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enter image description here I had to find the resistance of this circuit between A and B. My teacher said that since the blank wires have zero resistance, the potential difference across the ends of the blank wire is zero(according to v = ir) and thus 1,4 are equipotent and 2,5 are equipotent and the circuit is thus simplified.The equivalent resistance comes out to be 2r/5.

MY QUESTION:If the points 1,4 and 2,5 are equipotent how can current flow through the horizontal circuit.

MY REASONING: The potential difference across each branch of a parallel circuit is same and in this case it is 0 and hence no current can flow through the circuit.

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3 Answers 3

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You can explicitly compute the currents to see this.

Let us conventionally assume the currents are positive when they flow from left to right, let $V_i$ denote the potential at node $i$, and $I_1, I_2, I_3$ the currents respectively between the nodes 1-2, 2-4 and 4-5.

The blank wires shortcircuit some of the nodes, so that we have $$ V_1 = V_4 \qquad \text{ and } \qquad V_2 = V_5. $$

Applying the usual Ohm's law we see that: $$ V_2 - V_1 = r I_1, \\ V_4 - V_2 = V_1 - V_2 = 2r I_2, \\ V_5 - V_4 = V_2 - V_1 = r I_3 $$ from which it follows that $ I_1 = I_3 = - 2 I_2. $

As you can see the shortcircuits in the example do not prevent a flow of current in the circuit.

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  • $\begingroup$ I think you mean V1=V4 and V2=V5. $\endgroup$
    – MrAP
    Commented Dec 14, 2016 at 11:59
  • $\begingroup$ So will charge flow along 1-4 and 1-2 both, or along 1-4 only, or along 1-2 only? $\endgroup$
    – MrAP
    Commented Dec 16, 2016 at 16:01
  • $\begingroup$ if $A$ is at higher potential than $B$ the current will flow in the directions $1\to2, 4\to2$, and $4\to5$. That it flows in the direction $4\to5$ you can see it because due to the shortcircuit 1 and 4 are essentially the same, so $A$ being at an higher voltage means 4 being at an higher voltage than $B$, therefore the directionality of the flow. The calculations in the post also tell you that $I_1=I_3=-2I_2$ so that the other directions follow $\endgroup$
    – glS
    Commented Dec 16, 2016 at 16:45
  • $\begingroup$ Does negative current exist?I am not aware of negative current. Can you please explain it briefly. I have googled it and am not able to find something valuable. $\endgroup$
    – MrAP
    Commented Dec 16, 2016 at 17:26
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    $\begingroup$ you can think that between equipotent points charges can always "instantaneously" distribute themselves equally, because there is (ideally) zero resistance to the flow. I think the best way to think of equipotent points is to imagine that there is actually a resistence between them, just a very small one with respect to the others in the circuit. A good exercise in the present case would for example to put a resistence $r'$ between 1 and 4 and work out all the currents and voltage in this case. Than see what happens when $r'$ is small and how you recover the answers you got with your circuit $\endgroup$
    – glS
    Commented Dec 16, 2016 at 18:51
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It does not matter whether you have the same potential somewhere in your circuit. The main point is that you have a potential difference between A and B. Current will flow between A and B split over all possible paths, in your case you have three paths: 1-2-5, 1-4-5, 1-4-3-2-5.

The network of resistors can be replaced by an equivalent resistance which you correctly calculated to be 2R/5.

If you reshape the funny circuit in your question you will see that it is completely equivalent to the following more familiar circuit

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  • $\begingroup$ I said in the question that points having same potential have zero potential difference. I did not talk about zero potential. $\endgroup$
    – MrAP
    Commented Dec 14, 2016 at 11:09
  • $\begingroup$ Corrected zero->same. Does not change anything on the result. $\endgroup$
    – user1583209
    Commented Dec 14, 2016 at 11:10
  • $\begingroup$ You said"It does not matter whether you have the same potential somewhere in your circuit". What do you mean by this? $\endgroup$
    – MrAP
    Commented Dec 14, 2016 at 11:24
  • $\begingroup$ I mean that this is certainly not a sufficient condition for having no current through the whole circuit. As I tried to make clear in my answer there are three paths along which there is a potential difference so that current will flow along those paths. Also take a look at the link in my answer which is exactly the same circuit you have here. $\endgroup$
    – user1583209
    Commented Dec 14, 2016 at 11:33
  • $\begingroup$ You wrote that "Current will flow between A and B split over all possible paths, in your case you have three paths: 1-2-5, 1-4-5, 1-4-3-2-5" but {2,5},{1,4},{1,4},{2,5} respectively are all equi-potent. I do not see why current should flow in the whole circuit. $\endgroup$
    – MrAP
    Commented Dec 14, 2016 at 11:57
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enter image description here

This can surely help. As you can see potential drop across different resistors is V. Therefore current should flow in each resistors.

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  • $\begingroup$ You mean that charge will not flow along 1-4 and 2-5 but 1-2,2-3,3-4,4-5? $\endgroup$
    – MrAP
    Commented Dec 16, 2016 at 16:21
  • $\begingroup$ Yes. Because for current to flow through 1-4 and 2-5 there should be potential drop. $\endgroup$
    – user399511
    Commented Dec 16, 2016 at 16:23
  • $\begingroup$ I have another question in addition to the above. I have read that current chooses the least resistive path and according to this principle all the current should flow along the blank wire. What do you think? $\endgroup$
    – MrAP
    Commented Dec 16, 2016 at 16:24
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    $\begingroup$ No. You have read wrong. Current does not follow least resistive path. If it were to, in in parallel combination of resistor all the current should flow through the resistor of least resistance. Correct is: Current flow through the shortest path across which there is a potential drop. $\endgroup$
    – user399511
    Commented Dec 16, 2016 at 16:28
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    $\begingroup$ I cannot reconcile the opposing facts. You agree that current will not flow along 1-4 and 2-5 but the answers say that the short circuit draws a large amount of current. $\endgroup$
    – MrAP
    Commented Dec 16, 2016 at 19:17

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