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Here, the question asks me to work out the resistance between X and Y. The answer is 1.2 ohms.

In a normal circuit, the resistance between two points is simply whatever resistor value(s) is between the two points (and parallel circuit rules apply). So by this reasoning, why can't I say that the resistance between X and Y is 2 ohms, since there is ONLY a 2 ohms resistor between X and Y? I've seen these square resistors circuits before and they tend to confuse me, so how do they work, and how do I calculate the resistance between X and Y?

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    $\begingroup$ The path from X to Y can be done in 3 different ways. The direct X to Y, which has resistance 2 Ohms. The path from X down, and then across the diagonal. Then finally the path from X down, then along the bottom and back up. You can calculate the effective resistance of the second path & third paths (turn them both into one new path), and then calculate the effective resistance of the new path and the first path. Look up resistors in parallel and series. $\endgroup$ – Tweej May 15 '16 at 15:10
  • $\begingroup$ @Tweej why can't I calculate the effective resistance between all 3 paths at the same time, instead of doing path 2 and 3 first, and then path 1? I am thinking it's because path 1 isn't parallel to path 2 or 3, but then the new path from path 2 and 3 combined isn't parallel to the first path either $\endgroup$ – Logan545 May 15 '16 at 15:17
  • $\begingroup$ Because it's relatively simple to turn n parallel resistors into 1 resistor, or n series resistors into 1 resistor, but turning n parallel resistors and series resistors into 1 in 1 go is hard and we don't have definite rules for it. If we label the lower 2 corners as W for the left, and Z for the right, we can reduce the paths |WY| and |WYZ| into 1 resistor, which we'll just call |WY2|. We can then reduce |XW| and |WY2| into 1 resistor called |XY2|, then we can reduce |XY| and |XY2| into 1 resistor. $\endgroup$ – Tweej May 15 '16 at 15:27
  • $\begingroup$ @Tweej Ok. another thing: since there is only one way of a current to around between the two resistors adjacent to X (i.e. there is no junction, like the one vertically below X), why can't I assume that those resistors are in series? $\endgroup$ – Logan545 May 15 '16 at 15:35
  • $\begingroup$ Cause you're trying to find the resistance from X to Y. There's no point combining them into 1 resistor and finding the resistance from a point midway in a resistor. If you were trying to find the resistance from W to Y, then yes, you'd combine the resistors |WX| and |XY| into 1, but right now, that's not very helpful. $\endgroup$ – Tweej May 15 '16 at 15:40

The best way to solve these problems is to use the rules of resistors in series or parallel, and reduce it down into much simpler systems.

When we have resistors in parallel, we can use the rule $$ \frac{1}{R_{Total}} = \sum_i \frac{1}{R_i} $$

And when we have the resistors in series, we can use the rule $$R_{Total} = \sum_i R_i$$

We cannot make one general rule for resistors in series and parallel, as that seriously complicates things. Instead we just have to slowly work back, simplifying the problem. In this instance, labelling the lower corners as W for the left, and Z for the right, we have to reduce the resistors |WY| and |WZY| into one resistor, with the parallel rule - calling the new resistor |WY2|.

We then reduce the resistors |XW| and |WY2| to one resistor using the series rule - calling the new resistor |XY2|.

Finally we can turn |XY| and |XY2| into one resistor with the parallel rule again.

So what we have done is proven that this collection of resistors is equivalent to a single resistor, with resistance of 1.2 $\Omega$ when you fully do the maths.

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  • $\begingroup$ How to calculate resistors in series and parallel was already known to me, it was just the square formation with a diagonal resistor that threw me off $\endgroup$ – Logan545 May 16 '16 at 18:04

Try redrawing it as so - should be easier to figure out.

EDIT the top R2 in the second diagram should be R4. enter image description here

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  • $\begingroup$ Oops yes, the resistor on top labelled R2 should be R4. $\endgroup$ – M. Enns May 15 '16 at 16:22

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