# Self-Inductance of an Inductor

1. May 1, 2012

### forestmine

1. The problem statement, all variables and given/known data

An inductor has a current I(t) = (0.480 A) cos[(280 s-1)t] flowing through it. If the maximum emf across the inductor is equal to 0.490 V, what is the self-inductance of the inductor?

2. Relevant equations

ε = -L*di/dt

3. The attempt at a solution

I would have that this would be as easy as using the above equation, and taking a derivative as necessary. The one thing that throws me is the value of t and the fact that it is unknown.

So, what I did was,

ε = LdI(t)/dt

Taking the derivative of I(t), I get

dI(t)/dt = -(.48)(1/280)sin(1/280*t)

If I even did that correctly, I still don't have a value of t in order to solve for L. I know that the maximum emf across the inductor will occur right at t=0, right? As the current reaches a steady value, dI/dt goes to 0, at which point the inductor acts as a wire and there is no emf across it.

But if that's the case, and this max emf of .49 V occurs at t=0, then sin(0) = 0, and L = 0.

I think I'm completely missing something, here.

Any help in the right direction would be great.
Thanks!

2. May 1, 2012

### phyzguy

For what value of t with sin(t/280) have a maximum? What is the maximum value of sin(anything)?

3. May 1, 2012

### Staff: Mentor

The circuit is being driven by a sinusoidal source so DC steady-state behavior is not applicable. It's AC steady state that's of interest.

You can forget about the particular instant of time since you are given the maximum emf. What's the maximum value that sin(θ) can ever be? (And in a related factoid, remember that voltage and current for an inductor have a mutual phase shift with respect to each other).

4. May 1, 2012

### forestmine

Alright, so the maximum value that sin(x) can have is 1. So t would have to be 280.

I guess I don't understand how we move from knowing the maximum value of the emf to knowing that sin(x) needs to be a maximum. I'm having a hard time picturing this graphically.

I know that the voltage for an inductor leads the current by 90 degrees, but when I think about this graphically, if the voltage is max and right along the y-axis at 90 degrees, then the current is 0, along the x-axis.

5. May 1, 2012

### Staff: Mentor

You don't need to know the particular time. All you need to know is that at some instants
$E = L \frac{dI}{dt}$ will have a maximum (and a symmetrical minimum). Since L is a constant, you just need to know the maximum value that $\frac{dI}{dt}$ can take on, which occurs when its sin(whatever) = +/- 1. So just replace sin(whatever) with +/- 1 and get on with it

When you're dealing with maximums and minimums of a periodic signal you usually don't have to be concerned about phases or times. Just determine what maximum or minimum values the function can take on.

6. May 1, 2012

### forestmine

Ok, thank you! That makes sense. I was over-thinking this one.

So, I should be left with

dI(t)/dt = -(.48)(1/280)

ε = LdI(t)/dt

And solving for L, I come up with about 285, which doesn't seem to be the correct answer.

The correct answer is 3.65mH. Apparently I can't even use my calculator this afternoon?! I have no idea how they're getting this answer...

7. May 1, 2012

### Staff: Mentor

Your original current function had I(t) = (0.480 A) cos(280s-1t). What's ω for that cosine function?

8. May 2, 2012

### forestmine

Omega for the cosine function is 280. Got it, thanks again!

9. May 2, 2012

### truesearch

If max V = 0.49 and the max current is 0.48 then the reactance Xl = V/I = 1.02 ohms
Xl = ωL and you know ω
So you should be able to find L.