I'm trying to learn a little bit more about three phase current, power and KVA. We currently have some hi-tech UPS power systems at work and I'm trying to ensure that they are adequately provisioned to cope with a power outage. The UPS's are rated to about 200KVA. Power System
For simplicity imagine that two distribution boards are connected to the UPS. Each distribution board is capable of recording the current of each of the three phases.

Switch Board 1
I1=62.62 Amps
I2=67.4 Amps
I3=53.74 Amps

Switch Board 2
I1=60.52 Amps
I2=69.2 Amps
I3=54.44 Amps

We are in Aus where the devices typically run at 230-240V. Take the higher voltage please 240.

How do I use the known currents of each phase and switchboard to calculate:
A) The KVA of Each switchboard
B) The Power of each switchboard in KW

Please show any calculations in the solution. If I'm missing any vital information please ask and I'll attempt to find out more.

  • \$\begingroup\$ I assume the loads are Wye connected? \$\endgroup\$ Apr 11, 2014 at 7:32
  • \$\begingroup\$ @freyyr I'm really not sure, this isn't really my area of expertise but for now assume its a WYE connection not delta. If your keen you can put me on the right path to solve for both. \$\endgroup\$
    – Peter H
    Apr 11, 2014 at 10:02
  • \$\begingroup\$ What is the output of the UPS? 240 V AC? \$\endgroup\$ Apr 14, 2014 at 5:17

1 Answer 1


a) The KVA, or the apparent power is the product of the current and the voltage. For three-phase systems you can imagine it like a superposition. You have three currents comming in(or out) and each is at 240V(phase to ground). So the total apparent power would be: $$S=U_{phase}I_1+U_{phase}I_2+U_{phase}I_3$$ Where \$U_{phase}=240V. \$

b) The power is calculated as: $$P=S\cos\varphi$$ Where the phase difference between voltage and current is the angle \$\varphi\$, and \$\cos\varphi\$ is called power factor.

It is not possible to calculate the power \$P\$ (in kW) from the given data. See what other readings you can get.

There are far more details to it, now we are just looking at the first harmonic. You can look into the other stuff when you get this done.


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