EE MCQ

SSC JE Electrical 2019 with solution SET-1
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If the copper loss of a transformer at half full load is 400 W, then the copper loss corresponding to full load is

Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains the same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss.

At full load copper Loss = I2R

At half load copper Loss = (I/2)2 × R = I2/4 × R

400 = I2/4 × R

I2R = 4 × 400

I2R = Full load copper Loss = 1600 W

Related Question

SSC JE Electrical 2019 with solution SET-1
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Find the mutual inductance between two ideally coupled coils of 2H and 8H

Mutual inductance between the two coils is K = √L1.L2

K = √2 × 8

k = √16

k = 4 H

SSC JE Electrical 2019 with solution SET-1
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A coil is wound with 50 turns and a current 8A produces a flux of 200μWb. Calculate the inductance of the coil

The inductance of the coil is given by the relation

L = Nφ/I

Where

N = number of turns = 50

φ = flux = 200μWb

I = current = 8 A

L = 50 × 200 × 10−6 ⁄ 8

L = 1.25 mH

SSC JE Electrical 2019 with solution SET-1
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Please check out below figure.

From the figure it can be concluded that the voltmeter reads 5 volts as shown in the figure below. Based on the voltmeter and ammeter readings in the measuring network, determine the value of the resistor R Here Current I = 1/2 A = 0.5 A Voltage V = 5 V R = V/I = 0.5/5 R = 10Ω

SSC JE Electrical 2019 with solution SET-1
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Lamp efficiency is measured in

The luminous efficiency of an electric lamp is the ratio of the luminous flux (in lumen) emitted by the lamp to the electric power (watt) given to it i.e.

Luminous efficiency η = Luminous flux/Watt

Note:- Lumens is the unit of measurement for luminous flux, which is the total amount of visible light emitted by a source.

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