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 450 kW 300 kW 600 kW 150 kW 450 kW 300 kW 600 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) 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. 1270 mH 932 mH 730 mH 1135 mH 1270 mH 932 mH 730 mH 1135 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
MGVCL Exam Paper (30-07-2021 Shift 1) નીચે પૈકી 'અચરજ'નો પર્યાયી શબ્દ ક્યો છે? વિસ્મય જગત રજ શાશ્વત વિસ્મય જગત રજ શાશ્વત ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Dena Bank and Vijaya Bank recenlty got merged with which of the following banks? State Bank of India Indian Overseas Bank Punjab National Bank Bank of Baroda State Bank of India Indian Overseas Bank Punjab National Bank Bank of Baroda ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) In the following question, one part of the sentence may have an error. Find out which part of the sentence has an error and select the option corresponding to it. If the sentence contains no error, Select "No error'' option. (Avoid punctuation errors)(A) Bubin together with / (B) her parents are coming / (C) to attend the party. / (D) NO ERROR B A D C B A D C ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 15 Ω and the resistance of the resistor connected to the sound core was 45 Ω. Calculate the distance of the fault point from the test end. 225 m 150 m 75 m 450 m 225 m 150 m 75 m 450 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (15/60)*600= 150 m
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