MGVCL Exam Paper (30-07-2021 Shift 2) Operating system is a program that runs during log off a layer between the user and the hardware a program that runs during log off a program that runs for short duration a program that runs during log off a layer between the user and the hardware a program that runs during log off a program that runs for short duration ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 930 kVA 1240 kVA 620 kVA 2480 kVA 930 kVA 1240 kVA 620 kVA 2480 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
MGVCL Exam Paper (30-07-2021 Shift 2) Fill in the blanks with suitable Article from the given alternatives.I heard ____interesting story yesterday No article the an a No article the an a ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In transformer, bushing failure occurs due to(i) transformer vibrations(ii) dielectric losses(ii) partial discharge (i), (ii) and (iii) (i) and (iii) only (i) and (ii) only (ii) and (iii) only (i), (ii) and (iii) (i) and (iii) only (i) and (ii) only (ii) and (iii) only ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.3 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. 1.42 Hz 1.52 Hz 1.32 Hz 1.62 Hz 1.42 Hz 1.52 Hz 1.32 Hz 1.62 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))= (11/05)*cosδ= (11/0.5)*0.5M = (H*s)/(πf)= 4/(50π)fn = (1.76*(3/50π))= 9.4 rad/sec = 1.32 Hz
MGVCL Exam Paper (30-07-2021 Shift 2) In a short circuit test on circuit breaker, time to reach the peak re-striking voltage is 50 µs with the peak restriking voltage of 100 kV. Determine average RRRV (Rate of rise re-striking voltage)? 1 x 10⁶ kV/sec 2 x 10⁶ kV/sec 3 x 10⁶ kV/sec 4 x 10⁶ kV/sec 1 x 10⁶ kV/sec 2 x 10⁶ kV/sec 3 x 10⁶ kV/sec 4 x 10⁶ kV/sec ANSWER EXPLANATION DOWNLOAD EXAMIANS APP RRRV = Peak value of restriking voltage/time to reach the peak re-striking voltageRRRV = 100*1000/(50 μs)RRRV = 2*10⁶ kV/sec
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