# 5.4: Multiphase Power

- Page ID
- 9964

The electrical service to your home is a two-phase service.^{1} This means that two 110 volt, 60 Hz lines, plus neutral, terminate in the panel. The lines are π radians (180) out of phase, so we can write them as

\[ \begin{align*} x_1(t) &=110\cos[2π(60)t+φ] \\[4pt] &=\mathrm{Re}\{110e^{j[2π(60)t+φ]}\} \\[4pt] &=\mathrm{Re}\{X_1e^{j2π(60)t}\} \end{align*}\]

\[X_1=110ej^{φ}\]

\[ \begin{align*} x_2(t) &=110\cos[2π(60)t+φ+π] \\[4pt] &=\mathrm{Re}\{110e^{j[2π(60)t+φ+π]}\} \\[4pt] &=\mathrm{Re}\{X_2ej^{2π(60)t}\} \end{align*}\]

\[X_2=110e^{j(φ+π)}\]

These two voltages are illustrated as the phasors \(X_1\) and \(X_2\) in the Figure.

You may use \(x_1(t)\) to drive your clock radio or your toaster and the difference between \(x_1(t)\) and \(x_2(t)\) to drive your range or dryer:

\[x_1(t)−x_2(t)=220\cos[2π(60)t+φ]\]

The phasor representation of this difference is

\[X_1−X_2=220e^{jφ}\]

The breakers in a breaker box span the \(x_1\)−to-neutral bus for 110 volts and the \(x_1-\mathrm{to}-x_2\) buses for 220 volts.

Exercise \(\PageIndex{1}\)

Sketch the phasor \(X_1−X_2\) on the Figure.

Most industrial installations use a three-phase service consisting of the signals \(x_1(t)\),\(x_2(t)\), and \(x_3(t)\):

\(x_n(t)=110\mathrm{Re}\{e^{j[ω_0t+n(2π/3)]}\}↔X_n=110e^{jn(2π/3)},n=1,2,3\)

The phasors for three-phase power are illustrated in the Figure.

Exercise \(\PageIndex{2}\)

Sketch the phasor \(X_2−X_1\) corresponding to \(x_2(t)−x_1(t)\) on Exercise. Compute the voltage you can get with \(x_2(t)−x_1(t)\). This answer explains why you do not get 220 volts in three-phase circuits. What do you get?

## Constant Power

Two- and three-phase power generalizes in an obvious way to N-phase power. In such a scheme, the N signals \(x_n(n=0,1,...,N−1)\) are

\[ \begin{align*} x_n(t) &=A\cos(ωt+\frac {2π} {N} n) \\[4pt] &=\mathrm{Re}[Ae^{j2πn/N}e^{jωt}]↔X_n=Aej^{2πn/N} \end{align*}\]

The phasors \(X_n\) are \(Ae^{j2π(n/N)}\). The sum of all \(N\) signals is zero:

\[\begin{align*} \sum^{N−1}_{n=0}x_n(t) &= \mathrm{Re}\{A∑^{N−1}_{n=0} e^{j2πn/N} e^{jωt} \} \\[4pt] &=\mathrm{Re}{A\frac {1−ej^{2π}} {1−e^{j2π/N}} e^{jωt}} \\[4pt] &=0 \end{align*}\]

But what about the sum of the instantaneous powers? Define the instantaneous power of the n^{th} signal to be

\[ \begin{align*} p_n(t) &=x^2_n(t) &=A^2\cos^2(ωt+\frac{2π}{N} n) \\[4pt] &=\frac {A^2}{2} + \frac{A^2}{ 2} \cos(2ωt + 2\frac {2π} {N} n) \\[4pt] &=\frac {A^2} {2} + \mathrm{Re}\{\frac {A^2} 2 e^{j(2π/N)2n} e^{j2ωt}\} \end{align*}\]

The sum of all instantaneous powers is

\[P=\sum_{n=0}^{N−1}p_n(t) = N\frac {A^2} 2 \]

and this is independent of time!

Exercise \(\PageIndex{3}\)

Carry out the computations of Equation 5.4.16 to prove that instantaneous power P is constant in the N-phase power scheme.

## Footnotes

- It really is, although it is said to be “single phase” because of the way it is picked off a single phase of a primary source. You will hear more about this in circuits and power courses