Capacitance: given V waveform, find max. power

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courteous
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Homework Statement


A capacitance of [tex]60 \mu F[/tex] has the voltage waveform shown in Fig. 2-36. Find [tex]P_{max}[/tex].
[PLAIN]http://img828.imageshack.us/img828/5255/unled2copy.jpg

Homework Equations


[tex]p(t)=i(t)u(t)=\left(C\frac{du(t)}{dt}\right)u(t)[/tex]

The Attempt at a Solution



When is power at maximum?
Is it the time [tex]t[/tex] when the derivative of power [tex]p'(t)=C\left(\frac{du(t)}{dt}u(t)\right)'[/tex] is equal to zero?

If yes, well ... how do you differentiate this (piecewise) equation for [tex]v(t)[/tex] I came up with looking at Fig. 2-36:
[tex]v(t)=\begin{cases}<br /> \frac{50}{2}t-50k & \text{for $2k < t < 2(k+1) AND k_{even}$} \\<br /> -\frac{50}{2}t+50(k+1) & \text{for $2k<t<2(k+1) && k_{odd}$} <br /> \end{cases}[/tex]

Anyway, I must be over-complicating ... help me solve this "problem".* Help me with TEX: in the conditions for piecewise v(t) it should read "2k < t < 2(k+1) AND k_{even}". What am I doing wrong?
 
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courteous said:

Homework Statement


A capacitance of [tex]60 \mu F[/tex] has the voltage waveform shown in Fig. 2-36. Find [tex]P_{max}[/tex].
[PLAIN]http://img828.imageshack.us/img828/5255/unled2copy.jpg

Homework Equations


[tex]p(t)=i(t)u(t)=\left(C\frac{du(t)}{dt}\right)u(t)[/tex]


The Attempt at a Solution



When is power at maximum?
Is it the time [tex]t[/tex] when the derivative of power [tex]p'(t)=C\left(\frac{du(t)}{dt}u(t)\right)'[/tex] is equal to zero?

If yes, well ... how do you differentiate this (piecewise) equation for [tex]v(t)[/tex] I came up with looking at Fig. 2-36:
[tex]v(t)=\begin{cases}<br /> \frac{50}{2}t-50k & \text{for $2k < t < 2(k+1) AND k_{even}$} \\<br /> -\frac{50}{2}t+50(k+1) & \text{for $2k<t<2(k+1) && k_{odd}$} <br /> \end{cases}[/tex]

Anyway, I must be over-complicating ... help me solve this "problem".


* Help me with TEX: in the conditions for piecewise v(t) it should read "2k < t < 2(k+1) AND k_{even}". What am I doing wrong?

A quicker way to solve this would be to just graph the i(t) waveform, based on the v(t) waveform. You should be able to write the equation for the power based on the combined graphs...
 
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Yes, that is easier. Now I got the correct solution of [tex]P_{max}=1.5\text{ }A\times 50\text{ }V=75\text{ }W[/tex] ... but, I've got two more questions:

1) As the derivative [tex]\frac{du(t)}{dt}[/tex] at [tex]t=\{2,4,6,8\}[/tex] is undefined: are the end-points at those [tex]t[/tex] "empty" or "filled" (see picture)?
[PLAIN]http://img863.imageshack.us/img863/4064/dsc00992v.jpg

2) What would be a more general approach when the problem was more complex? I guess you would you help yourself with a computer, and then find the point where the derivative of [tex]p(t)=0[/tex] ... but what would you do with a piecewise function?
 
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Good work. On (1), you could just leave them empty, to signify undefined. It's not going to affect the power calculation. And for (2), I'd just use the equations for v(t) and i(t), and sum the power up in a piecewise fashion, leaving out the points that are undefined.