By123Neu
The power supply, in my case, is supposed to be a potentiometer that provides 20 volts at one output point and 10 volts at the other, both are brought to a current path, what happens in this case, as well as with the voltage itself.
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As long as you don’t reveal how the inner life with potentiometer looks, you can say exactly zero. A potentiometer is a resistance with three connections: it does not provide anything – only in the context of a concrete circuit, which you do not show. We don’t have a glass ball.
Why is that as important as the inner life of a potentiometer, there is only one grinding resistance, with which the output voltage is set?
then draw it up. I drew a picture of a potenziometer circuit. But you owe us these details, all where the second tension comes from. You can’t expect a well-founded answer to a half-hearted question. It is not about the inner life of the potenziometer, but how it is connected. You hid that in the box. How do you know how the potenziometer is connected if you don’t show it?
I suppose you have a center-tap pot. But you should see the data sheet, like here https://asset.conrad.com/media10/add/160267/c1/-/en/002996440DS00/datenblatt-2996440-bourns-ptv111-3415a-b103-dreh-potentiometer-005-w-10-k-1-st.pdf
There are also various types of potentiometers that have 4 connections.
I’m afraid you won’t go on without these informations. Take a picture. Or is there something?
Yes has 4 ports, two positives and two negative ports.
Your potenziometer has 4 connections?
I added a picture, right now?
So if I lock another potentiometer in parallel, would it work with the two different output voltages?
What do you want to do?
A compensating current flows at a connecting point between 2 voltages so that a common voltage is set.
The voltage which is set depends on the internal resistance of the voltage sources. Whether the circuit survives it is of course questionable.
The circuit should withstand this without any problems, because in the circuit a resistance, in that case a lamp is installed, the higher voltages are transmitted.
Then, of course, the differential voltage is present at this.
However, you must note that potentiometers are not ideal voltage sources, so the output voltage is extremely load-dependent.
Again: NOT EQUIPMENT. A circuit diagram must specify exactly (!) how a potenziometer is installed, i.e. how and with what it is connected, what resistance it has as the load looks. That’s exactly as if you ask, “Please, how do I repair my car – it doesn’t start? But you can’t open the hood, just tell me what to do.” Of course, you will get any good-meaning answers, but hardly one that solves the problem. You’ve been guilty of the schedule so far. If you had posted it, the question would have been answered in 5min.
What do you mean by the question?
As soon as you connect voltage sources, they are no longer galvanically separated with it makes no sense.
I only have one last question, is the balancing current always smaller than the current if the voltage sources were switched galvanically?
Only if U1 and U2 are load-independent otherwise you have the shape U1 -I*R1 and U2 -I*R2 instead of U1 and U2 terms, where I again depends on the external resistance.
The easiest thing is to draw the equivalent circuit and then rectify it or simulate it all.
In my case, if, for example, the total resistance is 30 ohms, a balancing current of 0.34 amperes would flow.
Formula: U1-U2/Rges
Providing the output voltage of a potentiometer is always dependent on the load resistance at the pot!
With open output terminals the pot is up to 410V!
In some potentiometers, different voltages can be set at two outputs.
These voltages depend on load and total resistance!
A total current source is formed from or one of the systems smokes, which always depends on the structure of the circuit.
I wouldn’t think of a better circuit than the one I hired.
But we can generally speak if one somehow manages to combine two different voltage-carrying conductors with different voltages with a load resistance, what would happen with the two different voltages at this point?
Put a picture like you imagine!
Your blackbox doesn’t work there!
Let’s say we have set the partial resistance at the pot differently, do we not have to have different output voltages?
Depending on the load resistance, a different voltage at the load resistance is set due to part and total resistance, and in the case of line interruption (high-resistance) there is possibly a the outer conductor voltage of 410V at the terminals.
So you want to say that no matter what voltage you set differently on both controllers, the same voltage is always provided depending on the load resistance.