A three-phase balanced wye-delta system has a line voltage of 208 V rms. The total real power absorbed by the load is 1200 W. If the power factor angle of the load is 20° lagging, we wish to determine the magnitude of the line current and the value of the load impedance per phase in the delta.

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Accepted Answer

The line current can be obtained from Eq. (11.26). Since the real power per phase is 400 W,

[latex]\begin{aligned}&P_{p}=\frac{V_{L} I_{L}}{\sqrt{3}} \cos heta \&Q_{p}=\frac{V_{L} I_{L}}{\sqrt{3}} \sin heta\end{aligned}[/latex] 11.26

[latex]\begin{aligned}400 &=\frac{208 I_{L}}{\sqrt{3}} \cos 20^{\circ} \\I_{L} &=3.54 \mathrm{~A} \mathrm{rms}\end{aligned}[/latex]

The magnitude of the current in each leg of the delta-connected load is

[latex]\begin{aligned}I_{\Delta} &=\frac{I_{L}}{\sqrt{3}} \\&=2.05 \mathrm{~A} \mathrm{rms}\end{aligned}[/latex]

Therefore, the magnitude of the delta impedance in each phase of the load is

[latex]\begin{aligned}\left|\mathbf{Z}_{\Delta} ight| &=\frac{V_{L}}{I_{\Delta}} \\&=\frac{208}{2.05} \\&=101.46 \Omega\end{aligned}[/latex]

Since the power factor angle is 20° lagging, the load impedance is

[latex]\begin{aligned}\mathbf{Z}_{\Delta} &=101.46 \underline{/ 20^{\circ}} \\&=95.34+j 34.70 \Omega\end{aligned}[/latex]

Question 11.5: A three-phase balanced wye-delta system has a line voltage o...

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