MGVCL Exam Paper (30-07-2021 Shift 2) What is the full form of UN in 'UN Human Rights Prize'? United Nations Uniformed Nations Unity Nation Uniform Nacional United Nations Uniformed Nations Unity Nation Uniform Nacional ANSWER DOWNLOAD EXAMIANS APP
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) If the velocity of electromagnetic wave in free space is 3 × 10⁸ m/s, then the velocity in a medium with εr of 4.5 and μr of 2 would be 9 × 10⁸ m/s 27 × 10⁸ m/s 3 × 10⁸ m/s 1 × 10⁸ m/s 9 × 10⁸ m/s 27 × 10⁸ m/s 3 × 10⁸ m/s 1 × 10⁸ m/s ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Speed of EM wave in medium = C/√(εr*μr)= (300000000)/√9= 300000000/3= 100000000= 1*10⁸ m/s
MGVCL Exam Paper (30-07-2021 Shift 2) હુ ગીત ગાઈશ' ને કર્મણીમાં ફેરવો. ગીતને હુ જ ગાઇશ ગીતને હુ ગાઈશ મારાથી ગીત ગવાશે હુ ગીતને ગાઈશ ગીતને હુ જ ગાઇશ ગીતને હુ ગાઈશ મારાથી ગીત ગવાશે હુ ગીતને ગાઈશ ANSWER DOWNLOAD EXAMIANS APP
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 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 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) 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 620 kVA 1240 kVA 2480 kVA 930 kVA 620 kVA 1240 kVA 2480 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
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