# Resistance of superconducting wire in parrallel with standard wire

The formula for parrallel resistors is $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}$

But how can you use this formula when one of the branches is a superconductor, eg:

Where the red resistor represents a wire with some resistance and the blue line represents a superconducting wire, how can the above equation be used to find the resistance between A and B, as it would mean dividing by zero?

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The effective resistance is zero here, because all electrons will flow through the blue wire. Your equation works when we are already given nonzero resistors... – Chris Gerig May 14 '12 at 20:16
However, beware of contact resistance... – user2963 May 14 '12 at 22:30

A more proper and simpler way to do this would be to use elementary algebra to obtain

$$R_T = \frac{R_1 R_2}{R_1 + R_2}$$

and putting $R_1 = 0$. The result is obvious :)

Or if you really like limits (completely unnecessary for your problem):

$$\lim_{R_1 \rightarrow 0} R_T = \lim_{R_1 \rightarrow 0} \frac{R_1 R_2}{R_1 + R_2} = 0$$

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s/calculus/algebra/ ? – dmckee May 14 '12 at 21:40
Ehm, English is not primary language, so I don't quite understand the difference between calculus and algebra in this case. Can you help me? – Pygmalion May 14 '12 at 21:47
Calculus would generally imply differentials ($\frac{\mathrm{d}}{\mathrm{d}x}$) or integrals ($\int \mathrm{d}x$). – dmckee May 14 '12 at 21:52
@dmckee OK, thanks, in my language term algebra usually refers to linear algebra and group theory. – Pygmalion May 14 '12 at 21:55
I don't see how this method is more proper/simpler than mine. It's basically the exact same thing - mathematically dealing with the singularity. – CHM May 14 '12 at 21:55

That formula is not in any way fundamental, the singularity at $R_i\to 0$ doesn't have much of a physical meaning, rather it means that the assumptions with which this formula was derived are violated. Namely that the voltage is nonzero.

The total resistance must be such that $$U = R_{\mathrm{tot}}\cdot I_{\mathrm{tot}}.$$ Since it's a parallel circuit, this is also $$U = R_1\cdot I_1 = R_2\cdot I_2.$$ But if one of the resistances is zero, this expression must be zero, and therefore (the total current being nonzero) we have $R_{\mathrm{tot}}=0$. Which is quite intuitive as said in Chris Gerig's comment: if there's a perfect shortcut, why would any electrons bother to take the path with nonzero resistance? They don't, so it wouldn't change anything to simply take the resistor away, in which case only the superconductor would be left, still with a resistance $0$.

Also note that a superconductor doesn't actually have zero resistance, just very much less than ordinary conductors. So you could still use the formula $$\frac1{\tfrac1{R_1}+\tfrac1{R_2}}$$ where, since $R_1\gg R_2$, $$\tfrac1{R_1}+\tfrac1{R_2}\approx \tfrac1{R_2}$$ and therefore $$\frac1{\tfrac1{R_1}+\tfrac1{R_2}} \approx \frac1{\frac1{R_2}} = R_2 \approx 0.$$

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Sorry, a superconductor does have zero resistance, by definition. – Chris Gerig May 15 '12 at 0:22
@ChrisGerig: no. A superconductor is defined by its magnetism (Meissner effect / "perfect diamagnetism"). The resistance happens to be zero for an ideal, infinitely stretched or topologically closed superconductor, but in any real circuit with superconducting elements this is just an approximation (albeit often a very good one). – leftaroundabout May 15 '12 at 7:37
Wrong: a superconductor is defined by Meissner effect IN ADDITIION to zero resistivity... This is basics of the legitimate theory, and your semantics goes against it. – Chris Gerig May 15 '12 at 23:19
I can't say I'm an expert on superconductivity. I learned that the Meissner effect is the correct criterium because perfect conductance doesn't do the trick. AFAIK, Meissner effect is also sufficient to derive the other characteristics of a superconductor – which do, put it any way you want, not include having in an electric circuit a superconducting part with zero resistance. You have to use a four-point probe to measure the zero resistance (thereby effectively removing the imperfect boundaries from the measurement), which is not what was asked for here. – leftaroundabout May 16 '12 at 12:10

If $$\lim_{R_1\to0} (\frac{1}{R_1}+\frac{1}{R_2}) = \infty$$

Then

$$\frac{1}{R_T} = \infty$$

And

$$R_T = 0$$

or more rigorously

$$R_T\to0$$

Note that this is a guess. I'm not a physicist.

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This is exactly what I would expect from a system containing a superconductor, but then again, I'd be happy to be proven wrong. – CHM May 14 '12 at 20:15