Electricity question

yonski · 2006-11-21 08:52:19

Okay, so this is definitely not maths, but it's really bugging me and i thought someone round here might be able to help :-) I've been given the following diagram and been asked to find the total resistance and the current in the circuit...

Apologies for the rubbish drawing (the jagged lines are supposed to be resistors btw lol). Each resistor has a resistance of 10 ohms. I think the idea is to simplify the circuit into something that can be worked with more easily, but i'm not quite sure how.

Thank you!

mathsyperson · 2006-11-21 09:26:47

This looks like a very interesting problem. I'll open it up to anyone who might want to try it but doesn't know the necessary formulae.

There are two things you need to know.

If there are resistors like this (in series): ------R------R------ , then the total resistance of that is determined by adding the resistances of the two resistors.

If there are resistors like this (in parallel):

-------R------
------| |----------
-------R------

Then the total resistance is determined by taking the reciprocal of all the resistances, adding all of those, then taking the reciprocal of the total. For example, for two resistances of 10 ohms, you'd do 1/10 + 1/10 = 1/5, and the reciprocal of that is 5, so the system's resistance is 5 ohms.

That's all you need to know to find the resistance, and once you have the resistance of the system, finding the current is fairly trivial.

I'll make another post if I manage to unravel this puzzle before anyone else does.

luca-deltodesco · 2006-11-21 10:19:29

according to crocodile clips, it would have a current of 426mA, which would mean those resistors have a resistance of ~117ohms

mathsyperson · 2006-11-21 10:23:22

I'm not *entirely* certain that this is right, but I'm pretty certain, so here goes:

The image shows the original circuit at a, and at each stage I have simplified it until we get f.

Going from a to b, I have just moved the points where the circuit branches to the start of the circuit. But it's the same thing topologically, so that's OK. The 2 curly lines each have a resistance of 20 ohms, as do the big straight lines on the left and bottom.

From b to c, I've recognised that 2 of the lines are in parallel, and so combined them. The diagonal line now has a resistance of 10 ohms.

From c to d, I've "unzipped" the circuit, so it branches at the start, rather than in the middle. But again, the topology remains the same. The two diagonal lines have a resistance of 20 ohms each.

From d to e, I've combined some more lines than were in parallel with each other. In both cases, the diagonal path has a resistance of 20 ohms and the other path has a resistance of 30, so combining them gives a resistance of 12.

From e to f, I've added the last bit of resistance to both diagonal paths, making them 22 ohms each, and then combined them because they were in parallel. Therefore, the circuit's total resistance is 11 ohms.

And if that's correct, then we can follow through to find that the circuit's current is 50/11 ≈ 4.5A.

If anyone thinks I've done something wrong or doesn't understand one of my steps, please do say so. I'm not completely sure that I'm right. Of course, if someone wants to say that they agree with everything that I've done and that I'm awesome in every way then that's fine too.:)

Edit: To Luca, there's no way that the resistance can be that high. Even if the electricity only went along one of the possible paths, the resistance would only total 40 ohms, and with all the alternative paths that run parallel to it, the resistance will definitely be lower than that. I don't know what Crocodile Clips did, but it did something wrong.

yonski · 2006-11-21 11:07:48

Yeah, i was kind of looking at it the same way but couldn't figure out how to combine them like you have. It all seems logical now though. We kept answering all the worksheets really quickly in class so my physics teacher was like 'right, this one'll sort you out' lol. Tomorrow i'm going in with a big grin on my face Thanks mathsyperson.

P.S. You are a little bit awesome, maybe.

mathsyperson · 2006-11-22 05:52:33

I think I've found a problem with my solution.

In one of the steps, I change this kind of setup:

-----------R-----------
-----R----| |-------
-----------R-----------

into this:

-------R-------R-------
----------| |---------
-------R-------R-------

by 'splitting' the first resistor up. Unfortunately, those two situations give different overall resistances, meaning that that reasoning doesn't work. Back to the drawing board...

yonski · 2006-11-27 08:57:38

Hmm yes, i see your problem.

Having looked at it again, I think the way to look at it is using symmetry and some PD logic. If you divide that bit of the circuit up symmetrically as follows

Now the PD at each of the identically coloured junctions is going to be the same, so as far as electricity is concerned they are the same point. This means that you can look at the circuit as 4 sections, each with resistors in parallel, and these 4 sections are themselves connected in series. Hence we have

R1 = (1/10 + 1/10)-¹ = 5
R2 = (1/10 + 1/10 + 1/10 + 1/10)-¹ = 5/2
R3 = R2 = 5/2
R4 = R1 = 5

So the total resistance is 5 + 5/2 + 5/2 + 5 = 15 ohms. And that gives the current as 50/15 = 3.333 amps.

Last edited by yonski (2006-11-27 08:59:32)

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#1 2006-11-21 08:52:19

Electricity question

#2 2006-11-21 09:26:47

Re: Electricity question

#3 2006-11-21 10:19:29

Re: Electricity question

#4 2006-11-21 10:23:22

Re: Electricity question

#5 2006-11-21 11:07:48

Re: Electricity question

#6 2006-11-22 05:52:33

Re: Electricity question

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