MGVCL Exam Paper (30-07-2021 Shift 1) Which of the following architectures is also known as "Five Principles of Peaceful Co-existence"? Panchayatana Manusmriti Vedanga Panchsheel Panchayatana Manusmriti Vedanga Panchsheel ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Case 1: The PV connected to the DC bus was injecting 20 kW of power constituting of 1 kW of lossCase 2: The same PV is connected to the AC bus constituting to 1.5 kW loss. Find out the change in efficiency from Case 1 to Case 2? 5% increase 5% decrease 2.5% decrease 2.5% increase 5% increase 5% decrease 2.5% decrease 2.5% increase ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In DC microgrid system,Efficiency α 1/PinFor 1 kW loss injecting power is 20 kWFor 1.5 kW loss injecting power will be proportional to 30 kWSo,Efficiency will be reduced 0.025
MGVCL Exam Paper (30-07-2021 Shift 1) માણસ: ઘર :: પશુ : આકાશ પાતાળ ગુફા અંતરિક્ષ આકાશ પાતાળ ગુફા અંતરિક્ષ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Determine maximum permissible load which a 15 MVA, ONAN cooled transformer to IS : 2026 can take for 6 hr, if the initial load on it was 12 MVA. The weighted ambient temperature is 20˚C. Assume the permissible load kVA as a fraction of rated kVA is 1.25. 12 MVA 18.75 MVA 15 MVA 9.6 MVA 12 MVA 18.75 MVA 15 MVA 9.6 MVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Maximum permissible load = Rated MVA rating*fraction of rated kVAMaximum permissible load = 15*1.25Maximum permissible load = 18.75 MVA
MGVCL Exam Paper (30-07-2021 Shift 1) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 15 Ω and the resistance of the resistor connected to the sound core was 45 Ω. Calculate the distance of the fault point from the test end. 150 m 75 m 450 m 225 m 150 m 75 m 450 m 225 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)= (15/60)*600= 150 m
MGVCL Exam Paper (30-07-2021 Shift 1) An alternator is supplying load of 300 kW at a power factor of 0.5 lagging. If the power factor is raised to unity, how many more kilowatts can alternator supply for the same kVA loading? 150 kW 450 kW 300 kW 600 kW 150 kW 450 kW 300 kW 600 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = S*cosφS = P/cosφS = 300/0.5S = 600 kVAFor same kVA rating and unity power factor,P = 600*1P = 600 kWPextra = 600 - 300Pextra = 300 kW
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