# What is the equivalent resistance in this part of the circuit?

I broke the circuit down into as basic as I can take it, and I know what X and Y are, but I can't figure out the equivalent resistance of the circuit in the middle. It seems to me like the current would flow around the resistor, but then there'd be no point of the resistor even being there...

Most basic:

It used to look like this (with a battery connected to two of the vertices):

• Which two vertices was the battery connected to? The equivalent resistance depends which two points you're measuring resistance across. To answer your basic question, though, you're correct in that the current flows around the resistor (hence no resistance), and that there is "no point of the resistor even being there." Well, except for pedagogical purposes, of course. Feb 13, 2015 at 17:17
• I'm confused - your top diagram has 3 resistors in total, but your bottom one has 6 (and none of them look like X or Y). How do these two diagrams go together? Feb 13, 2015 at 17:26
• Sorry. I was unclear. All of the resistors have the same value (100 Ohms), so the vertices don't matter. I added equivalent resistors by morphing the diagram into a 2-D model and then adding the equivalent resistors that I knew how to add. Feb 13, 2015 at 17:58

## 1 Answer

Current does flow around the resistor. Note that the resistor is completely shorted out (its ends are connected together with wires, meaning the potential difference across it is, by definition, zero).

Another way of looking at it is that the resistor is in parallel with a wire/short-circuit (i.e. $R_{wire} = 0$). You can calculate the total resistance of any resistor in parallel with a short circuit by using the basic simplified formula for parallel equivalent resistance of two resistances:

$$R_{eq} = \frac{R_1R_2}{R_1 + R_2}\\$$

With the values in your problem... $$R_{eq} = \frac{(100)(0)}{100 + 0} = \frac{0}{100} = 0$$