Solving 5 Unknown Resistor Values

WBahn

Joined Mar 31, 2012
30,242
Impedance of 10 kΩ WHERE?
So far I have
VA- (1/(1/r1)+(1/r2)) * (1/(1/r3)+(1/r4)+(1/r5))=0
So you have a voltage minus something that is very clearly anything but a voltage.

Would you expect the voltage of Vs2 to be a part of things?

Always, always, ALWAYS check the units.

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

What you have written is the following:

\(V_A \; - \; \( \frac{1}{ \( \frac{1}{r_1} \) } \; + \; \( \frac{1}{r_2} \) \) * \( \frac{1}{\( \frac{1}{r_3} \) } \; + \; \( \frac{1}{r_4} \) \; + \; \( \frac{1}{r_5} \) \) \; =\; 0\)
 

WBahn

Joined Mar 31, 2012
30,242
Va - ( (1/(1/r1)) + (1/(1/r2)) ) * ( (1/(1/r3)) + (1/(1/r4)) + (1/(1/r5)) ) = 0
Again, you have the first term is a voltage while nothing after that has a voltage anywhere. Does that make ANY sense at all? Is there ANY way that it could POSSIBLY be correct?

Leaving that aside, look at what you have now and simply it. You get

Va - ( r1 + r2 ) * ( r3 + r4 + r5 ) = 0

So you are subtracting something that has units of Ω² from something that has units of V.

Does that pass the "do the units work out" test?

Does that pass the "does the answer make sense" test?
 

WBahn

Joined Mar 31, 2012
30,242
Va - ( r1(I1) + r2(I2) ) * ( r3(I3) + r4(I4) + r5(I5) ) = Vs2
What are these new currents I1 through I5?

Assuming they are the currents through the corresponding resistors, they are without meaning unless you indicate their directions through the resistors.

Where Va= 3.3 Volts
and Vs2= 0.9 Volts
???????

Look at your schematic.

Va = 0.9 V.

Vs2 = 3.3 V.
 

MrAl

Joined Jun 17, 2014
11,565
Vcc is 3.3v
Node A is 0.9v
Node B is 0.85V
Impedance of 10K Ohms

View attachment 151442
Hello there,

What is that impedance of, where is it measured? You need to tell us that.

The typical way to solve this would be to analyze the circuit in the usual way, then plug in what is known and then solve for the resistor values. You may need 5 equations because there are 5 unknowns.
You also have to solve for whatever that "impedance" is and that will be one of the equations.
 

MrAl

Joined Jun 17, 2014
11,565
Vcc is 3.3v
Node A is 0.9v
Node B is 0.85V
Impedance of 10K Ohms

View attachment 151442

Hello again,

Is that circuit still considered valid?
I ask because when i first replied i assumed it was a reasonable circuit but after looking at it again i see that maybe it is not reasonable at all.

Look at the polarity of the source, positive on left, negative on right, and that is the ONLY source.
Now look at Va and Vb. The most positive Va is on the RIGHT and the less positive Vb is on the LEFT.
That does not make sense because if Vb is 0.85v and is closest to the positive terminal of the only source, then Va must be less than that because Va must be closer to the negative terminal than Vb.

The obvious thing to try is to reverse the source, but i dont think that works either because then it looks like Vb should be negative with Va positive.

Take another look and see what you think.
 

LesJones

Joined Jan 8, 2017
4,208
Another point is that I don't think it is possible to calculate values R10 aand R40 as they are effectivly in series. It may be possible to calculate the sum of these two resistor values.

Les.
 

MrAl

Joined Jun 17, 2014
11,565
Another point is that I don't think it is possible to calculate values R10 aand R40 as they are effectivly in series. It may be possible to calculate the sum of these two resistor values.

Les.
Hello there,

Yes good point, and to bring it home if we swap R1 and R4 we get the same voltages at the two nodes A and B. The only thing that changes is the more uninteresting voltages on either side of the source. The sum is all we can calculate, and to add to that, if we express a ratio as the sum over R3 we can calculate two resistors:
R2=-(v2*R41)/((v2-v1)*(R413+1)+E1)
R5=(v1*R41)/((v2-v1)*(R413+1)+E1)

where E1 is the REVERSED Vcc in the original problem (that means positive on the right not left), and
v1 is Va, v2 is Vb, and
R41=R1+R4, and
R413=R41/R3, and
everything else the same.
Note this is the same circuit with Vcc reversed, so it's not the original circuit. I didnt want to give too much away about the original circuit just yet.
So we kind of pick R3 at random until we find out what that "impedance" is supposed to be.
For this example with R3=3k and R41=R1+R4=3k and v1=0.924 and v2=-0.528 we get:
R2=4k and R5=7k exactly.
Note v2 was negative in this example and i think it always will be unless it reaches zero maybe with some strange values, but it must be impossible to go positive (remember E1 has positive right in these solutions too).

I have a feeling there is a controllability issue (from control systems theory) with this circuit but i'd have to look into that. We can do some things with this circuit but cant do some other things.

Just to note:
0.924=231/250
0.528=132/250
the numerators are backwards of each other :)
 
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