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Can you stop current flow with a resistor? And POTs?

  1. Apr 3, 2015 #1
    This thread will concern the workings of potentiometers, but I could use some help tackling something first. I will explain in my next post!



    Assume I have shorted out a battery with some resistor.

    I always thought that if you had a high enough resistor value, you could stop current from flowing. How? The battery has some potential difference in volts. I see this as an attractive force (well, when given a conducting path). So let's take this attractive force in volts, and convert it to joules. So let's say I have an attractive force of 10 Joules. Now onto the resistor. I see this as the inverse of the battery's attractive force, so it's a repelling force. So imagine I start swapping out that resistor with ones of higher and higher values. When the resistor reaches a repelling force that is equal to the battery's attractive force (so at 10 Joules), then current should cease to flow. It's analogous to taking two 9V batteries and wiring one's negative to the others positive and vise versa.



    I tried keeping that as elementary as possible so you can see where I'm coming from. I know it's incorrect, but I can't help to think of it that way. Why?

    If you were to theoretically short out a battery with a resistor value of 0 ohms (use your imagination), you could figure out the maximum current value that battery can supply. Just use the battery's internal resistance. But what happens when you go the other way with it? What is the least current a battery can supply? If I have a 1V battery, I can keep plugging in higher and higher resistor values -I'll just keep getting a smaller and smaller fraction for the current.

    Is there a limit?
     
  2. jcsd
  3. Apr 3, 2015 #2

    mfb

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    You need an "infinite resistance" for that.
    That does not work. Volts and Joules are different units for a good reason. You can have "10 Joules per Coulomb transferred", which is the same as 10 Volts.
    It is not. Otherwise we would call it battery, not resistor.
    Your models of batteries and resistors do not reflect reality.
    Right. And you can do that as long as you want (in theory, in reality even the battery without anything connected has a large but finite resistance, discharging it, due to the battery material, air and so on).
     
  4. Apr 3, 2015 #3

    berkeman

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    Just remember V=IR :smile:
     
  5. Apr 6, 2015 #4
    y3s.jpg

    This is my schematic of a POT, power source, and load. What's with the mediocre illustration of a POT? That's not the schematic symbol... I know. I will explain.

    In my initial post, I mentioned that you could figure out the maximum amount of current a chemical battery can deliver using the battery's voltage & internal resistance. Again, we assumed we shorted out the leads of the battery with a resistor of zero ohms just to get the battery to function. And then I went the other way with it and asked what the least current a battery could supply was. And well, there wasn't one. It just approaches zero.

    And?

    Assume that POT is 10K. Regardless of the load, there will always be .001 Amperes of current moving through the POT. That is wasted power. So why not just use two terminals of the POT? This would eliminate that wasted power issue and would still serve as a variable resistor.


    y4s.jpg



    Okay, now that I think of it, that isn't going to work. I put too much work into all of this to delete it though, lol.

    Is there a difference between a voltage divider and a variable resistor? I see people label POTs as variable resistors, but that last figure I drew is a variable resistor, but I can't think of any practical uses for it. You turn the dial and you'll change the resistance. Change the resistance and you'll modify the current drawn.
     
  6. Apr 6, 2015 #5

    mfb

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    There is. With a variable resistor the voltage for your load will depend a lot on its internal resistance. In the voltage divider setup, as long as the load has a resistance high enough, it will always have approximately the same voltage.
     
  7. Apr 6, 2015 #6

    Integral

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    There are 2 ways of wiring a variable resistor. First as a potentiometer, This is what you have drawn, what it accomplishes is to select a voltage out which can be anywhere between the terminal voltages. This is best used as a voltage source with small current draw.

    2nd It can be wired as a rheostat, with one end on the source and the output on the wiper, with the second end of the variable resistor floating, ie not connected. This is used for current control. I have seen huge rheostats used for stage lights.

    A voltage divider is a circuit used to pick a voltage between ground the source max it may or may not involve a variable resistor.
     
  8. Apr 6, 2015 #7
    But practically though, wouldn't you eventually reach a point where the current was so weak that it was cancelled out by currents induced by background radiation?
     
  9. Apr 6, 2015 #8

    mfb

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    That goes randomly in both directions, with some preferrence for discharging.
     
  10. Apr 6, 2015 #9

    sophiecentaur

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    This is something you will just have to deal with. You can't just make up your own rules. If you aren't prepared to accept and use the conventions then you will continue to struggle and get nowhere. It's not 'optional'.
    Also, the mathematical formulae are absolutely essential. Learn them and use them.
     
  11. Apr 6, 2015 #10
    The reason is that the whole point of inventing mathematics is so that you can be shown how to think about it rather than having to rely on physical intuition (which usually gets it dead wrong).

    The mathematics (in this case Ohm's Law) describe the situation- you need to start your thinking from there, not blindly use mathematics for your calculations while still relying on (very very weak) human physical intuition.
     
  12. Apr 7, 2015 #11

    sophiecentaur

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    The current would not be 'cancelled out" but subjected to variations in both directions. As you reduce the current the background noise makes it progressively harder and detect it but, given enough averaging time, you can chase that down into the noise. That principle applies to all measurements, though. That's the beauty of the Maths, though. It allows you to draw lines through the origin of graphs, where (low) numerical examples would produce nonsense.
    OF course the effect of noise has to be borne in mind in any mathematical model of a physical system but the Maths allows us to treat the noise as a bolt on, adding to the basic model.
     
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