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This question already has an answer here:

How do I calculate the equivalent resistance between A and B?

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marked as duplicate by Dave Tweed, Nick Alexeev, Anindo Ghosh, Kaz, placeholder Apr 7 '13 at 19:02

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    \$\begingroup\$ The short from top to bottom makes this an easy solve. Do you see why? \$\endgroup\$ – Andy aka Apr 6 '13 at 13:22
  • \$\begingroup\$ Possible duplicate: electronics.stackexchange.com/q/60977/17592 \$\endgroup\$ – user17592 Apr 6 '13 at 13:27
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    \$\begingroup\$ @Andy: Actually it would be just as easy to solve without the short. The difference is two parallel then series, versus two series then parallel. Same work either way. \$\endgroup\$ – Olin Lathrop Apr 6 '13 at 13:59
  • \$\begingroup\$ Andy possibly means that the short makes it easier to solve compared to if that short was replaced with a nonzero resistor. But if it is open, it is also easy to solve. \$\endgroup\$ – Kaz Apr 7 '13 at 7:47
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You calculate the equivalent resistance between two points by simplifying the circuit with respect to those two points. Combine parallel resistors and series resistors so long as they don't absorb one of your points.

Resistors are in parallel their terminals both share common nodes. For instance the 2 ohm and 6 ohm adjacent to node A are in parallel, and likewise for the 3 ohm and 6 ohm adjacent to terminal B. Start with that simplification and you're down to two resistors in the circuit - call them Rx and Ry. What does your picture look like then? Sometimes it helps to redraw the circuit to visualize this.

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Now I'm just going to move symbols around, but the following picture is exactly the same circuit:

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... which is also electrically identical to this:

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Does that make it more obvious what is in parallel? While R1 and R2 are in series I wouldn't combine them because node A would be absorbed by that simplification (it would be inside the simplified element).

Here's what it looks like when I collapse those elements using parallel resistance:

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It's obvious now that Rx and Ry are in series and can be combined right?

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A systematic way of determining resistors in parallel is to name every node in your circuit and associate with each resistor the Set of (two) nodes its terminals are connected to. Any other resistor that has Set Equivalence to that set is in parallel with it. A systematic way of determining resistors in series is if there is a node that appears in exactly two of the Sets those two resistors are necessarily in series. Sorry to get all computer-sciencey there. Hope it helps.

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  • \$\begingroup\$ @ vicatcu : Do you mean i let the voltage source tend to zero after the calculation. \$\endgroup\$ – HOLYBIBLETHE Apr 6 '13 at 13:22
  • \$\begingroup\$ its 7/2 ohms. The figure needed to be split along the short. \$\endgroup\$ – HOLYBIBLETHE Apr 6 '13 at 13:44
  • \$\begingroup\$ @ vicatcu : You have redrwan the figure which makes it is easy to solve. \$\endgroup\$ – HOLYBIBLETHE Apr 6 '13 at 13:49
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Look at the circuit carefully to see the real topology. Your circuit is drawn somewhat obfuscated and doesn't have component designators. Always put component designators on a schematic, especially if you are asking other people to talk about it. Here is the same circuit arranged more intuitively:

This should be easy to solve. If you have any problems figuring this out, ask what specifically you are stuck on.

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