# Modeling question

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"?

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

Regards

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

dlgoff
Gold Member
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).

Integral
Staff Emeritus
Gold Member
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.

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

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 $$v(t) =L\frac{di}{dt}$$.

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