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Lets say we have a circuit with a voltage source and a 5 ohm resistor. The voltage source is 20 v and so the current flowing is 4 amperes (according to ohms law). Now, we add a short wire in parallel to the resistor (you can think of it as having a resistor whose resistance is 0 cpne ted in parallel with the first resistor. Finding the current flowing in each wire is hard since as you can see, we first need the equivalent resistance. The equivalent resistance is (1/((1/5)+(1/0))) and it turns out to be 0. Now using the current divider formula, each current yields 0 ampere which is wrong. Thanks

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If the equivalent resistance is zero then there's also no voltage across it, per Ohm's Law. Then the current though the resistor is 0 V/ 5 Ω = 0 A. The current through the wire can't be calculated this way since 0 V/ 0 Ω is undefined.

Then the current will depend on the source's internal resistance. If that's 1 µΩ for instance the current will be 20 V/ 0.000001 Ω = 20 MA.
If the source has zero resistance current will be infinite.

Either way, applying this to the current divider formula gives for the resistor path:

\$ I_R = I_{tot} \dfrac{R_{tot}}{R} = I_{tot} \dfrac{0 \Omega}{5 \Omega} = 0 A \$

For the wire we get again

\$ I_W = I_{tot} \dfrac{R_{tot}}{R_W} = I_{tot} \dfrac{0 \Omega}{0 \Omega} = undefined \$

And we'll have to look at the external conditions to see how high the current is.

Edit

"Undefined sounds crazy to me"

It is, and mathematicians aren't happy with it either, but there's no other way. Any real thing you try leads to contradictions. Even the 0 volt that I claimed. (I know, I lied, but that was because otherwise I'd get dizzy. Ah, what the heck, let's go for dizziness.)

The voltage across the wire||resistor is

\$ V = \dfrac{R_{PSU}}{R || R_W + R_{PSU}} 20 V = \dfrac{0 \Omega}{0 \Omega + 0 \Omega} 20 V = undefined \$

I can't help it. But let's for the sake of argument say it's 10 V. 0 V didn't get us anywhere, and it has to be between 0 V and 20 V. Then the current through the resistor is 10 V/ 5 Ω = 2 A. The current through the wire is 10 V/ 0 Ω = \$\infty\$ A.
If we apply KCL:

\$ I_{tot} = I_R + I_W \$

That's

\$ \infty A = 2 A + \infty A \$

So far so good. But if we want t0 find \$I_R\$ from this we'll see that we can't! Despite the fact that we know it's 2 A. Let's try:

\$ I_R = \infty A - \infty A = undefined\$

Yeah, right, I always say undefined. Why would it be, if we know the result? OK, you're asking for it. So suppose

\$ \infty - \infty = 2 \$

Now we know that \$\infty\$ + \$\infty\$ = \$\infty\$, so

\$ (\infty + \infty) - \infty = 2 \$

or

\$ \infty + (\infty - \infty) = 2 \$

The value between the brackets is 2, that was our assumption. Then

\$ \infty + 2 = 2 \$

Subtract 2 from both sides, and

\$ \infty = 0 \$

which obviously isn't true. So our assumption was false. Now you can try with any number instead of 2 you'll always end up with a contradiction. So that's how we end up with undefined stuff and a dizzy head.

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  • \$\begingroup\$ What will the ammeter read on the second wire? Undefined sounds crazy to me \$\endgroup\$ – WantIt Jul 16 '12 at 10:49
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    \$\begingroup\$ @blackandyello With the second wire, you're breaking the mathematical model. The wire cannot have zero resistance. Since it doesn't, it's a resistor in parallel with the first resistor and you use the usual parallel resistance formula here. Also the meter has its own internal resistance that could be several ohms and that needs to be taken into account since the resistance of the wire may be lower than the resistance of the meter. \$\endgroup\$ – AndrejaKo Jul 16 '12 at 10:53
  • \$\begingroup\$ @blackandyello - I updated my answer in an attempt to explain that "undefined" is the least crazy of all possible solutions. \$\endgroup\$ – stevenvh Jul 16 '12 at 12:00
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The resistance of the second wire is not zero, unless you have added a superconductor. The resistance is probably just too low to measure on a multimeter.

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First, in the context of ideal circuit elements, a wire in parallel with a resistor is a wire, i.e.,

\$ 0 || R = 0\$ for any R.

Proof:

\$ 0||R = \displaystyle \lim_{r \rightarrow 0} \dfrac{1}{\frac{1}{r}+\frac{1}{R}} = r = 0\Omega\$

Now, a wire, by definition, is the circuit element that has zero volts across it for any current through it; a wire is a zero volt voltage source.

But, an elementary fact from circuit theory is that you cannot parallel differing voltage sources on pain of a contradiction. For example, if you parallel a 20V and 0V voltage source, by KVL you get: \$20V = 0V\$.

So, the most important problem here is that you're attempting to solve a circuit that has no solution when you place the wire in parallel with the voltage source.

Now, if instead you had a current source, then there's no problem. The current division works fine. There's zero current through the resistor and all of the source current is through the wire.

\$ I_R = I_S \dfrac{0}{0 + R}= 0\$

\$ I_0 = I_S \dfrac{R}{0 + R} = I_S\$

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  • \$\begingroup\$ "has zero volts across it for any current through it". Possible exception: infinite current. \$\endgroup\$ – stevenvh Jul 16 '12 at 13:04
  • \$\begingroup\$ and then all bets are off! ;^) \$\endgroup\$ – Alfred Centauri Jul 16 '12 at 13:06
  • \$\begingroup\$ Wouldn't a better proof be that \$\dfrac{1}{\frac{1}{a} + \frac{1}{b}} = \dfrac{a \cdot b}{a + b}\$? Then make a = 0 and you're there. \$\endgroup\$ – stevenvh Jul 16 '12 at 14:54
  • \$\begingroup\$ @stevenvh, the one I gave is more general since it extends straightforwardly to 3 or more parallel elements. You could also write \$\dfrac{1}{\frac{1}{r}+\frac{1}{R}} = \dfrac{r}{1+ \frac{r}{R}}\$ and this also extends easily. \$\endgroup\$ – Alfred Centauri Jul 16 '12 at 15:52

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