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JNTU B.Tech III MEASUREMENTS AND INSTRUMENTATION Previous Question Paper (September - 2021 - R18)

B.Tech | III | I | R18 | September - 2021

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

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

By utilizing attracted disc technique explain how electrostatic voltmeter works with help of neat diagram.

8 marks

Answer

The attracted disc electrostatic voltmeter operates on the principle of electrostatic attraction between two charged electrodes. It consists of a fixed plate and a movable disc (the attracted disc) suspended by a spring. When a potential difference is applied across the plates, an electrostatic force of attraction is generated, pulling the movable disc towards the fixed plate. This force is proportional to the square of the applied voltage.

The force of attraction F is given by:

F = (1/2) * V2 * (dC/dx)

where V is the applied voltage and dC/dx is the rate of change of capacitance with respect to the distance between the plates. As the disc moves, the spring provides a restoring force proportional to the displacement. The equilibrium position of the disc indicates the voltage value on a calibrated scale. This instrument is primarily used for high voltage measurements as the force is significant at higher potentials and it draws negligible current, making it ideal for high-impedance circuits.

b

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 corresponding to 100A (ii) The resistance for full-scale reading with 1000V. Calculate the power dissipation in each case.

7 marks

Answer

Given: Full-scale current Im = 10 mA = 0.01 A, Voltage drop Vm = 100 mV = 0.1 V. The internal resistance of the meter is Rm = Vm / Im = 0.1 / 0.01 = 10 Ω.

  1. Shunt resistance for 100 A: The total current I = 100 A. The shunt resistance Rsh = Rm / (I/Im - 1) = 10 / (100/0.01 - 1) = 10 / 9999 ≈ 0.0010001 Ω. Power dissipation in shunt Psh = Ish2 * Rsh = (99.99)2 * 0.0010001 ≈ 10 W.
  2. Multiplier resistance for 1000 V: The total voltage V = 1000 V. The multiplier resistance Rse = (V/Vm - 1) * Rm = (1000/0.1 - 1) * 10 = 9999 * 10 = 99,990 Ω. Power dissipation in multiplier Pse = Im2 * Rse = (0.01)2 * 99990 = 9.999 W.
215 marks
a

A PMMC ammeter has the following specification Coil dimension are 1cm x 1cm. Spring constant is 0.15 x 10-6 N m / rad, Flux density is 1.5 x 10-3 wb / m. Determine the number of turns required to produce a deflection of 900 when a current 2mA flows through the coil.

7 marks

Answer

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b

Describe the construction and working of PMMC instrument. Derive the equation for deflection if the instruments are spring controlled.

8 marks

Answer

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3

Differentiate between a C.T. and P.T. Discuss the theory of a P.T with phasor diagrams. Derive the expression for actual transformation ratio, ratio error and phasor angle error of a P.T.

15 marks

Answer

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4

Discuss the major sources of errors in the current transformer. What are the means to reduce errors in CT? Explain the design and constructional feature to reduce the error.

15 marks

Answer

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5

Explain the construction and working of induction type single-phase energy meter with a neat diagram and derive its equation.

15 marks

Answer

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615 marks
a

Explain the principle of working of LPF wattmeter.

7 marks

Answer

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b

Explain the operation of a three-phase dynamometer type wattmeter.

8 marks

Answer

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715 marks
a

Describe the circuit of Kelvin double bridge used for measurement of low resistance. Derive the conditions for balance.

7 marks

Answer

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b

Derive an equation for Wien’s bridge at balance with a circuit diagram and explain the measurement procedure for measuring unknown frequency using this bridge.

8 marks

Answer

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815 marks
a

How the transducers are classified based on the principle of operation?

7 marks

Answer

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b

Explain the construction and principle of working of an LVDT.

8 marks

Answer

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