MGVCL Exam Paper (30-07-2021 Shift 1) Match the following A = (iii), B = (i), C = (ii) A = (iii), B = (ii), C = (i) A = (ii), B = (iii), C = (i) A = (ii), B = (i), C = (iii) A = (iii), B = (i), C = (ii) A = (iii), B = (ii), C = (i) A = (ii), B = (iii), C = (i) A = (ii), B = (i), C = (iii) 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 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 - 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 Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) 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) 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? 450 kW 150 kW 600 kW 300 kW 450 kW 150 kW 600 kW 300 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) In computer networks, the nodes are the computer that routes data the computer that originates data the computer that terminates data all the mentioned above the computer that routes data the computer that originates data the computer that terminates data all the mentioned above ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) The operating time of an inverse definite minimum time (IDMT) overcurrent relay corresponding to plug-setting multiplier (PSM) = 1 and time-multiplier setting (TMS) = 1 will be infinite 1 sec zero 3 sec infinite 1 sec zero 3 sec ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of operating time for IDMT relay (t) = 0.14*TMS/[(PSM^0.02) - 1]Here,PSM = TMS = 1So,t = 0.14*1/(1 - 1)t = infinite
MGVCL Exam Paper (30-07-2021 Shift 1) Recently, India signed first ever term contract to import crude oil with which country? Japan China Nepal Russia Japan China Nepal Russia ANSWER DOWNLOAD EXAMIANS APP
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