# Equivalent Resistance

Discussion in 'Homework Help' started by gajoline, Aug 28, 2013.

1. ### gajoline Thread Starter New Member

Aug 28, 2013
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I am really confused , i can't figure out which of them are in parallel and which are in series !!!
Can you help me to find a way to figure this out ...

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2. ### Ridgerunner New Member

Jan 25, 2009
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Try Redrawing the Circuit.

hint: 4R and R are in parallel

3. ### studiot AAC Fanatic!

Nov 9, 2007
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It usually helps to redraw the network in a more easily recognisable form, preferably so that it falls into sections.

Here is one way.

What was the question, you will need to tell us your thoughts for more detailed help.

EDIT this diagram is incorrect see post#6

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4. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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Am I missing the 80Ω resistor?

5. ### WBahn Moderator

Mar 31, 2012
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Redrawing can definitely make things jump out at you. But you can also do it with the drawing as given.

If a node connects exactly two components, those components are in series since whatever current flows out of one component must flow into the other. If there are not exactly two components, they are not in series.

If you can put your two index fingers on the two modes on each side of a component and then, by sliding you fingers along those two nodes, get to the oopposite sides of another component, those two components are in parallel because they have the same voltage across them. If you can't do this, then they are not in parallel.

Of course, this description is assuming two-terminal components. It's a bit more complicated (and less well-defined) for components with more terminals.

6. ### studiot AAC Fanatic!

Nov 9, 2007
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Very good catch.

Here is the revised network I have redrawn it at 1 including the 80Ω resistor.

Fig2 highlights the first parallel R and 4R

Fig 3 shows how once you have the parallel combination of R & 4R this is in series with the 20 ohm resistor.

Fig4 shows that the series combination is in parallel with the 25 ohm resistor

and so on

Perhaps this exercise was meant to show how quickly the analysis by series & parallel can become complicated.

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7. ### gajoline Thread Starter New Member

Aug 28, 2013
10
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Ok , i think i solved it !!
Thank you all for your help especially studiot
It took me a while to figure it why the 4R resistor doesnt connect with the 25 ohm but i think i got it....
Do you guys have any other circuit that i can solve?? I really need the practice and i will post the answer here to confirm it !!

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8. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes it's easier with numbers than symbols.

Did you note how I redrew the network

Removing the diagonal in the top version of your circuit.

Then stretching the middle section to allow proper connectivity for the 25Ω in 1

Then expanding the lines around the 40Ω, 80Ω and R to bring out the next parallel combination?

I'm sure if you do a forum search you will find lots of worked examples, the issue comes up again and again.

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9. ### WBahn Moderator

Mar 31, 2012
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Often times, just a single change to get rid of a crossing wire will let you see how to simplify things. Consider

With that wire moved to the outside, we can quickly see how a minor redraw yields the second version. Basically, the junction of of the 4R and 25Ω resistors folds down to the bottom right node while the 20Ω resistor remains connected to the junction of the R and 4R resistors.

With this one, we can literally get the solution by inspection.

R || 4R = R * (4*1)/(4+1) = (4/5)R = 80Ω

This is in series with the 20Ω, giving us 100Ω.

This is in parallel with 25Ω, giving us (2500/125)Ω = 20Ω

This is in series with 80Ω, giving use 100Ω.

This is in parallel with 100Ω, giving us 50Ω.

This is in series with 10Ω and 40Ω, giving us a total of 100Ω.

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10. ### WBahn Moderator

Mar 31, 2012
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This problem provides a good example of where estimation can be very useful.

Let's say that you analyze the circuit and come up with a total resistance of 200Ω. Would you have caught that this was wrong?

A very, very, very good habit to get into is to look at bounding cases and put limits one what the correct answer has to be within. This is part of asking if the answer makes sense. If you actual answer is not within those bounds, then you KNOW the answer is wrong (though, in fairness, you do have to allow for the possibility that you computed a bound incorrectly, but in either case you have a conflict that must be resolved before you proceed).

Let's look at the original circuit and put some bounds on the final answer. Everthing to the right of the leftmost R resistor is in parallel with that R and will have an equivalent resistance of somewhere between 0 and ∞. At those two bounding cases, the R either goes away, leaving us 50Ω, or everything else goes away leaving us 150Ω. Thus, just a brief, casual inspection tells us that any answer we get that is outside the range 50Ω to 150Ω is wrong.

If all else fails and you get hopelessly mired down in finding an answer, you can always punt and say that the answer must like within this range and, therefore, the best estimate is 100Ω. You would express this as 100Ω±50Ω. Now, this turns out to be the correct answer for this particular problem, but that is sheer coincidence. But the fact remains that the actual correct answer is guaranteed to be within 50% of your best estimate, which in many cases in the real world would be good enough right there! Given that many people (in school and in the real world) frequently make mistakes that yield answers that are more than an order of magnitude off without batting an eye and scratching their head, this puts you a leg up on the competition already.

With a little more effort, we can bound this a lot tighter. Looking at the network that makes up the parallel resistance to that leftmost R, we can see that the lowest it can be is 80Ω because any path has to go through that resistor. Next we can quickly examine all of the paths from one end of R to the other looking for the single path with the lowest resistance, knowing that this sets a maximum value for out parallel resistance since all other paths will reduce it (and it doesn't matter if we don't actually find the best path, it just means that we don't get as good a bound as if we had). So this is clearly the path through the 25Ω resistor and down to 80Ω resistor and back for a total of 105Ω. So now we know that the parallel resistance must be between 80Ω and 105Ω. We could just note these off to the side and check the actual parallel resistance against them once we find it and before we finish up the circuit, which saves us some math (and actually makes for a better check because it narrows down what part of the computation has gone astray).

If we were to run the numbers real quick we would find that we have reduced our range to between 94.5Ω and 101.2Ω. Even without calculating these and only checking for the parallel resistance to be between the 80Ω and 105Ω, we have guaranteed that we cannot make a mistake that is going to result in an answer that is more than 5.5% off from the correct answer without catching it. This is almost always good enough in the real world, since these are likely 5% tolerance resistors to begin with!

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11. ### WBahn Moderator

Mar 31, 2012
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Huh? It DOES connect with the 25Ω resistor. Directly.

My guess is that you mean that it seemed to you that they should have been in series. If so, the key is to note that the node connecting them connects things other than just throse two resistors, thus they are not in series.

There are a lot of good things you can still do with this problem, too.

One would be to calculate what the maximum power that could be dissipated by this resistor network if every resistor had a limit of 1W. Which resistor is the limiting factor?

Another would be to ask what the total dissipation would be if the most likely loaded resistor were dissipating 1W and how much would the most heavily loaded resistor have to dissipate?

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12. ### gajoline Thread Starter New Member

Aug 28, 2013
10
0
Thanks for your time WBahn !!
I can't fully understand this: The 25 ohm is connected with the wire that leads to the 80 ohm and the 4R , how can we skip the 4R connection in the redrawn ?? and then connect both R and 4R to the 20 ohm (since in my mind the current in the original design flows to the R not the 4R but in the redrawn flows to both...)
I hope that you understand what i am trying to say

13. ### WBahn Moderator

Mar 31, 2012
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Let's color the key nodes and see that they all still connect to the same resistors.

First, let's look at my redrawn circuit.

Now let's identify the same nodes with the same colors in your redrawn circuit.

If you list which components are connected to each node, you will see that the lists are identical.

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14. ### gajoline Thread Starter New Member

Aug 28, 2013
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You are awsome !!!!
Thanks

15. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Here's a circuit from another forum.

Find the resistance between all possible combinations of terminals, such as Rab, Rac, Rad, Rae, Raf, Rag, Rah, Rbc, Rbd, Rbe, etc.

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16. ### WBahn Moderator

Mar 31, 2012
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That's a good one to practice on. Before you start solving it, bound the minimum and maximum values that each answer can be. That's quick and easy and there is a good chance that you will make a mistake in calculating at least one of these effective resistances.

When doing bounding, it can help to remember that given two parallel resistors, the combination must be no smaller than half the size of the smallest resistor nor any larger than the smallest resistor. But it also cannot be any larger than half the larger resistor, either,. So given 100Ω and 150Ω resistors in parallel, the effective resistance has to be between 50Ω and 75Ω.

Last edited: Aug 31, 2013
17. ### The Electrician AAC Fanatic!

Oct 9, 2007
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You think so?

18. ### The Electrician AAC Fanatic!

Oct 9, 2007
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This doesn't seem quite right. Suppose we have 10Ω and 100Ω in parallel. The parallel combination can be smaller than 50Ω.

19. ### WBahn Moderator

Mar 31, 2012
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That was a typo - typed smaller when I meant to type larger. My thought at the time was that the upper bound is the smaller of the smaller resistor or half the larger resistor, but that seemed a bit convoluted, so I put it into two sentences... and messed it up.

Thanks for the catch.

So if R1 < R2, then

$
\frac{R_1}{2} \, < \, R_{eq} \, < \, \text{min}\left( R_1, \frac{R_2}{2} \right)
$