EE MCQ

SSC JE Electrical 2019 with solution SET-1
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In the below circuit, X = ?

The circuit shown in the question is an AND GATE. In an AND gate has two or more inputs but it has only one output. An input signal applied to a gate has only two stable states, either 1 (HIGH) or 0 (LOW). In AND gate for any input A&B the output is A.B.

Related Question

SSC JE Electrical 2019 with solution SET-1
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The meter constant of an energy meter will be given in:

Meter constant = No. of revolution/kWh

Hence the number of revolutions made by the disc for 1 kWh of energy consumption is called meter constant.

The meter constant is always written on the name plates of the energy meters installed m homes, commercial and industrial establishments. If the meter constant of an energy meter is 1500 rev/kWh, it means that for the consumption of 1 kWh, the disc will make 1500 revolutions.

SSC JE Electrical 2019 with solution SET-1
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Two inductors of 4H and 6H are connected in series. The equivalent inductance of this combination is

Total inductance when inductor are connected in series

L = L1 + L2

L = 4 + 6 =10H

SSC JE Electrical 2019 with solution SET-1
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P = ρQHg is the water power equation. Head ‘H’ is measured in meter ‘g’ gravity constant is measured in meter/second square, then the measuring unit of ‘Q’, the flow rate of water is:

P = ρ QHg is the water power equation.

Where

H = heead = meter

g = gravity =m/s2

ρ =density = (kg/m3)

P = power = watt = N-m/sec

Q = water flow rate = m3/sec

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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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