JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year BASIC ELECTRICAL ENGINEERING Previous Question Paper (September - 2021 - R15)

B.Tech | II Year | II-I | R15 | September - 2021

Answer PaperPaper Code: 123BRMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

State and explain Maximum Power transfer theorem.

8 marks

Answer

The Maximum Power Transfer Theorem states that for a given DC network with a fixed internal resistance, the maximum power is transferred from the source to the load when the load resistance is equal to the internal resistance of the source.

Mathematically, if a source has an internal resistance Rth (Thevenin resistance), the power delivered to a load RL is given by:

PL = (Vth / (Rth + RL))2 × RL

To find the condition for maximum power, we differentiate PL with respect to RL and set it to zero:

dPL / dRL = 0

Solving this derivative yields the condition RL = Rth. At this point, the load receives the maximum possible power from the circuit. The efficiency of power transfer at this condition is exactly 50%, as half the power generated is dissipated as heat within the internal resistance of the source.

b

Find the current in the resistors R1 = 10Ω and R2 = 20Ω connected in parallel across a voltage source. The supply current is 50 A. Also, find power across each resistance.

7 marks

Answer

Given two resistors in parallel, R1 = 10Ω and R2 = 20Ω, with a total supply current I = 50 A, we use the current divider rule to find the individual currents.

The current through R1 is:

I1 = I × (R2 / (R1 + R2)) = 50 × (20 / (10 + 20)) = 50 × (20 / 30) = 33.33 A

The current through R2 is:

I2 = I × (R1 / (R1 + R2)) = 50 × (10 / (10 + 20)) = 50 × (10 / 30) = 16.67 A

The voltage across the parallel combination is V = I1 × R1 = 33.33 × 10 = 333.3 V. The power across each resistance is calculated using P = I2R:

P1 = (33.33)2 × 10 = 11108.89 W
P2 = (16.67)2 × 20 = 5557.78 W
2

Using Super position theorem, find the current through the 20Ω resistor for the network shown below:

15 marks

Answer

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315 marks
a

Explain the terms with respect to Alternating Quantity: i) Instantaneous value and ii) Time period.

8 marks

Answer

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b

If the form factor of a current waveform is 2 and the amplitude factor is 2.5, find the average value of the current if the maximum value of the current is 500 A.

7 marks

Answer

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415 marks
a

Explain the terms with respect to Alternating Quantity: i) average value ii) Root mean square value iii) Maximum value.

8 marks

Answer

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b

A 230 V, 50 Hz voltage is applied to a coil L = 5 H and R = 2Ω is in series with a capacitance C. What value must C have in order that the voltage across the coil is 400 V?

7 marks

Answer

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515 marks
a

Derive the expression of emf induced in a transformer.

8 marks

Answer

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b

Explain the working of transformer on lagging power factor load with a neat phasor diagram.

7 marks

Answer

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615 marks
a

Explain the various constructional features to be considered for a single phase transformer.

7 marks

Answer

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b

In a 25 KVA, 2000/200 V transformer the iron and full load copper losses are 350 W and 400 W respectively. Find the efficiency at 0.8 power factor lagging at i) Full load and ii) Half load.

8 marks

Answer

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715 marks
a

Derive the emf equation of a dc generator.

7 marks

Answer

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b

Explain the working of a 3-phase Induction Motor.

8 marks

Answer

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8

Explain the working of Attraction type moving iron instrument with a neat sketch.

15 marks

Answer

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About this paper

This page contains the question paper of BASIC ELECTRICAL ENGINEERING from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech IT II Year II-I, September - 2021, R15.

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