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) Find the word which is correctly spelt from the given options. Maximum Absurrd Sentimant Wrinkel Maximum Absurrd Sentimant Wrinkel ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which one of these is not a monitoring and diagnostic test for transformer oil? Volatility Level Analysis Furanic Compound Analysis Dissolved Metals Analysis Dissolved Gas Analysis Volatility Level Analysis Furanic Compound Analysis Dissolved Metals Analysis Dissolved Gas Analysis ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Volatility is a statistical measure of the dispersion of returns for a given security or market index.Analysis of a mineral oil sample for volatile compounds provides an indication of whether a fault or significant degradation has occurred.
MGVCL Exam Paper (30-07-2021 Shift 2) If the velocity of electromagnetic wave in free space is 3 × 10⁸ m/s, then the velocity in a medium with εr of 4.5 and μr of 2 would be 3 × 10⁸ m/s 9 × 10⁸ m/s 1 × 10⁸ m/s 27 × 10⁸ m/s 3 × 10⁸ m/s 9 × 10⁸ m/s 1 × 10⁸ m/s 27 × 10⁸ m/s ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Speed of EM wave in medium = C/√(εr*μr)= (300000000)/√9= 300000000/3= 100000000= 1*10⁸ m/s
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 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 kW W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 kW W1 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 kW W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 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) 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
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