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. 2.125 3.125 5.125 4.125 2.125 3.125 5.125 4.125 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) A coil having an inductance of 100 mH is magnetically coupled to another coil having an inductance of 900 mH. The coefficient of coupling between the coils is 0.45. Calculate the equivalent inductance if the two coils are connected in series aiding. 1135 mH 730 mH 932 mH 1270 mH 1135 mH 730 mH 932 mH 1270 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Mutual inductance (M) = k*√(L1*L2)k = 0.45M = 0.45*√(900*100*10⁻⁶) = 135 mHLeq = L1 + L2 + 2MLeq = 900 + 100 + 270Leq = 1270 mH
MGVCL Exam Paper (30-07-2021 Shift 1) જ્વાળા: ધુમાડો:: બરફ : ___ બાષ્પ ગરમી તાપમાન શીતળતા બાષ્પ ગરમી તાપમાન શીતળતા ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Case 1: The PV connected to the DC bus was injecting 20 kW of power constituting of 1 kW of lossCase 2: The same PV is connected to the AC bus constituting to 1.5 kW loss. Find out the change in efficiency from Case 1 to Case 2? 5% increase 2.5% decrease 5% decrease 2.5% increase 5% increase 2.5% decrease 5% decrease 2.5% increase ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In DC microgrid system,Efficiency α 1/PinFor 1 kW loss injecting power is 20 kWFor 1.5 kW loss injecting power will be proportional to 30 kWSo,Efficiency will be reduced 0.025
MGVCL Exam Paper (30-07-2021 Shift 1) Calculate the capacitance per meter of a 50 Ω load cable that has an inductance of 50 nH/m. 30 pF 10 pF 20 pF 40 pF 30 pF 10 pF 20 pF 40 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Z = √(L/C)C = L/Z²C = 50*10⁻⁹/(50*50)C = 0.02 nFC = 20 pF
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? 150 kW 450 kW 300 kW 600 kW 150 kW 450 kW 300 kW 600 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
EXAMIANS