Inverter Current Conundrum

[Cacahuete]

Hello:

We are taught by manufacturers that inverters have balanced outputs; the current on L1 is the same as that on L2. Thus, in Enphase (residential) solar systems and Tesla storage systems, for example, the software measures a single line (through CT's) and multiplies that number by two, to arrive at the actual solar production.

The question: Let's say we measure 7A on L1. Is the total production in Watts equal to 120V x 7A=840W x 2 for a total of 1.68kW or is the total production 7A x 2=14A x 240 for 3.36kW?

I suspect the answer has to do with phasing but, otherwise I'm perplexed...

Thanks.

 

[jaggedben]

First of all you can only assume that current on L1 and L2 is equal if there is no where else for it to go, i.e. no neutral conductor. This is true for Enphase inverters but I don't think it's true for Tesla storage systems.

Kirchoffs First Law is the principle at work here. All currents into a 'node' must equal the current out of the node. So for example if an inverter is a node with two circuit wires then the current on L1 and L2 must be equal.

The question: Let's say we measure 7A on L1. Is the total production in Watts equal to 120V x 7A=840W x 2 for a total of 1.68kW or is the total production 7A x 2=14A x 240 for 3.36kW?

I think it's really just 7A x 240V. Or in some cases it seems to me like metering equipment actually adds up the L1 and L2 voltage, so more like 7A x (120V x 2).

Note that 240V is a nominal voltage and when you measure voltage in the real world it make vary from that. (Anything from about 230 to 250 would be pretty normal, and sometimes even slightly outside of that.) Also L1 to neutral and L2 to neutral may be slightly different from each other. So a meter could actually measure something like 7A x (120V + 118V).

Finally we're assuming a power factor of 1 here, so that V x A = W, which isn't actually true with AC circuits although it's often 'close enough.'

 

[LarryFine]

To put it another way, current is not (directly) dependent on voltage, while power is.

That's why watt-meters must measure both voltage and current to read out power.

 

[Cacahuete]

Thank you gentlemen. This is actually much simpler than sussing out the physics of Ohms law. As I mentioned, the manufacturers have us place a CT around a single solar conductor and, in software, we tell the system to multiply that CT reading by two. From the example above, I was just trying to determine if the technician will see 1.68 or 3.36kW.

I will check it next time we have a tech on site and report what we find.

Thank you again.

 

[jaggedben]

If the software actually takes one L-N voltage and multiplies it by two that will be slightly less accurate than adding both L-N voltages. The manufacturers I'm familiar with do the latter.