Electronics > Beginners
Calculating value of resistors in parallel
<< < (6/7) > >>
paf:

--- Quote from: soldar on January 03, 2019, 10:01:35 am --- If you think you have equations that will resolve the problem you do not need the actual values; just give them names. Like "short here and there and we have this and that in parallel, let us call that value K", etc. then with the givens K,L,M,N you can go on to resolve and give us the final result.

But, furthermore, if you look carefully at the OP you can see I already short circuited some terminals and gave the values I obtained which, indeed, simplified things but not enough to make it easy. The values were right there all along.

--- End quote ---

Sorry, I don't have time to help you doing a simple calculation, to give results that are only useful to you. But I have time to help everybody who is facing a similar problem to yours.

In order to simplify the equations you can:

1) Short circuit some terminals in order to have the parallel of any two resistors in the circuit.
2) Write the equations in terms of conductance, not resistance, so the equations will be simpler.

Once again, more complete:

Calling the conductance G1=1/R1 and so on, you can do the following:

1)  Shorting 1 with 4 and 3 with 2, measure between those two terminals:   Gx=G1+G3
2)  Shorting 1 with 3 and 4 with 2, measure between those two terminals:   Gy=G2+G4
3)  Shorting 1 with 2 and with 3, measure between those  terminals and 4:  Gz=G3+G4
4)  Shorting 1 with 2 and with 4, measure between those  terminals and 3:  Gt=G2+G1
5)  Shorting 2 with 3 and with 4, measure between those  terminals and 1:  Gw=G4+G1
6)  Shorting 1 with 3 and with 4, measure between those  terminals and 2:  Gu=G2+G3







soldar:
I have several of these hot plate units and they all have the same three resistor configuration. Originally they had rotary switches which selected the power setting but the switches are gone.

I have set up one using four on/off switches as shown but that leaves several combination of switches with no function or duplicate power settings and it requires having a table at hand showing the power for each combination.

I have been thinking about the possibility of building or adapting a multi-position switch. The best idea I can come up with would be a rotary shaft with cams which would actuate four micro-switches. Maybe I will give it a try.

Maybe another possibility would be a 3 bit counter which could be moved up/down and the 3 bits converted to the necessary 4 bits which would control a relay each. Maybe simpler than building mechanical parts.
The Electrician:

--- Quote from: paf on January 02, 2019, 11:02:07 pm ---
Well, if you are not good at algebra, cheat a little with some short circuits. 

Calling the conductance G1=1/R1 and so on, you can do the following:

1)  Shorting 1 with 4 and 3 with 2, between those two terminals   Gx=G1+G3
2)  Shorting 1 with 3 and 4 with 2, between those two terminals   Gy=G2+G4
3)  Shorting 1 with 2 and with 3, between those  terminals and 4  Gz=G3+G4
4)  Shorting 1 with 2 and with 4, between those  terminals and 3  Gt=G2+G1


Measure Rx, Ry, Rz and Rt. Invert those and you got  Gx, Gy, Gz and Gt.

Solve the equations, and invert the found values to have the resistors.

--- End quote ---

This is a good idea, but the Gt you have chosen is a linear combination of Gx, Gy and Gz, namely: Gt=Gx+Gy-Gz

If you short 2 with 3 and with 4, between those terminals and 1, Gt=G1+G4
    then a solution is possible:



ArthurDent:
I had originally written out the 4 equations with 4 variable and thought about matrix reduction but being lazy quickly threw that method aside. I decided to make some assumptions using Occam’s Razor and that, plus a little insight, worked. It is interesting seeing how some still want to do it the long and hard way and are reluctant to accept another simpler way - not that the other ways aren’t correct. It reminded me of this short video.


The Electrician:

--- Quote from: ArthurDent on January 03, 2019, 04:07:49 pm ---I had originally written out the 4 equations with 4 variable and thought about matrix reduction but being lazy quickly threw that method aside. I decided to make some assumptions using Occam’s Razor and that, plus a little insight, worked. It is interesting seeing how some still want to do it the long and hard way and are reluctant to accept another simpler way - not that the other ways aren’t correct. It reminded me of this short video.

--- End quote ---

This is the 21st century, and we have computers to do the algebra nowadays.  There are only 4 unknowns in soldar's first post, and one would think that the proper 4 equations would yield a solution.  In reply #2, the first 4 equations mbest gave are sufficient, and those are the 4 equations I used in reply #19.  I could see right away that those 4 equations wouldn't be solvable by hand in the amount of time I was willing to devote.  But, I was certain that modern mathematical software probably wouldn't have a problem, and in fact Mathematica found a solution in a fraction of a second.

I then thought I would see what a symbolic solution looks like, so I substituted symbolic variables for the numeric values on the right side of the equals signs.  Mathematica was still crunching away after 1 minute, so I stopped it because I know from experience that when Mathematica runs that long, the result will be several pages of albegraic expressions.

paf's suggestion in reply #11 is a good one, and after fixing his small boo-boo, 4 equations result that can be solved by hand.  These equations can also be written as a matrix formulation:



 Nowadays there's no reason to shy away from a problem involving a lot of algebra when we have all this computer power at our disposal.

Navigation
Message Index
Next page
Previous page
There was an error while thanking
Thanking...

Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod