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Aug 27 at 2:42 comment added Ste Kulov @JohnD Oh, OK. I thought that was what TheAttack55 was trying to say, but on second look it sounds like they might have just made a mistake in the calculation??? I get the same values as you if I use the first and last data points (Vth=4.32 and Rth=22.7). By line, I mean the I-V curve. The I-V curve of this unknown circuit when you consider all the data points is slightly non-linear. So the Thevenin equivalent can only be an approximation to "best-fit" the points, as pointed out in qrk's answer. i.sstatic.net/9Qgn8UaK.png i.sstatic.net/pz0M6Yfg.png
Aug 24 at 16:57 comment added John D @SteKulov My answer didn't give EXACT results, but 0.78, 1.126, 1.41, 1.786, 2.127, 2.347, and 2.56. Well within what you might expect from experimental error. You wouldn't expect the points to lie on a line since a voltage divider is R2/(R1+R2).
Aug 24 at 7:11 comment added Ste Kulov @TheAttack55 My answer here is related to what’s going on im this question and gives the generic formulas. This is a very useful technique and am very glad it’s actually being taught in school. Many of my colleagues don’t know this at all. I actually just taught a class on this exact topic at my local hackerspace/makerspace last month to try and spread the word.
Aug 24 at 7:03 comment added Ste Kulov @JohnD I don’t understand how you got an answer that perfectly crosses all data points. From what I see in the table, the slopes between each data point is slightly different which means the system is non-linear. A decent approximation would be to do what TheAttack55 did and pick the first and last points to create a line from which would get close to most of the points. Is that what you did?
Aug 23 at 22:52 comment added John D @TheAttack55 I tried it and all the table values were consistent with my answers. So maybe try again?
Aug 23 at 21:27 comment added TheAttack55 @JohnD I did try this, and it worked fine for those 2 table values. I took the first table value as (2.56/2) = Vm[33/(33+Rth)] and the last one I worked by equating Vm = (0.78/2)/[5/(5+Rth)]. Solved to obtain what I thought would be the Rth value. It's when I analyzed the current of the other table values that I noticed this value wasn't true for the rest.
Aug 23 at 21:26 answer added qrk timeline score: 1
Aug 23 at 20:50 comment added John D Seems like you have a voltage divider with one unknown resistance, driven by a voltage source with an unknown value. But you have multiple data points for one resistance and the divider voltage. You only need two equations in two unknowns to find the answers.
Aug 23 at 20:41 comment added TheAttack55 @periblepsis That actually came up as I searched up similar approaches. Now that I think about it I was treading a similar path earlier until I got stuck. I used the average power formula (Vm x Im x Cos(0) x 1/2). I set Vm equal to a random constant 3 then came up with two separate equations that had separate values of Im which I substituted with Im = (Vpp/2)/(RL+Rth) where Vpp and RL were the table values. I equated them and thought I found the answer for Rth, but it only worked for 2 out of the 7 table values.
Aug 23 at 20:29 comment added TheAttack55 @pipe IKR? First time I can see a reward through a problem, but it's a tough one for me.
Aug 23 at 19:39 comment added periblepsis TheAttack, What do you know about the maximum power theorem?
Aug 23 at 19:27 comment added Voltage Spike @Theattack55 can you provide a source for the image?
Aug 23 at 19:27 history reopened Voltage Spike
Aug 23 at 19:26 history closed Andy aka
Voltage Spike
Needs details or clarity
Aug 23 at 19:23 comment added pipe I like it. This is a great exercise!
Aug 23 at 19:17 history edited Null CC BY-SA 4.0
added 29 characters in body; edited tags; edited title
S Aug 23 at 19:10 review First questions
Aug 23 at 19:17
S Aug 23 at 19:10 history asked TheAttack55 CC BY-SA 4.0