Where did the conclusion this is a bridge come from anyway?

It is five unknown resistors. The solution should be in terms of these five unknowns.

Because the circuit of this thread has the same topology as a Wheatstone bridge: https://en.wikipedia.org/wiki/Wheatstone_bridge

The Req seen by the voltage source can't be determined by ordinary parallel and series resistance formulas. This is the simplest circuit that can't be solved with those simple formulas, so it's often given to beginning students to confound them. It is often drawn in such a way as to make it even more tricky, as was done in the example in post #1.

The difficulty is discussed in this Wikipedia article: https://en.wikipedia.org/wiki/Y-Δ_transform

Under the heading "Simplification of networks", it is shown how to use the wye-delta transform to change the network to one where ordinary series and parallel resistor formulas can be used.

 

Then, instead of fomenting all of this folderol, and since you imply you have the wherewithal, Why didn't you just post a simple refutation which would prove your point?

I did. You couldn't understand it. I showed an explicit example. You couldn't understand it (but that's not surprising as it appears that you haven't yet learned what Thevenin equivalents are, so once you do you might be able to follow it).

I suspect that dannyf DID understand it as he stopped insisting that if you know the Thevenin equivalent you are golden.

 

Where did the conclusion this is a bridge come from anyway?

It is five unknown resistors. The solution should be in terms of these five unknowns.

I agree. It is worth noting that it is a Wheatstone bridge topology since drawing it that way does make it more familiar and easier to work with (for most people). But there is nothing magical about that topology, particularly with regards to finding the equivalent resistance of the network as seen by the source. In that sense, noting that it is a Wheatstone bridge works against many people (as amply demonstrated in this thread) because they want to immediately apply the kinds of techniques that they are used to applying to Wheatstone bridge circuits, which usually are concerned with finding the voltage across or the current through the bridge resistor. Just look at how many people near the beginning of the thread presented solutions for the current in R3, completely ignoring the fact that the TS wasn't even looking for a current at all!

At the end of the day it is just a network of five resistors that can be analyzed like any other network of (five or however many) resistors.

 

In other words, you gave him a single fish as opposed to teaching him how to fish. The "process" you showed him doesn't even apply to Wheatstone bridge circuits in general, let alone circuits in general. It only works on one specific, tiny, narrow, and generally uninteresting case. Real Wheatstone bridge circuits are generally only interesting when they are unbalanced. Yes, we usually null them initially, but they exist to measure things and those measurements are usually only possible because the measured quantity unbalances the bridge.

Well, that's a little harsh since what's remarkable about a Wheatstone bridge is that when it's nulled, after being unbalanced, the measurement of the unbalancing force can be made without regard to the load imposed by the measuring equipment just by monitoring what it takes to rebalance the bridge.

 

I did. You couldn't understand it. I showed an explicit example. You couldn't understand it (but that's not surprising as it appears that you haven't yet learned what Thevenin equivalents are, so once you do you might be able to follow it).

I suspect that dannyf DID understand it as he stopped insisting that if you know the Thevenin equivalent you are golden.

Dannyf may have stopped responding because he's afraid of bullies, but I'm not.

I've already demonstrated, in simple terms, why Thevenin's theorems aren't required to solve this problem, and yet you discount my efforts as irrelevant, spouting one of your victims' acquiescence to your self-serving blather as proof? At best, that's disingenuous.

 

Dannyf may have stopped responding because he's afraid of bullies, but I'm not.

I've already demonstrated, in simple terms, why Thevenin's theorems aren't required to solve this problem, and yet you discount my efforts as irrelevant, spouting one of your victims' acquiescence to your self-serving blather as proof. At best, that's disingenuous.

Huh? Did you not read ANYTHING I said right from the beginning? Where did I say that Thevenin's theorem was REQUIRED to solve this problem? In response to the suggestion to use Thevenin in Post #2, I stated very explicitly in Post #4, "One method that will NOT work (in general) is to use Thevenin equivalents...."

 

Dannyf may have stopped responding because he's afraid of bullies,

given all that has been said, I find anything more is fruitless.

 

Well, that's a little harsh since what's remarkable about a Wheatstone bridge is that when it's nulled, after being unbalanced, the measurement of the unbalancing force can be made without regard to the load imposed by the measuring equipment just by monitoring what it takes to rebalance the bridge.

Well, consider this point that hasn't been made yet.

The circuit posted by the TS is just five resistors. It is not even drawn in the classic layout of a Wheatstone bridge. Since not every circuit containing five resistors in this topology is a Wheatstone bridge, let alone a balanced Wheatstone bridge, doesn't it seem a bit presumptuous to just assume that THIS circuit MUST be intended to not only be a balance Wheatstone bridge, but to be one in which all five resistors are identically equal. For all you or anyone else knows that circuit represents five different heating elements in a kiln or the power routing resistances in an IC and the way the circuit is draw represents the physical orientation and connection of the resistances involved.

 

The "Wheatstone Bridge" is not a fun circuit to analyze by hand. Nice observation.

 

The "Wheatstone Bridge" is not a fun circuit to analyze by hand.

doesn't it depend on who's doing the analyzing?

 

Yea probably, My boss used to have "fun" prime factoring license plates in his head when he was driving or playing chess in his head with someone in the next room while he was doing dishes.

 

Well, consider this point that hasn't been made yet.

The circuit posted by the TS is just five resistors. It is not even drawn in the classic layout of a Wheatstone bridge.

It certainly is, with just a little massaging.

Since not every circuit containing five resistors in this topology is a Wheatstone bridge,

Wrong again.

Every circuit with the legs of the bridge connected to the load and the power supply connected as shown is a Wheatstone bridge.


let alone a balanced Wheatstone bridge, doesn't it seem a bit presumptuous to just assume that THIS circuit MUST be intended to not only be a balance Wheatstone bridge, but to be one in which all five resistors are identically equal. For all you or anyone else knows that circuit represents five different heating elements in a kiln or the power routing resistances in an IC and the way the circuit is draw represents the physical orientation and connection of the resistances involved.

Or not. You seem, because of your rhetoric, to want to lock everyone into your self-serving views and have everyone kow-tow to you regardless of the insanity of your edicts.

 

My boss used to have "fun" prime factoring license plates in his head when he was driving

he has a dangerous definition of "fun",

To turn the conversations around. We showed that superimposition + thevenin equivalency can be used to get one the answer.

Are there other approaches that will get to the answer faster and more intuitively?

 

Or not. You seem, because of your rhetoric, to want to lock everyone into your self-serving views and have everyone kow-tow to you regardless of the insanity of your edicts.

I simply took the circuit as given by the TS. You are the one that made huge, unwarranted, simplifying assumptions in order to get it reduced to something that you could handle regardless of whether it still bore any resemblance to the circuit as given by the TS.

 

I simply took the circuit as given by the TS. You are the one that made huge, unwarranted, simplifying assumptions in order to get it reduced to something that you could handle regardless of whether it still bore any resemblance to the circuit as given by the TS.

The circuit submitted by the TS was a simple Wheatstone bridge, and the only assumption I made (after my initial error, which I owned up to) was that - since there was no controverting evidence presented - in its simplest form it was balanced and that the resistances in its legs, and its load, were equal.

You, on the other hand, have been dancing around, posting poorly formulated trolls designed to elicit "damned if you do and damned if you don't" responses in order to what? Cement your position here as some kind of infallible guru?

Good luck with that, now that the cat's out of the bag.

BTW, what does "TS" mean?

 

Well, consider this point that hasn't been made yet.

The circuit posted by the TS is just five resistors. It is not even drawn in the classic layout of a Wheatstone bridge. Since not every circuit containing five resistors in this topology is a Wheatstone bridge, let alone a balanced Wheatstone bridge, doesn't it seem a bit presumptuous to just assume that THIS circuit MUST be intended to not only be a balance Wheatstone bridge, but to be one in which all five resistors are identically equal. For all you or anyone else knows that circuit represents five different heating elements in a kiln or the power routing resistances in an IC and the way the circuit is draw represents the physical orientation and connection of the resistances involved.

The drawing was kinda tricky to reduce but, once reduced, yielded a Wheatstone bridge which, in its simplest configuration, comprises four equally valued resistances for the legs of the bridge and another equally valued load across the junctions of the two legs.

Do you have a problem with that?

 

Huh? Did you not read ANYTHING I said right from the beginning? Where did I say that Thevenin's theorem was REQUIRED to solve this problem? In response to the suggestion to use Thevenin in Post #2, I stated very explicitly in Post #4, "One method that will NOT work (in general) is to use Thevenin equivalents...."

So, then, you agree with my procedure?

 

Well, consider this point that hasn't been made yet.

The circuit posted by the TS is just five resistors. It is not even drawn in the classic layout of a Wheatstone bridge. Since not every circuit containing five resistors in this topology is a Wheatstone bridge, let alone a balanced Wheatstone bridge, doesn't it seem a bit presumptuous to just assume that THIS circuit MUST be intended to not only be a balance Wheatstone bridge, but to be one in which all five resistors are identically equal. For all you or anyone else knows that circuit represents five different heating elements in a kiln or the power routing resistances in an IC and the way the circuit is draw represents the physical orientation and connection of the resistances involved.

Pie in the sky. You're grasping at straws since if the resistors are arranged in the configuration of a Wheatstone bridge and it looks lie a duck and quacks like a duck...

 

Ah, so if someone has five resistors, with unique reference designators, whose interconnection happens to match the topology of a Wheatstone bridge, then it is only reasonable to assume that all five resistors MUST be equal in value.

What garbage.

Oh, wait. I forgot. You claim that this "process" you used is valid even if the resistors are different values (see Post #64). So you are then claiming that the overall resistance is still just R even if all five resistors are different. But which R? There are five of them!

 

So, then, you agree with my procedure?

Correct me if I'm wrong.

Your "process" is to assume that all of the resistors are the same value, R, so that there is no current flowing in the bridge resistor. Then recognize that this results in the two vertical pairs being effectively isolated so that each vertical leg has resistance 2R and the two of these in parallel give a total resistance of R.

Then, because this "that process would be identical regardless of the values of the resistances", you claim to be able to use this "process" of yours regardless of what the actual values of the five different resistors turns out to be.

If that's your process, then I most definitely do NOT agree with it.