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I am trying to understand how current splits at nodes. I know that if two paths are available, each with R > 0, a current divider is used to find the split ratio. But what if one of the paths has R = 0, as in the circuit below?

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If I apply a current divider at the node just above I, I find all the current flows through the middle wire (and then through R2 and back to I), skipping R1 entirely. Is this correct?

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  • $\begingroup$ yes, it is basically a short circuit, current through $R_1$ is zero $\endgroup$
    – Courage
    Dec 7 '15 at 5:09
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Yes, under the conventions of schematics, the wire has zero resistance and no current flow in $R_1$. If some did, there would be a voltage difference across the wire, leading to infinite current. In practice, the wire will have some small resistance, so there will be some voltage drop and some current through $R_1$

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Simple. Your right, the current moves through the low resistance path, and negligible current through the resistor - barely noticeable. Think about this, though:

The resistance of wire is not TRULY zero. There is something called resistivity, which allows you to compute the resistance of a material based on length and surface area.

Now, if you have, say 1000 feet of thin copper wire, the resistance is 25 ohms or so; however, you'll notice from the equation of resistivity in the first link that we get that by multiplying by the length. If you use a more reasonable figure - for example, 1 foot of copper wire, you would find then you have a resistor of 0.025 ohms that is the wire itself. You use ideal wires with no resistance in your diagram, you account for it by simply adding a resistor that represents the wire resistance of that branch.

Now, if you were to redraw your diagram, you would have, in parallel: a resistor of $R_1$ and a resistor of 0.025 ohms.

If you choose $R_1$ to have even on ohm, that means there is more than 40 more times resistance through the branch with the actual resistor than the branch with just the wire! So very, very little current flows through even the a $\Omega$ resistance.

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