MGVCL Exam Paper (30-07-2021 Shift 3) The maximum value of power factor is ____ and it exists in a pure____ circuit. 1, resistive 1, inductive 0.866, inductive 0.866, resistive 1, resistive 1, inductive 0.866, inductive 0.866, resistive ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Value of power factor is varies from 0 to 1.R→PF = 1L→PF = 0 lagC→PF = 0 leadR-L→0 lag < PF < 1R-C→0 lead < PF < 1R-C-L→depends on value of R, L and C. PF = R/Z
MGVCL Exam Paper (30-07-2021 Shift 3) The process of peeling of rocks into layers is called Exfoliation Sublimation Delta Brachans Exfoliation Sublimation Delta Brachans ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Nairobi negotiations are related to which of the following Organisation? The United Nations Children's Fund World Trade Organisation World Economic Forum UNESCO The United Nations Children's Fund World Trade Organisation World Economic Forum UNESCO ANSWER DOWNLOAD EXAMIANS APP
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. 0.25 Ω 0.1 Ω 1 Ω 2.5 Ω 0.25 Ω 0.1 Ω 1 Ω 2.5 Ω ANSWER EXPLANATION DOWNLOAD EXAMIANS APP pu = actual/baserated phase consired at base value for alternator.ohmic pu = ohmic actual/ohmic basepu = ohmic actual*current base/voltage baseConsider all above phase valuesMVA (3phase) = 3 Vph IphIph = 500*10⁶/(3*28901.73) = 5766.66 Apu = ohmic actual*current base/voltage base0.5 = ohmic actual*5766.66/28901.73ohmic actual = 2.5 Ω
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. 300 m 400 m 200 m 100 m 300 m 400 m 200 m 100 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (20/60)*600= 200 m.
MGVCL Exam Paper (30-07-2021 Shift 3) A 3 pF capacitor is charged by a constant current of 2 pA for six seconds. The voltage across the capacitor at the end of charging will be 4 V 8 V 2 V 6 V 4 V 8 V 2 V 6 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation know that the charge stored in a capacitor is given as:Q = C*VQ = I*tV = (I*t)/C= (2μ*6)/(3μ)= 4 V
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