Why You Don't Teach Maximum Power With Calculus
Alright, so like, I was flipping through this calculus book (Calculus Seventh Edition, Larson, Hostetler, Edwards, if you must know; thick as a dictionary), and I came across this question:
The electric power P in watts in a direct-current circuit with two resistors R1 and R2 connected in series is
P = vR1R2 / (R1 + R2)^2
where v is the voltage. If v and R1 are held constant, what resistance R2 produces maximum power?
Deciding that I wanted to upload something other than horrible, horrible poetry or journal entries for a change, I figured that this would be as good a topic as any.
Now, I must admit that I'm not entirely sure where that formula comes from. Power equals VI, I^2R, and/or V^2/R. I wasn't able to come up with any algebraic manipulation to change the power formula into that, and I don't know why there is any multiplication of the resistances at all, since series resistances just add. But that's an exercise for another day, and is besides the point anyway. I'll assume that the equation works out somehow. I will also assume that the question means to ask what resistance R2 will provide it with maximum power, as opposed to the entire circuit. Reducing R2 to a short results in a lower total resistance, and thus a higher total current for the same source voltage. Given the relationship P = VI, the increase in I for a constant V increases the power in the circuit. So I could say that an R2 of 0 gives the ideal maximum power in the circuit without doing the slightest bit of calculation, which just doesn't seem right considering the purpose of the exercises in the book.
So, I assume that power does, in fact, equal vR1R2 / (R1 + R2)^2, and that maximum power is to be achieved for R2.
The maximum power theorum is nothing spectacular. It simply says that the greatest power will be dissipated in a component when its resistance equals that of a second series resistance, which can be a Thevenin equivalent resistance of a more complex circuit. Again, this makes sense without doing any calculation, for as resistance of a component increases, the voltage drop across the component also increases, and the total current decreases. If the resistance is decreased, then more current will flow, but there will be a smaller voltage drop across the component. Either way, not as much power gets to it. It's a trade-off, except when the resistances are equal.
In other words, R1 = R2 for maximum power transfer. Plain and simple.
This problem, though, would like to make everything difficult. Or, more difficult than it has to be. Which is fine, for the sake of practice. The purpose of the section is the application of differentiation, such as finding the maximum and minimum values of a function. Like maximum power dissipation.
Therefore, we must differentiate the power function P with respect to R2, since v and R1 are given as constants.
1. d/dx cu = c * u'
2. d/dx x = 1
3. d/dx q/p = ( pq' - qp' ) / (q)^2
4. d/dx u^n = nu^(n-1) * u'
5. d/dx ( u +- v) = d/dx u +- d/dx v
Thus,
P = vR1R2 / (R1 + R2)^2
Applying the quotient rule (3), we get:
P' = [ (R1 + R2)^2 * d/dR2(vR1R2) - (vR1R2) * d/dR2 (R1 + R2)^2 ] / [ (R1 + R2)^2 ]^2
Multiplying out the (R1 + R2)^2, we get R1^2 + 2R1R2 + R2^2.
We can apply rule 1 to the (vR1R2) to get simply (vR1), since R2 is our variable, and v and R1 are constants. The subtraction of vR1R2 is left alone. The derivative of (R1 + R2)^2 involves the chain rule (4). It becomes:
2 (R1 + R2) * d/dx (R1 + R2)
Using rule 5, d/dx (R1 + R2) becomes (0 + 1) = 1.
In the denominator, we still have [(R1 + R2)^2]^2. A power to a power is equal to the product of the powers. 2 * 2 = 4, and so the denominator becomes (R1 + R2)^4.
P' = [ (R1^2 + 2R1R2 + R2^2)(vR1) - (vR1R2)(2)(R1 + R2)(1) ] / [ (R1 + R2)^4 ]
Cleaning up the numerator a bit,
P' = [ (R1^2 + 2R1R2 + R2^2)(vR1) - (2vR1R2)(R1 + R2) ] / [ (R1 + R2)^4 ]
Next we distribute the vR1 to the R1^2 + 2R1R2 + R2^2, and the 2vR1R2 to the (R1 + R2).
P' = [ (vR1^3 + 2vR1^2R2 + vR1R2^2) - (2vR1^2R2 + 2vR1R2^2) ] / [ (R1 + R2)^4 ]
And then distribute the negative sign to the latter term.
P' = [ vR1^3 + 2vR1^2R2 + vR1R2^2 - 2vR1^2R2 - 2vR1R2^2 ] / [ (R1 + R2)^4 ]
To make things a little easier, gather all the like terms together.
P' = [ vR1^3 + 2vR1^2R2 - 2vR1^2R2 + vR1R2^2 - 2vR1R2^2 ] / [ (R1 + R2)^4 ]
Now just combine like terms. The vR1^2R2 terms cancel out entirely.
P' = [ vR1^3 - vR1R2^2 ] / [ (R1 + R2)^4 ]
There is a common factor of vR1 in the numerator:
P' = [ vR1 (R1^2 - R2^2) ] / [ (R1 + R2)^4 ]
And let's just leave it at that. That's enough to work with for the next part of the problem. We have the derivative function, and we set it to zero to find the horizontal tangents, hopefully at the maximums of the function. A fraction is zero if the numerator is zero, and so we can neglect the denominator entirely.
vR1 (R1^2 - R2^2) = 0
vR1 cannot be zero. Or divide both sides by vR1. Thus, R1^2 - R2^2 = 0
Adding R2^2 to both sides,
R1^2 = R2^2
Taking the square root of both sides,
R1 = R2
And done without NUMBERS even. How clever.
This is what the maximum power transfer theorum says in the first place, without all that calculus and algebra and whatnot. Not that it's the most difficult thing, but still. Enough things are a pain in the proverbial fanny without added basic circuit analysis to the mix. I don't know if I'd call this a proof of the theorum, or anything more than an example of it, because I don't know where that original equation came from, and it can be demonstrated other ways besides. Maybe I missed some obscure power formula somewhere in the bowels of the textbook, or maybe there's some very clever algebra involved that's just beyond me right now. If v is defined as R1R2 / (R1 + R2) somehow then it works, provided that it's substituted for only one of the Vs, but I don't know how V could be defined only in terms of R.
Feel free to prove V^2/R = vR1R2 / (R1 + R2)^2, if it's possible. Or don't. It...really doesn't matter. I go now.
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