# For a Y connection, why would a line-neutral current be the same as a line-line current?

For the following image of a Y-connection (taken from here)

How would a line-to-neutral phase current be the same as a line-to-line current?

Assuming I_{1N} is the phase current and I_{13} is the line current, how is it possible that I_{1N} = I_{N3} (as per original assumption),

as opposed to I_{1N} = I_{N3} + I_{N2}?

EDIT: From that same info. source,

"The terms line current and phase current follow the same logic: the former referring to current through any one line conductor, and the latter to current through any one component."

meaning for current, "line" =/= "line-to-line". So this question comes from a confusion with regards to what the terms actually mean.

• there is no "line to line" current. vector/phasor sum of the line currents is zero in that balanced system. – HelpMee Apr 12 '17 at 1:37
• But unbalanced load on reconnect big current and voltage swing. – Sunnyskyguy EE75 Apr 12 '17 at 3:33

In a wye connected 3-phase source, $I_{Line} = I_{Phase}$, because there is only one path. The current that flows in the line MUST flow in the phase.

There is no such thing as line-to-line current.

The phase voltages are out of phase by 120$^{\circ}$, which means phase currents will be out of phase by 120$^{\circ}$ for a balanced load.

The magnitude will be the same, so $I_1 = I_2 = I_3$, but $\vec{I_1} \ne \vec{I_2} \ne \vec{I_3}$, because of the phase angles.

If you do vector addition on these currents, the total is 0, which means there is no neutral current. $$\vec{I_N} = \vec{I_1} + \vec{I_2} + \vec{I_3} = 0$$

Because of the 120$^{\circ}$ phase shift, ac current will be flowing out and back to the sources at different times.

When maximum positive current flows out on $I_2$ (blue) (as shown in the drawing), $I_1$ (red) and $I_3$ (green) are 50% negative. Adding the instantaneous values at any instant in time will give 0.

If the loads are unbalanced, the unbalanced current will flow back to the sources over the neutral wire.

$$\vec{I_N} = \vec{I_1} + \vec{I_2} + \vec{I_3} \ne 0$$