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Find equivalent Resistance

In the given circuit, find the equivalent resistance between points A and B.

I solved this question in two different ways, out of which one gave the correct answer. The correct method is the most popular method: assuming the potential of given points and then calculating the others using the fact (or assumption) that in the absence of resistance, the potential difference is zero. This gives the correct answer: $R_{eq} = \frac{R}{3}$. I was confused regarding why the other method did not work.

Current follows the least resistance path when available. That is why you should not connect a conducting wire (R = 0) parallel to a resistor as it might lead to short circuit.

Using this, I can conclude that the current can flow in two possible paths: (1) from $V_A$ (leftmost) to $V_A$ (second from the right) to $V_B$ (rightmost) (2) from $V_A$ (leftmost) to $V_B$ (second from the left) to $V_B$ (rightmost).

Each of the paths has resistance R, so $R_{eq} = \frac{R}{2}$. But this is incorrect. Where is the mistake?


Update

Current follows the least resistance path when available.

The proof for this lies in the current divider rule. $$I_1 = I\frac{R_2}{R_1+R_2}$$ Putting $R_1 = 0$ we find that the entire current flows through $R_1$.

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    $\begingroup$ Scrub "path of least resistance" from your mind; that is a poor way of thinking about electricity. You can still be electrocuted by a high voltage even when your body is not along the path of least resistance to ground. This mistake kills people. $\endgroup$ – trentcl Jun 4 at 20:11
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    $\begingroup$ Your error is in the paragraph that starts with "Using this, I can conclude"... There are THREE equally valid paths, you omit the path of "VA (second from the right) to VB (second from the left)", using your terminology. $\endgroup$ – PcMan Jun 5 at 15:21
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I am used to smoothing out badly shaped circuits by pulling the wires:

enter image description here

Then I get a better circuit by cutting the extra wires:

enter image description here

So there are three resistors in parallel, indicating that the current flows through three possible paths.

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    $\begingroup$ This is awesome :) $\endgroup$ – Drake P Jun 5 at 17:18
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    $\begingroup$ @DrakeP - Thanks for your attention. $\endgroup$ – SG8 Jun 5 at 17:45
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    $\begingroup$ (+1) for the hands ;) $\endgroup$ – Nilay Ghosh Jun 6 at 3:30
  • $\begingroup$ @NilayGhosh - Thanks for your attention about that artistic part :) $\endgroup$ – SG8 Jun 6 at 16:34
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    $\begingroup$ This is surprisingly clear in very few words. Nice answer. +1 $\endgroup$ – Steeven Jun 8 at 8:10
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(continued from the previous answer by SG8)


Another way to think about this problem.

First:

enter image description here

Second:

enter image description here

Finally:

enter image description here

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Current follows the least resistance path when available.

This is wrong. Recall the Ohm's Law $$V=IR\Rightarrow I\propto \frac{1}{R}\ \ \ \ \ \ \text{if V constant.}$$ If there are two resistance one having resistance $1\ \Omega $ and the other $2\ \Omega$, connected in parallel then this doesn't mean that if the current will flow through $1\ \Omega$ resistance. Ohm's law says that there would be more current in the less resistant path.

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  • $\begingroup$ Okay, so a generalization of your statement is the current divider rule which states $I_1 = I \frac{R_2}{R_1+R_2}$ in case of 2 resistors. Putting $R_1 = 0$ we find that the entire current flows through $R_1$. This is the equivalent version of my statement. $\endgroup$ – S Das Jun 4 at 11:42
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    $\begingroup$ Your statement, assume one path to be a zero resistance path. It's not correct in general. You have to use Ohm's law. $\endgroup$ – Young Kindaichi Jun 4 at 12:22
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If you re-arrange the diagram you will find the you have three resistors in parallel. Another way to see this is to realise that (if we assume $V_A>V_B$) current flows from left to right through the first and third resistors, but from right to left through the middle resistor - so there are actually three paths from A to B, each with resistance $R$.

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  • $\begingroup$ The markings $V_A$ and $V_B$ are the part of method 1. Please explain using method 2 without assuming potential. And where is the mistake? $\endgroup$ – S Das Jun 4 at 12:44
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    $\begingroup$ @SDas If $V_B >V_A$ then you reach the same conclusion, only with current flowing in the opposite direction. Your mistake is where you say there are two possible paths from A to B - in fact, there are three paths, and each one has resistance $R$. $\endgroup$ – gandalf61 Jun 4 at 13:03
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Using this, I can conclude that the current can flow in two possible paths: (1) from $V_A$ (leftmost) to $V_A$ (second from the right) to $V_B$ (rightmost) (2) from $V_A$ (leftmost) to $V_B$ (second from the left) to $V_B$ (rightmost).

The critical mistake that you've made is that there's one more path that you've overlooked. Namely, the path from $V_A$ (leftmost) through the wire to $V_A$ (second from right), "backwards" through the middle resistor to $V_B$ (second from left), through the wire to $V_B$ (rightmost).

Each of these three paths has resistance $R$, so $R_{eq} = \frac{R}{3}$.

This whole argument about the three equal paths is equivalent to simply noticing that the three resistors are actually in parallel with each other.

(There's also the misconception about the path of least resistance, which is covered in Young Kindaichi's answer.)

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