A balanced three phase source with vL=240 V rms is supplying 8 kVA at 0.6 powder factor lagging to two wye connected parallel loads. If one load draws 3kW at unity powder factor , calculate impedence per phase of the second load.

2.35 + j8.34 Ω

Explanation:

Voltage = V = 240 V rms

supplying power = S = 8 kVA

power factor = pf = 0.6

Let P₁ represents one load draws 3kW at unity powder factor

The power angle is:

θ = cos⁻¹  pf = cos⁻¹  0.6 = 53.13°

Complex power supplied source is:

S =  S < θ = 8<53.13° kVA

Complex power for first load:

S₁ = P₁ = 3kVA

Since the power angle of first load is  θ₁ = 0°

According to principle of conservation of AC power, the power of second load is:

S₂ =  S - S₁

    = 8<53.13° - 3

    = 6.65<74.29° kVA

Since the second load is a Y connected load the phase voltage:

V =  V /

    = 240/1.732051

    = 138.564

    = 138.56 V

Complex power of second load:

S₂ = 3 V² / Z

impedance per phase of the second load:

Z =  3 V² / S₂

   = 3 (138.56)² /  6.65<74.29°

   = 3(19198.8736) / 6.65<74.29°

   = 57596.6208 / 6.65<74.29°

Z = 2.35 + j8.34Ω

answer:

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