MGVCL Exam Paper (30-07-2021 Shift 3) Name the application under MS Office software bundle, that we use to create audio- visual presentation. MS Access MS Word MS Excel MS Power point MS Access MS Word MS Excel MS Power point 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. 76.66% 50.55% 63.66% 28.33% 76.66% 50.55% 63.66% 28.33% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rectification efficiency = DC output power/AC input power
MGVCL Exam Paper (30-07-2021 Shift 3) Choose the word which expresses nearly the opposite meaning of the given word "DILIGENT " Active Busy Careful Lasy Active Busy Careful Lasy ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) If a cable of homogeneous insulation has maximum stress of 6 kV/mm, then the dielectric strength of insulation should be ____ 1.5 kV/mm 3 kV/mm 12 kV/mm 6 kv/mm 1.5 kV/mm 3 kV/mm 12 kV/mm 6 kv/mm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP If a cable have homogeneous insulation then the dielectric strength of insulation should be is equal to maximum dielectric strength.
MGVCL Exam Paper (30-07-2021 Shift 3) If the phase velocity of a plane wave in a perfect dielectric is 0.4 times its value in free space, then what is the relative permittivity of the dielectric? 1.25 2.5 4.25 6.25 1.25 2.5 4.25 6.25 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The equation is given by:Vp (Phase velocity) = (Velocity of light in free space/Phase velocity)²= (1/0.4)²= 6.25
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 = 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. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) 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
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