MGVCL Exam Paper (30-07-2021 Shift 1) નીચે પૈકી 'અચરજ'નો પર્યાયી શબ્દ ક્યો છે? વિસ્મય શાશ્વત જગત રજ વિસ્મય શાશ્વત જગત રજ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Insulation Power Factor testing on an oil-filled transformer is not helpful in determining ___ Whether the oil test indicates moisture Movement of the windings (change in capacitance) Turns ratio values Condition of paper insulation (change in watts losses) Whether the oil test indicates moisture Movement of the windings (change in capacitance) Turns ratio values Condition of paper insulation (change in watts losses) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP A form of AC testing that applies voltage and measures the leakage/loss current of electrical insulation.It is a type of insulation testing used to evaluate the integrity of electrical insulation.
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. 450 m 225 m 150 m 75 m 450 m 225 m 150 m 75 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) The miniature circuit breaker (MCB) shown in figure is three pole MCB five pole MCB two pole MCB four pole MCB three pole MCB five pole MCB two pole MCB four pole MCB ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Which of the following are the advantages of Aerial Bunched Cables (ABC)? Reduced height of supports All of these Elimination of theft of energy Higher safety and reliability Reduced height of supports All of these Elimination of theft of energy Higher safety and reliability ANSWER DOWNLOAD EXAMIANS APP
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? 450 kW 300 kW 600 kW 150 kW 450 kW 300 kW 600 kW 150 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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