Equivalent resistance in a circuit [duplicate]

I have this circuit and I need to find an equivalent resistance. I am very confused by the shape of the circuit because it appears like a more complex problem than I think it should be.

Ultimately, I think this is simply a series of 3 resistors and the total resistance should be 3R. Is this correct or do the extra wires matter?

• Hint: you can move everything around as long as you keep the topology the same. Try placing all points of the same voltage as $A$ at the top, and all points of the same voltage as $B$ at the bottom. Apr 17, 2017 at 4:33
• I gave wrong answer. I deleted it to avoid confusion. I agree it is R/3 now instead of R/2. Apr 17, 2017 at 5:01
• I suppose the first resistor has the same voltage as A and the third resistor has the same voltage as B but I don't understand how this shows it's parallel.
– M123
Apr 17, 2017 at 5:15

3 Answers

the answer should be R/3, the resistances are in parallel. you need to modify the circuit a little bit.

• Correct! I had just drawn that parallel circuit when your answer popped up. Apr 17, 2017 at 4:39
• I really can't figure out how this can be rearranged into a parallel circuit. What's the first step?
– M123
Apr 17, 2017 at 5:18
• @M123 I've uploaded a picture in gdrive: link . Hope it helps. Apr 17, 2017 at 5:36

Req = 1/(1/R + 1/R + 1/R) = 1 /3/R = R/3

The point after two R is A, and the point between two R is B. So just flip those points to make a single point of A and B, respectively. Then, you will see that the resistances are all in parallel. Thus, the total equivalent resistance is R/3.

• If you vote down this answer that means you don't understand it well. Apr 17, 2017 at 5:48

The extra wires absolutely do matter. There is a wire that shorts the first two resistors, meaning that there really is only one resistor in the circuit. So, the effective resistance is R.