# Finding Equivalent resistance

#### Sharon Alexander

Joined Sep 11, 2015
8
I tried to find it using star delta method but my answer always comes to be 5.04 ohms .

#### MikeML

Joined Oct 2, 2009
5,444
5.6 is correct

#### crutschow

Joined Mar 14, 2008
25,656
Show your calculations if you want more help.

#### WBahn

Joined Mar 31, 2012
26,301
I tried to find it using star delta method but my answer always comes to be 5.04 ohms .View attachment 92974
How are we supposed to be able to tell you why you are getting 5.04 Ω when you don't show us how you got 5.04 Ω?

Our crystal balls just aren't that good.

#### WBahn

Joined Mar 31, 2012
26,301
I tried to find it using star delta method but my answer always comes to be 5.04 ohms .View attachment 92974
One thing you should get in the habit of doing is putting bounds on what the answer has to be. That might or might now have helped in this case.

What would happen if you remove the 6 Ω resistor between A and D?

You know that adding that resistance back in can only make the resistance go down, so without it you end up with an upper bound on the total resistance. But with that resistor removed you have a circuit that is very easy to reduce and you get 5.65 Ω.

If short nodes B, C, and D we get another very simple network that is guaranteed to have a total resistance between E and F that is lower than the actual resistance and this works out to 5.16 Ω.

So with two simple steps we have made it possible to guess that the resistance is about 5.4 Ω and know that we are correct to within about 5%, which is probably more than good enough for most practical purposes. But more to the point we have established bounds that rule out 5.04 Ω as being a correct result.

#### Sharon Alexander

Joined Sep 11, 2015
8
this way i tried to solve the problem please look on to it i'm not able to find the mistake....

#### Sharon Alexander

Joined Sep 11, 2015
8
the formula used to
convert from delta abc to star
Ra=Rab*Rac/(Rab+Rac+Rbc)
same with Rb & Rc

formula used to convert star to delta
Rab=Ra+Rb+Ra*Rb/Rc
same with Rbc & Rca

#### Sharon Alexander

Joined Sep 11, 2015
8
One thing you should get in the habit of doing is putting bounds on what the answer has to be. That might or might now have helped in this case.

What would happen if you remove the 6 Ω resistor between A and D?

You know that adding that resistance back in can only make the resistance go down, so without it you end up with an upper bound on the total resistance. But with that resistor removed you have a circuit that is very easy to reduce and you get 5.65 Ω.

If short nodes B, C, and D we get another very simple network that is guaranteed to have a total resistance between E and F that is lower than the actual resistance and this works out to 5.16 Ω.

So with two simple steps we have made it possible to guess that the resistance is about 5.4 Ω and know that we are correct to within about 5%, which is probably more than good enough for most practical purposes. But more to the point we have established bounds that rule out 5.04 Ω as being a correct result.
One thing you should get in the habit of doing is putting bounds on what the answer has to be. That might or might now have helped in this case.

What would happen if you remove the 6 Ω resistor between A and D?

You know that adding that resistance back in can only make the resistance go down, so without it you end up with an upper bound on the total resistance. But with that resistor removed you have a circuit that is very easy to reduce and you get 5.65 Ω.

If short nodes B, C, and D we get another very simple network that is guaranteed to have a total resistance between E and F that is lower than the actual resistance and this works out to 5.16 Ω.

So with two simple steps we have made it possible to guess that the resistance is about 5.4 Ω and know that we are correct to within about 5%, which is probably more than good enough for most practical purposes. But more to the point we have established bounds that rule out 5.04 Ω as being a correct result.
I tried by removing 6 ohms i got 5.65 . and by shorting b,c,d got 5.165 ohms .Is this method valid in every circuit?
if yes, how did you identified that 6 ohms b/w A and D must be removed & B,C,D must be short , please explain both the steps in detail.

#### WBahn

Joined Mar 31, 2012
26,301
the formula used to
convert from delta abc to star
Ra=Rab*Rac/(Rab+Rac+Rbc)
same with Rb & Rc

formula used to convert star to delta
Rab=Ra+Rb+Ra*Rb/Rc
same with Rbc & Rca
Please show your actual calculations for converting from star to delta. I think you will find your problem.

Also, instead of converting the start back to delta, have you considered converting the remaining delta to a star?

#### WBahn

Joined Mar 31, 2012
26,301
I tried by removing 6 Ω i got 5.65 . and by shorting b,c,d got 5.165 Ω. Is this method valid in every circuit?
if yes, how did you identified that 6 Ω b/w A and D must be removed & B,C,D must be short , please explain both the steps in detail.
What you are looking for in the first case is a simplification to the circuit that will do two things: Make the resulting circuit easy to analyze AND result in a circuit that is guaranteed to result in a value that is too high. This is almost always accomplished by replacing some component with an open circuit. In the second case you also want to make the resulting circuit easy to analyze, but now you want the result to be guaranteed to be too low. This is almost always accomplished by replacing some component with a short circuit. The fewer components you alter, usually, the closer your bound will be. This is why the upper bound is so much closer than the lower bound in this case. But that rule's not set in stone by any means.

Instead of removing or shorting components you can also change values of some of them to make the analysis simple. For instance, in this case if we change all of the resistors in the inner delta to 6 Ω we know the resulting resistance will be too high and we can replace it with a star using 2 Ω resistors. In the bottom two legs we then have 9 Ω and 5.33 Ω, so call the series of those two 15 Ω. In the outer delta we can call those 18 Ω resistors so that our equivalent star uses 6 Ω resistors giving use 12 Ω across the bottom. Thus we have 12 Ω in parallel with 15 Ω and we know that that can be no more than 7.5 Ω even without living a calculator. So we have an upper bound by inspection. If we crank the math on that final parallel pair we have a tighter bound of 6.67 Ω. For the other bound we just choose lower values for resistors so in the inner delta we choose 2 Ω which gives use 2/3 Ω in the star. Then for the lower two branches we have 4 Ω and 7.67 Ω, so we will call the combined resistance 11 Ω. For the outer delta we will use 15 Ω which makes the star consist of 5 Ω resistors placing 11 Ω in parallel with 10 Ω which we know can be no less than 5 Ω and, by cranking the math, the lower bound is actually 5.24 Ω. Even though these bounds are quite a bit looser, they were very simple to arrive at and they eliminate your proposed answer as a possibility.

EDIT: Corrected typo in last resistance reported from 4.24 Ω to 5.24 Ω. Thank go to The Electrician for spotting it and pointing it out.

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#### The Electrician

Joined Oct 9, 2007
2,801
What you are looking for in the first case is a simplification to the circuit that will do two things: Make the resulting circuit easy to analyze AND result in a circuit that is guaranteed to result in a value that is too high. This is almost always accomplished by replacing some component with an open circuit. In the second case you also want to make the resulting circuit easy to analyze, but now you want the result to be guaranteed to be too low. This is almost always accomplished by replacing some component with a short circuit. The fewer components you alter, usually, the closer your bound will be. This is why the upper bound is so much closer than the lower bound in this case. But that rule's not set in stone by any means.

Instead of removing or shorting components you can also change values of some of them to make the analysis simple. For instance, in this case if we change all of the resistors in the inner delta to 6 ohms we know the resulting resistance will be too high and we can replace it with a star using 2 ohm resistors. In the bottom two legs we then have 9 ohms and 5.33 ohms, so call the series of those two 15 ohms. In the outer delta we can call those 18 ohm resistors so that our equivalent star uses 6 ohm resistors giving use 12 ohms across the bottom. Thus we have 12 ohm in parallel with 15 ohm and we know that that can be no more than 7.5 ohm even without living a calculator. So we have an upper bound by inspection. If we crank the math on that final parallel pair we have a tighter bound of 6.67 ohm. For the other bound we just choose lower values for resistors so in the inner delta we choose 2 ohm which gives use 2/3 ohm in the star. Then for the lower two branches we have 4 ohm and 7.67 ohm, so we will call the combined resistance 11 ohm. For the outer delta we will use 15 ohm which makes the star consist of 5 ohm resistors placing 11 ohms in parallel with 10 ohms which we know can be no less than 5 ohms and, by cranking the math, the lower bound is actually 4.24 ohm. Even though these bounds are quite a bit looser, they were very simple to arrive at and they eliminate your proposed answer as a possibility.
There's a single digit typo here; should be 5.24, otherwise the proposed answer (5.04) is not eliminated.

#### WBahn

Joined Mar 31, 2012
26,301
There's a single digit typo here; should be 5.24, otherwise the proposed answer (5.04) is not eliminated.
You are correct. Thanks for pointing it out.

#### Sharon Alexander

Joined Sep 11, 2015
8
finally i solved it...
but the method of upper bound and lower bound of Req is not yet clear
how & why did you chose 6 ohms to be removed not the 7 or 10/3?

#### Attachments

• 299 KB Views: 4

#### Sharon Alexander

Joined Sep 11, 2015
8
View attachment 92994 this way i tried to solve the problem please look on to it i'm not able to find the mistake....
the mistake happened while converting the star consisting 8,8,4 ohms to delta it's just due to calculation mistake.

#### WBahn

Joined Mar 31, 2012
26,301
finally i solved it...
but the method of upper bound and lower bound of Req is not yet clear
how & why did you chose 6 ohms to be removed not the 7 or 10/3?
It's pretty simple, really. By removing a resistor -- meaning that we are increasing it from it's actual value to infinity -- we are establishing an upper bound on the total resistance. That means that I want to not only simplify the circuit to make it easy to analyze, but I want to choose a simplification that will yield as small an overall resistance as I can obtain in order to get a tighter bound. When I remove one of the star resistors tied to one of the outer points {D,E,F}, I reduce the network to two parallel paths (due to the outer 16 Ω resistors) in which one of the 16 Ω resistors has the reduced inner circuit in parallel with it. If I remove either the 10/3 Ω or the 7 Ω resistor, I have a 16 Ω resistor in parallel with something that is between 16 Ω and 32 Ω which can be no smaller than 8 Ω. But if I remove the 6 Ω resistor then I have 32 Ω resistor in parallel with the parallel combination of 16 Ω and whatever the reduced inner circuit turns out to be, which by inspection is guaranteed to be less than 14 Ω meaning that the highest value the overall resistance can be is going to be less than 8 Ω. So that's the better choice.

#### Sharon Alexander

Joined Sep 11, 2015
8
I understood about the upper bound
Thank u

#### WBahn

Joined Mar 31, 2012
26,301
I understood about the upper bound
Thank u
That's good. Do you also understand how I came up with the lower bound?

Keep in mind that there is no single "right" answer in this process. There are many ways to go about bounding (or even just estimating) the result. In general there is a tradeoff between how tight the bound you get is and how much effort it took to get it. Since part of the whole point is to obtain a bound quickly, this tradeoff can be particularly tricky. For instance, in this circuit you could have removed the 10/3 Ω resistor and ended up with a circuit that, by inspection, reduces to a 16 Ω resistor in parallel with a 24 Ω resistor leaving you with an upper bound of 9.6 Ω. Not nearly as good a bound as the other choice, but definitely a lot easier to get.

#### Sharon Alexander

Joined Sep 11, 2015
8
the lower bound seems to be very tricky to understand..

#### WBahn

Joined Mar 31, 2012
26,301
the lower bound seems to be very tricky to understand..
What are you finding tricky about it?

The general idea is to make simplifications to the circuit that you know will lower the resistance. That way you know that the actual resistance is higher than the resistance of the simplified circuit.

In any passive resistor network, lowering the resistance of a resistor can never increase the overall resistance (it might lower it, but it can never raise it). Thus you lower resistance values to make the circuit simpler. That usually involves doing one of two things -- either lower a resistance value to create a balanced condition (such as making two parallel resistors equal so that they can be combined trivially or balancing a bridge so that the bridging element can be ignored) or shorting out the resistance altogether.

#### The Electrician

Joined Oct 9, 2007
2,801
Imagine that you had to find the equivalent resistance between the terminals A-B for this network:

You can easily see that if you increase the value of R56, the equivalent resistance would increase. Similarly, if you decrease the resistance of R56, the equivalent resistance would decrease. Of course, removing a resistor is the same as increasing its resistance to ∞, and shorting it out is the same as reducing its resistance to zero.

The same reasoning applies to R43, and to R30, etc. In fact, as a general proposition, the same thing applies to every resistor in the network. If you remove any resistor (with a caveat I'll explain), the equivalent resistance at A-B will increase (or at least, it won't decrease). If you short any two or more nodes together, the equivalent resistance will decrease (or at least, will not increase).

It's possible that changing a certain resistor won't make any change in the equivalent resistance at A-B. When will this be the case? Imagine that you apply a voltage to A-B, say one volt. Then you can measure the voltage across any other resistor in the network. If there is no voltage across a given resistor, then changing the value of that resistor will not change the equivalent resistance at A-B. But if a particular resistor does have a non-zero voltage across it, changing its value will change the equivalent resistance at A-B.

Consider your circuit from post #1. If the 7 ohm resistor is changed to 3 ohms, then removing the 6 ohm resistor between A and D (or shorting it) will have no effect on the equivalent resistance between E and F. This is because if you apply a voltage to E and F (when the 7 ohm resistor has been changed to 3 ohms), there will be no voltage across the 6 ohm resistor between A and D.

You should also be able to see why changing resistors that are distant from the A-B terminals will have less effect on the equivalent resistance than will changing resistors near to A-B.

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