Capacitance: given V waveform, find max. power

[courteous]

Homework Statement

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

Homework Equations

$$p(t)=i(t)u(t)=\left(C\frac{du(t)}{dt}\right)u(t)$$

The Attempt at a Solution

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

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

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?

 

[berkeman]

Homework Statement

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

Homework Equations

$$p(t)=i(t)u(t)=\left(C\frac{du(t)}{dt}\right)u(t)$$

The Attempt at a Solution

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

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

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...

 

[courteous]

Yes, that is easier. Now I got the correct solution of $$P_{max}=1.5\text{ }A\times 50\text{ }V=75\text{ }W$$ ... but, I've got two more questions:

1) As the derivative $$\frac{du(t)}{dt}$$ at $$t=\{2,4,6,8\}$$ is undefined: are the end-points at those $$t$$ "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 $$p(t)=0$$ ... but what would you do with a piecewise function?

 

[berkeman]

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.

 

[DeeNos]

Your approach to finding the maximum power is correct. To differentiate the piecewise equation for v(t), you can break it up into two separate equations for the even and odd cases, as follows:

v(t) = \begin{cases} \frac{50}{2}t-50k & \text{for } 2k < t < 2(k+1) \text{ and } k \text{ even} \\ -\frac{50}{2}t+50(k+1) & \text{for } 2k<t<2(k+1) \text{ and } k \text{ odd} \end{cases}

For the even case, we can differentiate with respect to t to get:

v'(t) = \frac{50}{2} = 25

And for the odd case:

v'(t) = -\frac{50}{2} = -25

So the derivative of v(t) is a constant value of 25 for both cases. Therefore, to find the maximum power, we just need to find the time t when the derivative of power p'(t) is equal to zero. This occurs when the derivative of voltage v'(t) is equal to zero, which happens at t = 0. This means that the maximum power occurs at t = 0, and we can find it by plugging in t = 0 into the equation for power:

P_{max} = C \frac{du(t)}{dt} u(t) = (60 \mu F)(25 V/s)(1) = 1500 \mu W

To fix the formatting issue, you can use the "&&" command instead of "AND" to separate the two conditions for the piecewise equation, like this:

v(t) = \begin{cases} \frac{50}{2}t-50k & \text{for } 2k < t < 2(k+1) && k \text{ even} \\ -\frac{50}{2}t+50(k+1) & \text{for } 2k<t<2(k+1) && k \text{ odd} \end{cases}

This will make the "AND" appear on the same line as the rest of the equation.