| Electronics > Beginners |
| Calculating value of resistors in parallel |
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| IanB:
OK, now I'm home from work, I've looked at the problem as promised. I see that you have a solution, and the problem is basically solved. I found that an algebraic solution was essentially intractable, and was indeed much harder than I suspected it to be. Instead I have attached an Excel workbook that shows how to tackle the problem using the solver add-in. Since you provided all possible measurements it is possible to look at this as an error minimization problem and find the most likely values of the resistors based on your measurements. Once it became clear that R2 was a very large value I eliminated it from the network and re-solved for the others. As you will see the results are that R1 = 61.7 Ω, R3 = 89.6 Ω and R4 = 180.9 Ω. It seems likely that your resistance measurements were not as accurate as they might be, since some residual errors remained after solving. |
| paf:
--- Quote from: soldar on January 02, 2019, 11:06:43 pm --- --- Quote from: paf on January 02, 2019, 11:02:07 pm --- Solve the equations, and invert the found values to have the resistors. --- End quote --- Easy to say isn't it? The problem is that several people come in and suggest a way but they don't do it. And I have thought of those ways and I always get stuck. So I would like anyone who thinks they can solve it to actually do the work of solving it and showing us how it is done. Because I just get stuck along the way. --- End quote --- Have you read all my message? I have said that short circuiting some terminals, and measuring the resistance you can have simpler equations. You can measure the parallel of any set of two resistors on the circuit. After doing that, the equations are simpler. I cannot measure the resistance in your circuit, from my home. But, will gladly do it if you show me how. |
| soldar:
--- Quote from: IanB on January 03, 2019, 03:31:14 am --- I found that an algebraic solution was essentially intractable, and was indeed much harder than I suspected it to be. --- End quote --- Yes, I have tried several times over the years and I always got stuck. It was just an academic exercise so I just dropped it each time. It never occurred to me to see if it might be only three elements. Once you think of that possibility it is easy to find out which element is missing because the reading between those two terminals becomes the sum of the other three individually. At that point the problem is trivial. I just had a look at another heater I got from the trash but this one has the elements exposed as the whole thing was covered with glass. You can see I have marked the three resistor elements. It has four connectors and the only problem and the reason it was discarded and replaced is that the connectors are failing due to the heat. The rest works fine and repairing the connectors is trivial. What I have found fails quite frequently is the rotary selector switch and those are expensive and difficult or impossible to find. In my case I fitted four switches as shown and it works well enough for me but I suppose I could fit three-position switches which would allow connecting each terminal to high/off/low rail. That would give more combinations for power. --- Quote from: IanB on January 03, 2019, 03:31:14 am --- Instead I have attached an Excel workbook that shows how to tackle the problem using the solver add-in. Since you provided all possible measurements it is possible to look at this as an error minimization problem and find the most likely values of the resistors based on your measurements. Once it became clear that R2 was a very large value I eliminated it from the network and re-solved for the others. --- End quote --- I know how to use Excel to solve for one unknown by doing successive iterations - approximations but with four unknowns I do not know how to do it. Thanks for that sheet. I will study it and learn. I always find Excel to be very useful but my knowledge is limited. --- Quote from: IanB on January 03, 2019, 03:31:14 am --- As you will see the results are that R1 = 61.7 Ω, R3 = 89.6 Ω and R4 = 180.9 Ω. It seems likely that your resistance measurements were not as accurate as they might be, since some residual errors remained after solving. --- End quote --- Yeah my readings / data was taken from a quick reading of power used in each position and then converted to ohms assuming a certain voltage. Just a rough approximation. But once you realize there is no fourth resistor then the value of the others are taken directly between terminals. The values you arrived at are the values I mention in my OP. The solution was right there staring at me and I just wasn't seeing it. Thank you all for your input. |
| soldar:
--- Quote from: paf on January 03, 2019, 09:14:46 am --- Have you read all my message? I have said that short circuiting some terminals, and measuring the resistance you can have simpler equations. You can measure the parallel of any set of two resistors on the circuit. After doing that, the equations are simpler. I cannot measure the resistance in your circuit, from my home. But, will gladly do it if you show me how. --- End quote --- 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. |
| soldar:
--- Quote from: soldar on January 03, 2019, 10:01:35 am --- --- Quote from: paf on January 03, 2019, 09:14:46 am --- Have you read all my message? I have said that short circuiting some terminals, and measuring the resistance you can have simpler equations. You can measure the parallel of any set of two resistors on the circuit. After doing that, the equations are simpler. I cannot measure the resistance in your circuit, from my home. But, will gladly do it if you show me how. --- End quote --- 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 for you all along. I know that with four unknowns you only need four equations to solve but I gave more redundant, information so that readers could use which data to use and so that it could be used to confirm any results obtained. --- End quote --- |
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