I haven't touched calculus in around a year and a half. However, this semester, I have to take Intro to Electrical Engineering as a requirement for my Computer Engineering major, and the first part of this class seems quite heavy on calculus. One of the problems I've been assigned is as follows:

The voltage and current at the terminals of the circuit element in Fig 1.5 are zero for $t < 0$. For $t \geq0$ they are:

v $= (16,000t + 20)e^{-800t}$ V

i $= (128t + 0.16)e^{-800t}$ A

At what instant of time is maximum power delivered to the element?

Now, I thought that all I had to do was multiply v times i and find the derivative of that equation, then find the maximum based on that, and then find the value of t for that. However, I tried that method on a practice problem and it didn't give me the correct answer. Additionally, the next part of this question is to actually find the maximum value for p, which makes it seem odd that I'd need to find the maximum value first, then the time at which it is at a maximum. So, please, help me figure out the actual method to solving this problem.

EDIT: Also, for clarity, Fig 1.5 simply shows an ideal circuit element, it isn't really necessary for solving the problem.

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    $\begingroup$ In principle, you should take the derivative of power. But it turns out that each of $v$ and $i$ is decreasing for positive $t$. $\endgroup$ – André Nicolas Aug 26 '12 at 20:23

Your approach is correct, multiply $vi$, take the derivative, set to zero. Without seeing your work, it is hard to say more. Having found the time, the value of the power is $vi$ at that moment. Usually I find the time first from the derivative and if you want the power you find it then. I plotted it in Alpha and it looks like the maximum is at $t=0$ as the equations are not valid before that.

  • $\begingroup$ Thanks for your help. I just realized I left my backpack with my work in it in a buddy's room, so I need to go track that down. I'll get back to you and see how it goes. I think I got messed up somewhere between multiplying the equations and then taking the derivative. $\endgroup$ – Sean Cogan Aug 26 '12 at 20:03
  • $\begingroup$ Thanks for your help again. I went through the process of finding the derivative step-by-step, and it is quite involved. I'm thinking my professor just wants us to graph the function, or be able to recognize that it decreases as t increases, where t >= 0. I'll check with him on what he expected from us, but in the meantime, you were a great help. $\endgroup$ – Sean Cogan Aug 26 '12 at 21:13

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