MGVCL Exam Paper (30-07-2021 Shift 2) A circuit breaker is rated 2500 A, 2000 MVA, 33 kV, 3 sec, 3-phase oil circuit breaker. Determine the breaking current. 350 kA 35 kA 25 kA 2500 A 350 kA 35 kA 25 kA 2500 A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Breaking capacity of circuit breaker = √3*VL*IbVL - line voltage in voltIL - Line current/breaking in ampereVL = 33 kVS = 2000 MVAIb = 30*1000 (in kVA)/(√3*33)Ib = 34.99 kA
MGVCL Exam Paper (30-07-2021 Shift 2) Nangal Wildlife Sanctuary is located in which state? Mizoram Tamil Nadu Punjab Sikkim Mizoram Tamil Nadu Punjab Sikkim ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In transformer, bushing failure occurs due to(i) transformer vibrations(ii) dielectric losses(ii) partial discharge (i) and (iii) only (i) and (ii) only (ii) and (iii) only (i), (ii) and (iii) (i) and (iii) only (i) and (ii) only (ii) and (iii) only (i), (ii) and (iii) ANSWER DOWNLOAD EXAMIANS APP
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) Fill in the blanks with suitable Article from the given alternatives.I heard ____interesting story yesterday a an No article the a an No article the ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 12 kV, 50 Hz and 300 kW load is operating at 0.8 pf lag. If the pf is to be improved to unity, using star connected capacitor, the per phase value of the reactive VAR required is: 100 kVAR 25 kVAR 50 kVAR 75 kVAR 100 kVAR 25 kVAR 50 kVAR 75 kVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The reactive VAR required = kW*(tanφ1 -tanφ2)Cosφ1 = 0.8φ1 = 36.87 deg.Cosφ2 = 0φ2 = 0 deg.kVAR required = kW*(tan(36.87) - tan(0))= 300*1000(0.75)= 225.00 kVARPer phase value = 225.00/3= 75 kVAR
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