MGVCL Exam Paper (30-07-2021 Shift 1) જ્વાળા: ધુમાડો:: બરફ : ___ શીતળતા બાષ્પ ગરમી તાપમાન શીતળતા બાષ્પ ગરમી તાપમાન ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) An alternator is supplying load of 300 kW at a power factor of 0.5 lagging. If the power factor is raised to unity, how many more kilowatts can alternator supply for the same kVA loading? 300 kW 150 kW 450 kW 600 kW 300 kW 150 kW 450 kW 600 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = S*cosφS = P/cosφS = 300/0.5S = 600 kVAFor same kVA rating and unity power factor,P = 600*1P = 600 kWPextra = 600 - 300Pextra = 300 kW
MGVCL Exam Paper (30-07-2021 Shift 1) Two-wattmeter method is used to measure the power taken by a 3-phase induction motor on no load. The wattmeter readings are 375 W and –50 W. Calculate the reactive power taken by the load. 325/√3 VAR 425√3 VAR 325√3 VAR 425/√3 VAR 325/√3 VAR 425√3 VAR 325√3 VAR 425/√3 VAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Reactive power taken, Q = √3*(W1 - W2)Q = √3*(375 + 50)Q = √3*425 kVAR
MGVCL Exam Paper (30-07-2021 Shift 1) The current flowing to a balanced delta connected load through line 'a' is 15 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 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(15 + 15∟-120°)= 7.5 + j*4.33 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(15 + 15∟120°)= 7.5 - j*4.33 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
MGVCL Exam Paper (30-07-2021 Shift 1) Choose the best option from the given alternatives which can be substituted for the given word/sentenceA disease prevailing in a locality Host Habitat Endemic Epic Host Habitat Endemic Epic ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Calculate the capacitance per meter of a 50 Ω load cable that has an inductance of 50 nH/m. 10 pF 30 pF 40 pF 20 pF 10 pF 30 pF 40 pF 20 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Z = √(L/C)C = L/Z²C = 50*10⁻⁹/(50*50)C = 0.02 nFC = 20 pF
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