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I'm learning electronics and this confused me. In this circuit it shows that current flows through both wires with resistors and lower only through one wire where there's no resistor (even though one is with really low resistance). I thought current should still flow, what's happening here?

enter image description here

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  • \$\begingroup\$ In "ideal" land wires have 0 resistance. \$\endgroup\$ – Tyler May 26 '16 at 17:52
  • \$\begingroup\$ Your diagram is a bit confusing with all the red ink applied. Could you please separate your observations from your questions. \$\endgroup\$ – Sparky256 May 26 '16 at 17:53
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    \$\begingroup\$ Models are not reality - in this case the model assumes zero resistance. BTW you are correct - current would flow in the 600 and the 1 ohm in the real world. \$\endgroup\$ – JIm Dearden May 26 '16 at 17:53
  • \$\begingroup\$ Oh, I see. And I thought I'll be able to use these simulators for learning.. \$\endgroup\$ – bah May 26 '16 at 17:57
  • \$\begingroup\$ If there is a voltage across a resistor, the current through it is a function of that voltage independent of the existence of other paths. However the voltage cross the resistor will usually be, at least to a degree in the real world, influenced by the total load presented by all the paths. As the intermediate node in your circuit has resistors both above and below, such dependence would quite strongly be the case here - in this case the presence of the "wire" path means the voltage is effectively zero. \$\endgroup\$ – Chris Stratton May 26 '16 at 17:57
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The old saying that "electricity follows the lowest resistance path" or whatever the case may be, is misleading. In true fact, parallel resistances form a current divider with each "leg" of the circuit carrying a current inversely proportional to the resistance. So, while a lower value resistance will carry more current than it's higher resistance neighbor, both will still carry some current. In the case of ideal circuit analysis, wires (and switches) have zero resistance. By evaluating the current divider formula as a resistance approaches zero, you can see that that leg will conduct a share of current that approaches 100%.

In summary, you are correct that in the real world, nothing is ideal and zero resistance is impossible. This means that some tiny current will be flowing through the 600 ohm and 1 ohm resistors, with the majority of it flowing through the closed switch. However, the current in the resistors is going to be so small it doesn't matter in practically any circuit.

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When drawing a schematic like you have done in your simulator, connecting one point to another point directly is you telling the simulator that these are actually the same point in the circuit.
You are not telling it, "I have a wire of some length and non-zero resistance connecting these points".

For the circuit you've drawn, the simulator sees something more like this:

schematic

simulate this circuit – Schematic created using CircuitLab

Its entirely possible that the simulator just optimizes away those 2 resistors because it knows that they will have no effect on the circuit, but even if it doesn't - just apply some simple Ohm's law to the situation.
The 2 ends of either of those resistors are connected to the same point. You need two different points to have a voltage difference, but you don't have that.
So you have 0V across the resistors and the current through them must also be 0A.

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  • \$\begingroup\$ The problem with this solution is that you have physically changed the circuit connections. The negative of the battery is now connected to the other side of the short circuit (a switch in the original circuit). If the switch in the original circuit is now opened it would produce a different set of current values but not in yours so the two circuits are not equivalent. A better approach would be to set the resistance of the closed switch nearer to its real world value (say 0.1 ohms) by adding a series resistor to the switch. The simultator would then do the rest. \$\endgroup\$ – JIm Dearden May 27 '16 at 9:39
  • \$\begingroup\$ @JImDearden - yes, but I was only trying to illustrate that particular configuration in order to explain it. \$\endgroup\$ – brhans May 27 '16 at 13:37
  • \$\begingroup\$ I fully appreciate that and I can see what you are trying to do, hence no down vote, but you fall into the same trap as the simulator taking the short circuit as zero ohms, in which case (as you correctly point out) it would make no difference which side of the short circuit you connected to ground. However, the OQ shows this particular 'short circuit' as being produced by a switch rather than a wire (difficult to see but it is there). I merely point out that your connection above the switch would produce a different answer to the OQ if the switch was opened. \$\endgroup\$ – JIm Dearden May 27 '16 at 13:52

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