Solve v(t) from Power and Current Plots

  • Thread starter Thread starter EvLer
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on deriving the voltage function v(t) from given power p(t) and current i(t) plots, utilizing the relationship p(t) = i(t) * v(t). It is established that if p(t) is a parabolic function and i(t) is linear, the voltage can be calculated by performing polynomial long division of p(t) by i(t). The participants confirm that if both functions start at the origin, the resulting voltage function will also be linear, simplifying the plotting process. The discussion also addresses the implications if one of the functions is a trigonometric function.

PREREQUISITES
  • Understanding of polynomial functions and their properties
  • Knowledge of basic calculus, specifically differentiation and integration
  • Familiarity with the concept of power in electrical circuits
  • Experience with graphing functions and interpreting their intersections
NEXT STEPS
  • Study polynomial long division techniques for function analysis
  • Explore the implications of trigonometric functions in power calculations
  • Learn about software tools for graphing and analyzing mathematical functions
  • Investigate the physical significance of power, current, and voltage relationships in electrical engineering
USEFUL FOR

Electrical engineers, physics students, and anyone involved in circuit analysis or mathematical modeling of electrical systems will benefit from this discussion.

EvLer
Messages
454
Reaction score
0
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?

Thanks in advance.
 
Physics news on Phys.org
EvLer said:
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?

Thanks in advance.

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.
 
EvLer said:
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?

Thanks in advance.

I am assuming that p(t) is a parabola starting at the origin, right? Then itt is of the form [itex]a t^2[/itex]. And if i(t) also starts at the origin, then it is of the form [itex]b t[/itex]. So the ratio is of the form [itex]c t[/itex]. 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
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
Replies
9
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
44
Views
4K
Replies
14
Views
2K
Replies
4
Views
1K
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
23
Views
2K
Replies
5
Views
1K