# RMS value calculation of 3 phase line voltages for power calculations

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: 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; 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. 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 \$$.)

• 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.
– Berk
Apr 27 at 11:56
• 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. 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).