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 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 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 circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field of 0.1 Wb/m². When the radius of the loop is 2 m, the emf induced in the loop will be 0.4π V 0.8π V zero 0.2π V 0.4π V 0.8π V zero 0.2π V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP emf = dΨ/dt = change of flux linkageemf = dф/dtemf = d(B*Area)/dtemf = B*d(area)/dtemf = B*d(πr²)/dtemf = Bπ*2r*dr/dtemf = Bπ*2r*change of radius per unit timeemf = 0.1*π*2*2emf = 4π
MGVCL Exam Paper (30-07-2021 Shift 2) નીચે પૈકી કયો 'દાદુર' નો સમાનર્થી નથી મેડક દર્દુર દેડકો સુમાર મેડક દર્દુર દેડકો સુમાર ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the committee on poverty estimation? Ad hoc committe S.S.Tarapore committee Tendulkar committee Karve committee Ad hoc committe S.S.Tarapore committee Tendulkar committee Karve committee 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? None of these 75˚C column 90˚C column 60˚C column None of these 75˚C column 90˚C column 60˚C column 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) A short-shunt dc compound generator supplies 50 A at 300 V. If the shunt field resistance, series field resistance and armature resistance are 30 Ω, 0.03Ω, and 0.05 Ω respectively, determine the e.m.f. generated. 304.5 V 306.5 V 308.5 V 302.5 V 304.5 V 306.5 V 308.5 V 302.5 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Vt = 300 VRse = 0.03 Ω, Rsh = 30 Ω, Ra = 0.05 ΩI_L = 50 AVsh = V + I_L*Rse = 300 + 50*0.03 = 301.5 VIsh = Vsh/Rsh = 301.5/30 = 10.05 AIa = Ish + I_L = 60.05 AEg = Vt + I_L*Rse + IaRaEg = 300 + 50*0.03 + 60.05*0.05 = 304.5025 V
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