Electric systems with more than one phase are called polyphase systems. Polyphase systems have multiple applications, particularly in power delivery, in which the transport of power is performed more efficiently over polyphase power lines than single phase lines. Among all the polyphase systems, two-phase and three-phase systems are the most common, but there are certain specialized applications in which a higher number of phases are used. Below, we introduce two-phase and three-phase systems by focusing on their general properties and mathematical description.

Two-phase systems use a pair of voltage sources with four separate wires to transport power. The phases of the voltage sources are shifted by $90^\circ$ from each other and, if the system is balanced (i.e., the magnitudes of the voltage sources are equal) $$\begin{equation}V_1(t)= V \cos(\omega t)\end{equation}$$ $$\begin{equation}V_2(t)= V \cos\left(\omega t - \frac{\pi}{2}\right)\end{equation}$$ or, in complex form $$\begin{equation}V_1 = V \angle{0^\circ} \end{equation}$$ $$\begin{equation}V_2 = V \angle{-90^\circ}\end{equation}$$ If the loads are also balanced, the currents produced by the sources are $$\begin{equation}I_1(t)= I \cos(\omega t-\psi)\end{equation}$$ $$\begin{equation}I_2(t)= I \cos\left(\omega t - \frac{\pi}{2}-\psi\right)\end{equation}$$ where $\psi$ is the angle difference between voltage and current.

The instantaneous power in a two-phase system is equal to $P(t)=P_1(t)+P_2(t)=V_1(t)I_1(t)+V_2(t)I_2(t)$, which can be shown to equal $$\begin{equation}P(t)= VI \cos\psi \end{equation}$$ Note that the instantaneous power in a balanced two-phase system is constant; the system therefore avoids the power flow variations caused by AC voltages and currents.

Three-phase AC circuits are widely used to transport power from generation plants (for example, nuclear, hydro, or wind facilities) to consumers. To decrease Ohmic losses in the wires, power is transmitted at high voltages and low currents, and transformers are often used to step the voltage up or down.

A typical arrangement of the voltage sources in a three-phase system is the wye configuration shown in Fig. 2. If the sources are balanced $$\begin{equation}V_{an}(t) = V \cos(\omega t)\end{equation}$$ $$\begin{equation}V_{bn}(t) = V \cos(\omega t-120^\circ)\end{equation}$$ $$\begin{equation}V_{cn}(t) = V \cos(\omega t+120^\circ)\end{equation}$$ where subscript $n$ stands for the neutral node. In complex form $$\begin{equation}V_{an} = V \angle{0^\circ} \end{equation}$$ $$\begin{equation}V_{bn} = V \angle{-120^\circ} \end{equation}$$ $$\begin{equation}V_{cn} = V \angle{120^\circ} \end{equation}$$ If the load is also balanced, the currents produced by the sources are $$\begin{equation}I_{an}(t)= I \cos(\omega t-\psi)\end{equation}$$ $$\begin{equation}I_{bn}(t)= I \cos\left(\omega t - 120^\circ-\psi\right)\end{equation}$$ $$\begin{equation}I_{cn}(t)= I \cos\left(\omega t + 120^\circ-\psi\right)\end{equation}$$ where $\psi$ is the angle difference between voltage and current, which depends on the load attached to the system.

Another typical arrangement of the voltage sources in a three-phase system is the delta configuration shown in Fig. 3. If the sources are balanced $$\begin{equation}V_{ab}(t) = V \cos(\omega t)\end{equation}$$ $$\begin{equation}V_{bc}(t) = V \cos(\omega t-120^\circ)\end{equation}$$ $$\begin{equation}V_{ca}(t) = V \cos(\omega t+120^\circ)\end{equation}$$ or, in complex form $$\begin{equation}V_{ab} = V \angle{0^\circ} \end{equation}$$ $$\begin{equation}V_{bc} = V \angle{-120^\circ} \end{equation}$$ $$\begin{equation}V_{ca} = V \angle{120^\circ} \end{equation}$$ Notice that in both the wye and delta configurations, the voltages of the sources are shifted by $120^\circ$ from each other.

The instantaneous power in a three-phase system is equal to $P(t)=P_{an}(t)+P_{bn}(t)+P_{cn}(t)=V_{an}(t)I_{an}(t)+V_{bn}(t)I_{bn}(t)+V_{cn}(t)I_{cn}(t)$, which can be shown to be equal to $$\begin{equation}P(t)= \frac{3VI}{2} \cos\psi \end{equation}$$ Although this equation was obtained for a delta-connected source, a similar result holds for wye-connected sources. Finally, the current through the neutral wire of a balanced three-phase system is zero $$\begin{equation}I_n = I_{an}+I_{bn}+I_{cn} = I\cdot \left(\angle{(-\psi)} + \angle{(120^\circ-\psi)} + \angle{(-120^\circ-\psi)}\right) = 0 \end{equation}$$ (the last equation can be proved using simple trigonometric formulas).

Comparing the last equation with the equation of the instantaneous power in two-phase systems, we see that three-phase systems are able to transport about 50% more power than two-phase systems using the same wire mass at the same voltage. For this reason, three-phase systems have replaced two-phase systems in most commercial distribution applications of electrical energy; however, two-phase circuits are still found in certain specific applications.

• Basic calculations in three-phase systems
  Three-phase (Y) system with given phase voltages
  Three-phase (D) system with given line voltages