MGVCL Exam Paper (30-07-2021 Shift 3) સારું:નરસું::શુક્લ: ? અમર અવિશાની પરમ કૃષ્ણ અમર અવિશાની પરમ કૃષ્ણ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The International boundary 'Radcliffe Line’ lies between India and China Afghanistan Japan Pakistan China Afghanistan Japan Pakistan ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The magnetizing inrush current in a transformer is rich in the harmonic component. third fifth second seventh third fifth second 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) If a cable of homogeneous insulation has maximum stress of 6 kV/mm, then the dielectric strength of insulation should be ____ 6 kv/mm 3 kV/mm 12 kV/mm 1.5 kV/mm 6 kv/mm 3 kV/mm 12 kV/mm 1.5 kV/mm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP If a cable have homogeneous insulation then the dielectric strength of insulation should be is equal to maximum dielectric strength.
MGVCL Exam Paper (30-07-2021 Shift 3) Which state is the largest producer of wind energy in India? Tamil Nadu Andhra Pradesh Maharashtra Madhya Pradesh Tamil Nadu Andhra Pradesh Maharashtra Madhya Pradesh 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) A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.4 p.u. and j0.2 p.u. respectively. The generator no load voltage is 1.1 p.u. and that of infinite bus is 1.0 p.u. The inertia constant of the generator is 4 MW-sec/MVA. Determine the frequency of natural oscillations if the generator is loaded to 60% of its maximum power transfer capacity and small perturbation in power is given. 1.8 Hz 1.6 Hz 1.2 Hz 0.8 Hz 1.8 Hz 1.6 Hz 1.2 Hz 0.8 Hz ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Frequency of natural oscillation is given by,fn = {((dPe/dδ)at(δo))/M)}dPe/dδ = ((V1*V2)/X*(cosδ))= (1.1/0.6)*cosδ= (1.1/0.6)*0.5= 0.91M = (H*s)/(πf)= 4/(50π)fn = (0..91/(4/50π))= 8.9 rad/sec= 1.2 Hz
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