# How to find the apparent power absorbed by a three phases load?

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\$$

• 300^2 / |sqrt(2)(1-j)| * 3 = 135 KVA ! – X J Jan 22 at 17:07
• It seems I'm having a little bit trouble getting to 135kVA. Where does the immaginary term go? – Kevin Jan 22 at 17:09
• Magnitude of (1-j) is sqrt(2), then the denominator is 2. Does this make sense? – X J Jan 22 at 17:13
• Yes, totally. Thanks! – Kevin Jan 22 at 17:17

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: $$P_{total, ap} = 3 \cdot P_{ap}=135~\text{kVA}$$