Power Systems The per unit impedance of a circuit element is 0.30. If the base kV and base MVA are halved, then the new value of the per unit impedance of the circuit element will be 0.30 0.60 0.0060 0.0030 0.30 0.60 0.0060 0.0030 ANSWER DOWNLOAD EXAMIANS APP
Power Systems The function of reflector is to bounce back most of the neutrons that escape from the fuel core. none of above. reduce the speed of the neutrons. stop the chain reaction. bounce back most of the neutrons that escape from the fuel core. none of above. reduce the speed of the neutrons. stop the chain reaction. ANSWER DOWNLOAD EXAMIANS APP
Power Systems If the percentage reactance of an element is 20 % and the full load current is 50 Amp, the short circuit current will be 350 Amp. 200 Amp. 300 Amp. 250 Amp. 350 Amp. 200 Amp. 300 Amp. 250 Amp. ANSWER DOWNLOAD EXAMIANS APP
Power Systems In a Gauss Seidel load flow method, the number of iterations may be reduced if the correction in voltage at each bus is multiplied by Lagrange multiplier Gauss constant Blocking factor Acceleration factor Lagrange multiplier Gauss constant Blocking factor Acceleration factor ANSWER DOWNLOAD EXAMIANS APP
Power Systems Which of following generating plants has the minimum operating cost Nuclear plant. Diesel plant. Hydroelectric plant. Steam plant. Nuclear plant. Diesel plant. Hydroelectric plant. Steam plant. ANSWER DOWNLOAD EXAMIANS APP
Power Systems Consider a three phase, 50 Hz, 11 kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self capacitance of the insulators is 5 times the shunt capacitance between the link and ground. Find the voltage across the two insulators V1 and V2? (V1 = Voltage of disc nearest to the conductor) V1 = 3.46 kV and V2 = 2.89 kV V1 = 6 kV and V2 = 5 kV V1 = 5 kV and V2 = 6 kV V1 = 2.89 kV and V2 = 3.46 kV V1 = 3.46 kV and V2 = 2.89 kV V1 = 6 kV and V2 = 5 kV V1 = 5 kV and V2 = 6 kV V1 = 2.89 kV and V2 = 3.46 kV ANSWER DOWNLOAD EXAMIANS APP
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