House with Delta mains connection

Exactly that. Two, on the two lines, one of which is not connected to the a.c. adapter and one that is. Going back to post 15, and assuming L3 (grey) is the one we are treating as the neutral, the a.c. adapter is connected L1 - L3 (brown - grey), ct1 is on L1 (brown) and ct2 is on L2 (black).

The adapter (“programmer”) can be ordered from the shop: Programmer - USB to serial UART - Shop | OpenEnergyMonitor

All the details for setting up the Arduino IDE are here: Learn→Electricity Monitoring→Using the Arduino IDE
I can’t help you with Platformio - It destroyed my system on my computer and it took me hours to restore everything, so it had to go.

Sorry Robert,

In my previous message, I asked, or rather hinted, at the question

Your explanation seems to imply that across L1-L2, the indicated currents I1 and I2 will be “double measured”, but that the corresponding power consumptions P1 and P2 will be halved, so that the measured total power consumption should be the same when the same load is connected across L1-L2, L1-L3 and L2-L3.

Is this the case? That across L1 - L2, the current is measured twice, ie I1 = I2 = 4.4A, but that the corresponding power consumption per CT is halved, ie P1 = P2 = 0.5kW? Are these the power reading that will be displayed?

Thanks,
Paul

I guess you’re talking about the current you’d measure with a meter. You must realise that the phase of the current comes into play - in the case of a load that’s connected L1 - L2, the current is at an angle of 60° with respect to the voltage - in terms of the diagram in post no.15, the real power is the product of voltage and current - the current being the current projected onto the voltage arrow, and that’s clearly half the length between the tips of the arrows. The same happens when you project it onto the shifted voltage (the other dashed line). Add the two together and you have the total power.

If you want the maths proof, I’m quite certain it’s going to be on the Web somewhere. It will probably be for a standard 3-phase set-up, so you’ll need to work though it changing the angles where necessary.

Robert,
I had forgotten about the phase angle, ie cos(60°) = 0.5. Thus

P = V * I * cos(60°)

I don’t think I have any more questions at this stage.
Thanks for your patience. I will place the order now. If I have any more questions I will come back.
Many thanks.