Timeline for Will any charge flow in this circuit?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2016 at 18:08 | comment | added | MrAP | Let us continue this discussion in chat. | |
| Dec 18, 2016 at 18:06 | comment | added | glS | 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 | |
| Dec 18, 2016 at 18:04 | comment | added | MrAP | R=rr'/(r+r'). When r'->0, R->0. Right? | |
| Dec 18, 2016 at 17:59 | comment | added | glS | as I said before, put a small resistance $r'$ in those wires, solve the circuit, and see what happens when $r'\to 0$ | |
| Dec 18, 2016 at 17:58 | comment | added | MrAP | 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? | |
| Dec 18, 2016 at 12:55 | comment | added | glS | 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 | |
| Dec 18, 2016 at 8:57 | comment | added | MrAP | So in conclusion what do you think(Y/N):Will charge be flowing along equipotent points or not? | |
| Dec 16, 2016 at 18:51 | comment | added | glS | 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 | |
| Dec 16, 2016 at 18:39 | comment | added | MrAP | 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. | |
| Dec 16, 2016 at 17:28 | comment | added | glS | 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. | |
| Dec 16, 2016 at 17:26 | comment | added | MrAP | 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. | |
| Dec 16, 2016 at 16:45 | comment | added | glS | 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 | |
| Dec 16, 2016 at 16:01 | comment | added | MrAP | So will charge flow along 1-4 and 1-2 both, or along 1-4 only, or along 1-2 only? | |
| Dec 15, 2016 at 18:07 | vote | accept | MrAP | ||
| Dec 14, 2016 at 12:08 | history | edited | glS | CC BY-SA 3.0 |
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| Dec 14, 2016 at 11:59 | comment | added | MrAP | I think you mean V1=V4 and V2=V5. | |
| Dec 14, 2016 at 11:07 | vote | accept | MrAP | ||
| Dec 14, 2016 at 11:07 | |||||
| Dec 14, 2016 at 10:53 | history | answered | glS | CC BY-SA 3.0 |