# Line voltage and phase voltage in three phase electricity

This page gives a definition of phase and line voltage in three phase electricity, and gives the relationship between the rms values of phase voltage and line voltage.

In three phase electricity, there are three separate voltages oscillating at the same frequency, but with each one oscillating out of step with the other two. If the first voltage is $V_0\cos\theta$, the second voltage oscillates as $V_0\cos(\theta+120^\circ)$, and the third voltage oscillates as $V_0\cos(\theta+240^\circ)$.

The **phase voltage** is the voltage difference between neutral and
a single phase. The **line voltage** is the voltage difference
between two of the phases.

If the phase voltage oscillates as $V_0\cos\theta$ then its root mean square (rms) voltage over one cycle, from $\theta=0$ to $2\pi$, is

$$ \begin{align} V_{\rm rms} &=\sqrt{{1\over2\pi}\int_{0}^{2\pi} \left(V_0 \cos\theta\right)^2d\theta} \\ &=V_0\sqrt{{1\over2\pi}\int_{0}^{2\pi}{1\over2}(1+\cos2\theta)d\theta} \\ &={V_0\over\sqrt2} \end{align} $$The integral is simply ${\pi}$ because the term $\cos2\theta$ evaluates to zero over this range.

If the phase voltage of line A is $V_0\cos\theta$, and the phase voltage of line B is $V_0\cos(\theta+2\pi/3)$, then the voltage between lines A and B is

$$ V_{\rm AB} = V_0\cos\theta - V_0\cos(\theta+2\pi/3) $$and the rms value of $V_{\rm AB}$ is

$$ \begin{align} V_{\rm AB\ rms} &= \sqrt{{1\over2\pi}\int_{0}^{2\pi} V_{\rm AB}^2 d\theta}, \\ &= V_0\sqrt{{1\over2\pi}\int_0^{2\pi}\left(\cos\theta-\cos(\theta+2\pi/3)\right)^2d\theta}. \end{align} $$Now

$$ \begin{align} \cos(\theta+2\pi/3) &= \cos\theta\cos(2\pi/3)-\sin\theta\sin(2\pi/3) \\ &= -1/2\cos\theta+\sqrt3/2\sin\theta \end{align} $$so the integral becomes

$$ \begin{align} \int_0^{2\pi} \left({9\over4}\cos^2\theta -{\sqrt3\over2}\sin\theta\cos\theta +{3\over4}\sin^2\theta\right) d\theta &=\int_0^{2\pi} \left({3\over4}+{3\over4}(1-\cos2\theta)-{\sqrt3\over4}\sin2\theta\right)d\theta \\ &=\int_{0}^{2\pi}{3\over2}d\theta \end{align} $$since the trigonometric terms in $\cos2\theta$ and $\sin2\theta$ vanish over the range, giving

$$ V_{\rm AB\ rms} = V_0\sqrt{3\over2} $$Thus the RMS line voltage is $\sqrt3$ times the RMS phase voltage.

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