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 ____ 12 kV/mm 1.5 kV/mm 6 kv/mm 3 kV/mm 12 kV/mm 1.5 kV/mm 6 kv/mm 3 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) A three phase four pole 50 Hz induction motor has a rotor resistance of 0.02 Ω/phase and stand-still reactance of 0.5 Ω/phase. Calculate the speed at which the maximum torque is developed. 1500 rpm 1525 rpm 1475 rpm 1440 rpm 1500 rpm 1525 rpm 1475 rpm 1440 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Condition for maximum torque,Sm = R2/X2= 0.04Ns = (120*f)/P= (120*50)/4= 1500 rpmS = (Ns - Nr)/NsNr = Ns(1 - Sm)= 1500*(1 - 0.04)= 1440 rpm
MGVCL Exam Paper (30-07-2021 Shift 3) સારું:નરસું::શુક્લ: ? કૃષ્ણ અમર પરમ અવિશાની કૃષ્ણ અમર પરમ અવિશાની ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Nairobi negotiations are related to which of the following Organisation? UNESCO World Economic Forum The United Nations Children's Fund World Trade Organisation UNESCO World Economic Forum The United Nations Children's Fund World Trade Organisation 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% 28.33% 50.55% 63.66% 76.66% 28.33% 50.55% 63.66% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rectification efficiency = DC output power/AC input power
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 = 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 = 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 = 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 = 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
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