MGVCL Exam Paper (30-07-2021 Shift 2) Match the following shown in figure B = (ii) & (1) A = (iii) & (2) C = (ii) & (1) A = (i) & (3) B = (ii) & (1) A = (iii) & (2) C = (ii) & (1) A = (i) & (3) ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a synchronous motor with field under excited, the power factor will be leading unity zero lagging leading unity zero lagging ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For synchronous motor,Under excitation - Lagging power factorOver excitation - Leading power factorFor synchronous generatorUnder excitation - Leading power factorOver excitation - lagging power factor
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is a national organisation monitoring the doping control programme in sports in India? Primary Anti-Doping Agency International Anti-Doping Protocol Indian Anti-Doping Agency National Anti-Doping Agency Primary Anti-Doping Agency International Anti-Doping Protocol Indian Anti-Doping Agency National Anti-Doping Agency ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Choose the best option from the given alternatives which can be substituted for the given word/sentence.A person who kills somebody especially for political reasons Pauper Stoic Alien Assassin Pauper Stoic Alien Assassin ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Encrypted security payload extension header is new in IPv6 IP IPv4 IPv5 IPv6 IP IPv4 IPv5 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 1.62 Hz 1.52 Hz 1.32 Hz 1.42 Hz 1.62 Hz 1.52 Hz 1.32 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