MGVCL Exam Paper (30-07-2021 Shift 3) સાબરમતી: અમદાવાદ:: મૂસી: લંડન ગોવા હૈદરાબાદ વેનિસ લંડન ગોવા હૈદરાબાદ વેનિસ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The most commonly used method for the protection of three phase feeder is None of these Differential protection Time graded protection Reverse power protection None of these Differential protection Time graded protection Reverse power protection ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Differential Pilot Wire Protection is simply a differential protection scheme applied to feeders.Several differential schemes are applied for protection of line but Mess Price Voltage balance system and Translay Scheme are most popularly used.
MGVCL Exam Paper (30-07-2021 Shift 3) Consider a solar PV plant with the following specific conditions:Analysis period: 1 yearMeasured average solar irradiation intensity in 1 year: 150 kWh/m²Generator area of the PV plant: 10 m² Efficiency factor of the PV modules: 15%Electrical energy actually exported by plant to grid: 135 kWhCalculate the performance ratio. 75% 50% 60% 80% 75% 50% 60% 80% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,A - total solar panel area in meter square.r - efficiency of solar panelh - solar installation per meter square area.Solar panel performance ratio = Actual energy (kWh)/(A*r*h)= (135/(0.15*10*150))= 0.60= 60 % (in percentage)
MGVCL Exam Paper (30-07-2021 Shift 3) Which feature is used in Ms Word to make the selected sentence to All Capital Letters or All Small Letters? Change latter Change sentence Change case Change capital Change latter Change sentence Change case Change capital ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The current flowing in to a balanced delta connected load through line 'a' is 10 A when the conductor of line ’b’ is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of ’abc’. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(10 + 10∟-120°)= 5 + j*2.88 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(10 + 10∟120°)= 5 - j*2.88 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
MGVCL Exam Paper (30-07-2021 Shift 3) In which of the following situations, there is no need to provide directional overcurrent protection. single end fed, single feeder double end fed, single feeder ring main single end fed, parallel feeder single end fed, single feeder double end fed, single feeder ring main single end fed, parallel feeder ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Directional overcurrent relay protection is used for parallel feeder and ring main system.It is not used for the protection of radial system.
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