MGVCL Exam Paper (30-07-2021 Shift 3) Message ____ means that the data must arrive at the receiver exactly as sent. None ot these Confidentiality Authentication Integrity None ot these Confidentiality Authentication Integrity ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Replace the braketed phrase grammatically and conceptually with the help of the given opation. If the given sentence is correct then select the option ’The given sentence is correct’.The corono virus continued to spread, with the development of infections advancing quickly continue to spread, with the development of infections continued for spread, with the development of infections The given sentence is correct continued to spreading, with the development of infections continue to spread, with the development of infections continued for spread, with the development of infections The given sentence is correct continued to spreading, with the development of infections ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The complex power injections at a bus are shown in figure Two lines connected drawing the powers as shown in figure. The complex power in the line-2 "S2" is 2-j1 2+j1 2-j5 2+j5 2-j1 2+j1 2-j5 2+j5 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Total power generate (Sg) = S1 + S2S2 = Sg - S1= 5 + 3j - 3 + 2j= 2 + 5j
MGVCL Exam Paper (30-07-2021 Shift 3) The most commonly used method for the protection of three phase feeder is None of these Reverse power protection Time graded protection Differential protection None of these Reverse power protection Time graded protection Differential protection ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Differential Pilot Wire Protection is simply a differential protection scheme applied to feeders.Several differential schemes are applied for protection of line but Mess Price Voltage balance system and Translay Scheme are most popularly used.
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 = (-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 = 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 = 10 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 = 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. 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 series RLC circuit, the supply voltage is 230 V at 50 Hz. The resonant current is 2 A at the resonant frequency of 50 Hz. Under the resonant condition, the voltage across the capacitor is measured to be equal to 460 V. What are the values of L and C? 0.53 mH and 15.83 pF 0.73 mH and 13.85 pF 0.73 H and 13.85 pF 0.53 H and 15.83 pF 0.53 mH and 15.83 pF 0.73 mH and 13.85 pF 0.73 H and 13.85 pF 0.53 H and 15.83 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Vc = I*Xc460 = 2*(1/(2*ᴨ*50*C))C = 2/(460*2*50*ᴨ)= 1.385*10⁻⁵ F= 13.85 FV_L = I*X_LL = V/(I*2*ᴨ*f)= 460/(2*ᴨ*50*2)= 0.7345 H
EXAMIANS