MGVCL Exam Paper (30-07-2021 Shift 2) Nangal Wildlife Sanctuary is located in which state? Sikkim Punjab Mizoram Tamil Nadu Sikkim Punjab Mizoram Tamil Nadu ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 930 kVA 2480 kVA 1240 kVA 620 kVA 930 kVA 2480 kVA 1240 kVA 620 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
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 generating station having 'n' section busbars each rated at Q kVA with ×% reactance is connected on the tie-bars systems through busbar reactances of b%. Determine the short circuit kVA if a 3-phase fault takes place on one section. Q[(1/x) + (n+1)/(bn+x)] Q[(1/x) + (n+1)/(bn-x)] Q[(1/x) + (n-1)/(bn+x)] Q[(1/x) - (n+1)/(bn-x)] Q[(1/x) + (n+1)/(bn+x)] Q[(1/x) + (n+1)/(bn-x)] Q[(1/x) + (n-1)/(bn+x)] Q[(1/x) - (n+1)/(bn-x)] ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Short circuit kVA of any bus is given by:Skva = Sbase*(100/% X)Where,Sbase - base rating in kVA% X - value of reactance in %
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 = 25 kW W1 = -5 kW and W2 = 35 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 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)
MGVCL Exam Paper (30-07-2021 Shift 2) The expression for voltage regulation of a short transmission line for leading power factor is given by: ____, where, 'I' is the line current, 'V_r' is the receiving end voltage, 'R' is the line resitance and 'X' is the line reactance. (IXcosφ_r + IRsinφ_r)/V_r (IRcosφ_r - IXsinφ_r)/V_r (IRcosφ_r + IXsinφ_r)/V_r (IXcosφ_r - IRsinφ_r)/V_r (IXcosφ_r + IRsinφ_r)/V_r (IRcosφ_r - IXsinφ_r)/V_r (IRcosφ_r + IXsinφ_r)/V_r (IXcosφ_r - IRsinφ_r)/V_r ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For lagging power factor = (IR*cosφ + IXsinφ)/V_rFor leading power factor = (IR*cosφ - IXsinφ)/V_rZero voltage regulation occurs on leading power factor only.
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