Given a symmetric and balanced three-phases circuit, I would like to calculate the apparent power absorbed a three-phases delta load. I’m given the rms value of the line voltage applied to the load, which is 300V, and \$Z_\Delta = \sqrt{2}(1-j)\$. How can I find the apparent power absorbed by the three phases load? I suppose the line voltage is equal to the phase voltage, so I calculate the power as \$P=\frac{V^2}{Z_\Delta}\$ and then I multiply it by 3, but I don’t think it’s right since the solution is \$135kVA\$
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\$\begingroup\$ 300^2 / |sqrt(2)(1-j)| * 3 = 135 KVA ! \$\endgroup\$– X JJan 22, 2020 at 17:07
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\$\begingroup\$ It seems I'm having a little bit trouble getting to 135kVA. Where does the immaginary term go? \$\endgroup\$– KevinJan 22, 2020 at 17:09
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\$\begingroup\$ Magnitude of (1-j) is sqrt(2), then the denominator is 2. Does this make sense? \$\endgroup\$– X JJan 22, 2020 at 17:13
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\$\begingroup\$ Yes, totally. Thanks! \$\endgroup\$– KevinJan 22, 2020 at 17:17
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1 Answer
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The current in each phase will be: $$I = \frac{V_{rms}}{Z_{\Delta}} = 150 \angle{45°}~ \text{A}\longrightarrow I_{rms} = 150~\text{A} $$
Apparent power is: $$ P_{ap} = V_{rms} \cdot I_{rms} = 45~\text{kVA} $$
And the total apparent power is: $$ P_{total, ap} = 3 \cdot P_{ap}=135~\text{kVA}$$
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\$\begingroup\$ P_ap can (maybe should?) be designated by the letter S. Active power is P, apparent is S \$\endgroup\$ Dec 29, 2021 at 18:39