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MGVCL Exam Paper (30-07-2021 Shift 1)
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For sea water with σ = 5 mho/m and ε_r = 80, what is the distance for which ratio signal can be transmitted with 90% attenuation at 25 kHz?

Condition of good conductor:

σ/ϵω >> 1

Equation is given by:

The constant ratio = σ/ϵω

= 44913.6

σ/ϵω >> 1.

hence sea water is a acting as a good conductor Since attenuation is 90%, so transmission is 10%

So,

e^(-αx) = 0.1

Attenuation constant = √(σμϵ/2)

= 0.702

-αx = ln(0.1)

x = 3.27 m

MGVCL Exam Paper (30-07-2021 Shift 1)
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What is the largest thermal power plant in India?

Tank - Name of power station - capacity

1 - Vindhyachal Thermal Power Station - 4,760MW.

2 - Mundra Thermal Power Station - 4620 MW

3 - The Sasan Ultra Mega power plant - 3960 MW

4 - The Tiroda thermal power plant - 3300 MW

MGVCL Exam Paper (30-07-2021 Shift 1)
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At what range of wind speed is the electricity from the wind turbine is generated?

For small turbine starts generating power 12.6 kph (3.5 m/s) is the typical cut-in speed.

At 36–54 kph (10–15 m/s) produces maximum generation power.

At 90 kph (25 m/s) maximum, the turbine is stopped or braked.