Can this be solved?

Discussion in 'General Electronics Chat' started by Llama, Jan 16, 2014.

1. Llama Thread Starter New Member

Jan 16, 2014
3
0
Hi,

I'm a mechatronics engineer working on 3 phase PMSMs. I'm trying to implement temperature compensation for the winding resistance, and am having some trouble.

V1, V2, I1 and I2 are known. I'm trying to work out Ra, Rb, Rc

I can do the wye-delta transformation, but I can't seem to figure this out using algebra.

2. #12 Expert

Nov 30, 2010
16,261
6,770
I1 Rc + I2 Rb = (I1 +I2) Ra

It's hard for me to think at that level.
Put up some real numbers and you'll get the answers.

3. Llama Thread Starter New Member

Jan 16, 2014
3
0
V1 = V2 = 20V.
I1 = 0.5
I2 = 0.45

Those are some typical values I guess.

Resistances should be somewhere just above 20Ohms, I'm just trying to automate the measurement without adding any more instruments to a rather convoluted setup.

4. #12 Expert

Nov 30, 2010
16,261
6,770
I max = .95
Sum of all R's = 21.0526

20V-.95 Ra = Vx
Vx/Rb + Vx/Rc = .95
Vx + VxRb/Rc = .95Rb
VxRc + VxRb = .95 RbRc
Vx (Rc + Rb) = .95RbRc
Vx = .95RbRc/(Rc + Rb)
.95RbRc/(Rc + Rb) = 20-.95Ra

Can you finish from there?

5. Llama Thread Starter New Member

Jan 16, 2014
3
0

Does adding in Vx as a variable really help in solving the equations?

I still can't get absolute variables from there. (I'm just getting one resistance in terms of the others)

6. #12 Expert

Nov 30, 2010
16,261
6,770
You could write out "20V - .95Ra" instead of Vx in each equation, but substitution is a common and harmless practice. (Look at the equations for active filters.)

As long as Rb and Rc have no voltage or current attached to either one of them, the only answer is in terms of each other. As far as I can tell, you can't get there from here.

7. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
If a real solution exists for the stated values then there would be an infinite range of real value solutions for Ra, Rb & Rc.

This may be reasoned as follows:

Take the values given

V1=V2=20 V
I1=0.5 A
I2=0.45 A

Suppose Ra=0Ω

Then Rb & Rc are readily found.

Suppose Ra ≠ 0Ω - what then ?

Let the unknown node voltage Vx be some value other than 0V - say 5V.

In that case Rc=(20V-5V)/0.5A [Ω], Rb=(20V-5V)/0.45A [Ω], Ra=5V/0.95A [Ω]

One can go on proposing any arbitrary real value for Vx less than 20V in this case and a solution will exist.

#12 likes this.
8. #12 Expert

Nov 30, 2010
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6,770
Well, that's a confirmation of what I said. Thanks for spelling out what I could only do intuitively.

9. Papabravo Expert

Feb 24, 2006
10,137
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If the original question was: does this problem have a unique solution, which was implied, then the answer is no. There are many cases where we expect a unique solution, but sometimes it just doesn't happen. Too many unknowns and not enough equations. The general rule is that for each unknown variable you need an independent equation. I see three unknowns and one equation after specifying V1=V2, I1, and I2.

10. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
It's possible to be "seduced" by the creation of three apparently independent equations. The third is a result of consideration of sourced & dissipated power balance.

$\text{(I_1+I_2)R_a+I_1R_c=V_1}$
$\text{(I_1+I_2)R_a+I_2R_b=V_2}$
$\text{{(I_1+I_2)}^2R_a+I_2^2R_b+I_1^2R_c={V_1I_1}+{V_2I_2}}$

or in matrix form

$\left( \begin{array}{l c r} (I_1+I_2) & 0 & I_1 \\ (I_1+I_2) & I_2 & 0 \\ {(I_1+I_2)}^2 & I_2^2 & I_1^2 \end{array} \right ) \left( \begin{array}{c}R_a \\ R_b \\ R_c \end{array} \right)= \left( \begin{array}{c}V_1 \\ V_2 \\ {V_1I_1}+{V_2I_2} \end{array} \right)$

A little reflection shows that the left hand 3x3 matrix is singular and therefore no (unique) solution exists for the unknown resistors.

Last edited: Jan 23, 2014