EE MCQ

MGVCL Exam Paper (30-07-2021 Shift 1)
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Which type of connection is employed for current transformers for the protection of delta- star connected 3-phase transformer?

The connection of CT on primary and secondary side is always reverse to the connection of transformer.

To protect the power transformer with star-delta against fault, the type of connection of CTs should be in the delta–star connection otherwise the relay will operate when an external fault occurs due to zero sequence currents

Related Question

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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A given solar cell has the following specifications: Isc = 4 A and Voc = 0.7 V. Consider that 20 identical cells with the above specifications are to be interconnected to create a PV module. What is the open-circuit voltage of the PV module if all the solar cells are connected in a series configuration?

For M no. of series configuration of solar cell,

Open circuit voltage of PV module= M*Voc

= 20*0.7

= 14 V

Short circuit current of PV module = M*Isc

= 20*4

= 80 A

MGVCL Exam Paper (30-07-2021 Shift 1)
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A balanced delta connected load of 8+j6 Ω per phase is connected to a 415 V, 50 Hz, 3-phase supply lines. If the input power factor is improved to 0.9 by connecting a star connected capacitor bank, then the required kVAR of the bank is

Current flow through the circuit,

I = V/Z

= 400/10

= 41.5 A

Power factor = 8/10

= 0.8

After connecting a capacitor bank, the reactive power component of current changes but the active power component of current is unchanged.

I1*cosφ1 = I2*cosφ2

41.5*0.8 = I2*0.9

I2 = 36.88 A

Q1 (before connecting capacitor) = √3*VL*IL*sinφ1

= √3*415*41.5*sin(36.86)

= 26.847 kVAR

Q2 (after connecting capacitor) = √3*VL*IL*sinφ2

= √3*415*36.88*sin(25.84)

= 15.867 kVAR

kVAR supplied by capacitor bank = Q1 - Q2

= 26.847 - 15.867

= 10.98 kVAR

MGVCL Exam Paper (30-07-2021 Shift 1)
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A given solar cell has the following specifications:

Open-circuit voltage, Voc: 0.7 V

Short-circuit current, Isc: 3.5 A

Short-circuit current, Isc: 3.5 A

Short-circuit current, Isc: 3.5 A

If the solar cell operates at maximum power point, calculate the efficiency.

Open-circuit voltage, Voc: 0.7 V

Short-circuit current, Isc: 3.5 A

Short-circuit current, Isc: 3.5 A

Short-circuit current, Isc: 3.5 A

If the solar cell operates at maximum power point, calculate the efficiency.

Equation is given by:

Efficiency = (F.F*Voc*Isc)/Pin

= (0.6*3.5*0.7)/10

= 0.14702

% efficiency = 14.7 %