Why does RL circuit equation use RI not R(dI/dt)?

[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"?

 

[dlgoff]

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

Regards

 

[asdf1]

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

 

[dlgoff]

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]

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.

 

[asdf1]

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

 

[asdf1]

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