Three equal star-connected inductors take 8 kW at 0.8 power factor when connected to a 460 V three - phase 3-wire supply. If one of the inductor in phase – Y is short circuited, then the line current in phase-Y is ___ Assume RYB sequence.

• A. 37.6 A
• B. 31.7 A
• C. 27.6 A
• D. 21.7 A

Answer: (A)

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