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I'm using an online circuit simulator found here.

When I have a single path through both LEDs, they each illuminate:

enter image description here

However, when I add a wire to create a cycle, one of the LEDs loses power:

enter image description here

Why does this happen? If I were to write my own simulator, I would check to see if a particular path leads to ground. If it leads to ground, it gets power, if doesn't lead to ground, it doesn't complete the circuit and doesn't get power.

In this case, I would say both paths lead to ground, and thus they should both get power. I thought the simulation might be correct if "electricity takes the path of least resistance".

But then I read that this is not true.

Is the simulation correct?
Would only one LED light up in real life?

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  • \$\begingroup\$ Remember that in this simulator, wires have zero resistance and will therefore create zero voltage drop, as there's no voltage drop across the wire, there's no voltage drop across the LED so it won't light up. When you hover over the LED with the mouse it will show you the voltage across it in the bottom right hand corner. In real life you would need a wire that caused the same voltage drop as the LED, which would be a really long bit of wire. Unlike incandescent bulbs, LEDs do almost nothing until you hit their threshold voltage and they suddenly fire up (more or less) \$\endgroup\$ – Sam Jul 11 '16 at 0:31
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One property of circuit theory is that all things connected to the same node are at the same potential (voltage).

If there is no voltage across your LED, no current will flow, and no light will come out.

To do what you are wanting to do, you should put the second LED on a different branch entirely.

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The resistance through the "off" LED is so much higher than the short that not enough current flows to light it up. So even though some current does flow, it is in the picoamperes or less.

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  • \$\begingroup\$ @TrevorHickey: With the fact that one branch is a short-length conductor and that the other is a semiconductor, specifically a P-N junction with associated depletion region. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 11 '16 at 0:35
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When a wire and LED is in parallel, voltage going in is the same for both. By ohm's law, V=I/R The resistance at LED is higher than the wire, therefore, lesser current will flow through the LED and more will flow through the wire.

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