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EMF and internal resistance

  1. Sep 4, 2015 #1
    Hello,

    We know that the electric field generated by a changing magnetic field is not conservative. But, for example let' say we have an EMF source that has an internal resistor, the voltage drop across the source is equal to EMF-rI where r is the internal resistance. My question is, inside the EMF source, we have non-conservative electric field but also do we have conservative electric field due to its internal resistor? Please if someone didn't understand the question tells me.

    Thank you.
     
  2. jcsd
  3. Sep 4, 2015 #2

    sophiecentaur

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    Hi.
    You have brought up two different issues. An emf source does not, by definition, have an internal resistor. A real source of emf, if it is located in a specific place, (often) has an equivalent circuit of a true emf in series with a series resistor. All the same rules apply to that resistor as the other resistors in the circuit.
    The issue of the non-conservative field is a separate one, I think and this video deals with it quite well. He presents it in an entertaining way.
     
  4. Sep 4, 2015 #3
    Yes, Sorry I meant about an emf source having a non-conservative electric field plus an internal resistor like in a coil. we can say that inside the coil we have a conservative electric field due to its resistor?
     
  5. Sep 4, 2015 #4

    sophiecentaur

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    Doesn't that video describe exactly the situation you describe (in effect)?
     
  6. Sep 4, 2015 #5
    Okay, you mentioned before that an internal resistor is like a normal resistor. So yes it describes the situation. Just I was confused about the internal resistor and the difference between normal one because physically the coil is one object that has ideal coil + resistor.
    Thanks anyway
     
  7. Sep 5, 2015 #6

    sophiecentaur

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    But with one difference. You can't measure the voltage drop across it - and it's only an 'equivalent' component in as far as the loss mechanism can be anywhere and anything that's 'inside' the component. In your though experiment, you couldn't put a meter where the one in the video appears so you wouldn't get any paradoxical results. This stuff is quite confusing because of the temptation to assume that you really would get different results just by re-connecting a meter.
    I'm not sure about this one because the resistor in question is distributed within the coil. Would the conservative field idea still apply?
     
  8. Sep 5, 2015 #7
    Then why there's drop in potential if there's no conservative electric field?
     
  9. Sep 5, 2015 #8

    sophiecentaur

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    Could there not be a combination of conservative and non-conservative field, perhaps?
     
  10. Sep 5, 2015 #9

    sophiecentaur

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    Perhaps someone else could chime in, here?
     
  11. Sep 7, 2015 #10

    sophiecentaur

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    Having read around, I now find (why has it taken me so long?) that any vector field will be a combination of a conservative field and a solenoidal (non-conservative) field. Wikkers is a way into this. It makes good enough sense, when you think about it.
     
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