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
MGVCL Exam Paper (30-07-2021 Shift 2) A 3-phase motor load has a p.f. of 0.397 lagging. The two wattmeter method used to measure power show the input as 30 kW. Find the reading on each wattmeter. W1 = 10 kW and W2 = 20 kW W1 = 5 kW and W2 = 25 kW W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW W1 = 10 kW and W2 = 20 kW W1 = 5 kW and W2 = 25 kW W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Cosφ = 0.397P = 30 kW = √3*V_L*I_L*cosφV_L*I_L = 30kW/(√3*0.397) = 43628.48 VAReadings of wattmeter W1 = V_L*I_L*cos(30+φ)Readings of wattmeter W2 = V_L*I_L*cos(30-φ)
MGVCL Exam Paper (30-07-2021 Shift 2) Recently, India and Benin signed MoUs on education, health and ____ e-visa facilities Trading Against Corruption Defence e-visa facilities Trading Against Corruption Defence ANSWER DOWNLOAD EXAMIANS APP
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? 75˚C column 60˚C column 90˚C column None of these 75˚C column 60˚C column 90˚C column None of these 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) In a synchronous motor with field under excited, the power factor will be lagging zero leading unity lagging zero leading unity ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For synchronous motor,Under excitation - Lagging power factorOver excitation - Leading power factorFor synchronous generatorUnder excitation - Leading power factorOver excitation - lagging power factor
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 1135 mH 1270 mH 932 mH 730 mH 1135 mH 1270 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
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