MGVCL Exam Paper (30-07-2021 Shift 3) Job, in operating system refers to System software Process Application Software Program System software Process Application Software Program ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A single phase full converter bridge, connected to 230 V, 50 Hz source is feeding a load R =10 Ω in series with a large inductance that makes the load current ripple free. For a firing angle of 450, calculate the rectification efficiency. 28.33% 50.55% 63.66% 76.66% 28.33% 50.55% 63.66% 76.66% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rectification efficiency = DC output power/AC input power
MGVCL Exam Paper (30-07-2021 Shift 3) What is the value of the characteristic impedance of a transmission line with impedance and admittance of 15 and 5? 1.732 1.414 0.577 0.7.7 1.732 1.414 0.577 0.7.7 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Characteristic impedance = √(impedance/admittance) = 1.73
MGVCL Exam Paper (30-07-2021 Shift 3) What items should be considered when looking at a site for a new transformer installation? Site Survey All of these Soils Report Environmental impact Site Survey All of these Soils Report Environmental impact ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Points should be consider for new transformer installation:LocationEnvironmental impactSoils conditionSurvey of landVentilationGroundingWiringFluid checkInspection
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 = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - 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) Fill in the blanks with suitable Preposition from the given alternatives.BCCI receives____1000 applications for Team India head coach position against for since over against for since over ANSWER DOWNLOAD EXAMIANS APP
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