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I've been trying to learn electronics stuff from this video series on YouTube: https://youtu.be/m4jzgqZu-4s

At 1:07 in the video above, a switch is closed, SO there is zero resistance across the switch BUT current flows in the switch despite it having no voltage across it. ie R = 0 but I <> 0. With the switch closed, both sides of the switch seem to be at the same voltage. That means that the potential difference between them is 0, so the current flowing seems to contradict ohm's law.

Sorry I couldn't phrase this question better, but what am I missing?

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    \$\begingroup\$ We don't like watching videos here. Please reproduce the circuit schematics using the embedded circuit editor. \$\endgroup\$ – Eugene Sh. Jul 20 at 1:28
  • \$\begingroup\$ I thought the video was helpful for understanding my intuition, since the question is fundamentally about what I'm misunderstanding. \$\endgroup\$ – JacKeown Jul 20 at 1:43
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    \$\begingroup\$ I'm voting to close this question as off-topic because this site does not accept questions that so fundamentally depend on an external link as to be entirely meaningless if one does not click on the link, or if it breaks. External links may only supplement, the statement of the question here on this site must be self sufficient \$\endgroup\$ – Chris Stratton Jul 20 at 1:54
  • \$\begingroup\$ An ideal zero ohm wire connecting a resistor and a battery in a circuit will have I = V/R flowing in it. Ohms law is not contradicted by having current flow in it. The drop across the wire is V = IR. If the R is 0 the V is 0. You'd be more upset if you got a voltage drop across a zero Ohm wire as since I = V/R then V/0 = infinite current ! :-) \$\endgroup\$ – Russell McMahon Jul 20 at 14:02
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    \$\begingroup\$ @RussellMcMahon strictly speaking if R=0 you can't rearrange the formula like that as you can't divide by 0. V=IR => V = 0 × I => V=0, I=anything. \$\endgroup\$ – Tom Carpenter Jul 20 at 14:34
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This is because the video is dealing with 'Ideals' to make it easier to visualise. In theory, a switch has no resistance, and when closed, it becomes a short circuit. If you were to connect a battery to a resistor, then measure from the positive terminal of the battery to where it connects to the resistor, there would be no voltage drop. If you measure across the resistor, you get a drop, and can measure the difference between the 2 points.

Essentially, the switch in the example is just seen as part of the wire with no resistance. The current is being drawn by the bulb, not the switch.

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  • \$\begingroup\$ you say, "If you were to connect a battery to a resistor, then measure from the positive terminal of the battery to where it connects to the resistor, there would be no voltage drop." This makes sense to me, but then doesn't the current along this path violate ohm's law? \$\endgroup\$ – JacKeown Jul 20 at 1:41
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    \$\begingroup\$ @JacKeown No, it doesn't violate Ohm's Law. The resistance of an ideal wire is zero, so there is no voltage drop across an ideal wire regardless of the current flowing through it. \$\endgroup\$ – Elliot Alderson Jul 20 at 2:24
  • \$\begingroup\$ Thank you Elliot. \$\endgroup\$ – JacKeown Jul 20 at 2:26
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The battery has two ends: a positive end and a negative end. The positive end is at a higher potential than the lower end. The current can flow only when the charge carrier leaving the positive end can return back to the negative end (conventional current).

Note: * It's not the charge carrier that should return actually. For more info, read this post in this site.. I don't know whether you can understand what electric field waves are or not, so I didn't mention the term here. Once you have understood what electric fields are, you should say its "electric field waves that should return: not the charge carriers".*

With this in mind, let’s come to your question. When the switch is opened, there is no way for the current to return back to the battery. So the light doesn't glow.

But once the switch is closed, there is a complete path for the current to return back to the battery. The current is flowing through the bulb because there is a potential difference between the ends of the bulb.

schematic

simulate this circuit – Schematic created using CircuitLab

I have modeled the bulb to be a simple resistor (ignore the resistance). The -ve end of the battery is always at 0V potential. So the resistor has a potential difference of 1.5V across it, when the switch is closed.

To clarify what the switch actually does, consider the bulb is removed. So, there is no way for a charge carrier to return back. So just closing the switch doesn't allow the current flow. What actually happens is, the switch connects two of its ends, so both of its ends are at the same potential. The flow of current is determined by the condition I have mentioned above. How much current flows? Well, that depends on the total resistance offered by the whole circuit. For ideal cases, you can ignore the wire's resistance, battery's internal resistance. But for the practical case, they must be taken into account.

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The “ideal” wire has 0 ohms resistance, so no delta V no matter how much current flows. We model this as all part of the same node. But since it’s an idealized model, it doesn’t account for all real-world effects. I think your confusion here maybe stems from that limitation of the model. A real wire might have something like 0.1 ohm to maybe 1 ohm. Another real-world effect omitted by that ideal model is that a battery has some internal source resistance — like the wire, it’s often small enough that we can omit that detail from the model. But if you’re trying to model how a circuit behaves differently with a fresh battery versus a nearly dead battery, you would need to account for the battery internal resistance. Another thing to watch out for is that ohm’s law doesn’t apply to everything, for example LEDs and other diodes are governed by a different physical effect.

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The voltage over the switch exists when the switch is at non-contacting position. Just because there's the voltage over the switch and the switch is metallic like the wires, the electrons rush along the closed switch just like they do in the wires (=meeting only a little resistance). This neutralizes the voltage over the switch to microvolt or millivolt range in practical switches. But there's still some voltage needed to make the current flow continuously. As said, that's in practice micro- or millivolts, the video rounds it to zero, because practical light bulbs need thousands of times more voltage for the same current.

A man named Georg Ohm noticed that every piece of metal have resistance R, which depends on material and dimensions. And that R can be used in formula I=U/R or as well U=IR very accurately. Practical electric cabling wires and switches have thousands of times lower resistance than practical light bulbs used with those wires and switches.

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Consider the relationship between the variables of Ohms Law

schematic

simulate this circuit – Schematic created using CircuitLab

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Very interesting question. Makes you think.

The answer lies in the fact that if there IS a current flowing, then it's value is controlled by elements outside your 0 ohm resistor.

Now, if resistor is zero, then VOLTAGE DROP ACROSS YOUR RESISTOR due to this current WILL be zero. There's no flaw.

On another note, in normal (i.e. non-superconductor) reality, 0 ohm doesn't exist. Even a switch.

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