MGVCL Exam Paper (30-07-2021 Shift 3) In which of the following situations, there is no need to provide directional overcurrent protection. single end fed, single feeder double end fed, single feeder ring main single end fed, parallel feeder single end fed, single feeder double end fed, single feeder ring main single end fed, parallel feeder ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Directional overcurrent relay protection is used for parallel feeder and ring main system.It is not used for the protection of radial system.
MGVCL Exam Paper (30-07-2021 Shift 3) What is the value of capacitance of a capacitor which has a voltage of 4 V and has 16 C of charge? 4 F 16 F 2 F 8 F 4 F 16 F 2 F 8 F ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation is given by:C = Q/V= 16/4= 4 F
MGVCL Exam Paper (30-07-2021 Shift 3) Arunachal Pradesh does not share its borders with which one of the following nations? Myanmar Bangladesh Bhutan China Myanmar Bangladesh Bhutan China ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Match the following shown in table A = (ii), B = (i), C = (iii) A = (iii), B = (ii), C = (i) A = (iii), B = (i), C = (ii) A = (ii), B = (iii), C = (i) A = (ii), B = (i), C = (iii) A = (iii), B = (ii), C = (i) A = (iii), B = (i), C = (ii) A = (ii), B = (iii), C = (i) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Bus Type - Known Parameter - Unknown ParameterLoad Bus -P, Q - V, phase angleGenerator Bus - P, V (magnitude) - Q, Voltage phase angleSlack Bus Voltage - magnitude and phase angle - P, Q
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) The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil. 0.75 H 0.5 H 0.25 H 1.25 H 0.75 H 0.5 H 0.25 H 1.25 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of self inductance:V = Ls*(di/dt)Ls = V/(di/dt)= 100/((6 - 2)/(0.01))= 0.250 H= 250 mH
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