MGVCL Exam Paper (30-07-2021 Shift 2) જે ક્રિયા દર્શાવે પણ વાક્યનો અર્થ પૂર્ણ કરે એ એને ____કહેવાય. - આપેલ વિકલ્પોમાંથી યોગ્યનું ચયન કરી વાક્ય પૂર્ણ કરો ક્રિયા-વિભક્તિ કૃદંત નીપાત વિભક્તિ ક્રિયા-વિભક્તિ કૃદંત નીપાત વિભક્તિ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Match the following shown in figure B = (ii) & (1) C = (ii) & (1) A = (iii) & (2) A = (i) & (3) B = (ii) & (1) C = (ii) & (1) A = (iii) & (2) A = (i) & (3) ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A THHN conductor will have a 60˚C termination on one end and a 75˚C termination on the other, which ampacity column will be used? 60˚C column None of these 75˚C column 90˚C column 60˚C column None of these 75˚C column 90˚C column ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Conductor with higher temperature ratings, provide the ampacity is determined based on the 60°C ampacity of the conductor.
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) A three-phase, 12 kV, 50 Hz and 300 kW load is operating at 0.8 pf lag. If the pf is to be improved to unity, using star connected capacitor, the per phase value of the reactive VAR required is: 100 kVAR 25 kVAR 50 kVAR 75 kVAR 100 kVAR 25 kVAR 50 kVAR 75 kVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The reactive VAR required = kW*(tanφ1 -tanφ2)Cosφ1 = 0.8φ1 = 36.87 deg.Cosφ2 = 0φ2 = 0 deg.kVAR required = kW*(tan(36.87) - tan(0))= 300*1000(0.75)= 225.00 kVARPer phase value = 225.00/3= 75 kVAR
MGVCL Exam Paper (30-07-2021 Shift 2) In a test by Murray loop for ground fault on 600 m of cable having a resistance of 1.5 Ω/km, the faulty cable is looped with a sound cable of the same length and area of cross section. If the ratio of the other two arms of the testing network at balance is 3 : 1, find the distance of the fault from the testing end of cables. 400 m 200 m 300 m 100 m 400 m 200 m 300 m 100 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,d is distance in mQ is value of variable resistor in ohmP is variable resistance in ohmHere Q/P = 1/3 is given,d = Q/(P + Q)*Loop length= 1/4*(2*600)= 300 m
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