MGVCL Exam Paper (30-07-2021 Shift 1) માણસ: ઘર :: પશુ : પાતાળ અંતરિક્ષ ગુફા આકાશ પાતાળ અંતરિક્ષ ગુફા આકાશ 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. 9.6 MVA 18.75 MVA 15 MVA 12 MVA 9.6 MVA 18.75 MVA 15 MVA 12 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) The operating system is the most common type of ___ Communication Application Software Firmware System Software Communication Application Software Firmware System Software 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 600 kW 300 kW 150 kW 450 kW 600 kW 300 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
MGVCL Exam Paper (30-07-2021 Shift 1) In Microsoft Word, which of the following menu is used to add Headings that are displayed automatically in the top margin of all pages? View Menu Tools Menu Format Menu Insert Menu View Menu Tools Menu Format Menu Insert Menu ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) A coil having an inductance of 100 mH is magnetically coupled to another coil having an inductance of 900 mH. The coefficient of coupling between the coils is 0.45. Calculate the equivalent inductance if the two coils are connected in series aiding. 932 mH 1270 mH 1135 mH 730 mH 932 mH 1270 mH 1135 mH 730 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Mutual inductance (M) = k*√(L1*L2)k = 0.45M = 0.45*√(900*100*10⁻⁶) = 135 mHLeq = L1 + L2 + 2MLeq = 900 + 100 + 270Leq = 1270 mH
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