MGVCL Exam Paper (30-07-2021 Shift 3)
In transformer testing, Sweep Frequency Response Analysis (SFRA) testing is a tool to help locate ____

Core movement
Winding damage
All of these
Short circuits

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MGVCL Exam Paper (30-07-2021 Shift 3)
In a 3-phase, 4-wire, 400/230 V system, a lamp (L₁) of 100 W is connected to one phase and neutral and a lamp (L₂) of 150 W is connected to the second pahse and neutral. If the neutral wire is disconnected accidently, what will be the voltage across a 150 W lamp (L₂)?

120 v
180 V
160 V
240 V

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MGVCL Exam Paper (30-07-2021 Shift 3)
Consider the following statements and choose the correct option.
Statement 1: The speed regulation of a synchronous motor is always unity.
Statement 2: In a synchronous motor, at no-load condition, and with normal excitation the armature current drawn by a synchronous motor is in phase with applied voltage.

Both Statement 1 and Statement 2 are TRUE
Both Statement 1 and Statement 2 are FALSE
Statement 1 is FALSE and Statement 2 is TRUE
Statement 1 is TRUE and Statement 2 is FALSE

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MGVCL Exam Paper (30-07-2021 Shift 3)
The miniature circuit breaker (MCB) with ____ trip characteristics trips instantaneously when the current flowing through it reaches between 10 to 20 times the rated current.

Class B
Class Z
Class D
Class C

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MGVCL Exam Paper (30-07-2021 Shift 3)
In a single phase energy meter, the pressure / voltage coil is wound on

both the limbs of the laminated core with same turns
both the limbs of the laminated core with different turns
the edges of the limb on the laminated core
the centre limb of the laminated core / the middle limb of the shunt magnet

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MGVCL Exam Paper (30-07-2021 Shift 3)
The current flowing in to a balanced delta connected load through line 'a' is 10 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’.

Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A.
Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A.
Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A.
Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A.

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