MGVCL Exam Paper (30-07-2021 Shift 3) Message ____ means that the data must arrive at the receiver exactly as sent. None ot these Integrity Authentication Confidentiality None ot these Integrity Authentication Confidentiality ANSWER DOWNLOAD EXAMIANS APP
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) વિષમ' ની સંધી છૂટી પાડો. - કયો વિકલ્પ સાચો છે? વી + સમ વિ + સમ વીસ + અમ વિસ્ + અમ વી + સમ વિ + સમ વીસ + અમ વિસ્ + અમ ANSWER DOWNLOAD EXAMIANS APP
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 ____ 3 kV/mm 12 kV/mm 6 kv/mm 1.5 kV/mm 3 kV/mm 12 kV/mm 6 kv/mm 1.5 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 star connected synchronous generator rated at 500 MVA, 50 kV has a reactance of 0.5 pu. Find the ohmic value of the reactance. 2.5 Ω 0.1 Ω 0.25 Ω 1 Ω 2.5 Ω 0.1 Ω 0.25 Ω 1 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) A 3 pF capacitor is charged by a constant current of 2 pA for six seconds. The voltage across the capacitor at the end of charging will be 6 V 4 V 2 V 8 V 6 V 4 V 2 V 8 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation know that the charge stored in a capacitor is given as:Q = C*VQ = I*tV = (I*t)/C= (2μ*6)/(3μ)= 4 V
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