three phase
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??
Any cross phase wiring of a WYE connection will yield sqrt(3)* higher voltage. You have shown this in your simulation. This can also be determined via phasor analysis.
The current flow in these phases will no longer be balanced, which may or may not be a concern.
As you note the motor will be running at 208V instead of it's designed 220 (which BTW is actually 110 @ 0degrees and 110 at 180 degrees). This motor is designed to take two active phases so insulation issues should be safe.
Yes, you are right.
No guarantees as to correctness, but this shows the general method.
By convention, phase rotation is anti-clockwise. An inductive load will make the current lag behind the voltage. (Remember, inductors resist changes in voltage dv/dt.)
Diagrams below are not to scale.
Best Answer
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:
$$V_{AB} = 2\times230\times\frac{\sqrt{3}}{2} = 230\sqrt{3} = V_{AN}\sqrt{3}$$