MGVCL Exam Paper (30-07-2021 Shift 3) Match the following shown in table A = (iii), B = (i), C = (ii) A = (ii), B = (i), C = (iii) A = (iii), B = (ii), C = (i) A = (ii), B = (iii), C = (i) A = (iii), B = (i), C = (ii) A = (ii), B = (i), C = (iii) A = (iii), B = (ii), C = (i) 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) Which of the following medium is used between CPU and RAM to speed up the processing power of a CPU? DRAM Cache memory Virtual memory Flash memory DRAM Cache memory Virtual memory Flash memory ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Nairobi negotiations are related to which of the following Organisation? World Trade Organisation World Economic Forum UNESCO The United Nations Children's Fund World Trade Organisation World Economic Forum UNESCO The United Nations Children's Fund ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Two-wattmeter method is used to measure the power taken by a 3-phase induction motor on no load. The wattmeter readings are 400 W and -50 W. Calculate the reactive power taken by the load 450/√3 VAR 450√3 VAR 350/√3 VAR 350√3 VAR 450/√3 VAR 450√3 VAR 350/√3 VAR 350√3 VAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Reactive power taken (Q) = √3*(W1 - W2)= √3*(450 + 50)= √3*450 kVAR
MGVCL Exam Paper (30-07-2021 Shift 3) A single phase motor connected to 400 V, 50 Hz supply takes 25 A at a power factor of 0.75 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.95 lagging. 72.55 pF 92.55 pF 62.55 pF 82.55 pF 72.55 pF 92.55 pF 62.55 pF 82.55 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For single phase induction motor,P = V*I*cosφ= 400*25*0.75= 7500 WQ1 = P*tan(φ) = 7500*0.80= 6015 VARNow power factor change to 0.95So, φ' = 25.84 degreeQ = 7500*tan(18.19)= 2464.424Qnet = Q - Q'= 6015 - 2464.424= 4500.8C = V²/(2*ᴨ*f*Q)= (400*400)/(2*ᴨ*50*3575.76)= 8.255*10^(-5)= 82.55 μF
MGVCL Exam Paper (30-07-2021 Shift 3) In a power network, 375 kV is recorded at a 400 kV bus. A 45 MVAR, 400 kV shunt reactor is connected to the bus. What is the reactive power absorbed by the shunt reactor? 49.55 MVAR 39.55 MVAR 69.55 MVAR 59.55 MVAR 49.55 MVAR 39.55 MVAR 69.55 MVAR 59.55 MVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Q = V²/XX = 400²/45= 3555.55 ohmFor 375 kV,Q = 375²/3555.55= 39.55 kVAR
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