Hello,

for 3-Phase Y-Connection the Line Voltage is,

VL=\(\sqrt{3}\)xVp

This can be proved using this equation,

Vab = Van + Vnb , looking at this circuit, how this equation is derived ?

please help

 

Hello,

for 3-Phase Y-Connection the Line Voltage is,

VL=\(\sqrt{3}\)xVp

This can be proved using this equation,

Vab = Van + Vnb , looking at this circuit, how this equation is derived ?

please help

\(V_{ab} = V_{a} - V_{b}\)

\(\text{Because, } V_{a} = V_{m} \angle 0^{o} \text{ and } V_{b} = V_{m} \angle -120^{o}, \text{ we have,}\)

\(V_{ab} = V_{m} - V_{m}(-1.5 - j0.866)\)

\( = V_{m}(1.5 - j0.866)\)

\(V_{ab} = \sqrt{3}V_{m} \angle 30^{o}\)

Similarly,

\(V_{bc} = \sqrt{3}V_{m} \angle -90^{o}\)

\(V_{ca} = \sqrt{3}V_{m} \angle -210^{o}\)

Thus the line-to-line voltage is √(3) times the phase voltage and is displace 30° in phase.

EDIT: After reading your post, I may have misinterpreted your question.

If you are asking where the equation,

Vab = Va-Vb comes from,

see the figure attached.

 

the attached figure solved my problem

thanks alot