- 734

- 0

## Main Question or Discussion Point

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"?

but why doesn't the "I" in "RI" have to be "dI"?

- Thread starter asdf1
- Start date

- 734

- 0

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"?

but why doesn't the "I" in "RI" have to be "dI"?

dlgoff

Science Advisor

Gold Member

- 3,734

- 1,639

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

Regards

Regards

- 734

- 0

no, but in that case there is an inductor,

so shouldn't the "I" be "dI"?

so shouldn't the "I" be "dI"?

dlgoff

Science Advisor

Gold Member

- 3,734

- 1,639

Think of two circuits; one with just a resistor and one with just an inductor (ideal with no resistance).asdf1 said:no, but in that case there is an inductor,

so shouldn't the "I" be "dI"?

Integral

Staff Emeritus

Science Advisor

Gold Member

- 7,184

- 55

- 734

- 0

still a little confused~

but isn't the current always changing?

so there isn't supposed to be "I"?

SGT

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].

- 734

- 0

thanks~

i think i thought too much...

:P

i think i thought too much...

:P

- Last Post

- Replies
- 11

- Views
- 5K

- Last Post

- Replies
- 1

- Views
- 1K

- Replies
- 1

- Views
- 460