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.

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Since You specify neither the resistances nor the connections between the resistances in the three-terminal devices 'A' and 'B', and since you've specified that there's no current through R0, the only information which can be gleaned from that is that the voltages on both ends of R0 are equal.

One would assume, however, that since this thread is about Wheatstone bridges, 'A' and 'B' would be, essentially, 2 resistor voltage dividers connected across Vs, with their taps connected to opposite ends of R0.

Now, if we call the resistors in 'A' R1 and R2, and the resistors in 'B' R4 and R5, all that's required for the voltage across R0 to be zero is that R1/R2 be equal to R4/R5.

So, when the bridge is balanced, (when there's no current through R0) R1 could be 1000 ohms, R2 could be 10000 ohms and R4 could also be 1000 ohms, in which case R5 would have to be 10000 ohms.

But, another valid solution for balance would be with R4 equal to 1 milliohm and R5 equal to 10 milliohms, so there are an infinite number of solutions.

Fact of the matter is that the resistance the balanced bridge presents to the supply will be the sum of R1 and R2 in parallel with the sum of R4 and R5.

Your bottom drawing seems to be refuting a post you disagree with, but in a nebulous kind of way. Would you make it more concrete, please?

 

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.

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.

If the voltages presented to each end of a resistor are equal, then the charge transferred through the resistor will be zero.

 

If the voltages presented to each end of a resistor are equal, then the charge transferred through the resistor will be zero.

Again, so what? That has absolutely ZERO bearing on the fact that that equation is WRONG!

Let's go ahead and assume that all of the resistors are the same value, R. That equation says that the equivalent resistance (as seen by the voltage source) is 2R, when in fact the equivalent resistance is simply R. The equation is WRONG!

 

Since You specify neither the resistances nor the connections between the resistances in the three-terminal devices 'A' and 'B', and since you've specified that there's no current through R0, the only information which can be gleaned from that is that the voltages on both ends of R0 are equal.

One would assume, however, that since this thread is about Wheatstone bridges, 'A' and 'B' would be, essentially, 2 resistor voltage dividers connected across Vs, with their taps connected to opposite ends of R0.

Now, if we call the resistors in 'A' R1 and R2, and the resistors in 'B' R4 and R5, all that's required for the voltage across R0 to be zero is that R1/R2 be equal to R4/R5.

So, when the bridge is balanced, (when there's no current through R0) R1 could be 1000 ohms, R2 could be 10000 ohms and R4 could also be 1000 ohms, in which case R5 would have to be 10000 ohms.

But, another valid solution for balance would be with R4 equal to 1 milliohm and R5 equal to 10 milliohms, so there are an infinite number of solutions.

Fact of the matter is that the resistance the balanced bridge presents to the supply will be the sum of R1 and R2 in parallel with the sum of R4 and R5.

Your bottom drawing seems to be refuting a post you disagree with, but in a nebulous kind of way. Would you make it more concrete, please?

Once again you are making absolutely unwarranted assumptions. The whole idea of a black box is precisely that you DON'T know what is in there. The whole idea of a Thevenin equivalent is that it doesn't MATTER that you don't know what is in there because, at the terminals, the Thevenin equivalent is indistinguishable from whatever is in the black box. I gave you the Thevenin equivalent for the combination of the black box and the voltage source. The claim that was made (by dannyf) is that "if you know the Thevenin equivalent then you are golden" and that you can then find the overall equivalent resistance of the network. Well, have at it.

 

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

I don't think I ever said that; what I said is that in a balanced bridge with R1=R2=R4=R5, then Req will be equal to R1.

 

Once again you are making absolutely unwarranted assumptions. The whole idea of a black box is precisely that you DON'T know what is in there. The whole idea of a Thevenin equivalent is that it doesn't MATTER that you don't know what is in there because, at the terminals, the Thevenin equivalent is indistinguishable from whatever is in the black box. I gave you the Thevenin equivalent for the combination of the black box and the voltage source. The claim that was made (by dannyf) is that "if you know the Thevenin equivalent then you are golden" and that you can then find the overall equivalent resistance of the network. Well, have at it.

Thanks, but no thanks.

The point which you seem to be avoiding is that there's a connection between the black boxes and ground in the top two instances which is missing in the third.

In fact, if Va is equal to Vb, then in the third instance the current through Ra, R0, and Rb will be zero, there will be no load for either Va or Vb to feed, and Req will be infinite, regardless of the values of the resistances.

In the first two instances however, (assuming each are Wheatstone bridges, since I think that's what we're talking about) if the bridges are balanced, Req will simply be equal to IA + IB / Vs, where IA is the current through A and IB is the current through B.

 

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.

 

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

because that's what it is?

 

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

My guess is that you are leading up the fact that a Thevenin equivalent is not a total, component-level equivalence. For example, in the original schematic in post #1, if all 5 resistors are of equal value, then there is zero current through R3 (the "center" resistor) even though there could be significant current through all of the other 4. A Thevenin equivalent circuit is how a source *appears* to a load, not how a source appears to itself.

ak

 

My guess is that you are leading up the fact that a Thevenin equivalent is not a total, component-level equivalence. For example, in the original schematic in post #1, if all 5 resistors are of equal value, then there is zero current through R3 (the "center" resistor) even though there could be significant current through all of the other 4. A Thevenin equivalent circuit is how a source *appears* to a load, not how a source appears to itself.

ak

How can I possibly be "leading up" to that fact when I started the last paragraph of Post #4 with the following sentence: "One method that will NOT work (in general) is to use Thevenin equivalents since the equivalent circuit is NOT equivalent as far as anything inside the equivalent is concerned."?

 

I don't think I ever said that; what I said is that in a balanced bridge with R1=R2=R4=R5, then Req will be equal to R1.

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.

 

Thanks, but no thanks.

The point which you seem to be avoiding is that there's a connection between the black boxes and ground in the top two instances which is missing in the third.

Well, at least we have firmly established that you have no clue what a Thevenin equivalent circuit is.

In fact, if Va is equal to Vb, then in the third instance the current through Ra, R0, and Rb will be zero, there will be no load for either Va or Vb to feed, and Req will be infinite, regardless of the values of the resistances.

So now you are saying that Req is infinite, when if the circuit in each of the boxes happens to be two series-connect resistors, each equal to Ro, between the top and bottom terminals and then tapped (meaning they are no longer in series) in the middle and brought out to the side terminal, we have exactly the circuit you insist on working with and in which you claim Req = Ro. Which is it?

In the first two instances however, (assuming each are Wheatstone bridges, since I think that's what we're talking about) if the bridges are balanced, Req will simply be equal to IA + IB / Vs, where IA is the current through A and IB is the current through B.

So what are the currents IA and IB? The claim has been made several times (though not by you, I don't think) that if you know the Thevenin equivalent that you have all the information you need to find Req. So I'm simply asking someone that believes that to back it up by finding Req for a trivially simple case.

 

How can I possibly be "leading up" to that fact when I started the last paragraph of Post #4 with the following sentence: "One method that will NOT work (in general) is to use Thevenin equivalents since the equivalent circuit is NOT equivalent as far as anything inside the equivalent is concerned."?

If you're not leading up to something, then why don't you just make a simple statement which explains what you mean instead of (it seems) digging potholes with conflicting arguments for folks to fall into?

 

because that's what it is?

The Wheatstone Bridge circuit I am familiar with has a galvanometer and 4 resistors.

One of these things is not like the other...

 

The Wheatstone Bridge circuit I am familiar with has a galvanometer and 4 resistors.

The bridges I drive over everyday all are made of steel and concrete, heavy as hell and look and feel unlike anything from a wheastone bridge circuit.

Why anyone calls that circuit a bridge beyond me,

 

If you're not leading up to something, then why don't you just make a simple statement which explains what you mean instead of (it seems) digging potholes with conflicting arguments for folks to fall into?

What I'm trying to "lead up to" is for you and a couple of others to grasp the very simple statement that explained what I meant.

 

Well, at least we have firmly established that you have no clue what a Thevenin equivalent circuit is.

Hmm... I've been here for only 9 days and you already have me all figured out and relegated to a particular box of your choosing? How outrageously arrogant!

So now you are saying that Req is infinite, when if the circuit in each of the boxes happens to be two series-connect resistors, each equal to Ro, between the top and bottom terminals and then tapped (meaning they are no longer in series) in the middle and brought out to the side terminal, we have exactly the circuit you insist on working with and in which you claim Req = Ro. Which is it?

I don't mean to be unkind, but it seems to me that if you realize that the top two circuits are identical and vastly different from the bottom circuit, then you're intentionally trying to muddy the water in order to save face.

What I tried to point out but which you didn't seem to comprehend (giving you the benefit of the doubt) is that in the bottom circuit, since Va = Vb, there'll be no voltage drop across the three series resistors (RA, RO, and RB). Such being the case, Ohm's law dictates that since no current is taken from VA, its load impedance must be infinite. The same argument applies to VB and, consequently, Req for the bottom circuit will be infinite.

So what are the currents IA and IB?

I gave their values in an earlier post.

The claim has been made several times (though not by you, I don't think)

Double negative...

that if you know the Thevenin equivalent that you have all the information you need to find Req. So I'm simply asking someone that believes that to back it up by finding Req for a trivially simple case.

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?

 

How ... ?

Because this thread is so long that I forgot that part.

ak

 

Because this thread is so long that I forgot that part.

ak

I can sympathize!

 

The bridges I drive over everyday all are made of steel and concrete, heavy as hell and look and feel unlike anything from a wheastone bridge circuit.
Why anyone calls that circuit a bridge beyond me,

Because if you draw it in a rectilinear fashion like any other circuit rather in the diamond-shape that now is common for bridge circuits, it looks like two river banks with a bridge spanning the gap between them.

ak