Electronic – Single LED connected in parallel to conducting wire does not light up

breadboardledparallel

LED connected in series to resistor. All is well.

Led connected in series to resistor. All is well.

This doesn't work.

This does not work.

Neither does this.

Neither does this.

This would be the schematic diagram.

schematic

simulate this circuit – Schematic created using CircuitLab

Why doesn't it light up? Is this the absence of a load, even though the breadboard connection has a minimal resistance?

Conclusion

Straightforward answer: Wire without load sucks up all the current because its resistance is approximately 0 (see Ohm's law V = RI). Points which have approx
0 resistance between them are called electrically common.

A bit more elaborate: The voltage on the wire is equal to 0. The LED leads are connected to this wire, hence no current can flow through the LED (analogous to a bird sitting on a high voltage power supply wire).
Why is V = 0 in the wire? Because there is no resistance. If you measure the pressure of a flowing river without any obstacles between two points of
not inmense distance between them, they are practically the same. The water can keep on flowing though, because
the initial pressure from the potential difference between the mountain top and sea level (the power source) produce the current. If there's an
obstacle in the way, say a small dam, water can accumulate (as well as electrons can do) on one side of the obstacle
(the resistor) creating a potential difference on each side of the obstacle. Therefore a voltage builds up. Resistors can be thought of as narrowings
of a pipe, but as such, the water analogy would fail, because Bernoulli's principle would come into consideration. Even though the voltage drops just like the
as pressure drops in the narrowing of a pipe, in a circuit it happens before and after the resistor, not only within it. With Bernoulli, current
(mass/time) is equal everywhere. That is why the water molecules are accelerated in the narrowing in order to get the same amout of mass through.
In circuits it is more like closing of lanes in a road due to an accident. The road gets narrower and technically the drivers would have to
accelerate in order to get the same amount of cars through in the same time interval. In reality, they slow down, producing a traffic jam, and with so
a "car potential difference" is built between before and after the accident site.

Best Answer

Ok, let's take this step-by-step up Wittgenstein's ladder.

Step 1:

  • Current is lazy and always takes the path of least resistance.

Given the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

For the current to get from point A to point B it's going to go straight down the simple wire rather than take the more difficult route through the LED. So no current flows through the LED, it just goes straight past.

Step 2:

  • Kirchhoff's Current Law:

At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node

Ok, but on the circuit above point A and point B are connected directly together, so they are in effect the same point. The circuit is basically the same as:

schematic

simulate this circuit

(imagine the little link bit isn't there - the editor won't let you do diagonal lines).

The current I that flows in must equal the current I that flows out of point A. So if all that's going in is going straight out, there's none left to go up to the LED.

Step 3:

  • Wires aren't perfect.

No wire has absolute zero resistance. The same with breadboards. So the actual circuit is more like this:

schematic

simulate this circuit

Ok, so we have an fixed voltage \$V_{CC}\$. Let's say this is, for simplicity. 5V. The LED has a fixed forward voltage drop. Let's assume for the sake of argument that it's 2V.

Ok. Let's take the LED out of the circuit initially and just work out the voltages dropped across the resistors R1, R4 and R5.

The total resistance for that section will be 100.002Ω (simply add them together). So the current through them would be \$\frac{5}{100.002} = 49.999mA\$.

Therefore the voltage dropped across R4 would be \$0.049999 \times 0.001 = 49.999{\mu}V\$.

Now if you attach the LED across that resistor it's only going to be getting 49.999µV, which is considerably less than the required forward voltage needed to turn it on. So it won't be conducting as it's not on, so the current through the resistors R2 and R3 will be zero.

Now there are more potential steps in Wittgenstein's ladder, but from here we're getting into the realms of subatomic physics, and even quantum theory, so we'll leave it there for now.