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 ?

Thanks !

  • 1
    \$\begingroup\$ Welcome to EE.SE. "a hypothetical LED of 1.5 ohm"??? At 1 amp the LED may become a laser, for a few uS. Your mind is looping. Please post concise, well thought out analysis and then your questions. \$\endgroup\$
    – user105652
    Feb 20, 2018 at 0:33
  • \$\begingroup\$ The GFCI works because the neutral/ground connection is on the other side of the GFCI. So the live connection passes through the GFCI and the neutral connection does not pass through the GFCI. Thus the GFCI reads 0 amperes on the neutral because the current is flowing around it not through it. \$\endgroup\$
    – user253751
    Feb 20, 2018 at 0:35
  • \$\begingroup\$ @immibis, could you please expand on that a bit ? are the statements before my last paragraph correct ? \$\endgroup\$
    – Suresh
    Feb 20, 2018 at 0:47
  • \$\begingroup\$ @immibis, thank you .. now i understand your answer, after reading Selvek's answer below. \$\endgroup\$
    – Suresh
    Feb 20, 2018 at 1:27
  • \$\begingroup\$ Hypothetical LEDs don't have a resistance, they have a maximum current and a typical operating voltage, and woe betide you if you control voltage, not current. Not to mention disposing of heat.... \$\endgroup\$
    – Ecnerwal
    Feb 20, 2018 at 2:44

1 Answer 1


Keep in mind that an "ideal ammeter" is the equivalent of a wire for understanding what the circuit does, and that current with only flow if you have a loop.

Also, note that the built-in schematic tool will let you draw and simulate these circuits quickly and easily to answer questions like this! You can start with my examples below.

Final Note: A "1.5 ohm LED" is not a real thing. I have substituted 1.5 ohm lamps in the schematics below. I won't go into details on the LED/lamp issue here because the lamp is just a hypothetical component filling the role of whatever load is connected to this circuit. I'm sure a little research will clarify how LEDs work.

Case 1: Yes, the ammeter reads 1A.


simulate this circuit – Schematic created using CircuitLab

Case 2: Here you have disconnected the negative terminal of the battery. There's no loop through which current can flow, and the ammeter reads 0A.


simulate this circuit

Case 3: This case is EXACTLY the same as case 1, because for all practical purposes, an ideal ammeter acts the same as a wire. Because ammeter 1 and ammeter 2 are part of the same current loop, they must read the same value - although here, ammeter 2 reads -1A because the current flows into its negative terminal instead of its positive terminal (clockwise around the loop).


simulate this circuit

So, what about that ground fault interrupter?

The key concept here is that the neutral wire and the Ground wire are both ~0V. However, the neutral wire is INTENDED to carry the return current. The ground wire is NOT intended to carry any current.

The circuit at left is the "intended" operation. Notice that the two ammeters on the Hot and Neutral wires read the same 1A.

In the circuit at right, a "person" (shown as a 100mOhm resistor) makes a connection from the Hot wire to the Ground wire (or, the grounded metal case of the circuit). Current is now flowing along the Ground wire, and the Hot and Neutral wires now have different currents. This triggers the GFI circuit and turns off power to the lamp.


simulate this circuit

  • \$\begingroup\$ Wow, perfect and super-helpful in multiple ways. Thank you !! Is there a way to accept this answer/upvote this etc ? \$\endgroup\$
    – Suresh
    Feb 20, 2018 at 1:08
  • \$\begingroup\$ Glad it's helpful! At the upper left of every post there are up/down arrows and a number for up/downvoting. Right below that is a check mark button for accepting an answer. That should all have shown up in the tutorial it gives you when you create an account - if you haven't read through that tutorial, I highly recommend it! \$\endgroup\$
    – Selvek
    Feb 20, 2018 at 1:12

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