# What is the difference between a rupture and the neutral conductor not existing from the start in star three phased system?

I have a pretty basic question.

In three phased systems, I know that if the loads are equal, the neutal conductor has no role and can be removed, my question concerns unequal loads.

My question is, In case of the rupture of the neutral conductor, When calculating the new voltage in each load, we apply the formula $$\ V_i=Vs_i-U_n \$$ where $$\U_n \$$ has its own formula with $$\ z_n\rightarrow +\infty \$$, However, If the neutral conductor didn't exist from the start, I would consider each load in series with another load of the system and apply the voltage divider to calculate the actual voltages with the difference of two phases as my main voltage (Am I right ?).

My question is, What is the difference between it rupturing and it not existing from the start? I understand that, before the rupture, $$\ U_n \$$ existed while it didn't at all in the second situation, but in case of a rupture, why don't I consider them in series just like I did with the second situation, I know both situations are dangerous since they give wrong voltages and can be dangerous however, They don't give the same voltage values so they obviously aren't the same(?).

The problem may lie in my severe weakness in basic electrical concepts.

• I'm neither following your formulas nor following your unclear question. I follow your last sentence and here possibly. lies the truth! – Andy aka Nov 23 '18 at 18:26

rupture
/ˈrʌptʃə/
verb
1. (especially of a pipe or container, or bodily part such as an organ or membrane) break or burst suddenly.
"if the main artery ruptures he could die"

Rupture usually refers to pipes or pressure vessels. I suspect this is a translation issue (but there is no location information in your user profile) and that a better word would be "break"

My question is, What is the difference between it rupturing and it not existing from the start?

There is none. They would give the same result.

I understand that, before the rupture, Un existed while it didn't at all in the second situation, ...

If the neutral is connected then Un should be 0 V.

... but in case of a rupture, why don't I consider them in series just like I did with the second situation, ...

Without a diagram it is not possible to understand this part. simulate this circuit – Schematic created using CircuitLab

Figure 1. The effect of unbalancing the load on a star (wye) connected 3-phase load with broken neutral.

I know both situations are dangerous since they give wrong voltages and can be dangerous however, They don't give the same voltage values so they obviously aren't the same(?).

My Figure 1 may help.

• Once the neutral connection is lost the star point can be pulled in the direction of the biggest load. In (a) the load is balanced so the star point will stay close to neutral voltage which should be 0 V.
• In (b) I have halved the value of R1 so it "pulls" the star point towards L1.
• Now L1-n (we'll use a little 'n' to signify the broken neutral point) is reduced so that load will not receive full power.
• Meanwhile R2 and R3 now have a higher voltage on them and, as you say, this may be dangerous - for the loads at least.

Does that help?