Question

3-A balanced wye-connected load of (4 + j3) Ω is connected across a three-phase source of 173 V (line-to-line) Find the transmission line current. Find the power factor, P and Q

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Answer #1

Given the line voltage is Vi-, = 1731.

The balanced wye-connected load having an impedance on each phase is Z=(4+j3)Omega

since the load is balanced,

the line to neutral voltage across the load on each phase is

VL-N-VL-L/V/3-173/V3 = 99.88 16V

Let the voltage on phase-A load is having a zero phase.

Vi-N (a) = 99.8816200

The transmission line current on phase-A is is

Vi-N(a) _ 99.881620 ° 1, (a) = 15.9811-jl 1.9858A

I(a)iNa IL (a) = -= 19.9763<-36.870A

So, the magnitude of transmission line current in each phase is 19.9763A

The power factor is

p.fcos-36.87°0.8lagging

The apparent power on phase-A is

Sa = VL-N(a)12(a)* == 99.88 16(15.98 11+jl 1.9858) = (15962+jl 197.2)VA

The total apparent power in the three phases is

S_{3phi}=3 imesS_a=(4788.6+j3591.5)VA

The 3-phase active power is the real part of 3-phase apparent power

P_{3phi}=Re[S_{3phi}]=4788.6W

The 3-phase reactive power is the imaginary part of 3-phase apparent power

Q_{3phi}=Im[S_{3phi}]=3591.5VAr

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