MGVCL Exam Paper (30-07-2021 Shift 1) An RLC series circuit consists of R = 16 Ω, L = 5 mH and C = 2 μF. Calculate the quality factor. 4.125 2.125 3.125 5.125 4.125 2.125 3.125 5.125 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For series RLC circuit,Q = (1/R)*√(L/C)Q = 1/16*√(5 mH/2 μF)Q = 50/16Q = 3.125
MGVCL Exam Paper (30-07-2021 Shift 1) The necessary equation to be solved during the load flow analysis using Fast-Decoupled methid is given below:where B' and B'' are formed using the imaginary part of the bus admittance matrix Ybus, n is the total number of buses in the system, m is the number of voltage regulated buses. The size of matrix B' is∆P/|Vi|= -B' ∆δ∆Q/|Vi| = -B'' ∆|Vi| (m-1) x (m-1) (n-1) x (n-1) (n-m-1) x (n-m-1) (n-m) x (n-m) (m-1) x (m-1) (n-1) x (n-1) (n-m-1) x (n-m-1) (n-m) x (n-m) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In FDLF method,B' corresponds to susceptance of unknown of PQ and PV busOrder of B' matrix is = (n -1)*(n - 1)Where n is total no. of buses.Order of B'' = (n - m - 1)*(n - m - 1)
MGVCL Exam Paper (30-07-2021 Shift 1) ___ is the process of actuating equipment operation at remote locations. Supervisory control Data processing Supplementary control Data acquisition Supervisory control Data processing Supplementary control Data acquisition ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) નીચેના રુઢિપ્રયોગોને અર્થ સાથે સાંકળી જોડી બનાવો.1. કાન કરડવા - 1. બાતમી મળી જવી.2. કાન ફૂંકવા - 2. વાત પર ધ્યાન ન આપવું.૩. કાને વાત પહોચવી - ૩. કાન 1-42-33-14-2 1-32-23-44-1 1-32-43-14-2 1-22-33-44-1 1-42-33-14-2 1-32-23-44-1 1-32-43-14-2 1-22-33-44-1 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) An alternator is supplying load of 300 kW at a power factor of 0.5 lagging. If the power factor is raised to unity, how many more kilowatts can alternator supply for the same kVA loading? 450 kW 150 kW 600 kW 300 kW 450 kW 150 kW 600 kW 300 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = S*cosφS = P/cosφS = 300/0.5S = 600 kVAFor same kVA rating and unity power factor,P = 600*1P = 600 kWPextra = 600 - 300Pextra = 300 kW
MGVCL Exam Paper (30-07-2021 Shift 1) A given solar cell has the following specifications:Open-circuit voltage, Voc: 0.7 VShort-circuit current, Isc: 3.5 AShort-circuit current, Isc: 3.5 AShort-circuit current, Isc: 3.5 AIf the solar cell operates at maximum power point, calculate the efficiency. 14.70% 15.70% 13.70% 12.70% 14.70% 15.70% 13.70% 12.70% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation is given by:Efficiency = (F.F*Voc*Isc)/Pin= (0.6*3.5*0.7)/10= 0.14702% efficiency = 14.7 %
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