a) Calculate the total complex power delivered to the -connected load of Example 11.2.

b) What percentage of the average power at the sending end of the line is delivered to the load?

Question

a) Calculate the total complex power delivered to the -connected load of Example 11.2.

b) What percentage of the average power at the sending end of the line is delivered to the load?

Accepted Answer

a) Using the a-phase values from the solution of Example 11.2, we obtain

[latex]V_{\phi} =\pmb{V}_{AB} = 202.72 \angle 29.04^{\circ}[/latex] V,

[latex]I_{\phi} = \pmb{I}_{AB} = 1.39 \angle -6.87^{\circ}[/latex] A.

Using Eqs. 11.52 [latex]S_{\phi} = p_{\phi} + jQ_{\phi} = V_{\phi}I^{*}_{\phi}[/latex],and 11.53[latex]S_{T}=3S_{\phi}=\sqrt{3}V_{L}I_{L}\angle heta_{\phi}[/latex], we have

[latex]S_{T} = 3\left(202.72 \angle 29.04^{\circ} ight)\left(1.39 \angle 6.87^{\circ} ight)[/latex]

= 682.56 + j494.21 VA.

b) The total power at the sending end of the distribution line equals the total power delivered to the load plus the total power lost in the line;

therefore

[latex]p_{input}= 682.56+ 3\left(2.4 ight)^{2}\left(0.3 ight)[/latex]

= 687.74 W.

The percentage of the average power reaching the load is 682.56/ 687.74 , or 99.25% Nearly 100% of the average power at the input is delivered to the load because the impedance of the line is quite small compared to the load impedance.

Question 11.4: a) Calculate the total complex power delivered to the -conne...

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