Unknown resistor solving

In message , dated Wed,

If I've understood your description properly, you are correct. R4 and R5 are in series and it impossible to determine their individual values. Vm depends only on the sum of their values.

So you have only two variables, R6 and (R4+R5).

If Vm is an open-circuit voltage, then R4 and R5 can take on any values. Vm will be equal to Va-Vb.

It's obviously a *balanced* circuit- leave the troll to work his own boring homework.

I think you've understood however in this case Vm is fixed (as is Vr and Vs) so therefore Va and Vb are dependant on R4, R5 and R6.

wombat

It's probably just my poor drawing but Vm is actually a fixed, known voltage.

hmmm... if only you knew. As it happens it's not homework, in fact it's got very little to do with electronics at all. It's just that the analysis used in circuits is analogous to my area of interest.

The way you have it drawn Vm is on a connection between R4 and R5, and therefore 0V. So I assume there is no connection there and the drawing is meant to indicate Vm measured between those two points. But then Vm is simply Va-Vb, R4 and R5 are indeterminate, and only ?R6 can be solved, this can't be right. Get your diagram right and we may be able to help.

In message , dated Wed,

Your diagram is confusing. You have a voltage Vm marked between the ends of R4 and R5 but those points are apparently connected by a wire, so Vm = 0, and Va = Vb.

I think you need to clarify the diagram.

Maybe you mean this, or maybe you mean that? There is no reasonable interpretation of your drawing:

View in a fixed-width font such as Courier.

Vr-Va Vr-Va Va-Vb ----- ----- - ----- R1 R1 R3||R6 -->

-->

R1 Va ?R4 Vr o--/\\/\\/\\/\\--+---o----+----------/\\/\\/\\/\\-+ | | | / \\ | \\ / | + R3 / \\ ?R6 --- \\ / - Vm / \\ | - \\ / | | | | Vs o--/\\/\\/\\/\\--+---o----+----------/\\/\\/\\/\\-+ R2 Vb ?R5

--> -->

Vs-Vb Vs-Vb Va-Vb ----- ----- + ----- R2 R2 R3||R6

Vr-Va Va-Vb 1) 0 =( ----- - ----- )*(R4 + R5) + Vm + Vb-Va R1 R3||R6

obtained by summing drops around the Va,R4,Vm,R5,Vb branch,

only establishes constraint on R6 and R4+R5.

Fair enough. All the voltage points marked are referenced to ground. Vm isn't the voltage between the ends of R4 and R5 it is in fact the voltage at the node formed by the ends of R4 and R5 which at the same potential (wired together).

The diagram could be drawn differently as shown below. Hopefully it's less confusing now:

R1 Va ?R4 Vr o--/\\/\\/\\/\\--+---o----+-/\\/\\/\\/\\--o Vm | | | / \\ | \\ / | R3 / \\ ?R6 | \\ / | / \\ | \\ / | | | | Vs o--/\\/\\/\\/\\--+---o----+-/\\/\\/\\/\\--+ R2 Vb ?R5

...or...

R1 Va ?R4 Vr o--/\\/\\/\\/\\--+---o----+-/\\/\\/\\/\\--o Vm | | / \\ \\ / R3 / \\ ?R6 \\ / / \\ \\ / | | Vs o--/\\/\\/\\/\\--+---o----+-/\\/\\/\\/\\--o Vm R2 Vb ?R5

Then you should have indicated a voltage source, not a measurement.

How about Vr and Vs? Are they also fixed by voltage sources?

View in a fixed-width font such as Courier.

Vr-Va Vr-Va Va-Vb Va-Vm ----- ----- - ----- = ----- R1 R1 R3||R6 R4 -->

-->

R1 Va ?R4 Vr o--/\\/\\/\\/\\--+---o----+-------/\\/\\/\\/\\ ---+ | | | / \\ | \\ / | R3 / \\ ?R6 +------+ \\ / | | / \\ | | + \\ / | --- | | | - Vm Vs o--/\\/\\/\\/\\--+---o----+-------/\\/\\/\\/\\----+ | - R2 Vb ?R5 ---

--> -->

Vs-Vb Vs-Vb Va-Vb Vb-Vm ----- ----- + ----- = ----- R2 R2 R3||R6 R5

Va-Vm Vm-Vb R4 Va-Vm 1) ----- = ----- or -- = ----- R4 R5 R5 Vm-Vb

and by inspection:

Vr-Va Va-Vb Va-Vb =(----- - ----- )*(R4+R5) R1 R3||R6

or

Va-Vb R4+R5= --------------- Vr-Va Va-Vb ----- - ----- R1 R3||R6

Va-Vb --------------- Vr-Va Va-Vb ----- - ----- R1 R3||R6

Making R5= -------------------

Va-Vm 1 + ----- Vm-Vb

Va-Vm and R4= ----- * R5 Vm-Vb

You need an additional constraint to fix R3||R6.

Measured? i.e. it's NOT a source?

...Jim Thompson

Didn't the OP say that all the voltages were known? If so, every voltage in the circuit is pinned, and any nonzero resistor values will work. I suspect that a certain small furry person didn't read the question carefully enough before posting.

Cheers,

Phil Hobbs

I presume VM to be a voltage measured to GROUND. Use your brain instead of your mouth, Fred ;-)

...Jim Thompson

Ok. First of all thanks for everyone's effort in a pretty boring post. Just to be absolutely clear:

-All voltages are known (because I can measure them).

-R4,5,6 are the only bits I don't know, but I do know they are arranged as shown.

If Fred's right with his most recent calcs then he came to the same conclusion I did which was that there's one too many unknowns.

wombat

Only in that R4 and R5 are paralleled by R6, so the solution causes R4 and R5 to be a function of R6.

Drop R6 and there will be a singular solution.

...Jim Thompson

R3 was given AFAIR - QED

Marte

Hmmm..singular usually means either infinite number of solutions or none at all, anything but unique, you might work on your terminology. Vm is worthless, it only sets ratio of R4/R5. The only determined solution is for R6||(R4+R5), the equivalent lumped resistor load.