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), (ii) and (iii) (i) and (iii) only (i) and (ii) only (ii) and (iii) only (i), (ii) and (iii) (i) and (iii) only (i) and (ii) only 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) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 930 kVA 1240 kVA 2480 kVA 620 kVA 930 kVA 1240 kVA 2480 kVA 620 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
MGVCL Exam Paper (30-07-2021 Shift 2) A wire pilot is generally economical for distances up to 10 km to 15 km 75 km to 100 km 30 km to 60 km 20 to 25 km 10 km to 15 km 75 km to 100 km 30 km to 60 km 20 to 25 km ANSWER EXPLANATION DOWNLOAD EXAMIANS APP A wire pilot is generally economical for distances up to 5 or 10 miles, beyond which a carrier-current pilot usually becomes more economical.
MGVCL Exam Paper (30-07-2021 Shift 2) 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 opposing. 730 mH 1270 mH 1135 mH 932 mH 730 mH 1270 mH 1135 mH 932 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP M = k√(L₁L₂)M= 0.45√(100*900)M = 135 mHL_eq for opposite connection of coils = L₁ + L₂ - 2ML_eq = 900 + 100 - 2*135L_eq = 730 mH
MGVCL Exam Paper (30-07-2021 Shift 2) If all the devices are connected to a central hub, then the topology is called Tree Topology Bus Topology Star Topology Tree Topology Tree Topology Bus Topology Star Topology Tree Topology ANSWER DOWNLOAD EXAMIANS APP
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