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? 798 rpm 882 rpm 1575 rpm 1425 rpm 798 rpm 882 rpm 1575 rpm 1425 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) A single phase full converter bridge, connected to 230 V, 50 Hz source is feeding a load R =10 Ω in series with a large inductance that makes the load current ripple free. For a firing angle of 450, calculate the rectification efficiency. 28.33% 50.55% 63.66% 76.66% 28.33% 50.55% 63.66% 76.66% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rectification efficiency = DC output power/AC input power
MGVCL Exam Paper (30-07-2021 Shift 3) Which state is the largest producer of wind energy in India? Andhra Pradesh Madhya Pradesh Tamil Nadu Maharashtra Andhra Pradesh Madhya Pradesh Tamil Nadu Maharashtra ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - State - Capaciy1 - Tamilnadu - 7.5 GW2 - Maharastra - 5 GW3 - Karnataka - 4.8 GW4 - Rajsthan - 4.3 GW
MGVCL Exam Paper (30-07-2021 Shift 3) In the following question, one part of the sentence may have an error. Find out which part of the sentence has an error and select the option corresponding to it. If the sentence contains no error, Select "No error" option. (Avoid punctuation errors).(A) The boys campaigned / (B) not only in / (C) Mumbai also in Chennai / (D) NO ERROR B D A C B D A C ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A three phase four pole 50 Hz induction motor has a rotor resistance of 0.02 Ω/phase and stand-still reactance of 0.5 Ω/phase. Calculate the speed at which the maximum torque is developed. 1475 rpm 1500 rpm 1440 rpm 1525 rpm 1475 rpm 1500 rpm 1440 rpm 1525 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Condition for maximum torque,Sm = R2/X2= 0.04Ns = (120*f)/P= (120*50)/4= 1500 rpmS = (Ns - Nr)/NsNr = Ns(1 - Sm)= 1500*(1 - 0.04)= 1440 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% 40% 25% 20% 30% 40% 25% 20% 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 %
EXAMIANS