Mathematics of three-phase electric power

An important property of three-phase power is that the instantaneous power available to a resistive load, ${\displaystyle \scriptstyle P\,=\,VI\,=\,{\frac {V^{2}}{R}}}$, is constant at all times. Indeed, let

${\displaystyle {\begin{aligned}P_{Li}&={\frac {V_{Li}^{2}}{R}}\\P_{TOT}&=\sum _{i}P_{Li}\end{aligned}}}$

To simplify the mathematics, we define a nondimensionalized power for intermediate calculations, ${\displaystyle \scriptstyle p\,=\,{\frac {1}{V_{P}^{2}}}P_{TOT}R}$

${\displaystyle p=\sin ^{2}\theta +\sin ^{2}\left(\theta -{\frac {2}{3}}\pi \right)+\sin ^{2}\left(\theta -{\frac {4}{3}}\pi \right)={\frac {3}{2}}}$

Hence (substituting back):

${\displaystyle P_{TOT}={\frac {3V_{P}^{2}}{2R}}.}$

Since we have eliminated ${\displaystyle \theta }$ we can see that the total power does not vary with time. This is essential for keeping large generators and motors running smoothly.

Notice also that using the root mean square voltage ${\displaystyle V={\frac {V_{p}}{\sqrt {2}}}}$, the expression for ${\displaystyle P_{TOT}}$ above takes the following more classic form:

${\displaystyle P_{TOT}={\frac {3V^{2}}{R}}}$.

The load need not be resistive for achieving a constant instantaneous power since, as long as it is balanced or the same for all phases, it may be written as

${\displaystyle Z=|Z|e^{j\varphi }}$

so that the peak current is

${\displaystyle I_{P}={\frac {V_{P}}{|Z|}}}$

for all phases and the instantaneous currents are

${\displaystyle I_{L1}=I_{P}\sin \left(\theta -\varphi \right)}$

${\displaystyle I_{L2}=I_{P}\sin \left(\theta -{\frac {2}{3}}\pi -\varphi \right)}$

${\displaystyle I_{L3}=I_{P}\sin \left(\theta -{\frac {4}{3}}\pi -\varphi \right)}$

Now the instantaneous powers in the phases are

${\displaystyle P_{L1}=V_{L1}I_{L1}=V_{P}I_{P}\sin \left(\theta \right)\sin \left(\theta -\varphi \right)}$

${\displaystyle P_{L2}=V_{L2}I_{L2}=V_{P}I_{P}\sin \left(\theta -{\frac {2}{3}}\pi \right)\sin \left(\theta -{\frac {2}{3}}\pi -\varphi \right)}$

${\displaystyle P_{L3}=V_{L3}I_{L3}=V_{P}I_{P}\sin \left(\theta -{\frac {4}{3}}\pi \right)\sin \left(\theta -{\frac {4}{3}}\pi -\varphi \right)}$

Using angle subtraction formulae:

${\displaystyle P_{L1}={\frac {V_{P}I_{P}}{2}}\left[\cos \left(\varphi \right)-\cos \left(2\theta -\varphi \right)\right]}$

${\displaystyle P_{L2}={\frac {V_{P}I_{P}}{2}}\left[\cos \left(\varphi \right)-\cos \left(2\theta -{\frac {4}{3}}\pi -\varphi \right)\right]}$

${\displaystyle P_{L3}={\frac {V_{P}I_{P}}{2}}\left[\cos \left(\varphi \right)-\cos \left(2\theta -{\frac {8}{3}}\pi -\varphi \right)\right]}$

which add up for a total instantaneous power

${\displaystyle P_{TOT}={\frac {V_{P}I_{P}}{2}}\left\{3\cos \varphi -\left[\cos \left(2\theta -\varphi \right)+\cos \left(2\theta -{\frac {4}{3}}\pi -\varphi \right)+\cos \left(2\theta -{\frac {8}{3}}\pi -\varphi \right)\right]\right\}}$

Since the three terms enclosed in square brackets are a three-phase system, they add up to zero and the total power becomes

${\displaystyle P_{TOT}={\frac {3V_{P}I_{P}}{2}}\cos \varphi }$

or

${\displaystyle P_{TOT}={\frac {3V_{P}^{2}}{2|Z|}}\cos \varphi }$

showing the above contention.

Again, using the root mean square voltage ${\displaystyle V={\frac {V_{p}}{\sqrt {2}}}}$, ${\displaystyle P_{TOT}}$ can be written in the usual form

${\displaystyle P_{TOT}={\frac {3V^{2}}{Z}}\cos \varphi }$.

For the case of equal loads on each of three phases, no net current flows in the neutral. The neutral current is the inverted vector sum of the line currents. See Kirchhoff's circuit laws.

${\displaystyle {\begin{aligned}I_{L1}&={\frac {V_{L1-N}}{R}},\;I_{L2}={\frac {V_{L2-N}}{R}},\;I_{L3}={\frac {V_{L3-N}}{R}}\\-I_{N}&=I_{L1}+I_{L2}+I_{L3}\end{aligned}}}$

We define a non-dimensionalized current, ${\displaystyle i={\frac {I_{N}R}{V_{P}}}}$:

${\displaystyle {\begin{aligned}i&=\sin \left(\theta \right)+\sin \left(\theta -{\frac {2\pi }{3}}\right)+\sin \left(\theta +{\frac {2\pi }{3}}\right)\\&=\sin \left(\theta \right)+2\sin \left(\theta \right)\cos \left({\frac {2\pi }{3}}\right)\\&=\sin \left(\theta \right)-\sin \left(\theta \right)\\&=0\end{aligned}}}$

Since we have shown that the neutral current is zero we can see that removing the neutral core will have no effect on the circuit, provided the system is balanced. Such connections are generally used only when the load on the three phases is part of the same piece of equipment (for example a three-phase motor), as otherwise switching loads and slight imbalances would cause large voltage fluctuations.

With linear loads, the neutral only carries the current due to imbalance between the phases. Devices that utilize rectifier-capacitor front ends (such as switch-mode power supplies for computers, office equipment and the like) introduce third order harmonics. Third harmonic currents are in-phase on each of the supply phases and therefore will add together in the neutral which can cause the neutral current in a wye system to exceed the phase currents.[3][4]