# Why can't I find Phase to phase voltage like this

I know that that in 3phase ac, Vll = sqrt(3) * Vln

I got confused about the sqrt(3) part so I draw phaser diagram and the resultant vector will give me the phase -phase voltage But it only gives me 230V , why? Can anyone explain how it got sqrt(3) in equation??

• Sure, draw a straigth line from point A to B and then calculate the resultant vector. – Marko Buršič Oct 15 '16 at 8:51

What you are calculating (the black line) isn't actually $V_{AB}$, it is rather the vector sum of $V_{AN}$ and $V_{BN}$ which is actually just $-V_{CN}$.

$V_{AB}$ is the voltage of $V_A$ referenced to $V_B$. To calculate that, you have to subtract one from the other. So the vector sum is actually:

$$V_{AB} = V_{AN} - V_{BN}$$

Doing the calculation, we get:

\begin{align}\\ V_{AB} = V_{AN} - V_{BN} &= 230\angle0^\circ - 230\angle120^\circ\\ &=398\angle30^\circ\\ &=(230\sqrt{3})\angle30^\circ\\ \end{align}

Now lets look at it in a vector diagram: Notice how the calculation you did (left) and the correct form (right) differ. You can see from the diagram that the line drawn for $V_{AB}$ is actually equal to the vector that takes us from the point $V_B$ to $V_N$ (-$V_{BN}$), and then from $V_N$ to $V_A$ ($V_{AN}$).

If you want to calculate that it is exactly $\sqrt{3}$, we can do a bit of trigonometry on the newly formed triangle: From that we can see that:

$$\frac{V_{AB}}{2} = 230\times\sin{60}$$

We know that $\sin(60) = \frac{\sqrt{3}}{2}$, so we can say directly that:

$$V_{AB} = 2\times230\times\frac{\sqrt{3}}{2} = 230\sqrt{3} = V_{AN}\sqrt{3}$$

• VAB=VAN−VBN, isn't this actually vector difference.??? Thinking back vector sum is the term that confused me – Athul Sep 7 '18 at 5:39
• @Athul, correct, all terms in that are vectors. – Tom Carpenter Sep 7 '18 at 6:14