Calculating 3 phase input power

Thread Starter


Joined Aug 8, 2019
Hi just wondering if anyone can help me / reassume myself I understand three phase input power correctly.

Say I have a power supply which has a three phase input. Say 100V as example.
And this PSU is required to produce 100W (ignore efficency).
To calculate the input current per phase line.
Is it simply: I = P (100W) / sqrt(3) x V (100V)?

Does give me the Input current per phase line?

thank you


Joined Oct 6, 2013
Based on:
Power = Voltage (V) x Current (I) x Power Factor (PF) x square root of three

I come up with:
I = P / (V x Sqrt(3))

Nevermind... It's not my area and I thought I understood what I looked up, but that seems like it would be a total current instead of by phase.
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schmitt trigger

Joined Jul 12, 2010
Your equation is correct. The stated current is per phase.
And the voltage is from phase to phase.
In a balanced system, of course.
Only the power is the TOTAL power.


Joined Aug 7, 2020
You are almost correct. Your equation is correct provided that you are using the phase-to-phase voltage AND your power supply has a power factor of unity. (I'm assuming that you intended to put brackets around "sqrt(3) x V (100V")
If your power factor is not that good, the current will be higher.
If the value you have for P is actually the VA rating of the power supply, not the Power then you are correct.
Your answer will the the total current i.e. the scalar sum of the current in all three phases. The current in each phase will be a third of that.
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Joined Aug 21, 2017
Just see behind eyes the three SEPARATE lines making three separate sinks from the phase line to the GND. In such configuration You are free to add all three brutally.
Say, Your line amperage is 10 A and that is 220/380 V standard network. Then, in no respect is the load switched as 380 V "lamps" between phase of the load are 220 V "lamps" from each phase to GND, but in calculation take 220V*10A=2200W each phase and such loads are 3, thus You consume 6.6 kW. Yet the inside the black box may be used indeed the line to line load, thus the current in load will be sqrt(3) times lesser but the voltage be 380, the factual POWER will be as You calculated in beginning.