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.
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 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
Let, E→EMF in volt φ→Flux in weber A→Parallel path (for wave winding = 2) Z→No. of armature conductors N→speed in rpm P = Number of poles E = PφNZ/(60A) E = 8*0.020*625/(60*2) E = 500 V
EXAMIANS