MGVCL Exam Paper (30-07-2021 Shift 2) A delta-connected balanced three-phase load is supplied from a three-phase, 400 V supply. The line current is 20 A and the power taken by the load is 10 kW. Find the power factor. 0.52 0.62 0.72 0.82 0.52 0.62 0.72 0.82 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = 10 kWVL = 400 VIL = 20 A3- phse power, P = √3*VL*IL*cosφcosφ = 10*1000/(√3*400*20)= 0.7216
MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which expresses nearly the opposite meaning of the given word "WHIMSICAL" Curious Funny Regular Fantastic Curious Funny Regular Fantastic 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 synchronous motor with field under excited, the power factor will be unity leading lagging zero unity leading lagging zero ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For synchronous motor,Under excitation - Lagging power factorOver excitation - Leading power factorFor synchronous generatorUnder excitation - Leading power factorOver excitation - lagging power factor
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is not a Font style in Microsoft Word? Italic Superscript Bold Regular Italic Superscript Bold Regular ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A MOSFET rated for 15 A, carries a periodic current as shown in figure. The ON state resistance of the MOSFET is 0.15 Ω. The average ON state loss in the MOSFET is 15 W 3.8 W 7.5 W 33.8 W 15 W 3.8 W 7.5 W 33.8 W ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Ip = 10 ATon = πT = 2πIavg = Ip*√(Ton/T)Iavg = 10*√(1/2)= 7.07 AThe average ON state loss in MOSFET, P = Iavg²*RP = 7.07²*0.15P = 7.5 W
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