I came up with the following question:

The vertices of tetrahedron ABCD are all connected by 1-ohm resistors. What is the total resistance between A and B?

I haven't the faintest idea how to solve it. Does it involve infinite path summations at some point?

[edit: how about from A to C?]


closed as off-topic by Kyle Kanos, Brandon Enright, David Z Apr 3 '14 at 5:49

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  • $\begingroup$ What's the definition of "total resistance"? $\endgroup$ – Ana S. H. Apr 2 '14 at 22:49
  • $\begingroup$ Welcome to Physics SE! This question might find a new home at electric engineering? $\endgroup$ – Stefan Bischof Apr 2 '14 at 23:16
  • $\begingroup$ @StefanBischof topically yes, but they've told me they don't take no-effort homework questions either. $\endgroup$ – David Z Apr 3 '14 at 5:49
  • $\begingroup$ ... why is it put on hold as off topic? $\endgroup$ – oink Apr 3 '14 at 14:55

By symmetry, C and D are at the same potential. You can merge them to one point, called CD. Then A to CD is $\frac 12 \Omega$, as is CD to B. You have two parallel $1 \Omega$ resistors, so the total is $\frac 12 \Omega$

  • $\begingroup$ How can C and D be at the same potential if there is a resistor between them? $\endgroup$ – LDC3 Apr 3 '14 at 4:06
  • $\begingroup$ @LDC3 Wheatstone bridge : ever heard of it? $\endgroup$ – evil999man Apr 3 '14 at 4:08
  • $\begingroup$ @LDC3 What does this condition imply about the current? So it's as though the said resistor isn't there. $\endgroup$ – WetSavannaAnimal Apr 3 '14 at 4:18
  • $\begingroup$ ABCD goes around the outside of the figure, not ACBD as you think it does. Where does it say a wheatstone bridge? It says tetrahedron. $\endgroup$ – LDC3 Apr 3 '14 at 4:33
  • $\begingroup$ @LDC3 If you connect 2 points at same potential with a resistor, will they cease to be at the same potential? $\endgroup$ – evil999man Apr 3 '14 at 4:46

Ross Millikan got you half the way there. The effective 1/2 ohm path through CD is also in parallel with the direct AB resistor, so the total resistance is 1/3 ohm.

  • $\begingroup$ Forgot CD to B to be in series with that 1/2? $\endgroup$ – evil999man Apr 3 '14 at 4:10

I think everyone forgot how to do this:

$\frac {1}{R_{eff}} = \frac {1}{R1} + \frac{1}{R2} + ...$

$\frac {1}{R_{eff}} = \frac {1}{1} + \frac{1}{3}$

$\frac {1}{R_{eff}} = \frac {4}{3}$

$R_{eff} =$ 0.75 ohms

  • $\begingroup$ I don't follow this answer. Could you please explain it more? $\endgroup$ – Brian Moths Apr 3 '14 at 4:11
  • $\begingroup$ You have serious misconceptions about words series and parallel. $\endgroup$ – evil999man Apr 3 '14 at 4:11
  • $\begingroup$ OK, I took a short cut. We have two parallel paths, one path is one 1-ohm resister and the other path is three 1-ohm resistors. So the 2 parallel paths are 1/1 + 1/3, which is what I have above. $\endgroup$ – LDC3 Apr 3 '14 at 4:36
  • $\begingroup$ Well, there is ACB also. You can either apply symmetry or kirchoff's laws $\endgroup$ – evil999man Apr 3 '14 at 4:48

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