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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

enter image description here

But it only gives me 230V , why? Can anyone explain how it got sqrt(3) in equation??

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

Vectors for -VCN and VAB

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:

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:

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

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

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