given as:

R = V/IF = 150/30 × 10−3 = 5000V

Hence the value of resistance to be connected in series

R− R = 5000 − 4 = 4996Ω

 

Ques.33. A 15-volt moving iron voltmeter has a resistance of 300 ohms and an inductance of 0.12 Henry.  The instrument reads correctly on DC and on AC at 36 Hz when it shows a voltage of 14.75 V. What will its reading for the same voltage at 100 Hz?

  1. 15.4 V
  2. 15 V
  3. 14.85 V
  4. 14.5 V

Answer.4. 4996 ohms

Explanation:-

Impedance at 100 Hz

On D. C., only ohmic resistance is involved and the voltmeter reads correctly. But on AC it is the impedance of the instrument which has to be taken into account. 

When the frequency is 100 Hz

Z2 = R2 + XL2 = R2 + 2πfL2

= 3002 + (2 × π × 100 × 0.12)2

Z = 300.6Ω

Voltmeter reading on A.C = Voltmeter voltage × voltmeter resistance ⁄ Impedance

= (15 × 300) ⁄ (309.3) = 14.55V

 

Ques.34. If the torque/weight ratio of an instrument is low, then it can be concluded that

  1. The meter will have a uniform scale
  2. The meter will have the non-uniform scale
  3. The sensitivity of the meter will be high
  4. The sensitivity of the meter will be low

Answer.4. The sensitivity of the meter will be low

Explanation:-

Torque/weight ratio an instrument determine the sensitivity. The frictional torque is dependent upon the weight of the moving parts and exerts a considerable influence on the performance of an indicating instrument and the deflection of the instrument will depend upon the frictional torque to a considerable extent. The ratio of the deflecting torque to frictional torque is, therefore, a measure of the reliability of the instrument indication. As the weight of the moving system is proportional to the frictional torque, therefore the (deflecting torque/weight) ratio of an instrument is an index of its performance and for better performance, this ratio should be higher. It has been observed that this ratio should not be less than 0.1 for the satisfactory performance of the instrument.

The low torque/weight ratio signifies heavy weight of the moving system due to which the instrument incurs frictional losses hence the sensitivity will be low.

 

Ques.35. Hot wire instrument reads

  1. Peak value
  2. R.M.S value
  3. Average value
  4. None of the above

Answer.2. R.M.S value

Explanation:-

Working Of Hot Wire Instrument

Working of the hot-wire instrument is based on the heating effect of electric currents. When the current to be measured passes through the hot wire, the wire gets heated and then expands. Since the wire is fixed between two points, it sags due to expansion which is magnified by the phosphor-bronze wire and silk thread. This expansion is taken up by the spring and the silk thread, which causes the pulley to rotate and the pointer is deflected.

Hot wire instrument

Deflecting Torque

The deflection of the pointer of the hot-wire instrument is proportional to the extension of the hot wire, which is itself proportional to the square of the current.

Hence

Deflecting Torque. Td ∝ I2

If spring control is used, then

Controlling torque T∝ θ (deflection)

For balanced condition

TC = Td

∴ θ ∝ I2

Thus, these instruments have a square-law type scale. They read the r.m.s value of current and are independent of its frequency.

 

Ques.36. A moving coil instrument gives full deflection with 15 mA. The instrument has the resistance of 5 ohms. If a resistance of 0.80 ohms is connected in parallel with the instrument, the instrument will be capable of reading upto

  1. 150 mA
  2. 600 mA
  3. 750 mA
  4. 1087 mA

Answer.4. 1087 mA

Explanation:-

The resistance of instrument Rm = 5 ohms

Shunt resistance Rs = 0.80 ohms

Full scale deflection current Im = 15mA = 0.015A

Instrument Current I =?

The shunt resistance is given as

Rs = ImRm ⁄ (I − Im)

0.80 = (0.015 × 5) ⁄ (I − 0.015)

0.80I − 0.012 = 0.075

I = 0.10875

I = 10875mA

 

Ques.37. Which one of the following is the main cause of magnetic decay in PMMC type instrument?

  1. Variation in the resistance of the moving coil
  2. Quality of spring
  3. Aging of the spring
  4. Aging of the magnets

Answer.4. Aging of the magnets

Explanation:- 

Permanent Magnets Moving Coil “PMMC” Instrument 

These instruments are used either as ammeters or voltmeters and are suitable for d.c. work only. This type of instrument is based on the principle that when a current carrying conductor is placed in a magnetic field, the mechanical force acts on the conductor. The coil placed in the magnetic field and carrying the operating current is attached to the moving system. With the movement of the coil, the pointer moves over the scale to indicate the electrical quantity being measured. This type of movement is known as the D’Arsonval movement.

Errors In PMMC Instrument

The basic sources of errors in PMMC instruments are friction, temperature, and aging of various parts. To reduce the frictional errors ratio of torque to weight is made very high.

The most serious errors are produced by the heat generated or by changes in the temperature. This changes the resistance of the working coil, causing large errors. In case o ~ voltmeters, a large series resistance of very low-temperature coefficient is used. This reduces the temperature errors.

The aging of permanent magnet and control springs also cause errors. The weakening of the magnet and springs cause opposite errors. The weakening of the magnet cause less deflection while weakening of the control springs cause large deflection, for a particular value of current. The proper use of material and pre-aging during manufacturing can reduce tt errors demo weakening of the control spring.

 

Ques.38. Which instrument has a necessarily the “square law” type scale?

  1. The permanent magnet moving the coil
  2. Hotwire instruments
  3. Moving iron repulsion
  4. None of the above

Answer.2. Hot wire Instrument

Explanation:-

Working Of Hot Wire Instrument

Working of the hot-wire instrument is based on the heating effect of electric currents. When the current to be measured passes through the hot wire, the wire gets heated and then expands. Since the wire is fixed between two points, it sags due to expansion which is magnified by the phosphor-bronze wire and silk thread. This expansion is taken up by the spring and the silk thread, which causes the pulley to rotate and the pointer is deflected.

Deflecting Torque

The deflection of the pointer of the hot-wire instrument is proportional to the extension of the hot wire, which is itself proportional to the square of the current.

Hence

Deflecting Torque. Td ∝ I2

If spring control is used, then

Controlling torque T∝ θ (deflection)

For balanced condition

TC = Td

∴ θ ∝ I2

Thus, these instruments have a square-law type scale. They read the r.m.s value of current and are independent of its frequency.

 

Ques.39. Which of the following cannot be described as the advantages of moving coil permanent magnets type instrument?

  1. They have uniform scale
  2. They have low torque/weight ratio
  3. They have efficient eddy-current damping
  4. They have no hysteresis loss

Answer.2. They have a low torque/weight ratio

Explanation

Advantages of PMMC Type instruments

  1. In moving-coil instruments, the power consumption is very small.
  2. Torque to weight ratio is very high.
  3. Moving-coil instruments have uniformly graduated scales.
  4. The same instrument can be used both as an ammeter and a voltmeter by a suitable shunt or series-resistance (multiplier).
  5. Moving-coil instruments are free from errors due to external magnetic fields since the working magnetic field strength itself is very high.
  6. Very large scale deflection (300°) is possible using the “circular scale “type of instruments.
  7. Damping employed in moving-coil instruments is perfect since an eddy current damping is used.
  8. Due to the application of intense polarized or unidirectional field, the stray magnetic field has no effect.

Disadvantages of PMMC Type Instrument

  • PMMC type of instrument can be operated in direct current only. In alternating current, the instrument does not operate because in the positive half the pointer experiences a force in one direction and in the negative half, the pointer experiences the force in the opposite direction. Due to the inertia of the pointer, it remains in its zero position.
  • Compared to moving iron instruments, these instruments are costlier,
  • Friction error due to Jewel and pivot suspension, temperature, aging of control springs, and permanent magnets might introduce errors.

 

Ques.40.  A moving coil ammeter having a resistance of 10 ohms gives full-scale deflection when a current of 5 mA is passed through it. The instrument can be used for the measurement of voltage upto 5 V by

  1. Connecting a resistance of 990 ohm parallel to the ammeter
  2. Connecting a resistance of 990 ohms parallel to the load
  3. Connecting a resistance of 990 ohms series with the instrument
  4. Connecting a resistance of 990 ohms in series with the load

Answer.3. Connecting a resistance of 990 ohms series with the instrument

Explanation:-

img.4

The resistance of instrument Rm = 10 ohms

External resistance R =?

Full scale deflection current Im = 5mA = 0.005A

Voltage across ammeter Vm = ImRm = 10 × 0.005 = 0.05A

Voltage across external resistance R =  ImR = 0.005R

Total voltage = ImRm + ImR

5 = 0.05 +  0.005R

R = 990 Ω

Note:- Direct formula

R = V/I −Rm

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