Yes, it is commonly known as "current", but the physical phenomenon is "Intensity", that is why the "I".
As for what you tell me about: RMS, you're right. This took importance because: At that time, there was first the continuous voltage and then the alternating voltage. Then to see the equivalence between the two, they were compared.
They looked for the exact value of a certain "current" (as you know) produced the same calorific effects as an alternating current over a given resistance. From there they simplified and came to postulate the mathematical equation. However, the 0.707 is due to the fact that the voltage coming from the outlets has no DC component, ie, the sine wave perfectly oscillates between "Vp, 0, -Vp". Of this last I do not remember much, perhaps I am mistaken, but I am almost sure that it is so.
Thanks for reply, I hope to see you in my next posts. The idea is also to exchange knowledge.
Thanks for clarifying. You can measure AC using peak-to-peak, peak, or RMS. What you described is peak-to-peak. That is where you measure the most positive and the most negative points, add the together (ignoring the negative sign). Peak is just one side, either the positive or negative. You already know the RMS. To convert, you divide peak-to-peak by 2 and multiply by .707 = RMS. I'm sure you can calculate RMS from peak. And this is based on if you are generating a pure sine wave. The calculation is different if you have a square or a saw tooth wave. You are correct that for certain things you need to evaluate the circuit in a DC rather then an AC value. But if you deal in radio transmission, you need to understand how DC is used to set a 'Q' point so you can correctly bias a transistor. The AC is the signal you are trying to amplify. It sounds somewhat convoluted but if you understand the basics (e.g. ohm's law, kirchhoff's and thevenin's theorem) it becomes easier.
Yes, I enjoy bantering with you and hope to do more in the future. Have a good evening.