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Modeling question

  1. Aug 12, 2005 #1
    if you model a RL-circuit, the correct answer should be L(dI/dt)+RI=V
    but why doesn't the "I" in "RI" have to be "dI"?
  2. jcsd
  3. Aug 12, 2005 #2


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    If ther were no inductor, would you need (dI/dt)?

  4. Aug 13, 2005 #3
    no, but in that case there is an inductor,
    so shouldn't the "I" be "dI"?
  5. Aug 13, 2005 #4


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    Think of two circuits; one with just a resistor and one with just an inductor (ideal with no resistance).
  6. Aug 13, 2005 #5


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    Such terms are not arbitrarily assigned. The voltage drop across an inductor is directly proportional to the CHANGE in current. The voltage drop across a resistor is proportional to the CURRENT. This is reflected in the modeling equation which must be based on the physics of the system.
  7. Aug 14, 2005 #6
    still a little confused~
    but isn't the current always changing?
    so there isn't supposed to be "I"?
  8. Aug 15, 2005 #7


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    In a resistor the instant voltage is proportional to the instant current:[tex] v(t) = R i(t)[/tex] (Ohm's law).
    In an inductor, the instant magnetic flux is proportional to the instant current: [tex]\phi(t) = L i(t)[/tex].
    But by Faraday's law, the voltage in a circuit is the derivative of the flux: [tex]v(t) =\frac{d\phi}{dt}[/tex], so [tex]v(t) =L\frac{di}{dt}[/tex].
  9. Aug 17, 2005 #8
    i think i thought too much...
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