JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year FUNDAMENTALS OF ELECTRICAL ENGINEERING Previous Question Paper (March - 2022 - R13)

B.Tech | II Year | I Semester | R13 | March - 2022

Answer PaperPaper Code: 113AVMaximum Marks: 75Time: 3 Hours

Answer any five questions. Each Carries Equal Marks.

Main

115 marks
a

What is dot convention? Explain with an example.

8 marks

Answer

The dot convention is a graphical method used to determine the polarity of the induced voltage in mutually coupled coils. When current enters the dotted terminal of one coil, the induced voltage in the other coil is positive at its dotted terminal.

Rules for dot convention:

  • If current enters the dot of one coil, the potential at the dot of the coupled coil is positive.
  • If current leaves the dot of one coil, the potential at the dot of the coupled coil is negative.

For example, in two coils with mutual inductance M, if a time-varying current i1 enters the dot of coil 1, an induced voltage v2 = M(di1/dt) appears across coil 2, with the dotted terminal being positive relative to the undotted terminal.

b

Two resistors of value 1000 Ω and 4000 Ω are connected in series across a constant voltage supply of 150 V. Find:

(i) potential difference across 4000 ohm resistor

(ii) calculate the change in supply current when a 12,000 ohm resistance is connected across the larger resistor.

7 marks

Answer

(i) The total resistance R_total = 1000 + 4000 = 5000 Ω. Using the voltage divider rule, the potential difference across the 4000 Ω resistor is:

V_4000 = 150 × (4000 / 5000) = 120 V

(ii) Initially, the supply current I1 = 150 / 5000 = 0.03 A. When a 12,000 Ω resistor is connected in parallel with the 4000 Ω resistor, the equivalent resistance of the parallel combination is:

R_p = (4000 × 12000) / (4000 + 12000) = 48,000,000 / 16,000 = 3000 Ω

The new total circuit resistance is R_new = 1000 + 3000 = 4000 Ω. The new supply current is:

I2 = 150 / 4000 = 0.0375 A

The change in supply current is:

ΔI = I2 - I1 = 0.0375 - 0.03 = 0.0075 A (or 7.5 mA)
215 marks
a

Draw a parallel magnetic circuit and explain the procedure in solving it with an example.

7 marks

Answer

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b

In the circuit shown in figure 1 find IL/IS.

8 marks

Answer

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315 marks
a

Obtain the expression for the RMS value and average value of a sinusoidal waveform.

8 marks

Answer

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b

A sine wave of v(t) = 200 sin 50t is applied to a 5 Ω resistor in series with a coil. The reading of a voltmeter across the resistor is 120 V and across the coil, 75 V. Calculate the power and reactive volt-amperes in the coil and the power factor of the circuit.

7 marks

Answer

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415 marks
a

Derive the expression for the bandwidth in parallel RLC resonant circuit.

7 marks

Answer

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b

230-V rms 50-Hz supply serves a load that is 10 kW (resistive), 15 kVAR (capacitive), and 22 kVAR (inductive). Find:

(i) the apparent power

(ii) the current drawn from the supply.

8 marks

Answer

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515 marks
a

State and explain Tellegen’s theorem.

6 marks

Answer

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b

In the circuit shown in figure 2 find the voltage ‘V’ using Thevenin’s theorem.

9 marks

Answer

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615 marks
a

State and explain compensation theorem.

6 marks

Answer

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b

Determine the voltage ‘V’ in the circuit shown in figure 3 using super position theorem.

9 marks

Answer

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715 marks
a

Draw the characteristics of DC shunt generator and explain.

7 marks

Answer

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b

Explain the principle of operation of DC motor.

8 marks

Answer

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815 marks
a

Explain the principle of operation of synchronous generator.

7 marks

Answer

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b

Discuss the constructional details of single phase transformer.

8 marks

Answer

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About this paper

This page contains the solved question paper with answers of FUNDAMENTALS OF ELECTRICAL ENGINEERING from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech EIE II Year I Semester, March - 2022, R13.

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