MGVCL Exam Paper (30-07-2021 Shift 3) કામકાજમાં ધ્યાન કન્દ્રિત કરજો'.આ આજ્ઞાર્થ વાક્યની સંભાવનાર્થ વાક્યરચના કઈ હોઈ શકે? કામકાજમાં ધ્યાન કેન્દ્રિત નથી કરવું? હવે તો કામકાજમાં ધ્યાન કેન્દ્રિત કરશો ને? હવે તો તમે કામકાજમાં ધ્યાન કેન્દ્રિત કરો. કામકાજમાં ધ્યાન કેન્દ્રિત કરી લીધું છે. કામકાજમાં ધ્યાન કેન્દ્રિત નથી કરવું? હવે તો કામકાજમાં ધ્યાન કેન્દ્રિત કરશો ને? હવે તો તમે કામકાજમાં ધ્યાન કેન્દ્રિત કરો. કામકાજમાં ધ્યાન કેન્દ્રિત કરી લીધું છે. ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Choose the best option from the given alternatives which can be substituted for the given word/sentence.Code of diplomatic etiquette and precedence Secular Sheath Wardrobe Protocol Secular Sheath Wardrobe Protocol ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A star connected synchronous generator rated at 500 MVA, 50 kV has a reactance of 0.5 pu. Find the ohmic value of the reactance. 0.1 Ω 0.25 Ω 1 Ω 2.5 Ω 0.1 Ω 0.25 Ω 1 Ω 2.5 Ω ANSWER EXPLANATION DOWNLOAD EXAMIANS APP pu = actual/baserated phase consired at base value for alternator.ohmic pu = ohmic actual/ohmic basepu = ohmic actual*current base/voltage baseConsider all above phase valuesMVA (3phase) = 3 Vph IphIph = 500*10⁶/(3*28901.73) = 5766.66 Apu = ohmic actual*current base/voltage base0.5 = ohmic actual*5766.66/28901.73ohmic actual = 2.5 Ω
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 = 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 = 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 = 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 = 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) Continuous and rapid variations in the load current magnitude which causes voltage variations is known as Harmonics Voltage sag Voltage distortion Flicker Harmonics Voltage sag Voltage distortion Flicker ANSWER EXPLANATION DOWNLOAD EXAMIANS APP When voltage changes occur in rapid succession, with magnitudes large enough to cause lighting level variations.The human eye-brain response is most sensitive to periodic r.m.s. voltage changes that occur at around 8 - 10 cycles per second.
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? 1.25 2.5 6.25 4.25 1.25 2.5 6.25 4.25 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
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