MGVCL Exam Paper (30-07-2021 Shift 3) The magnetizing inrush current in a transformer is rich in the harmonic component. fifth second third seventh fifth second 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) Rearrange the following to form a meaningful sentence and find the most logical order from the given options.P: globally, out of which over 200 millionQ: are now in India, andR: flipkart has 700 million usersS: growing exponentially RPQS RPSQ RSQP PQRS RPQS RPSQ RSQP PQRS 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. 60% 75% 50% 80% 60% 75% 50% 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) Message ____ means that the data must arrive at the receiver exactly as sent. Integrity Authentication None ot these Confidentiality Integrity Authentication None ot these Confidentiality ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A rectangular conductor is 1.6 inches wide and 0.25 inch thick. What is its area in square mils? 400000 square mils 400 square mils 4000 square mils 40000 square mils 400000 square mils 400 square mils 4000 square mils 40000 square mils ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 1.6 inches = 1.6 inches x 1,000 mils per inch = 1,600 mils0.25 inch = 0.25 inch x 1,000 mils per inch = 250 milsArea = 1,600 x 250 = 400,000 square mils
MGVCL Exam Paper (30-07-2021 Shift 3) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 20 Ω and the resistance of the resistor connected to the sound core was 40 Ω. Calculate the distance of the fault point from the test end. 300 m 400 m 200 m 100 m 300 m 400 m 200 m 100 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (20/60)*600= 200 m.
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