Physics Ninja shows you to obtain the transform between Wye and Delta Resistor networks. Here’s a link to a more general proof drive.google.com/file/d/130gN...
Found this while searching an explanation for a homework, but I probably found the best way to properly getting introduced to the circuit part of electrical engineering, will definitely go through your videos! Thank you!
For those confused by the proof and its assumptions, reading the following may help. First step: we assume that there is a solution. By this I mean that we assume (very important) that there is a function that gives you R1 in terms of Ra, Rb and Rc only. We assume there is a function that gives you R2 in terms of Ra, Rb and Rc only. We do the same thing for R3; and, conversely, we assume there is a function that gives you Ra in terms of R1, R2 and R3 only; the same being true for Rb and Rc. This is no triviality, being these assumptions what gives sense to the whole proof. Now, if we assume all this, then whatever is true to one choice of voltages must be true to all choices of voltages. Let's clarify this: Let Vn be the voltage at the Y center. If we choose Vn3 = Vn, there is no current on R3 (Y circuit) and Rc is in parallel with Rb and Ra (no current going through node N3 on Delta circuit). By this you derive the first equation (N1-N2 in 4:09). Now, this equation makes no mention of the voltage we chose, so it must be true for all choices of voltage. If it was not the case, then there would be no solution If the previous paragraph sounds weird, all I can say is that it may take some time to digest the argument (rereading the second paragraph may help), but, nevertheless, it's a truthful argument. Once you have derived the Delta to Y equations, you have proved nothing (we assumed there was a solution to start with). All you have to do, though, is to check if they really are a solution by direct substituion (for any given three voltages at the nodes, the currents entering the nodes must be the same on the Delta and Y configurations) and the theorem is rigorously proven. No deep circuit theory needed, no superposition; the central argument is purely mathematical, logical. Now, once you've checked the solutions, then to manipulate them to obtain the Y to Delta conversion is legitimate, allowing you to prove rigorously the second theorem. I hope this helps whoever was, like me at the beginning, quite confused by the video and its assumptions.
Found this while searching an explanation for a homework, but I probably found the best way to properly getting introduced to the circuit part of electrical engineering, will definitely go through your videos! Thank you!
I knew about the first part, but I have to admit I did not know the second part: It is a very ingenious trick!
Thanks a lot.
Very great job , i appreciate yout effort...
You rock! Thanks a million!!!
Very nice explained
Brilliant Proof!
Thank you very much
Perfect!
Very useful indeed
Perfect
How they can be parallel in delta network
Best proof
Thanks indeed , but i suggest that you label different Resistans value to each resistance
Calculation on the transitor network
Sorry but I do not understand how you can simply ignore a terminal of the network when calculating the equivalent resistance between the other two.
Great question. Will get back to you early next week. Ninja is on vacation.
Here’s a more general proof
drive.google.com/file/d/130gNaUIwi_HZYXLLy_pG-38Okxu0oMrK/view?usp=drivesdk
@@PhysicsNinja Thanks!
@@PhysicsNinja Hey man , answer this question please
Boom Bap which question?
whats the point if this? why is this taught to electricians when all this is theoretical?
For those confused by the proof and its assumptions, reading the following may help.
First step: we assume that there is a solution. By this I mean that we assume (very important) that there is a function that gives you R1 in terms of Ra, Rb and Rc only. We assume there is a function that gives you R2 in terms of Ra, Rb and Rc only. We do the same thing for R3; and, conversely, we assume there is a function that gives you Ra in terms of R1, R2 and R3 only; the same being true for Rb and Rc.
This is no triviality, being these assumptions what gives sense to the whole proof.
Now, if we assume all this, then whatever is true to one choice of voltages must be true to all choices of voltages. Let's clarify this:
Let Vn be the voltage at the Y center. If we choose Vn3 = Vn, there is no current on R3 (Y circuit) and Rc is in parallel with Rb and Ra (no current going through node N3 on Delta circuit). By this you derive the first equation (N1-N2 in 4:09). Now, this equation makes no mention of the voltage we chose, so it must be true for all choices of voltage. If it was not the case, then there would be no solution
If the previous paragraph sounds weird, all I can say is that it may take some time to digest the argument (rereading the second paragraph may help), but, nevertheless, it's a truthful argument.
Once you have derived the Delta to Y equations, you have proved nothing (we assumed there was a solution to start with). All you have to do, though, is to check if they really are a solution by direct substituion (for any given three voltages at the nodes, the currents entering the nodes must be the same on the Delta and Y configurations) and the theorem is rigorously proven. No deep circuit theory needed, no superposition; the central argument is purely mathematical, logical.
Now, once you've checked the solutions, then to manipulate them to obtain the Y to Delta conversion is legitimate, allowing you to prove rigorously the second theorem.
I hope this helps whoever was, like me at the beginning, quite confused by the video and its assumptions.
Thanks a lot
How Ra is parallel with RB and Rc
😂