Related Question
Here, Y phase is shorted,
Equation of 3-phase power = √3*VL*IL*cosφ
IL = P/(√3*VL*cosφ)
= 8000/(√3*400*0.8)
= 12.55 A
Phase impedance = Vp/Ip
= 460/(√3*12.55)
= 21.16 ohm
cosφ = 0.8, So φ = 37 degree.
Here Impedance of Y phase is shorted,
So, Neutral point and Y phase at same potential,
Three line voltages are given by,
Vry = 460∟0°
Vyb = 460∟120°
Vbr = 460∟-120°
Here N and Y phase at same potential,
Ir = (460∟-120°)/(21.16∟37°)
= -21.7∟83° A
Ib = (-460∟120°)/(21.16∟37°)
= 21.7∟-157° A
Ib = -(Ir + Iy)
= -21.7∟83° + 21.7∟-157°
= 37.6∟53° A
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