RL Circuit Differential equation

[IBY]

Homework Statement

Basically, I am deriving the following equation:
I=I_0(1-e^{-\frac{R}{L}t})

Homework Equations

1) L\frac{dI(t)}{dt}+I(t)R=V
2) \frac{V}{R}=I_0

The Attempt at a Solution

In relevant equation 1), R was divided both sides, and by using 2), I turned it into:

\frac{L}{R}\frac{dI(t)}{dt}+I(t)=I_0

By solving the homogenous part, I got:

I(t)=I_0e^{-\frac{R}{L}t}

If so, with the non-homogenous part, I got:

I(t)=I_0+I_0e^{-\frac{R}{L}t}

Obviously, something went wrong here because that plus sign is supposed to be a minus. The question is, what is the mistake?

 

[vela]

The constant in the homogeneous part isn't I₀. It's an arbitrary constant C that you solve for by using the initial condition.

 

[IBY]

But at initial condition, isn't C=I_0?

After all, that is the current right when the switch closes in the circuit, causing the current to flow. Because all I am left with after solving the homogenous part is:

I=Ce^{-\frac{R}{L}t}

 

[vela]

No, the initial condition has to be satisfied by the complete solution, so you have

I(t) = V/R + Ce^{-\frac{R}{L} t}

and you want to solve for C by setting I(0)=0 because the current through an inductor I(t) must be continuous. Note that the current isn't I₀ at t=0 because there was no current flowing just prior to the switch closing.

 

[IBY]

Okay, so based on that, C=I(t)-I_0?

Which means, substituting that into: I(t)=I_0+(I(t)-I_0)e^{-\frac{R}{L}t}

 

[IBY]

Uh oh, when I do the math above, I get I_0=I(t)

 

[vela]

When you set t=0, you get

I(0) = V/R + C

You don't still have I(t) floating around.

 

[IBY]

But what is I(0), then? If I(0) is current at initial condition, then that is I_0. That makes C=0.
Okay, I am doing this whole math logic thing really wrong somewhere, and I can't grasp where it is.

 

[vela]

Reread what I said in post 4.

 

[IBY]

Oh, if I(0)=0 because current doesn't exist at t=0, then C=-I_0.
Then I(t)=I_0-I_0e^{-\frac{R}{L}t}
Thanks!

 

[IBY]

D'oh! I have been assuming the graph was a decay all along just because the e was there. I forgot, this RL circuit is one with a battery. :)