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JNTU B.Tech III MEASUREMENTS AND INSTRUMENTATION Previous Question Paper (February - 2022 - R18)

B.Tech | III | III-I | R18 | February - 2022

Answer PaperPaper Code: 155CBMaximum Marks: 75Time: 3 hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

A moving coil instrument gives a full-scale deflection of 10mA when the potential difference across its terminal is 100mV. Calculate,

i) The shunt resistance for a full-scale deflection of 100A

ii) Find the resistance for full-scale reading with 100V

Also, calculate the power dissipation in each case.

8 marks

Answer

First, determine the internal resistance of the instrument (Rm):

Rm = Vm / Im = 100 mV / 10 mA = 10 Ω

i) For a shunt resistance (Rsh) to measure 100 A:

Rsh = (Im * Rm) / (I - Im) = (0.01 * 10) / (100 - 0.01) ≈ 0.0010001 Ω

Power dissipation in the shunt: Psh = Ish2 * Rsh ≈ (99.99)2 * 0.0010001 ≈ 10 W.

ii) For a series resistance (Rse) to measure 100 V:

Rse = (V / Im) - Rm = (100 / 0.01) - 10 = 9990 Ω

Power dissipation in the series resistor: Pse = Im2 * Rse = (0.01)2 * 9990 = 0.999 W.

b

Derive an equation for the torque developed in PMMC instrument.

7 marks

Answer

In a PMMC instrument, the deflecting torque (Td) is produced by the interaction of the magnetic field and the current-carrying coil. The force on one side of the coil is F = BIlN, where B is flux density, I is current, l is length of the conductor, and N is the number of turns. For two sides of the coil, the torque is:

Td = F * d = (B * I * l * N) * d

Since the area of the coil A = l * d, the equation becomes:

Td = B * A * N * I

Where:

  • B = Flux density in the air gap (Wb/m2)
  • A = Area of the coil (m2)
  • N = Number of turns
  • I = Current in the coil (A)

Thus, Td = G * I, where G = BAN is the displacement constant, showing that the deflecting torque is directly proportional to the current.

215 marks
a

A PMMC ammeter has following specification, coil dimension of 1cm×1cm, spring constant is 0.15×10-6 N-m/rad, Flux density is 1.5×10-3wb/m2. Determine the number of turns required to produce a deflection of 90°, when a current of 2mA flows through the coil.

7 marks

Answer

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b

With the help of a neat diagram, explain the working of attracted disc type voltmeter.

8 marks

Answer

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315 marks
a

Discuss how AC potentiometer can be used for calibration of wattmeter.

8 marks

Answer

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b

Explain the procedure to calibrate voltmeter and ammeter using DC potentiometer.

7 marks

Answer

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415 marks
a

With help of a neat diagram explain the working of coordinate type potentiometer.

8 marks

Answer

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b

Draw the equivalent circuit diagram and phasor diagram of the current transformer.

7 marks

Answer

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515 marks
a

Discuss the construction and Working Principle of Electrodynamometer type 1-φ wattmeter with help of a neat diagram.

9 marks

Answer

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b

Explain any two errors that occur in electrodynamometer type 1-φ wattmeter and its compensation.

6 marks

Answer

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615 marks
a

With help of a neat diagram, explain the construction and working of a three-phase energy meter.

9 marks

Answer

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b

Two-watt meters are connected to measure the input to a balanced 3 phase circuit indicating 2000W and 500W respectively. Find the power factor of a circuit,

i) When both the reading is positive and

ii) When the latter reading is obtained after reversing the connections to the current coil of the first instrument.

6 marks

Answer

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715 marks
a

Draw the circuit of Kelvin double bridge used for measurement of low resistance. Explain its working principle

8 marks

Answer

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b

An AC bridge is balanced at 2KHz with the following components in each arm: Arm AB=10KΩ, Arm BC=100µF in series with 100KΩ, Arm AD=50KΩ. Find the unknown impendence R ± jX in the arm DC, if the detector is between BD.

7 marks

Answer

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815 marks
a

Derive an equation for gauge factor in strain gauge.

7 marks

Answer

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b

With help of a neat diagram, explain the principle and working of LVDT.

8 marks

Answer

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