Question 2.39: A balanced star-connected load of (8 + j6) Ω per phase is co...
A balanced star-connected load of (8 + j6) Ω per phase is connected to a balanced three-phase, 400 V supply. Find the line current, power factor, power, and total volt-amperes.
Phase voltage, V_{P}=\frac{Line Voltage, V_{L}}{\sqrt{3}}=\frac{400}{\sqrt{3}}=231 V
Impedence per phase, Z_{P}=\sqrt{R²+X_{L}²}=\sqrt{8²+6²}=10 Ω
Phase current, I_{P}=\frac{V_{P}}{Z_{P}}=\frac{231}{10}=23.1 A
Line current I_{L} = I_{P} = 23.1 A
Power factor \cos Φ=\frac{R}{Z}=\frac{8}{10}=0.8 (lagging)
Total power, P = \sqrt{3} V_{L} I_{L} \cosΦ
=\sqrt{3} × 400 × 23.1 × 0.8
= 12,800 W
Total volt amperes =\sqrt{3} V_{L} I_{L}
=\sqrt{3} × 400 × 23.1 =16,000 VA