MGVCL Exam Paper (30-07-2021 Shift 3) The magnetizing inrush current in a transformer is rich in the harmonic component. second fifth third seventh second fifth third seventh ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 2nd harmonics contains - 60 to 55 %3rd harmonics contains - 40 - 50%
MGVCL Exam Paper (30-07-2021 Shift 3) In a power network, 375 kV is recorded at a 400 kV bus. A 45 MVAR, 400 kV shunt reactor is connected to the bus. What is the reactive power absorbed by the shunt reactor? 69.55 MVAR 59.55 MVAR 39.55 MVAR 49.55 MVAR 69.55 MVAR 59.55 MVAR 39.55 MVAR 49.55 MVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Q = V²/XX = 400²/45= 3555.55 ohmFor 375 kV,Q = 375²/3555.55= 39.55 kVAR
MGVCL Exam Paper (30-07-2021 Shift 3) The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil. 0.25 H 0.5 H 0.75 H 1.25 H 0.25 H 0.5 H 0.75 H 1.25 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of self inductance:V = Ls*(di/dt)Ls = V/(di/dt)= 100/((6 - 2)/(0.01))= 0.250 H= 250 mH
MGVCL Exam Paper (30-07-2021 Shift 3) સાબરમતી: અમદાવાદ:: મૂસી: હૈદરાબાદ વેનિસ ગોવા લંડન હૈદરાબાદ વેનિસ ગોવા લંડન ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Consider a solar PV plant with the following specific conditions:Analysis period: 1 yearMeasured average solar irradiation intensity in 1 year: 150 kWh/m²Generator area of the PV plant: 10 m² Efficiency factor of the PV modules: 15%Electrical energy actually exported by plant to grid: 135 kWhCalculate the performance ratio. 50% 60% 75% 80% 50% 60% 75% 80% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,A - total solar panel area in meter square.r - efficiency of solar panelh - solar installation per meter square area.Solar panel performance ratio = Actual energy (kWh)/(A*r*h)= (135/(0.15*10*150))= 0.60= 60 % (in percentage)
MGVCL Exam Paper (30-07-2021 Shift 3) A three phase four pole 50 Hz induction motor has a rotor resistance of 0.02 Ω/phase and stand-still reactance of 0.5 Ω/phase. Calculate the speed at which the maximum torque is developed. 1475 rpm 1500 rpm 1440 rpm 1525 rpm 1475 rpm 1500 rpm 1440 rpm 1525 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Condition for maximum torque,Sm = R2/X2= 0.04Ns = (120*f)/P= (120*50)/4= 1500 rpmS = (Ns - Nr)/NsNr = Ns(1 - Sm)= 1500*(1 - 0.04)= 1440 rpm
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