Which is not what you have in post #6. Does what you say here mean you agree that what you have in post #6 is not the circuit shown in post #1.

Yes.

 

since I was only 99.9999999% sure, I did a simulation in ltspice and it turned out that I was wrong. The current throught he middle resistor is not 0.226 but 0.2264.



So hooray to thevenin equivalency.

The exact value is 12/53 amps, which is approximately .2264150943396 amps.

 

However, I'll still stick to my guns about all of the resistors being equal valued since there's no hard evidence proving otherwise.

You are correct that there is no hard evidence but completely askew in your conclusion.

With a lack of hard evidence one must assume the general case of any arbitrary value for the resistors.

 

You are correct that there is no hard evidence but completely askew in your conclusion.

With a lack of hard evidence one must assume the general case of any arbitrary value for the resistors.

Au contraire, since the OP was asking for a process, not an answer, and that process would be identical regardless of the values of the resistances.

For example, consider the simple case of two resistors in parallel. The process used to determine the equivalent resistance of the pair is to use the formula: Rt =(R1 * R2)/(R1 + R2). Note that the process is identical for any pair of resistors in parallel, regardless of their values.

 

So, if given a circuit with two resistors in series, R1 and R2, you would be comfortable assuming that the total resistance is 2*R1?

That would depend on the context, of course.
In this instance the assumption is warranted becausef the OP was asking to be taught how to fish (and still may be) instead of just begging for a fish.

Since the OP is asking how to solve a circuit having that topology, wouldn't it seem more reasonable to show them how to solve any circuit with that topology instead of a technique that only works under one very special and rare set of circumstances? Instead of teaching them how to fish in any body of water, you are teaching them only how to fish in a bathtub.

So, if given the classic inverting amplifier op-amp circuit with feedback resistor Rf and input resistor Ri, you would be comfortable assuming that the gain of the amplifier is -1?

If the non-inverting input was connected to 0 volts, and if the opamp supply was bipolar and provided enough headroom for the output not to be railed, and if Rf and Ri were the same resistance, then yes, but I fail to see the relevance with respect to this thread.

It's relevance is your claim that it is reasonable to assume that resistances are the same based only on the fact that they were given with different reference designators and not with actual values. Well, as I said before, if all we have are the reference designators, then all we know is that we don't know the value of the resistor at all, which means that we have zero basis for assuming that they are the same value as any other resistor.

 

Au contraire, since the OP was asking for a process, not an answer, and that process would be identical regardless of the values of the resistances.

For example, consider the simple case of two resistors in parallel. The process used to determine the equivalent resistance of the pair is to use the formula: Rt =(R1 * R2)/(R1 + R2). Note that the process is identical for any pair of resistors in parallel, regardless of their values.

Indeed, that formula for resistors in parallel works for any two values of resistance. But it does so precisely because it was developed without assuming that the two resistors have the same value. If the person finding the "process" of determining the equivalent resistance had made that assumption, they would have concluded that Rt = R1/2 = R2/2, which clearly does NOT work if the two resistors have different values.

Similarly, the "process" you presented:

However, I'll still stick to my guns about all of the resistors being equal valued since there's no hard evidence proving otherwise.

In that case, R1 and R2 will be in series, as will R4 and R5, since regardless of the value of R3 there'll be no current through R3 and the series string R1R2 will be in parallel with the series string R4R5, yielding a total resistance of R across the power supply.

This does NOT work for arbitrary values of resistance. Just look at all main caveat that your "process" relies on -- that there be no current in R3. Well, if the resistors aren't all the same value then there probably IS current in R3 and that blows the rest of your "process" right out of the water.

For instance, let's say that R1 = R3 = R5 = 1 kΩ and that R2 = R4 = 99 kΩ. According to your "process", the total resistance would be 50 kΩ, yet it is trivially easy to see that the total resistance MUST be strictly less than 3 kΩ.

 

Okay, so what is the equivalent resistance, Req, seen by the voltage source, Vs, in the following circuit?

View attachment 107710

Let's keep the math as simple as possible:

Vs = 10 V
Ro = 10 kΩ
Va = Vb = 5 V
Ra = Rb = 5 kΩ

What is Req?

Whatever answer you get, I will absolutely guarantee you that you are off by more than an order of magnitude.

.

I see what you're getting at.

There is no unique answer, but here are 3 example cases for the content of box A:



Box B could be any of those 3 example cases, or others.

The pattern is that the two vertically oriented resistors are both the same value, which can be any value R. The horizontally oriented resistor is 5000-R/2.

If the A and B boxes contain the same arrangement and values, we will have a balanced Wheatstone circuit, and Req will be R ohms.

If the A and B boxes aren't the same then we have this:



To determine Req, set R3 = RA+Ro+RB and use the general formula I gave in post #32, or use any other convenient method to solve for Req.

 

Or, even more simply:



The 10 Ω resistor is completely masked from the bridge resistor by the power supply. But it also completely dominates the resistance seen BY the power supply. So whatever answer someone gives I need only pick a value for that resistor that is an order of magnitude smaller to make their answer more than an order of magnitude too large. Or, if they say that Req is anything less than 1 kΩ, I can just remove that resistor entirely and now their answer is more than an order of magnitude too small. In all cases, however, the bridge resistor sees absolutely no difference.

 

so Req is Rthevenin + Rload


To get a single equivalent resitance you can thevenize and add the R3 to the Thevenin resistance result, this should give the single equivalent resistor asked for. Req = ( R1+R2 || R4 +R5 ) + R3


This also may help..
http://ion.chem.usu.edu/~sbialkow/Classes/564/Thevenin/Thevenin.html

I could start a whole thread on this I suppose. The OP's question was about a procedure more than the value, and it is in fact a bridge circuit and should be treated that way..
Thanks for the link !

 

Indeed. I misread the original post and failed to realize that the topology shown in the schematic is that of a Wheatstone bridge. Mea culpa.

However, I'll still stick to my guns about all of the resistors being equal valued since there's no hard evidence proving otherwise.

In that case, R1 and R2 will be in series, as will R4 and R5, since regardless of the value of R3 there'll be no current through R3 and the series string R1R2 will be in parallel with the series string R4R5, yielding a total resistance of R across the power supply.

I think I see your point here, IF all the resistors are equal, then the bridge is balanced and no voltage appears across R3 and therefore no current through R3 ? ?

 

I could start a whole thread on this I suppose. The OP's question was about a procedure more than the value, and it is in fact a bridge circuit and should be treated that way..
Thanks for the link !

Just be sure to take note that both the process used by TheBigPablo as well as his solution are completely wrong.

That the result is wrong is obvious by inspection. He claims that the equivalent resistance is at least what the value of the bridge resistor is. So let's make that value infinite by removing it. That makes the equivalent resistance infinite. Does anyone buy that.

Always, always, ALWAYS ask if the answer makes sense!

 

I think I see your point here, IF all the resistors are equal, then the bridge is balanced and no voltage appears across R3 and therefore no current through R3 ? ?

But so what? It is a completely unwarranted assumption. It would be like calculating how far your car can coast by assuming that it is always downhill everywhere it goes -- and then claiming that your solution applies even when it is uphill.

 

Since the OP is asking how to solve a circuit having that topology, wouldn't it seem more reasonable to show them how to solve any circuit with that topology instead of a technique that only works under one very special and rare set of circumstances? Instead of teaching them how to fish in any body of water, you are teaching them only how to fish in a bathtub.



It's relevance is your claim that it is reasonable to assume that resistances are the same based only on the fact that they were given with different reference designators and not with actual values. Well, as I said before, if all we have are the reference designators, then all we know is that we don't know the value of the resistor at all, which means that we have zero basis for assuming that they are the same value as any other resistor.

And zero basis for assuming that they aren't, regardless of your pontification.

Instead of nit-picking, ad infinitum, why don't you just teach a man to fish by posting a universal solution for the resistance presented by an unbalalanced Wheatstone bridge across its power supply connections?

 

Indeed, that formula for resistors in parallel works for any two values of resistance. But it does so precisely because it was developed without assuming that the two resistors have the same value.

Well, sure, but so what?

If the person finding the "process" of determining the equivalent resistance had made that assumption, they would have concluded that Rt = R1/2 = R2/2, which clearly does NOT work if the two resistors have different values.

More nonsense; Apples and oranges.

 

And zero basis for assuming that they aren't, regardless of your pontification.

Instead of nit-picking, ad infinitum, why don't you just teach a man to fish by posting a universal solution for the resistance presented by an unbalalanced Wheatstone bridge across its power supply connections?

Posting an equation that can be used to solve a very specific topology is hardly teaching someone to fish -- it is just giving them a fish. When faced with even a slight modification to the topology they won't have the tools to solve it and must come asking for another fish. Teaching them to fish involves helping them learn how to solve ANY circuit. That involves learning how to apply tools such as mesh and nodal analysis. Those tools are essentially universal in their applicability. Additional tools with wide applicability include delta-wye transforms. You can also apply other concepts such as equating power of the actual network with the power in the equivalent network -- this is the defining concept for such things as RMS values of waveforms. Gee, seems like someone suggested the use of these approaches clear back in Post #4.

 

I think I see your point here, IF all the resistors are equal, then the bridge is balanced and no voltage appears across R3 and therefore no current through R3 ? ?

Yes, but in truth what really matters is that the ratio of R1:R2 be equal to the ratio of R4:R5, in which case the voltage difference across R3 will be zero and no charge will flow through R3, regardless of its resistance.

 

Posting an equation that can be used to solve a very specific topology is hardly teaching someone to fish -- it is just giving them a fish. When faced with even a slight modification to the topology they won't have the tools to solve it and must come asking for another fish. Teaching them to fish involves helping them learn how to solve ANY circuit. That involves learning how to apply tools such as mesh and nodal analysis. Those tools are essentially universal in their applicability. Additional tools with wide applicability include delta-wye transforms. You can also apply other concepts such as equating power of the actual network with the power in the equivalent network -- this is the defining concept for such things as RMS values of waveforms. Gee, seems like someone suggested the use of these approaches clear back in Post #4.

I don't understand your analogy, since the only topology being discussed is a Wheastone bridge and giving someone a fish would be equivalent to giving them the answer to a problem while teaching them how to fish would be equivalent to giving them a rod and reel and showing them how to use it.

Something like this:

 

I think I see your point here, IF all the resistors are equal, then the bridge is balanced and no voltage appears across R3 and therefore no current through R3 ? ?

Yes, but in truth what really matters is that the ratio of R1:R2 be equal to the ratio of R4:R5, in which case the voltage difference across R3 will be zero and no charge will flow through R3, regardless of its resistance.

I agree. All resistors are equal and there is no current through R3.
Then how Req = ( R1+R2 || R4 +R5 ) + R3 this matters?

 

Yes, but in truth what really matters is that the ratio of R1:R2 be equal to the ratio of R4:R5, in which case the voltage difference across R3 will be zero and no charge will flow through R3, regardless of its resistance.

And that has no effect whatsoever on the fact that this approach doesn't work for finding the equivalent resistance seen by the source. Go back and answer the question posted in Post #57

http://forum.allaboutcircuits.com/threads/equivalent-resistance-help.124949/page-3#post-1009357

There is no current in the bridge resistor, so you should be able to answer it very quickly. And your answer will be wrong. Since you won't be able to answer it correctly, that means that your "process" doesn't work. It really is that simple.

 

I agree. All resistors are equal and there is no current through R3.
Then how Req = ( R1+R2 || R4 +R5 ) + R3 this matters?

That equation for Req is simply wrong. As the saying goes, this is obvious to the most casual observer. It says that if we make R3 infinite, that the equivalent resistance seen by the voltage source is infinite. That is clearly incorrect.