EE MCQ

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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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The current flowing to a balanced delta connected load through line 'a' is 15 A when the conductor of line 'b' is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of 'abc'.

For the given question,

Ib = 0

Ia = -Ic

Calculation:

Ia1 = 1/3*(Ia + a*Ib + a²*Ic)

= 1/3*(15 + 15∟-120°)

= 7.5 + j*4.33 A

Ia2 = 1/3*(Ia + a²*Ib + a*Ic)

= (1/3)*(15 + 15∟120°)

= 7.5 - j*4.33 A

Iao = (1/3)*(Ia + Ib + Ic)

= (Ia - Ia)

= 0