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enter image description here

I'm trying to self-study this problem. This chapter explained the delta-to-y transformation but I can't seem to find one here.

My best guess is that both resistors on the right are short-circuited which gives them a value of 0. Then you add the two 3 ohm resistor on the left together, and thus the answer 6 ohms.

But I don't know if this is true.

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    \$\begingroup\$ To me, the answer is 7 Ohms. The 3 resistors to the right are in parallel with each other and can be redrawn as such if you move the two rightmost nodes to the left (over the vertical resistor terminals). \$\endgroup\$ – Ricardo Sep 16 '14 at 18:23
  • \$\begingroup\$ @Ricardo, sounds correct. I didn't picture them in parallel. \$\endgroup\$ – George Chalhoub Sep 16 '14 at 18:25
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    \$\begingroup\$ All 'internal' nodes are at the same potential, hence shorted. \$\endgroup\$ – copper.hat Sep 16 '14 at 18:30
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    \$\begingroup\$ @copper.hat I can see it now. \$\endgroup\$ – Ricardo Sep 16 '14 at 18:40
  • \$\begingroup\$ @Ricardo: I made the same mistake. \$\endgroup\$ – copper.hat Sep 16 '14 at 18:41
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You're right. The three resistors 3 \$\Omega\$, are shorted, so that the equivalent resistance is the sum 3 + 3 = 6.

enter image description here

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    \$\begingroup\$ @Null: They are in parallel and shorted. \$\endgroup\$ – copper.hat Sep 16 '14 at 18:32
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    \$\begingroup\$ @Ricardo: Look at the outer edges. All 'internal' nodes are shorted together. (I fell for it too!) \$\endgroup\$ – copper.hat Sep 16 '14 at 18:33
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    \$\begingroup\$ @copper.hat Ah, yes. Whoever draws schematics like this should be shot. \$\endgroup\$ – Null Sep 16 '14 at 18:34
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    \$\begingroup\$ @Null: I know, I was 'certain' about my 7 answer for a while. \$\endgroup\$ – copper.hat Sep 16 '14 at 18:35
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    \$\begingroup\$ @Null This is simply an educational schematic to practice on. \$\endgroup\$ – hryghr Sep 16 '14 at 18:36
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This answer is wrong, as pointed out by Tinchito, but I'm leaving it here shamelessly for didactic reasons, as it seems as the most common mistake when looking quickly at the circuit.

At first sight, you think you could redraw the schematic as follows (but you can't):

schematic

simulate this circuit – Schematic created using CircuitLab

If that was the case, then,

$$R3||R4||R5 = 1 \Omega$$

and

$$R_{eq} = R1 + R2 + (R3||R4||R5) = 7 \Omega$$

Looking closer you'll see that there's a zero resistance path that shorts out all three resistors, as pointed out by Tinchito, making the correct answer \$6 \Omega\$.

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    \$\begingroup\$ It fooled me for a while - I misread it like you did!! \$\endgroup\$ – Andy aka Sep 16 '14 at 20:36
  • \$\begingroup\$ @Andy - Yeah, it fooled many others here, too, it seems. That's probably the intention of the circuit's original author. I bet he or she could be a magician or illusionist. \$\endgroup\$ – Ricardo Sep 16 '14 at 21:11

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