Circuit Problem: How Many Ways to Light All the Bulbs?

[pixel01]

Hi everyone.

Now I ve got this problem, but not a homework. I hope some of you can help. There is a battery say B, and n small bulbs A1, A2...An. I can make a circuit as : B-> A1-> A2...An . I can also change it to B->A2-A1...An . Of course there are many ways to make a circuit so that all the bulbs will be lit.
The question here is: how many possible ways are there to make a circuit.

 

[mjsd]

do u just want to connecting in one loop or multiple loops etc... ? the number will be drastically different

 

[pixel01]

Thanks for your replying. There's no loop. It can be a tree or a series, but not a loop.

 

[olebuckeye]

Being from the old school I am having a bit of trouble trying to decide what you are describing, must be new math. But from what I can glean from your post then I would say that the answer would be close to infinity. You have not given the battery capacity, lamp watts, etc.

 

[pixel01]

In fact this is a math problem, not a electrical. The capacity of the battery is consider to light all the bulbs given. Here is my drawing for the case that there are only two lamps A1 and A2. In this case, there are 3 possible ways to make the circuits. If there are 3 bulbs, a tree circuit can be drawn.

 

[pixel01]

I hope this thread could be moved back to the maths box.

 

[olebuckeye]

Since I have to take my shoes off to count over ten then this is out of my experience in regards to math. But by looking at your drawing then it looks like you have two circuits in series and one in parallel. Since lamps are a resistive load then this changes the properties of the circuit.

 

[pixel01]

As I mentioned above, this one is not electrical. We have to consider the battery is infinitely strong ! and also all the bulbs are circuited parallelly meaning they all have one potentials without any voltage drop even if the number of bulbs could be very high.

 

[olebuckeye]

If you have a infinitely strong battery and a n number of lamps then I still don't fathom the question other than my first response to your post. The battery could be made of half of the material in the universe and the lamps made out of the other half. Then the brightness of the lamps would still be decided if the circuit was series or parallel.

 

[olebuckeye]

I wasn't trying to be funny but you posted it as an electrical problem and I thought that maybe the math was leading you to think that a electrical solution would find the answer. This is an electrical forum and you will receive electrical answers.

With an infinite power souce and infinite lamps then the answer would be infinite. In physics they say that when infinities are encountered then there is a problem with the math.

In the answer box I would draw a figure 8 sideways.

 

[pixel01]

Thanks any way. I will try to post it again in the maths box.

 

[mjsd]

mmm... sounds like you want to see how many ways (topologically inequivalent) you can build a tree with N distinct objects. A graph theory type problem i think...
but firstly, what is the definition of the bulb being lit? could there be a situation when battery is not connected to any bulbs? or are you really saying that all N distinct objects must be included in the tree graph?

anyway, it seems that a systematic way to do this is to first work out how many topological inequivalent graph can you have (assuming all items are the same first). then for each inequivalent graph work out the number of ways you can permute the N items. mm...not going to be easy, perhaps graph theory may help

 

[pixel01]

You seem to make the question more clear. The problem here is that each bulb is different from any other bulb. That's why I have numbered it A1, A2..An. For this reason, just when you make it a circuit in series, meaning B-A1-A2...An and you permute the knots, there are (n+1)j!/2 ways. But you can also make it trees and the number rises rapidly, I can not find the final solution. Hope anyone can help this.

 

[LM741]

pixel01>> do we assume : B-A1-A2 is the same as A2-A1-B?

 

[dlgoff]

Looks like (n+1)!/2 to me. The 2 would come from B-A1-A2... being the same as ...A2-A1-B. Or am I wrong?

 

[pixel01]

pixel01>> do we assume : B-A1-A2 is the same as A2-A1-B?

Yes, that's right.

 

[pixel01]

Looks like (n+1)!/2 to me. The 2 would come from B-A1-A2... being the same as ...A2-A1-B. Or am I wrong?

(n+1)!/2 is true only for the case of series: B-A1-A2..An. If n>2, we can make it trees and the final number must add up.

 

[Dlimer]

Hello, I am new to this forum so Hi everyone. I did not read all the replys to your question yet but the answer is 3. Series, Parallel and a combination thereof. I hope that is what you were looking for.

 

[Dlimer]

If the drawing given was expected to work, it won't. You must have a loop of some sort or at least a ground for electron flow. Your diagrams you gave are not good. Sorry.

 

[mjsd]

(n+1)!/2 is true only for the case of series: B-A1-A2..An. If n>2, we can make it trees and the final number must add up.

So that's why I suggested to first work out all inequivalent topology with same items for all nodes first, then work out how many permutation you can have for each topology,..eg: for 5 items

0-0-0-0-0

0-0-0-0
...0

...0
0-0-0
...0

0-0-0
...0
...0


then fill in the boxes... the first one 0-0-0-0-0, there are (n+1)!/2
the second one you probably get 5 x n!/2 (you will have to check these)

by the way, can you do a three or four way connection?
say

0--|--0
...|
...0

do a few for n small first then you may be able to find a pattern and extrapolate

 

[pixel01]

Hi Mjsd, the problem is the knots are not identical, they are numbered so the number goes up very fast when n increases. For n=2, I can find 3 ways. n=3 the number of ways is 12+3=15. The thing here is how to find a formula to calculate N=number of ways for a given n (bulbs).
To Dlimer: For simplicity, I do not draw the ground wire. All the bulbs and battery have been grounded, that's why they need only one line connecting to each other.

 

[mjsd]

ok, I guess I don't know you "circuit" well enough from just these descriptions.
for n=2 you only have one type of topology (right?) namely
0-0-0
but within this topology there are 3 different arrangements of nodes:
B-A1-A2
A1-B-A2 (or A2-B-A1)
A1-A2-B
so total is 3

now for n=3 there are two types of topology (right?) disregarding the content of each nodes for the moment... (are we on the same spectrum here?)
0-0-0-0

0-0-0
...|
...0

for first type you get 12 different arrangements
for 2nd type you get 4 different arranagements:
A1-A2-A3
...B

A1-A3-A2
...B

A2-A1-A3
...B

A1-B-A3
...A2

right? are these what you meant by possible different arrangements. Or is the battery somehow "special" which reduces the number of combinations you can get. say excluding the last one above (leaving you with 3 only)

my strategy was that get "shape" of circuit first then decide the arrangements for each shape... unless the battery is somehow special.. in that case you will have to specific the battery node when working out the "shapes".

 

[pixel01]

Oh, I made a mistake with n=3. There 4 cases when you make it a tree. So totally is 16. You have got to the point. Now that you can count with n=5 or 6 .. but n=10 or more it really difficult. I think there should be a formula to calculate the number of ways N according to n.
It's really difficult. Let say when n=5, you also have possibility to make trees with one and two branches!

 

[Dlimer]

When I look at your diagrams, they only show a pure parallel circuit (one not three). You are throwing with the discussion of a "tree circuit". Even if you add a "branch" onto any of the diagrams you presented, the new component or bulb is still parallel to the others. So, either I am missing something or ,I suggest, you really do not have enough information for the question.

 

[mjsd]

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

 

[pixel01]

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.

 

[mjsd]

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?

 

[pixel01]

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.

 

[mjsd]

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"

 

[pixel01]

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.