MGVCL Exam Paper (30-07-2021 Shift 2) In a synchronous motor with field under excited, the power factor will be zero lagging unity leading zero lagging unity leading 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) Which one of these is not a monitoring and diagnostic test for transformer oil? Dissolved Metals Analysis Furanic Compound Analysis Volatility Level Analysis Dissolved Gas Analysis Dissolved Metals Analysis Furanic Compound Analysis Volatility Level 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) 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 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 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) 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. 1240 kVA 2480 kVA 930 kVA 620 kVA 1240 kVA 2480 kVA 930 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) The line currents in amperes in phases a, b and c respectively are 500 +j150, 100 – j600 and – 300 + j600 referred to the same reference vector with phase sequence of abc. Find the symmetrical component of currents. Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Calculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 500 + j150 + (- 0.5 + j*0.866)*(100 – j6000) + (-0.5 - j*0.866)*(– 300 + j600)= (546.41 + j*156.47) AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= 500 + j150 + (- 0.5 - j*0.866)*(100 – j6000) + (-0.5 + j*0.866)*(– 300 + j600)= (-146.41 - j*56.47) AIao = 1/3*(Ia + Ib + Ic)= 1/3*(500 + j150 + 100 – j6000 – 300 + j600)= (100 + j*50) A
MGVCL Exam Paper (30-07-2021 Shift 2) What is the full form of the acronym 'AIDS' in diseases? Amino Intersect Deficiency Syndrome Acquire in Deficiency Syndrome Assert in Dander Syndrome Acquired Immune Deficiency Syndrome Amino Intersect Deficiency Syndrome Acquire in Deficiency Syndrome Assert in Dander Syndrome Acquired Immune Deficiency Syndrome ANSWER DOWNLOAD EXAMIANS APP
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