MGVCL Exam Paper (30-07-2021 Shift 3) Find the word which is correctly spelt from the given options. Profitable Clasification Idantical Dimnished Profitable Clasification Idantical Dimnished ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil. 0.25 H 0.5 H 1.25 H 0.75 H 0.25 H 0.5 H 1.25 H 0.75 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of self inductance:V = Ls*(di/dt)Ls = V/(di/dt)= 100/((6 - 2)/(0.01))= 0.250 H= 250 mH
MGVCL Exam Paper (30-07-2021 Shift 3) The sites where the capacity factor is ____ are not considered suitable for wind power generation. less than 24% greater than 12% less than 12% greater than 24% less than 24% greater than 12% less than 12% greater than 24% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The capacity factor is the ratio of an actual electrical energy output over a given period of time to the maximum possible electrical energy output over that period.Nuclear power plants are at the high end of the range of capacity factors
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 = (-5 + j2.44) A.Ia₂ = (5 - j2.44) 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 = 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 = 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 = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - 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
MGVCL Exam Paper (30-07-2021 Shift 3) The magnetizing inrush current in a transformer is rich in the harmonic component. second seventh third fifth second seventh third fifth ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 2nd harmonics contains - 60 to 55 %3rd harmonics contains - 40 - 50%
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? 2.5 1.25 4.25 6.25 2.5 1.25 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
EXAMIANS