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? 69.55 MVAR 59.55 MVAR 49.55 MVAR 39.55 MVAR 69.55 MVAR 59.55 MVAR 49.55 MVAR 39.55 MVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Q = V²/XX = 400²/45= 3555.55 ohmFor 375 kV,Q = 375²/3555.55= 39.55 kVAR
MGVCL Exam Paper (30-07-2021 Shift 3) What is the full form of the acronym 'SWIFT', in financial institution? Social Wide in Financial Transaction Society for Worldwide Interbank Financial Telecommunication Society website in Financial Terms Social World interconnection in Financial Tech process Social Wide in Financial Transaction Society for Worldwide Interbank Financial Telecommunication Society website in Financial Terms Social World interconnection in Financial Tech process ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A 10-pole induction motor is supplied by a 6-pole alternator, which is driven at 1400 rpm. If the motor runs with a slip of 5%, what is its speed? 1425 rpm 882 rpm 1575 rpm 798 rpm 1425 rpm 882 rpm 1575 rpm 798 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Slip of the induction motor,S = (Ns - Nr)/NsNs = (120*f)/PPoles (P) = 61400 = (120*f)/6f = 70 HzFor P = 10Ns = (120*70)/10= 840 rpmNr = Ns(1 - S)= 840*(1 - 0.05)= 798 rpm
MGVCL Exam Paper (30-07-2021 Shift 3) The sending end and receiving end voltages of the short transmission line are 150 kV and 120 kV respectively. Calculate its percentage voltage regulation. 30% 25% 20% 40% 30% 25% 20% 40% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,Vs - sending end voltage in kVVr - Receiving end voltage in kV.Voltage regulation = (Vs - Vr)/Vr= (150 - 120)/120= 0.25% R = 25 %
MGVCL Exam Paper (30-07-2021 Shift 3) Arunachal Pradesh does not share its borders with which one of the following nations? Bangladesh Myanmar China Bhutan Bangladesh Myanmar China Bhutan ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The current flowing in to a balanced delta connected load through line 'a' is 10 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 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(10 + 10∟-120°)= 5 + j*2.88 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(10 + 10∟120°)= 5 - j*2.88 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
EXAMIANS