EE MCQ

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

Related Question

MGVCL Exam Paper (30-07-2021 Shift 1)
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Determine the ohmic value of the current limiting reactor per phase external to a 30 MVA, 11 kV, 50 Hz, three phase synchronous generator which can limit the current on short circuit of 5 times the full load current. The reactance of the synchronous generator is 0.05 pu.

The ratio of full load current to short circuit current = 1/5

Xsc = j/(1/5)

External reactance required = j*((1/5) - 0.05))

= j*0.15 pu

Full load current = (30*1000)/(√3*11)

= 1574.6 A

Per unit reactance = j*0.15 = (I*Xr)/V

j*0.15 = (1574.6*Xb)/((11/√3)*1000))

= 0.60 ohm