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. :)