MGVCL Exam Paper (30-07-2021 Shift 1) An RLC series circuit consists of R = 16 Ω, L = 5 mH and C = 2 μF. Calculate the quality factor. 5.125 2.125 3.125 4.125 5.125 2.125 3.125 4.125 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For series RLC circuit,Q = (1/R)*√(L/C)Q = 1/16*√(5 mH/2 μF)Q = 50/16Q = 3.125
MGVCL Exam Paper (30-07-2021 Shift 1) Which of the following is an independent malicious program that spread copies of itself from computer to computer? Trojan horse Trap door Worm Virus Trojan horse Trap door Worm Virus ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Replace the underlined phrase in braket grammatically and conceptually with the help of the given options. If the given sentence is correct then select the option 'The given sentence is correct'.The new client had to pay money to initiates a account and activate their service The given sentence is correct have to pay money to initiate an had to pay money to initiate an had to pay money to initiate a The given sentence is correct have to pay money to initiate an had to pay money to initiate an had to pay money to initiate a 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? 150 kW 600 kW 450 kW 300 kW 150 kW 600 kW 450 kW 300 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 which of the following States, Tarapur Nuclear Power Plant is located? Uttar Pradesh Assam Maharashtra Bihar Uttar Pradesh Assam Maharashtra Bihar 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. 15 MVA 9.6 MVA 12 MVA 18.75 MVA 15 MVA 9.6 MVA 12 MVA 18.75 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
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