5
\$\begingroup\$

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.

\$\endgroup\$
5
  • 4
    \$\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, 2014 at 18:23
  • \$\begingroup\$ @Ricardo, sounds correct. I didn't picture them in parallel. \$\endgroup\$
    – My Name
    Sep 16, 2014 at 18:25
  • 2
    \$\begingroup\$ All 'internal' nodes are at the same potential, hence shorted. \$\endgroup\$
    – copper.hat
    Sep 16, 2014 at 18:30
  • 1
    \$\begingroup\$ @copper.hat I can see it now. \$\endgroup\$
    – Ricardo
    Sep 16, 2014 at 18:40
  • \$\begingroup\$ @Ricardo: I made the same mistake. \$\endgroup\$
    – copper.hat
    Sep 16, 2014 at 18:41

2 Answers 2

Reset to default
12
\$\begingroup\$

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

\$\endgroup\$
10
  • 1
    \$\begingroup\$ @Null: They are in parallel and shorted. \$\endgroup\$
    – copper.hat
    Sep 16, 2014 at 18:32
  • 1
    \$\begingroup\$ @Ricardo: Look at the outer edges. All 'internal' nodes are shorted together. (I fell for it too!) \$\endgroup\$
    – copper.hat
    Sep 16, 2014 at 18:33
  • 5
    \$\begingroup\$ @copper.hat Ah, yes. Whoever draws schematics like this should be shot. \$\endgroup\$
    – Null ♦
    Sep 16, 2014 at 18:34
  • 1
    \$\begingroup\$ @Null: I know, I was 'certain' about my 7 answer for a while. \$\endgroup\$
    – copper.hat
    Sep 16, 2014 at 18:35
  • 1
    \$\begingroup\$ @Null This is simply an educational schematic to practice on. \$\endgroup\$
    – hryghr
    Sep 16, 2014 at 18:36
3
\$\begingroup\$

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\$.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ It fooled me for a while - I misread it like you did!! \$\endgroup\$
    – Andy aka
    Sep 16, 2014 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, 2014 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.