# 3-ph current quest

Discussion in 'General Electronics Chat' started by DYLH, Aug 13, 2013.

1. ### DYLH Thread Starter New Member

Aug 13, 2013
28
1
Hypothetical scenario
380/220v 3-phase wye system.

1-phase load (line to line on phases A B) @ 1900 VA
1-phase load (line to ground on phase A) @ 549 VA

What is the expected current on each phase? How did you arrive at it?

2. ### BillB3857 Senior Member

Feb 28, 2009
2,400
348
So what have you done or what do you think would be a way to solve your homework problem?

3. ### DYLH Thread Starter New Member

Aug 13, 2013
28
1
I've been out of school for nearly 15 years... don't recall a problem like this. We tended to just take the total VA and divide by the 3 phase voltage to get a total current.. with the assumption that the phases were all relatively balanced.

Speaking w/ people I work with, there seem to be two ways this is done to simplify for a design to figure out feeder current.. so we're trying to reconcile the the rules of thumb with the theoretical application.

Option 1
---------
Divide load per phase by the line to neutral voltage
Phase A = 3693 / 220 = 16.8 A
Phase B = 3144 / 220 = 14.3 A
Phase C = 2194 / 220 = 9.97 A

Option 2
---------

Phase A = 3693 VA
Phase B = 3144 VA
Phase C = 2194 VA

6582 common across all three
6582 / (380 * sqrt(3)) = 10.0 A for A, B, & C

1900 common to A & B
1900 / 380 = 5.0 A additional for A & B

Finally, 549 only on A
549 / 220 = 2.5

Phase A = 10 + 5.0 + 2.5 = 17.5
Phase B = 10 + 5.0 = 15
Phase C = 10

If the total VA on the phases are the same, both give the same result.

Option 2 is more conservative, and 'works' in the scenario where there is only a 5.0 A (380V 1-ph 1900 VA) load on the 3-ph service... the load would be 1900 / 380 = 5.0 A

Option 1 would result in the 380V 1-ph 1900 VA load reporting as 950 / 220 = 4.3 A... that doesn't seem right. How can a 5.0A current turn into 4.3 A without a transformer?

My guess is that neither simplification is 'correct'. What is the real theory here.. how would it work out?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
The problem requires some attention to the phase displacements of the various currents.

Ignoring the current due to the A line to neutral load for the moment.....

The total VA connected directly across the A-B phases is 6582/3 + 1900 VA or 4094 VA.

If we take the the load A-B load current as the reference phase then the A-B current due to the 4094 VA load would be 4094/380= 10.774 Amps at 0°.

Between B-C and C-A there are equal loadings of 2194 VA equating respectively to currents of 5.774 A at -120° and 5.774 A at -240°.

The line A to neutral current of 549/220 or 2.5 A will lag the A-B load current by 30°. So the total current draw from line A would be

$\text{I_A=I_{AB}-I_{CA}+I_{AN}=10.774\angle{0^o} - 5.774\angle{(-240^o)}+2.5\angle{(-30^o)}=17.02\angle(-21.55^o) \ Amps}$

The other currents can be found by similar reasoning.