In my textbook in the power calculations section of the balanced three phase circuits, a part confused me a little bit, it is this part:

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Why are we dividing by \$\sqrt{3}\$ to find RMS values of \$V_{\phi}\$ and \$I_{\phi}\$? As far as I know, we are dealing with sinusoidal sources and in sinusoidal sources, RMS transformation was done by diving to \$\sqrt{2}\$. In the same textbook in the previous chapter, power calculation for sinusoidal sources were shown like this;

enter image description here

I didn't understand where this \$\sqrt{3}\$ come from in the 3-phase power calculation.


You're not doing an RMS transformation. You're converting L-N (RMS) to L-L (RMS).

You can use either but you have to remember that the phase to phase voltage is \$ \sqrt{3} \$ times the phase to neutral voltage.

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Figure 1. The phasor 3-phase and neutral diagram.

The \$ \sqrt{3} \$ term just comes from the trigonometric relationship between the voltages in Figure 1. (Remember that the \$sin(60) = \frac {\sqrt 3} 2 \$.)

  • \$\begingroup\$ Oh, so every source magnitude we use already are RMS values,I get it now. I already knew about the line to line and line to neutral conversion but I guess the explanation in the textbook confused me to think there was an RMS transformation, thanks. \$\endgroup\$
    – Berk
    Apr 27 at 11:56
  • 1
    \$\begingroup\$ Good. Wait a while to see if any other answers come in. They might give you some other insights. Upvote any useful ones and accept one that answers your original question. \$\endgroup\$
    – Transistor
    Apr 27 at 12:26

The \$\sqrt {3}\$ comes from the math when converting from line to phase quantities for wye or delta connected loads (or sources).

For any three-phase load, the total power is sum of the power consumed per phase.

$$P_T = P_{\phi_A} + P_{\phi_B} + P_{\phi_C}$$ $$P_T = V_{\phi} \ I_{\phi_A}\ cos \theta_{\phi_A} + V_{\phi} \ I_{\phi_B}\ cos \theta_{\phi_B} + V_{\phi} \ I_{\phi_C}\ cos \theta_{\phi_C}$$

For a balanced load, this simplifies to:

$$P_T = 3\ P_{\phi}$$

$$P_T = 3\ V_{\phi} \ I_{\phi}\ cos \theta_{\phi}$$

In a wye connected load, line current equals phase current \$(I_L = I_\phi)\$, but line voltage is larger \$(V_L = \sqrt {3}\ V_\phi)\$ due to two sources, which means:

$$P_T = 3\ \frac {V_{L}} {\sqrt {3}} \ I_{L}\ cos \theta_{\phi}$$ $$P_T = \sqrt {3}\ V_{L} \ I_{L}\ cos \theta_{\phi}$$

For a Delta connected load, line voltage equals phase voltage \$(V_L = V_\phi)\$. Line current comes from two phases, so it is larger \$(I_L = \sqrt {3}\ I_\phi)\$. This means:

$$P_T = 3\ V_{L} \ \frac {I_{L}} {\sqrt {3}}\ cos \theta_{\phi}$$ $$P_T = \sqrt {3}\ V_{L} \ I_{L}\ cos \theta_{\phi}$$

This proves the universality of \$P_T = \sqrt {3}\ V_{L} \ I_{L}\ cos \theta_{\phi}\$ or \$P_T = 3\ V_{\phi} \ I_{\phi}\ cos \theta_{\phi}\$ for wye or delta connected loads.

Some motors can be connected in wye/delta, so they have info on nameplates, but you usually don't know how a load or source is connected internally (and it doesn't really matter).

Education tends to approach it from a phase to neutral (wye) or phase to phase (delta) basis, but most calcs are done from a line-to-line perspective (cause that can be measured at the terminals of the device).


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