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

enter image description 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.

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

3-phase currents

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.

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

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