# Equivalent resistance of this circuit [closed]

The circuit below has a equivalent resistance but I couldn't find it, I found it interesting because it seems a trciky configuration, how would you solve it? The equivalent resistance gotta be $$R$$, (I know it because I used a simulator). Hint: This following circuit can be thought of as Wheatstone bridge, in which the middlemost resistor has no current flowing akin to your $$2R$$, in the question. • Thank you! I see your point, I didn't realise this was alike a Wheatstone bridge. Feb 9 at 13:30

This one does look tricky at first glance. However, you can use the symmetry of the circuit to come to a conclusion.

Because the outer paths (going through the top and right; and through the bottom and left resistors, respectively), are completely equivalent, the current through these two paths must be the same. But then the voltage drop across the top $$R$$ and the leftmost $$R$$ are also identical, meaning there is no potential difference across the central resistor with $$2R$$.

So this is like the two outermost paths connected in parallel, giving $$\frac{1}{R_\text{total}} = \frac{1}{2R} + \frac{1}{2R} = \frac{1}{R}\\ R_\text{total} = R$$

• Well, I see. thank you. It's difficult to me realising if a resistor has current or not. I guess I have to practice more Feb 9 at 13:29
• Shouldn't $R_{eq} = (R+R) \| (2R) \| (R+R) = \frac{2R}{3}$, since both $R$ pairs are connected in series, and both are connected in parallel to $2R$? Feb 9 at 14:23
• From symmetry there is no current across $2R$, so we might as well treat that as an open line. They are also not both in parallel with the $2R$, the current through $2R$ would have to pass through one of the $R$ on either side.
– noah
Feb 9 at 14:27