# Single phase circuit on 3 phase system calc

Discuss Single phase circuit on 3 phase system calc in the Electrical Forum area at ElectriciansForums.net

#### gib8319

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Hi,

I have a query about calculating current in a 3 phase system. So a 22kw load in 3 phase assuming a power factor of 0.8 would be 22000/(1.732x400x0.8) = 39.69A

and say a 7.2kw load on a single phase supply would be 31.30A if you simply use ohms law.

But what happens when you calculate the current in a single phase circuit on a 3 phase system. I.e a single phase load connected to L1 (or L2 or L3) and neutral? Do you use the single phase current calc or the 3 phase current calc?. and why?

#### Strima

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Arms
Esteemed
Use the single phase calculation.

#### Pete999

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Arms
Esteemed
Hi,

I have a query about calculating current in a 3 phase system. So a 22kw load in 3 phase assuming a power factor of 0.8 would be 22000/(1.732x400x0.8) = 39.69A

and say a 7.2kw load on a single phase supply would be 31.30A if you simply use ohms law.

But what happens when you calculate the current in a single phase circuit on a 3 phase system. I.e a single phase load connected to L1 (or L2 or L3) and neutral? Do you use the single phase current calc or the 3 phase current calc?. and why?
Most SP&N are derived from a TP4 wire supply so use SP &N calculations, out of interest are you a Trainee??

#### gib8319

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Hi Pete, yes new to the industry. It's one of those questions the textbooks don't really answer and I'm now studying 3 phase. So to clarify, if I have a 3 phase DB and put a single phase circuit with 5kw on L1, L2 or L3 it's exactly the same current calculation as if it was a single phase installation?.

Thanks for your response, new to the site and really appreciate your help.

#### Andy78

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Mentor
Arms
Esteemed
Hi Pete, yes new to the industry. It's one of those questions the textbooks don't really answer and I'm now studying 3 phase. So to clarify, if I have a 3 phase DB and put a single phase circuit with 5kw on L1, L2 or L3 it's exactly the same current calculation as if it was a single phase installation?.

Thanks for your response, new to the site and really appreciate your help.
The voltage would not be three phase 400V for the circuit you are doing the calc for so you wouldn't use the three phase 400V calc.

What you do need to pay attention to is load balancing across the DB when adding single phase circuits to three phase systems.

#### Pete999

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Arms
Esteemed
Hi Pete, yes new to the industry. It's one of those questions the textbooks don't really answer and I'm now studying 3 phase. So to clarify, if I have a 3 phase DB and put a single phase circuit with 5kw on L1, L2 or L3 it's exactly the same current calculation as if it was a single phase installation?.

Thanks for your response, new to the site and really appreciate your help.

#### Lucien Nunes

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Mentor
Arms
Esteemed
Yes. As mentioned above. a single-phase installation using one line and neutral, is simply one third of a three-phase installation. You have slightly added to the confusion by including power-factor in the 3-phase equation but omitting it from the single (it's the same for both) and mentioning Ohm's law, which deals with resistance, not power.

There is actually no difference in the way current and power are calculated for single phase and balanced three-phase. The 3-phase equation looks different because:
a) it deals with the power supplied by all three phases at once, and
b) when describing a 3-phase system it is conventional to quote the line-line voltage (400V in the UK) which is √3 (1.732) times the line-neutral voltage (230V in the UK). Those two voltages come from the same system, it's just a matter of where you measure.

So we can take the simple single-phase power equation P = I x V, where V refers to the line-neutral voltage, and convert it into the three-phase equation. We'll ignore power-factor for now, you can include it in either single or 3-phase calculations if it's applicable.

P = I x V (where P is the power supplied by one phase, V is the line-neutral voltage)
Now substitute V (line-line) /√3 for V (line-neutral)
P = I x V (line-line) / √3
And multiply by three so that P becomes the total power, not just of one phase
P = 3 x I x V (line-line) / √3
But 3 / √3 = √3, so that cancels:
P= √3 x I x V(line-line)
Which rearranges to:
I = P / (√3 x V (line-line))
Which on a UK supply becomes
I = P / (1.732 x 400)
I.e. what you stated to begin with for 3-phase systems.

So it's the same calculation, just altered to make more sense when dealing with three phases at once.

Mod