I don't know the resistance of my house when I am on the second floor, so I made up that resistance.
I don't know what it would be either. In anything but theory, it's close enough to infinity that it probably doesn't matter.
If I establish that circuit, will I actually induce a voltage / current between the Earth and A0?
That's beyond what I understand. I was kinda hoping someone with more electronics experience than me would jump in here. I understand some theory, but it's part physics, part chemistry, and part magic. I'm trying to design stuff that works. How the electrons make their trek from one place to another -- well, I can speculate based on what I know (or think I know). Sooner or later, I run out of answers.
So.. is it you conducting between earth and A0? I don't know. It might be static energy, or the electrons flowing through your body (we are partly electrical beings). I can assume the amount you measure will vary respective to ground when that ground is earth via USB, or an isolated battery.
Are you talking about a parallel circuit? In parallel circuits, current is shared between the paths, but the one with lower resistance will get more current.
Yes, right.. See the picture I've attached. This is more like how I pictured it. I chose more reasonable resistances of 100 and 3.6 ohms, rather than 100M and 3.6M, because with a 5v source, in reality, those relays will never turn on. The triggers are 5v square waves, at different frequencies, so they turn on and off independently. (This was an illustrative point I made in the last post, but otherwise not important.)
If the + terminal is 5V, the voltage across the relays won't be 5V due to the resistors, I think.
Right. The coil has its own resistance, so it's essentially a voltage divider.
But, the voltage in the parallel circuits should be the same, or am I wrong again?
If the trace on the - side of the coils, between the resistors, had 0 ohms resistance, then yes -- the voltage at the top end of both resistors would be the same (and current through each side would be relative to its resistance). But that trace does
have some resistance, so it will affect the voltage at each end. By how much is proportionate to how much current is flowing. In most cases, not much different, but if you're dealing with long / thin traces, high current, or very low voltages, it's significant enough that you need to know it's there.
Sorry, I prefer to work with graphics.
I'm visual too. Also lazy. ;-)
In Ground Loop 3.png, there would be a difference in earth reference between the left relay and the right relay, correct? So the voltage across the relays are not the same as each other.
Well, I didn't intend for the resistors to be parallel to the relays when I wrote my last post, so your diagram 3 has the advantage of having a low-impedance ground reference. My last post assumes the absurd 100M and 3.6M values between the coil - pins and ground. (Again, the relays would never actually work this way, it was just an example.) The voltage through the coils would be nearly exactly the same. The only thing that prevents them from being precisely the same is variance in the coil impedances (tolerance), and the imperfect nature of the ground bus beneath them. To help understand the last part, imagine those traces as very low-value resistors, not a straight line.
Since the connections between
the relay grounds is not perfect, there will be voltage on the ground bus between relay 1 and 2 if relay 3 is on, for instance. This affects relay 1 and 2's point of reference, because the + terminal is no longer with respect to 0v. It's with respect to 0v + the voltage across the ground bus.
The better way to design a PCB based on your latest diagram would be to put the ground connection under relay 2, so it's (closer to) equidistant from each of the relays. If you have a fourth relay, then you can align them in a square, with the ground lead in the middle. Or, just make that PCB trace fat enough that its impedance is so low that it doesn't affect your circuit.
We've gotten away from your original quandry (why the capacitive sensing doesn't work from a battery). I don't really know how to help with that specifically, since I've never tried to do it myself. All I have is theories, and I'm not sure all those are correct, or if I've explained everything right. I often find out how much I still have to learn by trying to explain my thoughts to someone else, so I might get schooled here soon. :-)