MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which best expresses the similar meaning of the given word "LUCID" Clear Sensible Cloudy Puzzled Clear Sensible Cloudy Puzzled ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A generating station having 'n' section busbars each rated at Q kVA with ×% reactance is connected on the tie-bars systems through busbar reactances of b%. Determine the short circuit kVA if a 3-phase fault takes place on one section. Q[(1/x) + (n+1)/(bn+x)] Q[(1/x) + (n+1)/(bn-x)] Q[(1/x) + (n-1)/(bn+x)] Q[(1/x) - (n+1)/(bn-x)] Q[(1/x) + (n+1)/(bn+x)] Q[(1/x) + (n+1)/(bn-x)] Q[(1/x) + (n-1)/(bn+x)] Q[(1/x) - (n+1)/(bn-x)] ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Short circuit kVA of any bus is given by:Skva = Sbase*(100/% X)Where,Sbase - base rating in kVA% X - value of reactance in %
MGVCL Exam Paper (30-07-2021 Shift 2) In a typical DC micro grid, hybrid storage system is a combination of None of these Lead acid battery and Lithium ion battery Lead acid battery and Capacitor Lead acid battery and ultra capacitor / super capacitor None of these Lead acid battery and Lithium ion battery Lead acid battery and Capacitor Lead acid battery and ultra capacitor / super capacitor ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Hybrid storage system is a combination of microturbines, fuel cells, photovoltaics (PV), Lead acid battery and ultra or super capacitors etc.
MGVCL Exam Paper (30-07-2021 Shift 2) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 1240 kVA 2480 kVA 620 kVA 930 kVA 1240 kVA 2480 kVA 620 kVA 930 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 4-pole, 50 Hz induction motor runs at 1425 rpm. Find out the percentage slip of the induction motor. 4% 6% 3% 5% 4% 6% 3% 5% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Slip, s = (Ns - Nr)/NsNs = 120f/PNs = 1500 rpms = (1500 - 1425)/1500s = 0.05%s = 5%
MGVCL Exam Paper (30-07-2021 Shift 2) A single phase 240 V, 1 kW heater is connected across single phase 240 V, 50 Hz supply through an SCR. For firing angle delay of 90˚, calculate the power absorbed in the heater element. 250 W 275 W 225 W 300 W 250 W 275 W 225 W 300 W ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Resistance of heater, R = V²/PR = 57.6 ohmRMS value of voltage through controlled SCR = (Vm/2π)*(1 + cosα)= 120 VPower absorb by heater = Vrms²/R= 120²/57.6= 250 W
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