A circuit problem.

  • Thread starter pixel01
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  • #26
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the real puzzling thing here for me is that the component at the junction say A in the following diagram
0-A-0
...0

it seems to be that A will have to be a 3-terminal device rather than just the usual 2-terminal (+/-).... so how do you actually connect them? Or the circuit concept is just a misnomer and has absolutely no relevance to the problem... in that case, it would simply be a graph theory type problem

The battery B and all bulbs An have only 2 terminals each. We must consider one terminal (say -) is grounded and it is not needed to draw in the circuit. So all the lines connecting knots are + terminal.
 
  • #27
mjsd
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mmm... that seems even more weird... because that would imply all three
A1-B-A2 , A1-A2-B, A2-A1-B are equivalent in the circuit sense, just two bulbs in parallel with the battery, all you are doing is exchange the order of the branches
eg. A1-B-A2 is really

.+-----+------+
.|........|........|
A1.......B.......A2
.|........|........|
GND....GND...GND

and A1-A2-B is really

.+-----+------+
.|........|........|
A1......A2.......B
.|........|........|
GND....GND...GND

since all connections are assumed to be ideal connecting wires, it doesn't matter whether A1 is further or closer to B than A2, circuit properties like voltage drop, current through, are the same. So,...I am confused?
 
  • #28
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Hi Mjsd,
You have drawn correctly. Electrically, all the circuits are equivalent, but mathematically they are different.
Even in the electrical aspect, they have some differences. For example considering the posibility of each bulb to be off, in case B-A1-A2, the possibility of A2 is off is double that of A1, provided that every line has the same posibility of disconnected.
As I said at first, this problem is more in mathematical than in electrical aspect.
 
  • #29
mjsd
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Hi Mjsd,
You have drawn correctly. Electrically, all the circuits are equivalent, but mathematically they are different.
Even in the electrical aspect, they have some differences. For example considering the posibility of each bulb to be off, in case B-A1-A2, the possibility of A2 is off is double that of A1, provided that every line has the same posibility of disconnected.
As I said at first, this problem is more in mathematical than in electrical aspect.

mmm..... B-A1-A2 can be arranged to look exactly the same as say A1-B-A2; all you need is have the common node connecting all three devices as the center of the circuit, then the three of them connects into that center node in a symmetric manner (ie. separated by 120 deg)...get some kind of a star shape. and via rotation by 120 or 240 deg you get the other two set up... I still don't get what you mean by "inequivalent" or "equivalent"
 
  • #30
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Here I draw the two circuits : B-A1-A2 and A1-B-A2.
In the first case, A2 is more probable to lose power than A1. In the second circuit, they are the same.
 

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  • #31
mjsd
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see my diagram when it is ready...hopefully it will convey what I mean before.

case I and II are effectively the same after redrawing (unless you have some weird rules governing your system)
 

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  • #32
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No, it is not like that. You can not stretch the wire that way. Imagine, the bulbs are far from each other, so in the case I, you connect from B to A1 and then from A1 to A2. It is completely different from your circuit.
You have created one more node in your drawing. The requirement here is that the wire can only be connected to the battery or a certain bulb, not to common node.
 
Last edited:
  • #33
mjsd
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tell me how you can connect two wires to one terminal without "making one more node". My point is coming from elementary circuit theory point of view.. if you know what I mean. you wanna connect two wires to the same terminal of the device, the "new node" that you have created is the point where the two wires and the terminal meet, before connection there is no (essential) node. You have a (essential, 3-way) node after connection because conservation of "stuff" holds at that junction (eg. KCL)...
you said that wires can only be connected to battery/device... true... but nothing forbids current to go from A1 to A2 without entering B in between, so there will be a direct link beween A1 and A2 even when u have B in between. This is the effect of connecting two wires together a same point.

I brought this issue up because there may be problems in defining what are equivalents and what are not when you have tree structure that may have some rotational symmetry in them.
 

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