# Equivalent resistance between C and D?

I hate to add to the list of equivalent resistance problems, but having browsed through the existing questions here and elsewhere online I am still unable to redraw the following circuit to find the equivalent resistance between points C and D. All the resistors have equal resistance R. Can this be solved by redrawing and applying Kirchoff's laws? Or must other techniques be used?

• If nothing is connected to the outgoing wires, it's just a bunch of resistors connected in parallel and serial manner. No Kirchoff is involved.. – Eugene Sh. Aug 9 '16 at 18:39
• 1st you can reduce the parallel resistors to a single equivalent, then do the same for the series ones. Remove the extra node labels you're not using - they're just there to confuse you (particularly since 2 of them are actually the same node). Redraw your circuit with the reduced/replaced equivalent resistors. Rinse & repeat. – brhans Aug 9 '16 at 18:40
• @SpehroPefhany This is the bit I can't intuitively understand - why are B and C the same node? Beyond this I have no issue with the problem, but was unsure how to justify B and C's equivalence. – user99755 Aug 9 '16 at 18:48
• In lumped-element circuit analysis (the kind you are doing) any nodes connected by a wire are exactly equivalent. That is a simplification, but a very useful one especially for low frequencies and low-ish currents. – Spehro Pefhany Aug 9 '16 at 18:52
• Re-draw the diagram with C on one side and D on the other, exclude A and B the answer should be clear. – Voltage Spike Aug 9 '16 at 19:01

$${{R_4 \cdot R_5} \over {R_4 + R_5} } + { {(R_2 + R_3)\cdot R_1} \over R_1 + R_2 + R_3} = {R^2 \over 2R} + {2R^2 \over 3R} = {7 \over 6} R$$ which goes in parallel with R6, so:
$$({7 \over 6}R^2) / ({13 \over 6}R) = {7 \over 13}R$$ should be the equivalent resistance.