Modeling question

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  • #1
asdf1
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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"?
 

Answers and Replies

  • #2
dlgoff
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If ther were no inductor, would you need (dI/dt)?

Regards
 
  • #3
asdf1
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no, but in that case there is an inductor,
so shouldn't the "I" be "dI"?
 
  • #4
dlgoff
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asdf1 said:
no, but in that case there is an inductor,
so shouldn't the "I" be "dI"?
Think of two circuits; one with just a resistor and one with just an inductor (ideal with no resistance).
 
  • #5
Integral
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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.
 
  • #6
asdf1
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@@a
still a little confused~
but isn't the current always changing?
so there isn't supposed to be "I"?
 
  • #7
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].
 
  • #8
asdf1
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thanks~
i think i thought too much...
:P
 

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