# Currents on a three phase system

1. Nov 5, 2014

### mariano54

Hi everyone,

is it possible to sum at any given moment in time L1 + L2 + L3, algebrically speaking? If so, what would this value represent?

Thanks

2. Nov 5, 2014

### zoki85

If L1,L2,L3 are currents than their sum will represent neutral current (if there is a neutral conductor in 3- phase system, that current flows through it).
Generally, one third of the phasors sum of all 3 phasors represent zero sequence phasor (in treatment by symmetrical system components)

3. Nov 5, 2014

### mariano54

Hi, thanks for answering. The neutral current, however, is a vector sum, am I correct? What if I sum L1 + L2 + L3, just the values?

4. Nov 5, 2014

### zoki85

Values must be taken at same instant of time, then the above holds.

5. Nov 5, 2014

### mariano54

The problem I'm having is this, the documentation of the device I'm monitoring is giving me "L1-N", "L2-N", "L3-N", and all three are positive values. Maybe I'm missing what the "-N" means here. That's why I wanted to know what would I be getting by adding them up. Additionally, the vector sum is also provided as a different value, which is clearly different.

6. Nov 5, 2014

### mariano54

I'm reading that in order to calculate how much power the system is using you actually use all three phases:

current L1 * voltage L1 +. ... + currentL3 * voltageL3 (the formula is more complex, but I'm interested only on the phases)

so the currents in the single phases are relevant. Logically speaking, if I can use all three phases to calculate how much power the system is using, shouldn't I be able to add them up and say how much current is going into the system and thus used by the system?

Thanks

7. Nov 5, 2014

### zoki85

Ok
This sound very fishy/wrong to me. Show me calculation you have on mind on certain example.

8. Nov 5, 2014

### mariano54

P1 = V1 x I1
P2 = V2 x I2
P3 = V3 x I3

Total power = P1 + P2 + P3

is this formula/approach correct?

9. Nov 5, 2014

### zoki85

P1 = V1 x I1 x cos φ1
P2 = V2 x I2 x cos φ2
P3 = V3 x I3 x cos φ3

Total power = P1 + P2 + P3

Now, it is correct

10. Nov 5, 2014

### mariano54

Perfect, thanks.

11. Nov 5, 2014

### mariano54

One more thing, regarding this point. What I meant was, what happens if I sum only the current values, without takng into account the phase shift, only the magnitudes.

I mean, even if th system is perfectly balances and the neutral is zero, there's still current going in, right?

12. Nov 5, 2014

### zoki85

Why would you do that? For a balanced 3-phase system you'll get 3x higher line current and do what with it?

The phases have currents of equal magnitude. Phase shift between any two is 1200 el.

13. Nov 5, 2014