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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.

MrAP
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3 Answers3

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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.

glS
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  • I think you mean V1=V4 and V2=V5. – MrAP Dec 14 '16 at 11:59
  • So will charge flow along 1-4 and 1-2 both, or along 1-4 only, or along 1-2 only? – MrAP Dec 16 '16 at 16:01
  • 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 – glS Dec 16 '16 at 16:45
  • 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. – MrAP Dec 16 '16 at 17:26
  • when you are dealing with currents in this context you have to conventionally decide what direction corresponds to "positive current". In the post I specified this. Once you do it "negative current" simply means current flowing in the other direction. – glS Dec 16 '16 at 17:28
  • I understood everything you wrote. I am having a doubt. I intuitively think that if two points are equipotent, no charge flow along them but some answers on this website and many websites state just the opposite,i.e., most of the current flows through the least resistant path,i.e., the short-circuit. – MrAP Dec 16 '16 at 18:39
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    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 – glS Dec 16 '16 at 18:51
  • So in conclusion what do you think(Y/N):Will charge be flowing along equipotent points or not? – MrAP Dec 18 '16 at 08:57
  • see related question on this. I think the concept of "current" in a circuit only makes sense when there is a steady flow with some average velocity of the charge carriers. When the points are equipotential the charges can (ideally) redistribute themselves istantaneously between the points. In this case the concept of "current" is simply not meaningful, as the current can have any value. It is a case in which the underlying approximations of Ohm's law break – glS Dec 18 '16 at 12:55
  • One more concern, though the correct answer is 2r/5, i think that the zero resistance wires make the equivalent resistance of the respective parallel combinations zero and reasoning this way, the answer comes out to be 0. Is there anything wrong with this approach? – MrAP Dec 18 '16 at 17:58
  • as I said before, put a small resistance $r'$ in those wires, solve the circuit, and see what happens when $r'\to 0$ – glS Dec 18 '16 at 17:59
  • R=rr'/(r+r'). When r'->0, R->0. Right? – MrAP Dec 18 '16 at 18:04
  • solve for the current not the resistance. Anyway, comments are not for extended discussions, you can ask another question asking specifically for that if you want – glS Dec 18 '16 at 18:06
  • Let us continue this discussion in chat. – MrAP Dec 18 '16 at 18:08
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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

user1583209
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  • I said in the question that points having same potential have zero potential difference. I did not talk about zero potential. – MrAP Dec 14 '16 at 11:09
  • Corrected zero->same. Does not change anything on the result. – user1583209 Dec 14 '16 at 11:10
  • You said"It does not matter whether you have the same potential somewhere in your circuit". What do you mean by this? – MrAP Dec 14 '16 at 11:24
  • 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. – user1583209 Dec 14 '16 at 11:33
  • 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. – MrAP Dec 14 '16 at 11:57
  • I only wrote this to make clear the direction of current flow. Yes, {2,5} and {1,4} are equi-potent in theory. However in practice the connector from 1 to 4 and the one from 2 to 5 does have some small resistance (much smaller than the ones you have in the circuit). Therefore in practice you will have a small voltage between 1 and 4; and between 2 and 5. – user1583209 Dec 14 '16 at 12:17
  • Do you agree that there is current flowing in the circuit which I linked in the answer? – user1583209 Dec 14 '16 at 12:18
  • I don't know.I am confused. – MrAP Dec 14 '16 at 15:37
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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.

user399511
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  • You mean that charge will not flow along 1-4 and 2-5 but 1-2,2-3,3-4,4-5? – MrAP Dec 16 '16 at 16:21
  • Yes. Because for current to flow through 1-4 and 2-5 there should be potential drop. – user399511 Dec 16 '16 at 16:23
  • 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? – MrAP Dec 16 '16 at 16:24
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    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. – user399511 Dec 16 '16 at 16:28
  • But 1-4,2-5 are short circuits and this article says that all current flows through it:http://physics.stackexchange.com/questions/74368/path-of-least-resistance-vs-short-circuit – MrAP Dec 16 '16 at 17:12
  • Read the answers given to that question ( link of which you provide in the comment). They have explained the rest. – user399511 Dec 16 '16 at 17:31
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    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. – MrAP Dec 16 '16 at 19:17
  • In a wire electron ar in random motion. Such that If we place an Area A normal to the wire, the number of electrons crossing the A,say, from left to right will be equal to the number of electrons crossing A from right to left. Hence there is no current in the wire. For current to flow, We must apply electric field across the wire so that free charges gets accelerated as per F=qE. Hence, There will be some number of electron crossing A in the direction opposite to the applied Electric Field Therefore, We get current. – user399511 Dec 17 '16 at 04:57
  • Is electric field present across the short-circuit? – MrAP Dec 18 '16 at 08:55
  • Where ever there is a potential difference, There is Electric Field. – user399511 Dec 18 '16 at 09:53