Modeling question

1. Aug 12, 2005

asdf1

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. Aug 12, 2005

dlgoff

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

Regards

3. Aug 13, 2005

asdf1

no, but in that case there is an inductor,
so shouldn't the "I" be "dI"?

4. Aug 13, 2005

dlgoff

Think of two circuits; one with just a resistor and one with just an inductor (ideal with no resistance).

5. Aug 13, 2005

Integral

Staff Emeritus
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. Aug 14, 2005

asdf1

@@a
still a little confused~
but isn't the current always changing?
so there isn't supposed to be "I"?

7. Aug 15, 2005

SGT

In a resistor the instant voltage is proportional to the instant current:$$v(t) = R i(t)$$ (Ohm's law).
In an inductor, the instant magnetic flux is proportional to the instant current: $$\phi(t) = L i(t)$$.
But by Faraday's law, the voltage in a circuit is the derivative of the flux: $$v(t) =\frac{d\phi}{dt}$$, so $$v(t) =L\frac{di}{dt}$$.

8. Aug 17, 2005

asdf1

thanks~
i think i thought too much...
:P