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?
 A: 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):
$$\lim_{R_1 \rightarrow 0} R_T = \lim_{R_1 \rightarrow 0} \frac{R_1 R_2}{R_1 + R_2} = 0$$
A: 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.
A: 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
$$
  \frac1{\tfrac1{R_1}+\tfrac1{R_2}} \approx \frac1{\frac1{R_2}} = R_2 \approx 0.
$$
