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. 1.25 H 0.75 H 0.5 H 0.25 H 1.25 H 0.75 H 0.5 H 0.25 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) 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. 75% 80% 50% 60% 75% 80% 50% 60% 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) Fill in the blanks with suitable Article from the given alternatives.Everyone is ready to accept ____ new rules framed by the executive body an No article the a an No article the a ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Which state is the largest producer of wind energy in India? Andhra Pradesh Maharashtra Madhya Pradesh Tamil Nadu Andhra Pradesh Maharashtra Madhya Pradesh Tamil Nadu ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - State - Capaciy1 - Tamilnadu - 7.5 GW2 - Maharastra - 5 GW3 - Karnataka - 4.8 GW4 - Rajsthan - 4.3 GW
MGVCL Exam Paper (30-07-2021 Shift 3) Message ____ means that the data must arrive at the receiver exactly as sent. Authentication Integrity None ot these Confidentiality Authentication Integrity None ot these Confidentiality ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 2 hours. What is the permissible overload for a duration of two hours. Assume the permissible load kVA as a fraction of rated kVA is 1.43. 700 kVA 1050 kVA 715 kVA 1430 kVA 700 kVA 1050 kVA 715 kVA 1430 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload for transformer = Full load kVA*factor for permissible overload= 1000*1.43= 1430 kVA
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