MGVCL Exam Paper (30-07-2021 Shift 2) Recently, which country has withdrawn the 'Generalized System of Preferences' with India? United States of America Russia Peru Pakistan United States of America Russia Peru Pakistan ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The operating time of an inverse definite minimum time (IDMT) overcurrent relay is ____ proportional to the plug-setting multiplier (PSM) and____proportional to the time-multiplier setting (TMS). directly, directly inversely, inversely directly, inversely inversely, directly directly, directly inversely, inversely directly, inversely inversely, directly ANSWER DOWNLOAD EXAMIANS APP
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? Indian Anti-Doping Agency National Anti-Doping Agency International Anti-Doping Protocol Primary Anti-Doping Agency Indian Anti-Doping Agency National Anti-Doping Agency International Anti-Doping Protocol Primary Anti-Doping Agency ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit of figure, the switch is closed to position-1 for a long time. At t = 0 the switch position is changed from postion-1 to postion-2. Find an expression for the current i(t). i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Standard eqution of current for this condition,i(t) = i(0+)*cos(wt)i(0+) = E/RFor LC circuit,w = 1/√(LC)i(t) = (E/R)*cos((1/√(LC))*t) A
MGVCL Exam Paper (30-07-2021 Shift 2) Encrypted security payload extension header is new in IPv6 IPv5 IP IPv4 IPv6 IPv5 IP IPv4 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A delta-connected balanced three-phase load is supplied from a three-phase, 400 V supply. The line current is 20 A and the power taken by the load is 10 kW. Find the power factor. 0.62 0.72 0.52 0.82 0.62 0.72 0.52 0.82 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = 10 kWVL = 400 VIL = 20 A3- phse power, P = √3*VL*IL*cosφcosφ = 10*1000/(√3*400*20)= 0.7216
EXAMIANS