Question

A balanced star load having an impedance of (20 – j30) Ω in each phase is supplied by a star-connected generator with line voltage of 400 V, 50 Hz. What is the total active power?

Answer 1

Answer

$Z=\sqrt{(20)^2+ (30)^2} =36\Omega$

Angle

$\phi=\tan^{-1}(-30/20)=-56^°$

V=400volts

Total active power

$P=\frac{V^2 \cos\phi}{Z}\\=\frac{(400) ^2\times cos(-56°) }{36} \\=\frac{160000\times0.56}{36}\\=2.5KW$