Quote:
Originally Posted by AdrianH
Hello kalee20, I am not going to disagree with you, it was just the formula given is V = L.2I/tr as if the 1 amp was in fact +1 amp to 1 amp?

OK  the general formula V = L x dI/dt is correct.
If the current changes from +I to I in tr, then the change in current is 2I, giving the formula stated. But they do clarify is as 1 amp peakpeak (which would have to be +0.5A to 0.5A).
The formula in your later clipping, energy = 1/2 x L x I^2, is also correct, this is the energy in an inductor when it is passing a current I. Nothing here about peakpeak, or times! It's analogous to a mass m moving at velocity v, having kinetic energy 1/2 x m x v^2, the factor 1/2 creeps in again.