EE MCQ

MGVCL Exam Paper (30-07-2021 Shift 1)
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The operating time of an inverse definite minimum time (IDMT) overcurrent relay corresponding to plug-setting multiplier (PSM) = 1 and time-multiplier setting (TMS) = 1 will be

Equation of operating time for IDMT relay (t) = 0.14*TMS/[(PSM^0.02) - 1]

Here,

PSM = TMS = 1

So,

t = 0.14*1/(1 - 1)

t = infinite

Related Question

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 %

MGVCL Exam Paper (30-07-2021 Shift 1)
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The power factor angle of a purely inductive circuit is

Power factor for purely inductive circuit is 0 and power factor angle is 90 degree.

Power factor for purely capacitive circuit is 0 and power factor angle is 90 degree

Power factor for purely resistive circuit is 1 and power factor angle is 0 degree

MGVCL Exam Paper (30-07-2021 Shift 1)
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A 16 A, Type B Miniature Circuit Breaker (MCB) will disconnect a fault in 0.1s when the fault current is

Class B trip curve:

The MCB with class B trip characteristics trips instantaneously when the current flowing through it reaches between 3 to 5 times rated current. These MCBs are suitable for cable protection.

Hence, For 16 A current Type B MCB disconnect fault at 3*16 = 48 A to 5*16 = 80 A.

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