MGVCL Exam Paper (30-07-2021 Shift 3) The International boundary 'Radcliffe Line’ lies between India and Japan China Pakistan Afghanistan Japan China Pakistan Afghanistan 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. 0.75 H 0.5 H 0.25 H 1.25 H 0.75 H 0.5 H 0.25 H 1.25 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) Choose the word which expresses nearly the opposite meaning of the given word "DILIGENT " Lasy Busy Careful Active Lasy Busy Careful Active ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A 200/400 V, 20 kVA, 2-winding transformer is connected as an auto-transformer to transform 600 V to 200 V. Calculate the kVA rating of the auto-transformer. 50 kVA 25 kVA 30 kVA 60 kVA 50 kVA 25 kVA 30 kVA 60 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here auto transformer change voltage from 600 V to 200 V.Voltage ratio (a) = 600/200 = 3kVA rating of the auto-transformer = (a/(a - 1))*(kVA rating of auto transformer)= (3/2)*20= 30 kVA
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? 0.7.7 1.414 1.732 0.577 0.7.7 1.414 1.732 0.577 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Characteristic impedance = √(impedance/admittance) = 1.73
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 = 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. 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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