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? None of these 90˚C column 60˚C column 75˚C column None of these 90˚C column 60˚C column 75˚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) Choose the word which expresses nearly the opposite meaning of the given word "WHIMSICAL" Regular Fantastic Funny Curious Regular Fantastic Funny Curious 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 transformer, bushing failure occurs due to(i) transformer vibrations(ii) dielectric losses(ii) partial discharge (ii) and (iii) only (i) and (ii) only (i), (ii) and (iii) (i) and (iii) only (ii) and (iii) only (i) and (ii) only (i), (ii) and (iii) (i) and (iii) only ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) 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) The Manager, with / (B) all his employee, was / (C) fired from the company /(D) NO ERROR B A C D B A C D ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field of 0.1 Wb/m². When the radius of the loop is 2 m, the emf induced in the loop will be 0.2π V 0.4π V zero 0.8π V 0.2π V 0.4π V zero 0.8π V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP emf = dΨ/dt = change of flux linkageemf = dф/dtemf = d(B*Area)/dtemf = B*d(area)/dtemf = B*d(πr²)/dtemf = Bπ*2r*dr/dtemf = Bπ*2r*change of radius per unit timeemf = 0.1*π*2*2emf = 4π
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