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If two resistors (5 Ohms & 10 Ohms) are connected in parallel, current will flow through both resistors. Then, assume a parallel circuit (5 Ohms & 0 Ohms), why does current flows to the path with 0 Ohms only and not through both of the paths?

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  • $\begingroup$ Are you certain this is true, and not that $V/R$ flows through both paths, but through the short path this just happens to be infinite? $\endgroup$
    – jacob1729
    Oct 30 '18 at 14:29
  • $\begingroup$ Do you have a source available that is able to produce a non-zero potential across a 0-ohm resistance? $\endgroup$
    – The Photon
    Oct 30 '18 at 16:11
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Because an ideal 0 Ohm resistance would offer no resistance to flow, while the 5 Ohm line would offer resistance. Energy travels from high to low potential through the path with the least resistance.

Assuming the resistance isn't a function of current for the 0 Ohm line (not realistic usually; but for a theoretical ideal situation it's what we're concerned about), there's no reason for the current to flow through the line with resistance. You have a path with no resistance, and infinite capacity; which is why all the energy wants to choose that path.

The reason it might not make sense intuitively is because in real situations, this doesn't really happen. With two non-superconducting wires, one having a 5 Ohm resistor and one having no resistor, there should still be some current through the 5 Ohm resistor. This is because regular wire has an internal resistance, and so all the energy cannot freely flow through as it would with an ideal 0 resistance path.

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  • $\begingroup$ But if you take Ohm's law and solve the problem with one the the resistors gooing to zero, you see that the current going trough the fixed resistor does not change. Why would it change if we take the limit r->0 $\endgroup$ Oct 30 '18 at 14:08
  • $\begingroup$ @Wolphramjonny You have to consider the current and the current source in that case as well. The current going to the fixed resistor doesn't change if you have an ideal perfect voltage source which could supply any current. The thing is, to keep the same current over the 5 Ohm resistor while the other line approaches 0 Ohms, you would need to put increasing amounts of current through the low resistance line as it's resistance drops. For a real voltage source with a limited current as well, you wouldn't be able to maintain the voltage drop and the required current... $\endgroup$
    – JMac
    Oct 30 '18 at 14:49
  • $\begingroup$ @Wolphramjonny ... (continued) When you say the ideal line has no resistance, it would also have to sustain any current put through it, and therefore everything would choose to flow through it, not just almost everything. It also creates a short circuit due to the lack of resistance, so there's no reason for current to flow through a resistor that has no potential difference anymore. $\endgroup$
    – JMac
    Oct 30 '18 at 14:51
  • $\begingroup$ right, there is no longer a voltage drop due to the shorcircuit, so taking the limit which assumes a voltage difference is wrong $\endgroup$ Oct 30 '18 at 14:55
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    $\begingroup$ @Wolphramjonny The limit only behaves like that when you have purely ideal systems with unphysical circuit elements. It doesn't make sense to take the limit as r -> 0 while leaving all other variables unchanged, because the situation doesn't represent anything physical anymore. That's why I said it's non-intuitive; we never would experience it that way. You can't apply Ohm's law in a vacuum here, you need to consider what the current source and voltage source would actually do. I'd say Kirchoff's laws are far more relevant than Ohm's law here. $\endgroup$
    – JMac
    Oct 30 '18 at 15:02
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You can assume they follow same way as 5&10 ohms, and the ratio between this two path is infinite, which basically means before you can observe this phenomenon your power supply already dead.

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  • $\begingroup$ You could however add a suitable resistor in series to the two resistors of this question. $\endgroup$
    – Jasper
    Oct 30 '18 at 14:59
  • $\begingroup$ @Jasper then since your total current is finite and the ratio between this two path is infinite, then it means no current is on the 5 ohm path. $\endgroup$ Oct 30 '18 at 16:41
  • $\begingroup$ Sure, I just wanted the power supply to survive :) $\endgroup$
    – Jasper
    Oct 30 '18 at 17:21