0
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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.

Wheatstone bridge

$\endgroup$
1
  • $\begingroup$ Thank you! I see your point, I didn't realise this was alike a Wheatstone bridge. $\endgroup$ Feb 9 at 13:30
1
$\begingroup$

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 $$\frac{1}{R_\text{total}} = \frac{1}{2R} + \frac{1}{2R} = \frac{1}{R}\\ R_\text{total} = R$$

$\endgroup$
3
  • $\begingroup$ 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 $\endgroup$ Feb 9 at 13:29
  • $\begingroup$ 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 $2R$? $\endgroup$
    – uriyaba
    Feb 9 at 14:23
  • $\begingroup$ 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. $\endgroup$
    – noah
    Feb 9 at 14:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.