I have an electrical question for a project I am working on. Let’s say, we have a heat pump drawing current, at an electrical power measured from the IoTa of 3kW, PF = 0.9, Voltage_rating = 240 V, and this HP is wired through a two-pole 60 A amp breaker. How do we go about computing the total real current draw from this heat pump? My first thought would be to use:

I = P/V *(1/PF) (which I presume IoTa does directly when you ask for current).

However, I am confused with how the two poles go into action. Does the heat pump in reality draw I = 2 * I_pole where each pole (or leg?) current is given from the above equation? Or is there some other way to compute this?

I don’t know what you mean by total “real” current. Current is current. It is measured directly by the CT as RMS Amps. When you create an output with units Amps or plot a circuit using Graph+ with units Amps, you get RMS Amps.

If you are measuring a 240V two-wire circuit (two-pole breaker) with one CT and “double” checked. The Amps indicated for the circuit are correct.

If you are measuring a 240V two-wire circuit using two CTs, one on each conductor, the Amps should be the same (current is constant throughout a circuit) and is the Amps of the circuit.

If you are measuring a 240V three-wire circuit using two CTs, you are actually measuring two 120V circuits. The current in each one is as measured by its corresponding CT.

RMS Amps do not take into consideration the direction of current flow with respect to voltage and as such includes the current that is both active and reactive. IoTaWatt measures voltage and current directly. It integrates voltage and current to determine Watts (real power). From those fundamental measurements, it calculates other metrics such as VA, VAR and PF.