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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.

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  • \$\begingroup\$ A generic red LED usually needs at least 0.5mA to visibly light up and the voltage drop at that current for it is about 1.5V. Since the breadboard connection has a low resistance, the 1.5V are never generated across the LEDs terminals and the current is shunted via the breadboard, due to the LED being "closed". \$\endgroup\$ – Linards Jan 3 '15 at 23:34
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    \$\begingroup\$ You need to CAREFULLY read the answers and understand the very useful facts that they are conveying. This is a good question BUT is based on an extremely major misunderstanding on your part. Investigating our misunderstandings is how we learn and you have done an excellent job of explaining what your understanding is. If you now learn from what people are telling you and keep on in the same investigatory vein you will do well. If you do not do as above you will be in trouble. \$\endgroup\$ – Russell McMahon Jan 3 '15 at 23:57
  • \$\begingroup\$ ALL of the answers are "rigorous enough". Maxwell will answer your question BUT you need to be comfortable with basic everyday approximations aka in this case - when the LED's leads are "shorted together" there is no voltage across the LED so no current can flow through it so it cannot light. In this case "no" and cannot" are extremely close approximations to "NO!" and CANNOT. | In Majenko's step3. "it's worse than that Jim"as the LEDs constant voltage approximation is in fact an inverse exponential model and the LED's effective resistance is "utterly immense" (at least :-) ). \$\endgroup\$ – Russell McMahon Jan 4 '15 at 1:29
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    \$\begingroup\$ Just crossed my mind that this is the exact same scenario as birds sitting on high-voltage power lines... \$\endgroup\$ – nomadStack Jan 4 '15 at 2:12
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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.

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    \$\begingroup\$ Wow. You have successfully blown my mind with the elaborateness of your answer. Thank you very much. Kirchhoff's law was what I was missing and indeed answers my question. You've also incited my curiosity with Wittgenstein's ladder in the following way: Kirchhoff's laws can be derived from Maxwells' equations of classical electrodynamics [1]. In order to answer my original question correctly, quantum field theory would be necessary. Correct? Do you know of a rigorous explanation? [1]: physics.stackexchange.com/questions/102458/… \$\endgroup\$ – nomadStack Jan 4 '15 at 1:00
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    \$\begingroup\$ Beg pardon, but I want to burn the phrase "current always takes the path of least resistance" out of the collective consciousness. In fact, current takes all paths of non-infinite resistance, and is divided among those paths in inverse proportion to each path's resistance. \$\endgroup\$ – Jamie Hanrahan Jan 4 '15 at 3:08
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    \$\begingroup\$ @JamieHanrahan Read the very first link in the answer about Wittgenstein's ladder!!! \$\endgroup\$ – Majenko Jan 4 '15 at 8:29
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    \$\begingroup\$ Very nice! Corollary to step 3 : using a bigger power supply, by the time you are passing about 1600 amps through R4 (1mOhm) there will be 1.6 volts across it, so you will indeed see D1 start to illuminate. Corollary to the corollary : don't try this at home! \$\endgroup\$ – Brian Drummond Jan 4 '15 at 12:55
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    \$\begingroup\$ @Majenko: That's fine. But the "path of least resistance" phrase is all too often given without further explanation, or even a note that it's a simplification. Even with further explanation it's a problem, because people remember best what they hear first. I think it would be better to drop it. \$\endgroup\$ – Jamie Hanrahan Jan 6 '15 at 18:50
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Let's look at that circuit again:

schematic

simulate this circuit – Schematic created using CircuitLab

I see a power supply, a resistor, and an LED. You know what all of these components do. But that's not all of the schematic. There are lines. You know, these things:

schematic

simulate this circuit

What do those things mean? Lemma: they mean that everything touching any one line has the same voltage.

"What, no! Those are wires!", I hear you thinking. OK, what is the difference between a wire and a resistor? Not much: a wire just has a very low resistance. You can find tables that give resistance per length for standard wire gauges. So, wires are just really small resistors.

Now consider Ohm's law:

$$ V = IR $$

Let's just say we have 1A going through a 10kΩ resistor. What's the voltage across that resistor?

$$ V = (1\:\mathrm A)(10\:\mathrm k\Omega) = 10\:\mathrm{kV} $$

What happens if the resistance gets a lot smaller? Say, 1Ω?

$$ V = (1\:\mathrm A)(1\:\Omega) = 1\:\mathrm{V} $$

The voltage gets smaller. The smaller the resistance, the smaller the voltage. As the resistance approaches 0Ω, the voltage approaches 0V, regardless of the current. Mathematically:

$$ \lim_{R\to 0\:\Omega} IR = 0\:\mathrm V $$

The lines in a schematic are idealized wires that have zero resistance. Thus, the voltage across them is always zero. You are actually using real wires which have some resistance, but it is negligibly small.

Look back at the schematic. Both sides of the LED are touching the same wire, so there can't be any voltage across the LED. So it can't light.

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  • \$\begingroup\$ If the voltage across the points of a given wire are always zero, how does the current keep on flowing? In order to have a current, a voltage is necessary (excluding the case of an free charged particle in space). \$\endgroup\$ – nomadStack Jan 4 '15 at 2:06
  • \$\begingroup\$ @nomadStack The power supply (V1) provides the difference in voltage that pumps the current through the wires. Perhaps a thought experiment will help resolve your confusion: what happens if you make the wires shorter? What happens if you make them so short they cease to exist? \$\endgroup\$ – Phil Frost Jan 4 '15 at 2:43
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    \$\begingroup\$ @nomadStack: as Phil Frost said, the voltage difference comes from the + and - terminals of the power supply. The voltage diff between two points on a wire will not quite be zero, but it will be nearly zero; the difference will likely be unmeasurable here due to the very low R of the wire, unless the current is very large. You might try connecting a voltmeter across your LED - even if you don't see why the V is zero (in the "doesn't work" case) you can at least demonstrate that that is the case. \$\endgroup\$ – Jamie Hanrahan Jan 4 '15 at 3:16
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    \$\begingroup\$ @nomadStack perhaps also consider that indeed, there must be voltage across something (see Kirchoff's voltage law for a formal explanation of what I think you might be thinking), but that thing is the resistor (R1), not the wires. \$\endgroup\$ – Phil Frost Jan 4 '15 at 3:18
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Wow !

Assuming the breadboard's contacts are perfect, there is no voltage between the LED terminals, so there is no current, no power, no light. I1=0.

Of course, there is a minuscule resistance so with a very high input voltage you may see a few photons emitted before frying the breadboard, but, meh !

In your third photo, you are short circuiting your power supply. Hope it is protected (not just a battery) !

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  • \$\begingroup\$ Why isn't there any voltage between the LED terminals? Do you agree with the schematic diagramm? In the third photo I'm short-circuiting my power source with I2 indeed. \$\endgroup\$ – nomadStack Jan 3 '15 at 23:29
  • \$\begingroup\$ The resistance of the small strip of metal is maybe in the order of 0.001 ohms. So there is a voltage (U=RI), it is minuscule, less than a millivolt. Due to the non linearity of the LED, the resistance of the LED terminals, the current across the breadboard is probably billions time larger than the current across the LED. \$\endgroup\$ – TEMLIB Jan 3 '15 at 23:35
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In the second and third photos, both pins of the LED are connected to the same metal strip in the breadboard, so there is no voltage across the LED - therefore no light.

In your schematic diagram, you show the two leads of the LED connected together, so they will be at the same voltage. Again, no voltage across the LED, so no light.

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even though the breadboard connection has a minimal resistance

It's important to recognize the difference between "no resistance" (a short circuit) and "infinite resistance" (an open circuit). They are not the same; in fact, they're opposites. "No resistance" does not mean "no current can flow".

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I have the same breadboard here. The first image makes sense, that's how an LED should be connected. In the second image, everything except for one end of the resistor and the positive are plugged into row 6 so both pins of the LED are touching the same connection. In the third image, everything is plugged into the positive connection, so basically you're shorting it in the 2 non working examples.

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