MGVCL Exam Paper (30-07-2021 Shift 2) Choose the best option from the given alternatives which can be substituted for the given word/sentence.A person who kills somebody especially for political reasons Pauper Assassin Alien Stoic Pauper Assassin Alien Stoic ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The necessary equation to be solved during the load flow analysis using Fast-Decoupled method is given below:where B' and B'' are formed using the imaginary part of the bus admittance matrix, Ybus. If the system consists of 6 buses of which one of the bus is slack bus and 2 bues are voltage regulated buses, then then the matrix B' is of order ____and the matrix B'' is of order____, respectivity.∆P/|Vi| = - B'∆δ∆Q/|Vi| = - B''∆|V| 5 and 3 3 and 5 6 and 3 3 and 6 5 and 3 3 and 5 6 and 3 3 and 6 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In Fast-decoupled method,Order of B' = No. of PV buses = 5Order of B'' = No. of PQ buses = 3
MGVCL Exam Paper (30-07-2021 Shift 2) A circuit breaker is rated 2500 A, 2000 MVA, 33 kV, 3 sec, 3-phase oil circuit breaker. Determine the breaking current. 2500 A 35 kA 25 kA 350 kA 2500 A 35 kA 25 kA 350 kA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Breaking capacity of circuit breaker = √3*VL*IbVL - line voltage in voltIL - Line current/breaking in ampereVL = 33 kVS = 2000 MVAIb = 30*1000 (in kVA)/(√3*33)Ib = 34.99 kA
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 12 kV, 50 Hz and 300 kW load is operating at 0.8 pf lag. If the pf is to be improved to unity, using star connected capacitor, the per phase value of the reactive VAR required is: 100 kVAR 75 kVAR 50 kVAR 25 kVAR 100 kVAR 75 kVAR 50 kVAR 25 kVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP The reactive VAR required = kW*(tanφ1 -tanφ2)Cosφ1 = 0.8φ1 = 36.87 deg.Cosφ2 = 0φ2 = 0 deg.kVAR required = kW*(tan(36.87) - tan(0))= 300*1000(0.75)= 225.00 kVARPer phase value = 225.00/3= 75 kVAR
MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit of figure, the switch is closed to position-1 for a long time. At t = 0 the switch position is changed from postion-1 to postion-2. Find an expression for the current i(t). i(t) = (E/R) sin√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Standard eqution of current for this condition,i(t) = i(0+)*cos(wt)i(0+) = E/RFor LC circuit,w = 1/√(LC)i(t) = (E/R)*cos((1/√(LC))*t) A
MGVCL Exam Paper (30-07-2021 Shift 2) A 3-phase motor load has a p.f. of 0.397 lagging. The two wattmeter method used to measure power show the input as 30 kW. Find the reading on each wattmeter. W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Cosφ = 0.397P = 30 kW = √3*V_L*I_L*cosφV_L*I_L = 30kW/(√3*0.397) = 43628.48 VAReadings of wattmeter W1 = V_L*I_L*cos(30+φ)Readings of wattmeter W2 = V_L*I_L*cos(30-φ)
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