I have an elementary question about current flow:
If I connect a simple 1.5 V battery to a hypothetical LED of 1.5 ohm and an ideal ammeter, the ammeter would show a current of 1 Ampere and the LED would glow (let us assume the power is enough).
Now, instead if:
- ---------- 1.5 ohm LED ----- ideal ammeter ----- Ground
- ------- free
That is, The positive terminal of the 1.5 V battery connects to one end of the LED, and the other end of the LED to the ideal ammeter which in turn goes to a thick copper wire connected to ground, then the LED will not glow and the ammeter will read zero, since the negative end of the battery is free. (Yes ?)
But if it is this:
- ---------- LED ----- ideal ammeter 1 ----- Ground
- ------- ideal ammeter 2 ---- ground
then I assume the LED will glow, the ideal ammeter 1 will show a current of 1 Ampere. What about ideal ammeter 2 ? Doesn't it also read 1 ampere ?
The reason I am asking is because I am trying to understand the principle of ground fault circuit interrupters, but then I got confused about the basic scenario above. So in a ground fault circuit interrupter, if there is a path to ground through a person touching a live wire, then there is a current difference between the return path and the forward path that triggers the interrupter... but if the negative end is also connected to ground (through the person or by being earthed), shouldn't the return path have the same current ?