For single phase induction motor, P = VIcosφ P = 400*30*0.7 P = 8400 W Q1 = P*tan(φ) = 8400*1.020 Q = 8568.82 VAR Now power factor change to 0.9 So, φ' = 25.84 degree Q = 8400*tan(25.84) Q = 4067.95 Qnet = Q - Q' Qnet = 8568.82 - 4067.95 Qnet = 4500.8 C = V²/(2ᴨf) C = (400*400)/(2ᴨ*50) C = 8.95*10⁻⁵ C = 89.5 μF
Bus Type - Known Parameter - Unknown Parameter Load Bus -P, Q - V, phase angle Generator Bus - P, V (magnitude) - Q, Voltage phase angle Slack Bus Voltage - magnitude and phase angle - P, Q
For the given circuit shown in figure, D1 is forward bias mode - On state D2 is in reverse bias mode - off state D3 is in reverse bias mode - off state.
The series electromagnet is energized by a coil known as current coil which is connected in series with the load so that it carry the load current. The flux produced by this magnet is proportional to and in phase with the load current.