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 Power point MS Word MS Access MS Excel MS Power point MS Word MS Access MS Excel ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) સારું:નરસું::શુક્લ: ? પરમ કૃષ્ણ અવિશાની અમર પરમ કૃષ્ણ અવિશાની અમર ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Two-wattmeter method is used to measure the power taken by a 3-phase induction motor on no load. The wattmeter readings are 400 W and -50 W. Calculate the reactive power taken by the load 350√3 VAR 450/√3 VAR 450√3 VAR 350/√3 VAR 350√3 VAR 450/√3 VAR 450√3 VAR 350/√3 VAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Reactive power taken (Q) = √3*(W1 - W2)= √3*(450 + 50)= √3*450 kVAR
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 = (-5 + j2.88) A.Ia₂ = (5 - j2.88) 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 = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) 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. 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) In a power network, 375 kV is recorded at a 400 kV bus. A 45 MVAR, 400 kV shunt reactor is connected to the bus. What is the reactive power absorbed by the shunt reactor? 39.55 MVAR 69.55 MVAR 59.55 MVAR 49.55 MVAR 39.55 MVAR 69.55 MVAR 59.55 MVAR 49.55 MVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Q = V²/XX = 400²/45= 3555.55 ohmFor 375 kV,Q = 375²/3555.55= 39.55 kVAR
MGVCL Exam Paper (30-07-2021 Shift 3) The maximum value of power factor is ____ and it exists in a pure____ circuit. 1, resistive 0.866, resistive 0.866, inductive 1, inductive 1, resistive 0.866, resistive 0.866, inductive 1, inductive ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Value of power factor is varies from 0 to 1.R→PF = 1L→PF = 0 lagC→PF = 0 leadR-L→0 lag < PF < 1R-C→0 lead < PF < 1R-C-L→depends on value of R, L and C. PF = R/Z
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