- #1

gruba

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## Homework Statement

Find active power of the given three-phase system.

[itex]U_{RS}=U_{ST}=U_{TR}=220\sqrt 3 V[/itex]

[itex]R=\omega L=\frac{1}{\omega C}=10\Omega[/itex]

**2. The attempt at a solution**

After reducing the circuit to one phase (see attachments), equivalent impedance is [itex]\underline{Z_e}=\sqrt{101}e^{-j84,3^{o}}[/itex].

[tex]U_{RS}=U_{ST}=U_{TR}=U_{L}=\sqrt{3}U_R\Rightarrow U_R=220 V\Rightarrow \underline{U_R}=220e^{-j\pi/3}\Rightarrow \underline{I_R}=\frac{220}{\sqrt{101}}e^{j54,3^{o}}[/tex]

Active power of each resistor is given by

[tex]P_1=P_2=P_3=\mathfrak{R}(\underline{U_L}\underline{I_R^{*}})=\sqrt 3 U_RI_R\cos\angle(\underline{U_L}\underline{I_R})\Rightarrow P=3P_1[/tex]

[tex]\cos\angle(\underline{U_L}\underline{I_R})\approx 0.099\Rightarrow P_1\approx 825.88 W\Rightarrow P=2477.64W[/tex]

Is this correct?