xconwing
- 20
- 0
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
How to Simplify Inductor Expressions Using Calculus?
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
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
- 42,895
- 983
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. [itex]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[/itex] which, if [itex]\lim_{\tau\to\infty} i(\tau)= 0[/itex] is [itex](1/2)i^2(\tau)[/itex]
If Li is a single function, then it is NOT always true.
If the latter this is easy. [itex]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[/itex] which, if [itex]\lim_{\tau\to\infty} i(\tau)= 0[/itex] is [itex](1/2)i^2(\tau)[/itex]
If Li is a single function, then it is NOT always true.