MGVCL Exam Paper (30-07-2021 Shift 3)

MGVCL Exam Paper (30-07-2021 Shift 3)
Data Encryption Standard (DES) was designed by

IBM

Microsoft

Apple

None of these

IBM

Microsoft

Apple

None of these

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MGVCL Exam Paper (30-07-2021 Shift 3)
A star connected synchronous generator rated at 500 MVA, 50 kV has a reactance of 0.5 pu. Find the ohmic value of the reactance.

2.5 Ω

0.25 Ω

1 Ω

0.1 Ω

2.5 Ω

0.25 Ω

1 Ω

0.1 Ω

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MGVCL Exam Paper (30-07-2021 Shift 3)
Which of the following statements are correct?
(i) Advanced metering infrastructure (AMI) is typically more automated and allows real- time, on-demand interrogations with metering endpoints.
(ii) An Automated Meter Reading (AMR) device only communicates from the customer to the energy supplier.
(iii) AMR meters only provide kWh information and possible peak kW demand for the month.

(ii) and (iii) only

(i) and (ii) only

(i), (ii) and (iii)

(i) and (iii) only

(ii) and (iii) only

(i) and (ii) only

(i), (ii) and (iii)

(i) and (iii) only

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MGVCL Exam Paper (30-07-2021 Shift 3)
A conductor having surface density is embedded in a dielectric medium of permittivity. The electric field in the medium is E. If it is known that the pressure p on the conductor surface is equal to the electric energy density in the medium, then p (in SI unit)is given by

σ²/(2ε)

σ²/(2πε)

σ²/(4π)

σ/(4πε)

σ²/(2ε)

σ²/(2πε)

σ²/(4π)

σ/(4πε)

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MGVCL Exam Paper (30-07-2021 Shift 3)
Which among these components connects to physical equipments of the distribution system?

Central control room with host computers

Remote Terminal Unit

None of these

Communication infrastructure

Central control room with host computers

Remote Terminal Unit

None of these

Communication infrastructure

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MGVCL Exam Paper (30-07-2021 Shift 3)
Find the current through the resistor and the capacitor for the current shown in figure The switch is closed at t = O and initial change in the capacitor is zero.

I(t) = 5e⁵⁰⁰⁰ᵗ A.

I(t) = 10(1 - e⁵⁰⁰⁰ᵗ) A

I(t) = 10e⁵⁰⁰⁰ᵗ A.

I(t) = 5(1 - e⁵⁰⁰⁰ᵗ) A

I(t) = 5e⁵⁰⁰⁰ᵗ A.

I(t) = 10(1 - e⁵⁰⁰⁰ᵗ) A

I(t) = 10e⁵⁰⁰⁰ᵗ A.

I(t) = 5(1 - e⁵⁰⁰⁰ᵗ) A

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