Another quick electronics quiz - Page 1
Forum section: electronics / general

author: KTP
author_rank: Frequent Contributor
timestamp: September 24, 2010, 07:39:41 pm
content: I stumped my professor with this little puzzle (but only for a minute or so).Hint: (You can solve this without touching pen or paper).

author: Zero999
author_rank: Super Contributor
timestamp: September 24, 2010, 08:03:45 pm
content: I admit I'm stumped. I couldn't even do it with a pen and paper.Nodal analysis wasn't my strongest area and I haven't done it since college as I've never had to do it in real life.

author: KTP
author_rank: Frequent Contributor
timestamp: September 24, 2010, 08:07:31 pm
content: Don't give up!You don't need to resort to nodal analysis.Another hint. Start analyzing by injecting a 1 amp test current at point A and pull it out at point B. KCL is your friend.

author: Bored@Work
author_rank: Super Contributor
timestamp: September 24, 2010, 08:32:07 pm
content: Ah, the cube. EE students are tortured with it since 1826. If you know how to solve it it is easy

author: jimmc
author_rank: Frequent Contributor
timestamp: September 24, 2010, 08:36:43 pm
content: Try thinking in 3 dimensions. Symmetry is your friend.Jim

author: Time
author_rank: Frequent Contributor
timestamp: September 24, 2010, 08:37:52 pm
content: 5 bananas

author: KTP
author_rank: Frequent Contributor
timestamp: September 24, 2010, 08:42:56 pm
content: Quote from: Time on September 24, 2010, 08:37:52 pmnice job. had you seen this puzzle before?

author: Time
author_rank: Frequent Contributor
timestamp: September 24, 2010, 08:45:29 pm
content: Quote from: KTP on September 24, 2010, 08:42:56 pmThis was just a freak coincidence. Back in college I put "5 bananas" as an answer to a lengthy physics problem on an exam and the prof. thought it was hilarious. Its just been my goto answer for every tricky problem I encounter. Sorry about that.

author: KTP
author_rank: Frequent Contributor
timestamp: September 24, 2010, 08:53:15 pm
content: Quote from: Time on September 24, 2010, 08:45:29 pmLOL! I thought you were actually answering the quiz!

author: Zero999
author_rank: Super Contributor
timestamp: September 24, 2010, 08:58:00 pm
content: Quote from: KTP on September 24, 2010, 08:07:31 pmI already did that, I can use a simulator, I'm not that stupid.Quote from: BoredAtWork on September 24, 2010, 08:32:07 pmI can now see it's a cube and the simulator told me it's 5 Ohms but I wouldn't know how to get there.

author: KTP
author_rank: Frequent Contributor
timestamp: September 24, 2010, 09:24:06 pm
content: Quote from: Hero999 on September 24, 2010, 08:58:00 pmSee if this helps. Mentally allow 1 amp to flow into node A and the same 1 amp to flow out of node B. The arrows show how the current divides up between the identical resistors. Pick a path and sum up the voltages across the resistors. Req = Vtotal / 1

author: Kiriakos-GR
author_rank: Super Contributor
timestamp: September 24, 2010, 09:32:31 pm
content: My brain uses always the " simple mode " ... and so I got 12 Ohm

author: Bored@Work
author_rank: Super Contributor
timestamp: September 24, 2010, 10:31:21 pm
content: Quote from: Hero999 on September 24, 2010, 08:58:00 pmIt is not only a cube, but A and B are on opposite ends. So everything is symmetrical. This is the key.Start at A. Because of the symmetry, the currents in R1, R4 and R12 (those directly connected to A) must be the same. We don't need to know how large they are, just that the three currents are the same. And because R1, R4 and R12 are each 6 Ohm, Ohm's law tells us there is the same voltage across them.Now for the next trick. Because there is the same voltage over R1, R4, and R12 their other ends (the ends not connected to A) are at the same voltage potential. Because of this one can connect these other ends together. It doesn't change a thing, because of the same voltage potential there will be no current over these added connections. But these connections parallel the three resistors, and one can replace them with a single one. Now comes into play that the professor did use 6 Ohm for each resistor. It makes calculation without a pocket calculator easy. 6 Ohm || 6 Ohm || 6 Ohm = 2 Ohm.Do the same for the resistors at B, R5, R7 and R8.Then you see that R2, R3 are also parallel. Replace them with R2 || R3 = 3 Ohm. Same for R9, R10.The rest should be straight-forward. Sum up the resistors and you should again get two parallel ones. I would now need to make a drawing, but I am to lazy to do it.Another way to solve this is to think of the cube, cut it along a plane through A-B into two pyramids, calculate the pyramids and parallel the resulting pyramid resistors. When cutting the cube, the resistors at the vertices you cut through need to be split into two 12 Ohm parallel resistors.

author: Mechatrommer
author_rank: Super Contributor
timestamp: September 25, 2010, 11:01:09 am
content: (5 minutes later) ooo ok... 5 bananas!

author: djsb
author_rank: Super Contributor
timestamp: September 25, 2010, 12:07:58 pm
content: 6 bananas.

author: jimmc
author_rank: Frequent Contributor
timestamp: September 25, 2010, 04:17:39 pm
content: For those who prefer pictures...http://www.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/64.42.htmlJim

author: Zero999
author_rank: Super Contributor
timestamp: September 25, 2010, 04:43:55 pm
content: That's better, thanks.

author: djsb
author_rank: Super Contributor
timestamp: September 25, 2010, 04:58:01 pm
content: So it's 5/6 x R = 5 Ohms.