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How to model internal resistance

  1. Dec 18, 2008 #1
    If I set up a coil with thin wire, push a magnet through it, and connect either end to a voltmeter (in series), then is this an accurate way to measure the total voltage? I might not be phrasing this right, so ask me to clarify if I should. I'm guessing that you can't just simplify a circuit like this by drawing a battery as the inductor and just add a resistor to the side which stands for the internal resistance. Or can you?
  2. jcsd
  3. Dec 19, 2008 #2


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    Why would you model a battery as an inductor? And I don't know what you're trying to ask.
  4. Dec 19, 2008 #3
    Are you trying to model an ac or a dc source?
    The usual model for a dc battery is an ideal voltage generator in parallel with a resistor...if you are trying to model an ac voltage source, typically a rotating piece of machinery such as an alternator, I'm sure there are standard models available....
  5. Dec 19, 2008 #4
    I might not be using the terms correctly, as I'm only a beginner with circuitry. I used the term inductor to describe a source of voltage. I'll assume that it's DC (a coil of wires that have magnetic flux go from A to B).

    Using equations for the area of the loops, the number of turns, and the magnetic field, I can get the EMF. Now, let's assume that using equations gives me an EMF of 50 V. If those coils of wire have an internal resistance known to be 20 ohms, then will connecting a voltmeter to the circuit reveal a 50 V difference between either end of the voltage source?
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