GR0877 Problem 68
PROBLEM STATEMENT: This problem is still being typed.
SOLUTION: (D) The total resistance of the circuit is equal to the resistance of three 
Found a typo? Comment!
Jump to the problem
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
| 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 |
| 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |
| 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 |
| 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70 |
| 71 | 72 | 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80 |
| 81 | 82 | 83 | 84 | 85 | 86 | 87 | 88 | 89 | 90 |
| 91 | 92 | 93 | 94 | 95 | 96 | 97 | 98 | 99 | 100 |
Want to get a full PDF with solutions? Read THIS.
The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.
Physicsworks Blog 
Because of the symmetry of the problem, there is no potential drop across any of the horizontal resistors. This means you can just yank them out of the circuit, which simplifies the calculations as noted above.
I’m confused as to how we know there’s no potential drop across the horizontal resistors. If current I flows through the whole circuit, then I/3 flows through each arm after the first junction. Then, on the left and right, the current splits into two equal parts because the resistances after those junctions are equal. Now, we have I/6 flowing from the right arm towards the center and I/6 flowing from the left arm to the center. At the center, there are three paths of equal resistance, so the current into that junction splits into three equal parts of I/9. To me, it looks like there’s a net flow of current toward the center. Where have I gone wrong?
Thanks!
I am not sure exactly where you went wrong, but I’m thinking about it like this: take one of the horizontal resistors — each is pulling the exact same amount of current from their left and right, so net current across either of the horizontal resistors is zero. In fact, it’s like they’re not even there (or the whole section of the circuit each is on). So the problem becomes three 2R resistors in parallel. …I hope this helps.
Makes sense now. How would you deal with it if there were no symmetry?
If there is no symmetry, than you should use Y-Δ transform: http://en.wikipedia.org/wiki/Y-Δ_transform
But this is unlikely to be on PGRE.