# Power to work of an inductor

1. Feb 1, 2013

### xconwing

How can you get from the first expression to the second. I've done calculus, just need a fresher. I don't know how to deal with the dt/dt

thanks

2. Feb 1, 2013

### HallsofIvy

Staff Emeritus
What you have written is not clear. Is "Li" a single function or is this the constant, L, times the function i?

If the latter this is easy. $L\int_{-\infty}^t i(\tau) (di/d\tau)d\tau= L\int_{-\infty}^t i di = L\left[(1/2)i^2(\tau)\right]_{\infty}^t$ which, if $\lim_{\tau\to\infty} i(\tau)= 0$ is $(1/2)i^2(\tau)$

If Li is a single function, then it is NOT always true.

3. Feb 1, 2013

### Staff: Mentor

Yeah, L is the constant value of the inductance. i(t) is the current as a function of time.