EE MCQ

SSC JE Electrical 2019 with solution SET-2
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Identify the correct statement regarding a nuclear power plant?

In nuclear power plant the graphite is used as a moderator. The job of the moderator is to absorb some of the kinetic energy of the neutrons to slow them down. This is because slow neutrons are more easily absorbed by uranium-235. A neutron slowed In this way can start the fission process.

Related Question

SSC JE Electrical 2019 with solution SET-2
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Observe the given figure. The maximum power to the load resistor is when Ri is =?

The maximum power transfer theorem states that the DC voltage source will deliver maximum power to the variable-load resistor only when the load resistance is equal to the source resistance. Similarly, this theorem states that the AC voltage source will deliver maximum power to the variable complex load only when the load impedance is equal to the complex conjugate of the source impedance.

Hence when Ri = RL the transfer of power will be maximum.

SSC JE Electrical 2019 with solution SET-2
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In a 4 pole. 20 kW, 200 V wave wound DC shunt generator, the current in each parallel path will be

Given

Power (P) = 20 kW = 20 × 103 W

Voltage (V) = 200 V

P = VI = I = P/V

I = (20 × 103)/200

I = 100A

For wave wound No. of Parallel Path = 2

Current in Each parallel Path for wave wound

I = 100/2 = 50 A

SSC JE Electrical 2019 with solution SET-2
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If a zero-centered voltmeter has a scale from 5V to − 5V, then the span of it is

Voltage Scale Span of instrument = Vmax − Vmin

Given

Vmax = 5 V

Vmin = −5V

Span = 5 −(−5) = 10 V

SSC JE Electrical 2019 with solution SET-2
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The input for a condenser in a steam power plant comes from a/an

Condenser – Condenser is a heat exchanger in which cooling water is circulated through the tubes. The exhaust steam from the turbine enters the condenser where it is cooled and converted to condensate (water).

SSC JE Electrical 2019 with solution SET-2
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Observe the figure and find the correct relation from the four given options?

According to Kirchhoff’s law states that the current entering a node is always equal to the current leaving a node.

From the above Diagram

Current Flowing towards the Point: I1, I2, I6

Current Flowing Away from the Point: I3, I4, I5

Hence I1 + I2 + I6 = I3 + I4+ I5