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JNTU B.Tech II ELECTRICAL TECHNOLOGY Previous Question Paper (August/September - 2022 - R16)

B.Tech | II | I | R16 | August/September - 2022

Answer PaperPaper Code: 133ANMaximum Marks: 75Time: 3 Hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a

Derive the torque equation of a DC motor.

8 marks

Answer

The mechanical power developed in a DC motor is the product of the electromagnetic torque (Ta) and the angular velocity (ω). From the electrical power equation, the power developed is P = EbIa, where Eb is the back EMF and Ia is the armature current.

The back EMF equation is given by:

Eb = (φZP N) / (60A)

Equating electrical power to mechanical power (Ta × ω):

Ta × (2πN / 60) = (φZP N / 60A) × Ia

Solving for Ta:

Ta = (φ Z P Ia) / (2πA)

Where φ is flux per pole, Z is the total number of armature conductors, P is the number of poles, Ia is the armature current, and A is the number of parallel paths.

b

A 6-pole wave connected DC generator having 60 slots on its armature with 6 conductors per slot, runs at 750 rpm and generates an open circuit voltage of 230 V. Find the useful flux per pole.

7 marks

Answer

Given parameters: P = 6, A = 2 (for wave connection), Z = 60 slots × 6 conductors/slot = 360 conductors, N = 750 rpm, Eg = 230 V.

The EMF equation for a DC generator is:

Eg = (φ Z P N) / (60A)

Substituting the values:

230 = (φ × 360 × 6 × 750) / (60 × 2)
230 = φ × 27000

Calculating the flux per pole:

φ = 230 / 27000 = 0.008518 Wb

The useful flux per pole is approximately 8.52 mWb.

215 marks
a

Explain in detail about the classification of DC generators.

8 marks

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b

A 200V DC shunt motor takes a total current of 100A and runs at 750 rpm. The resistance of the armature winding and shunt field winding is 0.1 ohms and 40 ohms respectively. Find the total copper losses.

7 marks

Answer

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315 marks
a

Derive the induced e.m.f equation of transformer.

8 marks

Answer

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b

A 15kVA, 1-phase, 50Hz, 500/250V transformer gave following test results: OC test (LV) side: 250V, 4.0A, 200W; SC test (HV) side: 25V, 30A, 300W. Calculate efficiency and regulation at full-load, 0.9 p.f lagging.

7 marks

Answer

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415 marks
a

Describe the construction details of a 1-φ transformer?

8 marks

Answer

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b

Draw the phasor diagram of 1-φ transformer on no-load.

7 marks

Answer

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515 marks
a

Draw and explain the Torque-Slip characteristics of a 3-phase Induction Motor.

9 marks

Answer

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b

The frequency of emf in the stator of a 4-pole induction motor is 50Hz, and that of rotor is 1.5Hz. Calculate slip and at what speed will the motor run?

6 marks

Answer

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615 marks
a

Compare slip ring and squirrel cage induction motors.

8 marks

Answer

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b

Explain any one of the starting methods of 3-phase Induction motor.

7 marks

Answer

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715 marks
a

What is distribution and coil span factors? Explain.

8 marks

Answer

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b

Explain the principle of operation of Alternator.

7 marks

Answer

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815 marks
a

Explain the working principle of a shaded pole motor.

8 marks

Answer

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b

Write short notes on Attraction type of Moving iron instrument.

7 marks

Answer

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This page contains the solved question paper with answers of ELECTRICAL TECHNOLOGY from JNTU for B.Tech ECE II I, August/September - 2022, R16.

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