### RMS and math question

Hello!

I've a question.  If I get an RMS voltage (sum of squares, and sqrt) proportional to the AC current (using YHDC CT) - how to do I turn this into a AC current...?  I am using a 100R resistor to give a 0-35amp scale based on 2.5v bias, and 5v max.

There doesn't appear to be a direct multiplication which works, i.e. peak of 2.5v x 14.14 gives 35amps, but this doesn't work.

I've obviously missed something....

James

### Re: RMS and math question

Provided that you are using a pure resistance, Ohms Law works just the same with alternating rms average values of voltage and current as it does for direct voltages and currents.

But I'm struggling to understand the real question that you are asking. Is it how to choose a burden resistor? Is that what your 100 Ω resistor is?

Have you looked here?

### Re: RMS and math question

Hiya Robert,

Thanks for the reply.  I have built the circuit as described in your link, and have a 100R as a burden resistor.  I was doing the code the simple way - i.e. read the maximum voltage, subtract the bias (2.5v) and then multiply to get an amps reading - as the voltages were proportional to the amps.

I've updated that code to use the RMS method as in the emmonlib code, and I get a voltage back - just can't get my head around how to convert it into an amps.  I can't see in the emmonlib any conversion code - i.e. rms volts x 2 = amps.

James

### Re: RMS and math question

It is the burden resistor that converts current to voltage, not the other way round. The current transformer is just that - it works on currents, so you need to switch your thinking round a little when dealing with c.t's.

You have a primary current of 35 A, you say. Let's assume you are using the YHDC SCT-013-000 which has a ratio of 100 A : 50 mA (a current transformer's ratio is always given in Amps). Therefore, it will generate a secondary current of 35 / 100 × 50 mA = 17.5 mA. That current will flow in the burden resistor and generate a voltage of 17.5 mA × 100 Ω = 1.75 V.

Because we have been dealing with rms values so far, this is the rms value. For a sine wave only (not quite true but true in practice) , the peak-peak value is 2√2 × the rms value, so your peak-peak voltage is 4.95 V. So that's just about OK for a 5 V range at the ADC input.

Going back a stage, the rms voltage is 1.75 V and you get 1024 counts per 5 V. Therefore the count (rms value - why not? !) is 1.75 / 5 × 1024 = 358.4

So now you know that 35 A gives you 358.4 counts, given that c.t. and burden resistor.

All I've actually done there is work through the input circuit quite logically. You have read CT and AC power adaptor installation and calibration theory, haven't you? In there, you'll see how all that relates to the calibration constant that is fed into the methods of emonLib.

### Re: RMS and math question

Hiya Robert,

I had read the pages on this site many times, but obviously not that page well enough.  Your worked example is most helpful.

I'll implement this tonight and see if that sorts it.

Many thanks,

James

### Re: RMS and math question

Hiya Robert,

Code updated, bugs fixed and that is working nicely now.

Very many thanks for your help!

James