studiot Posted August 25 Share Posted August 25 (edited) 21 hours ago, Z.10.46 said: A voltage source is a component that creates a potential difference between two points in a circuit, meaning it drives a certain amount of electrons to move from one point to another in a specific direction according to a time frequency. The internal resistance depends on the properties of the voltage source. In the quantum world, a small portion of electrons might even choose to move in the opposite direction of others, which can slightly reduce the voltage, and some electrons might go elsewhere, leaving the circuit and dissipating as heat. The quantum world is indeed strange, with some electrons following the rules, others doing the opposite of the rules, and some having no rules to follow at all... But this shows a misunderstanding of rectification. Yes, sorry, you should use diodes for rectification to keep only the positive sinusoidal signal, followed by filtering with capacitors to smooth the signal. You might also consider adding a regulator at the end. In any case, blocks 2 and 4 correspond to a device that transforms a sinusoidal signal into a constant signal. I wasn't sure how to say this in English. Are there any other points that need to be clarified in the schematic as well? This looks really promising, you have some idea about voltage sources. +1 Electrical engineering is an intensely practical subject. It is very fortunate that we are able to model the behaviour of electrical systems by carefully defining a few ideal or perfect components or elements. But we should always keep in mind that there are no such things as ideal or perfect elements in the real world. Note this is true in other disciplines as well as electrical enginering. When we measure the performance of real world voltage sources we find the graph in fig1 applies without exception whether we are talking about alternating or direct, pulsed or steady and whatever their source of EMF (electromotive force). So it applies to generators, dynamos, batteries thermogenerators etc. Looking at fig1 we see that the ideal voltage source output remains constant, whatever the current draw. But the output of the real world voltage source fall proportionately with increasing current from a maximum at zero current. So we construct a model of a real world voltage source by combining an ideal voltage source with an ideal resistor in series, as shown in fig2. The ouput of this real source appears between terminals A and B and is called the terminal voltage. In the next post or two I will go on to show how to use this extra information to determine the effect of combining two batteries of different voltages and capacities across the same load. We can also discuss conservation laws in the light of this new information. The situation there is somewhat more complicated than a simple answer. You will need to be clear upon the difference between an EMF and a Potential Difference (pd) and also the difference between voltage and current. There are circuits which are called voltage to current converters, but they are not rectifiers and act quite differently from what we are talking about. Edited August 25 by studiot Link to comment Share on other sites More sharing options...
Z.10.46 Posted August 27 Author Share Posted August 27 Thank you for your response. Could you please continue your valuable explanations? I would also like to know, for example, if any tests and real measurements have already been carried out to confirm that there is no voltage coming from the ground when U1 and U3 are different. If U1 and U3 are batteries, normally one or both batteries stop working, but before that, is there any voltage coming from the ground or not? The one from the Dark Web mentions that a high voltage can be observed coming from the ground, which is the main cause of battery destruction. In the case of a magnet rotating around a coil, there is no chemical reaction, just a disturbance in the magnetic fields. This voltage can light up a bulb and adapt to all loads, drawing a quantity of electrons from the grounding. Link to comment Share on other sites More sharing options...
studiot Posted August 27 Share Posted August 27 1 hour ago, Z.10.46 said: The one from the Dark Web mentions that a high voltage can be observed coming from the ground, which is the main cause of battery destruction. In the case of a magnet rotating around a coil, there is no chemical reaction, just a disturbance in the magnetic fields. This voltage can light up a bulb and adapt to all loads, drawing a quantity of electrons from the grounding. It's probably called the ~Dark Web because there is no electric voltage coming from the ground. So there are no electric lights in the Dark Web. Forget it. 1 hour ago, Z.10.46 said: Thank you for your response. Could you please continue your valuable explanations? I would also like to know, for example, if any tests and real measurements have already been carried out to confirm that there is no voltage coming from the ground when U1 and U3 are different. If U1 and U3 are batteries, normally one or both batteries stop working, but before that, is there any voltage coming from the ground or not? The one from the Dark Web mentions that a high voltage can be observed coming from the ground, which is the main cause of battery destruction. In the case of a magnet rotating around a coil, there is no chemical reaction, just a disturbance in the magnetic fields. This voltage can light up a bulb and adapt to all loads, drawing a quantity of electrons from the grounding. This suggests to me that you still don't appreciate the nature of electrical quantities. Since you haven't answered my comments. On 8/25/2024 at 11:31 PM, studiot said: You will need to be clear upon the difference between an EMF and a Potential Difference (pd) and also the difference between voltage and current. There are circuits which are called voltage to current converters, but they are not rectifiers and act quite differently from what we are talking about. Further I still have no idea what level to pitch the next bit at. At the end of this discussion I was going to point you at Millman's theorem (also known as the parallel generator theorem). This represents the general case and was developed for three phase elctrical supply where the three phases are often connected to a single load in what is known as a star connection. So help me to help you here Link to comment Share on other sites More sharing options...
Z.10.46 Posted August 27 Author Share Posted August 27 17 minutes ago, studiot said: It's probably called the ~Dark Web because there is no electric voltage coming from the ground. So there are no electric lights in the Dark Web. Forget it. This suggests to me that you still don't appreciate the nature of electrical quantities. Since you haven't answered my comments. Which ones? Further I still have no idea what level to pitch the next bit at. At the end of this discussion I was going to point you at Millman's theorem (also known as the parallel generator theorem). This represents the general case and was developed for three phase elctrical supply where the three phases are often connected to a single load in what is known as a star connection. So help me to help you here I understand your reasoning. So, you assume that an internal resistance would be so high that the voltages U1 and U3 would be equal? In Millman's theorem, Z1, Z2, and Z3 are fixed and carefully chosen so that U1, U2, U3, and VM are always equal. But in my circuit, U1 and U2 are not equal. Why assume that it is the internal resistance that will increase to make U1 and U2 equal? A voltage coming from the ground could also make U1 and U2 equal, and we might not realize it... Link to comment Share on other sites More sharing options...
studiot Posted August 27 Share Posted August 27 38 minutes ago, Z.10.46 said: I understand your reasoning. So, you assume that an internal resistance would be so high that the voltages U1 and U3 would be equal? In Millman's theorem, Z1, Z2, and Z3 are fixed and carefully chosen so that U1, U2, U3, and VM are always equal. But in my circuit, U1 and U2 are not equal. Why assume that it is the internal resistance that will increase to make U1 and U2 equal? A voltage coming from the ground could also make U1 and U2 equal, and we might not realize it... What is very clear is that you have not understood it. What I can't understand is why you would wish to argue rather than expand your knowledge and understanding. Firstly I haven't assumed anything (except linear circuit theory. Non linear theory gets exceedingly hairy) Secondly you haven't understood the most basic property of all real world voltage sources. I have already been through this. It is impossible to specify the output of a real world voltage source without reference to the load current, which is something you are trying to do. Thirdly U1 and U2 cannot be anything but equal since they are connected across the same pair of circuit terminals. Link to comment Share on other sites More sharing options...
Z.10.46 Posted August 27 Author Share Posted August 27 What I don't understand is why it is stated that the internal resistance increases to allow the two voltages to be equal. This is the standard explanation, but if we admit that this circuit acts as a sort of catalyst for a voltage coming from the grounding, we would see the same effect. So, we can't know which explanation is correct: is it the internal resistance that increases, or is it the voltage coming from the ground that makes U2 and U3 equal? Both could be physical explanations for what we actually observe. Link to comment Share on other sites More sharing options...
studiot Posted August 27 Share Posted August 27 (edited) 29 minutes ago, Z.10.46 said: What I don't understand is why it is stated that the internal resistance increases to allow the two voltages to be equal I never said it does, nor does millman's theorem. Forget the ground. The circuit analysis must also describe the situation where there is no ground. Edited August 27 by studiot Link to comment Share on other sites More sharing options...
Z.10.46 Posted August 27 Author Share Posted August 27 We can only determine this if precise energetic tests are conducted to measure the energies involved and identify the real cause: is it the increase in internal resistance, or is it a quantity of electrons coming from the ground? Have these tests been carried out to truly understand the real cause of this phenomenon? Link to comment Share on other sites More sharing options...
studiot Posted August 27 Share Posted August 27 23 minutes ago, Z.10.46 said: We can only determine this if precise energetic tests are conducted to measure the energies involved and identify the real cause: is it the increase in internal resistance, or is it a quantity of electrons coming from the ground? Have these tests been carried out to truly understand the real cause of this phenomenon? I keep telling you that the only way to correctly and accurately analyse such problems is to use measured properties or parameters. Those analysts with suffieient experience may be able to guess or estimate suitable values for at least some of the variables. Consider this simple question I have a length of rope specified at 100kg breaking stength. What force is rquired to break it ? In other words if I hang it up and add weights to the bottom what weight will I require to break it ? Link to comment Share on other sites More sharing options...
Z.10.46 Posted August 27 Author Share Posted August 27 Therefore, you would need to hang a weight of approximately 100 kg to break the rope. But if there are multiple forces greater than 100 kg applied, it wouldn’t be clear which force breaks the rope just by looking at the sum of the forces. A detailed analysis of the energy involved is needed. Link to comment Share on other sites More sharing options...
studiot Posted August 27 Share Posted August 27 26 minutes ago, Z.10.46 said: But if there are multiple forces greater than 100 kg applied, it wouldn’t be clear which force breaks the rope just by looking at the sum of the forces. This is nonsense as it is physically impossible to apply a force to the rope greater than required to break it. By definition. In the practical real world the real world breaking force will be greater than 100kg for statistical reasons. But statistical reasons don't enter into our consideration of circuits. I gave you a graph of the characteristics of a general voltage source. What do you think would be the correct circuit equation to describe it? It is important to use observable quantities. Link to comment Share on other sites More sharing options...
Z.10.46 Posted August 27 Author Share Posted August 27 (edited) Yes, a weight of 100 kg will break the rope. However, if you have two weights, one of 100 kg and one of 1000 kg, and you hang them on the cable, the rope will also break. If you don’t analyze the applied energy and how the rope was broken, it could be misleading to say that only the 100 kg weight caused the breakage. In the case of two batteries with voltages U1 and U2, where U1 is greater than U2, and if the battery with voltage U2 is destroyed, a detailed analysis of how U2 was destroyed is necessary to determine the true source of the energy that caused its failure. It’s important to assess whether the destruction was solely due to the voltage U1 or if it resulted from a higher voltage coming from the ground. Edited August 27 by Z.10.46 Link to comment Share on other sites More sharing options...
studiot Posted August 27 Share Posted August 27 26 minutes ago, Z.10.46 said: Yes, a weight of 100 kg will break the rope. However, if you have two weights, one of 100 kg and one of 1000 kg, and you hang them on the cable, the rope will also break. If you don’t analyze the applied energy and how the rope was broken, it could be misleading to say that only the 100 kg weight caused the breakage. In the case of two batteries with voltages U1 and U2, where U1 is greater than U2, and if the battery with voltage U2 is destroyed, a detailed analysis of how U2 was destroyed is necessary to determine the true source of the energy that caused its failure. It’s important to assess whether the destruction was solely due to the voltage U1 or if it resulted from a higher combined voltage or other factors. In what way do you think energy will cause a battery to fail ? Here's a challenge. I have just weighed a standard C cell at 71 grammes say 70 to make the math easy. How much energy do you say you need to add to 'destroy' it ? Please provide a figure, since you think it can be done. I will then tell you how to add more than that amount of energy without destuction, and then take destroy the battery by taking away said amount of energy. Link to comment Share on other sites More sharing options...
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