Answer to Question #110169 in Electric Circuits for xinyu

Question

Three identical coils, each of resistance 10 ohms and inductance 42 mH are connected to a 415 V , 50 Hz , 3-phase supply. Determine the total power dissipated when the loads are connected in a star-connected system

Answer 1

415 V is line voltage. The phase voltage will be $\sqrt{3}$ times lower, i.e., 240 V. Total (apparent) power in AC circuits is measured in VA and consists of true power (regular power measured in W) and reactive power (measured in VAr). Each phase has its own total power:

$S_1=\frac{U_\text{ph}^2}{Z}=\frac{U_\text{ph}^2}{\sqrt{R^2+(\omega L)^2}}.$

Three phases produce three times more power:

$S=3S_1=\frac{3U_\text{ph}^2}{\sqrt{R^2+(\omega L)^2}}=\\ \space\\=\frac{U_\text{line}^2}{\sqrt{R^2+(2\pi f L)^2}}=10.4\text{ kV⋅A}.$

We can also find the power factor:

$\text{cos}\phi=\frac{R}{\sqrt{R^2+(\omega L)^2}}=0.604.$

Using the power factor, we can find the real P and reactive Q power:

$P=S\text{cos}\phi=6.28\text{ kW},\\ Q=S\text{sin}\phi=8.29\text{ kVAr}.$

You can check that

$S=\sqrt{P^2+Q^2}.$