# How to get v(t)?

1. Jan 17, 2005

### EvLer

Hi everyone,
I am not sure how to solve this:
given a plot of power p(t) and current i(t), I am supposed to come up with a plot of voltage v(t). I know that p(t) = i(t)*v(t). The problem is that p(t) waveform is a parabola, while i(t) is linear. Does that mean I have to divide (long division) p(t) equation by i(t) equation and plot the result? Is there an easier way to do it just from looking at the graphs?
What if one of them is a trig. function? Is software the only way?

2. Jan 17, 2005

### stunner5000pt

are you given the two functions for p(t) and i(t)?? If so i suggest long division as the most direct route! There is NO way that V(t) is a sine function because if it was:
Since I(t) is linear I(t) = At so it is a straight line
Let s say V(t) = Sin (Ct+D) then P(t) = At Sin(Ct+D)

now i am sure in previous classes you have seen that if your function looks like that then P(t) would also wave because A(t) would serve as the amplitude for the waving function and thus P(t) would wave up and down as WELL (Which it doesnt).

How complex are the functions for P(t) and I(t)? if they are polynomials i am sure you don't need to seek alternative routes.

3. Jan 17, 2005

### nrqed

I am assuming that p(t) is a parabola starting at the origin, right? Then itt is of the form $a t^2$. And if i(t) also starts at the origin, then it is of the form $b t$. So the ratio is of the form $c t$. You just need to pick a point at a given time on both the p(t) and i(t) graphs to fix your constant "c".

So, are your graphs crossing the origin? That's the key point. I fthey do it's trivial. If they don't they must cross the point P=0 and I=0 at the same time and the ratio is still a linear graph that can be plotted using two points only. So still trivial. And if they do not equal zero at the same time, I am not sure they physically make sense (but I might be missing something).

Pat