The following is a question from a practice Physics GRE exam (found online at ETS's website).
The circuit shown in the figure consists of eight resistors, each with resistance R, and a battery with terminal voltage V and negligible internal resistance. What is the current flowing through the battery?
Here is the figure in question:
In case the figure doesn't load, here's the problem from ETS: https://www.ets.org/s/gre/pdf/practice_book_physics.pdf, page 54, problem 68.
Since this is a GRE question I figured there was a shorter approach than a brute-force, Kirchoff's Voltage Law around every loop. I've been trying to figure out what a simpler solution would be by reducing parallel/series resistors, which currently there aren't any that can be reduced that way. Then I tried a triangle-star conversion, where I basically reduced each closed square in the diagram (each square was essentially a triangle because each had a resistor-less wire on one of the sides) but that also didn't lend itself to parallel/series reduction. I couldn't figure out another approach after that.
The correct answer is 3/2**V/R*, and an abbreviated solution I saw someone else put online (http://physicsworks.files.wordpress.com/2012/09/gr0877_solutions.pdf, problem 68, if you're interested) said to treat the problem as 3 separate 2 R resistors in parallel, but I am not sure how that works, because there are still the two horizontal resistors. Could someone explain to me either why this approach to the problem is correct or an alternative approach to the problem? Thank you!