MGVCL Exam Paper (30-07-2021 Shift 3)
A 10 A, Type C Miniature Circuit Breaker (MCB) will trip at fault currents

between 50 A and 100 A
greater than 100 A
between 10 A and 50 A
less than 10 A

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

MGVCL Exam Paper (30-07-2021 Shift 3)
Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 20 Ω and the resistance of the resistor connected to the sound core was 40 Ω. Calculate the distance of the fault point from the test end.

200 m
100 m
300 m
400 m

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

MGVCL Exam Paper (30-07-2021 Shift 3)
Determine the ohmic value of the current limiting reactor per phase external to a 40 MVA, 15 kV, 50 Hz, three phase synchronous generator which can limit the current on short circuit of 6 times the full load current. The reactance of the synchronous generator is 0.06 pu.

0.6 Ω
0.2 Ω
0.8 Ω
0.4 Ω

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

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) = 10e⁵⁰⁰⁰ᵗ A.
I(t) = 10(1 - e⁵⁰⁰⁰ᵗ) A
I(t) = 5(1 - e⁵⁰⁰⁰ᵗ) A

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

MGVCL Exam Paper (30-07-2021 Shift 3)
The short circuit current of a solar cell is

constant and not affected by radiation
inversely proportional to radiation intensity
changes logarithmically with radiation
directly proportional to radiation intensity

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP

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

ANSWER EXPLANATION DOWNLOAD EXAMIANS APP