Maximum Power Theorem
How do you get the maximum power output across a component? As you increase it's resistance, it gets a larger share of the voltage, but the power across it also reduces from v^2/R. Now you might say, aha! The voltage increase is squared, so that will have a larger impact. But you have to keep in mind that the voltage does not increase 1 for 1 with an increase in resistance. This is an interesting puzzle, and it's not to my knowledge a part of the syllabus.
The overall question is this:
In a circuit composed of a cell wth internal resistance r and emf E (can't write epsilons on this) in series with a variable resistor, what resistance R should the resisistor be set to to maximise the power output of the variable resistor?
Good luck! I will post a worked solution tomorrow.
It's been a couple of days, so the first step is to find the power output of the resistor. Now, power can be expressed as P=IV, but in this case it's quicker to substitute in I=V/R to get power as P=V^2/R. You ought to be really quite familiar with power in that format. A quick units check gives 1V=1J/1C and 1Ohm=1V/1A=1J*1s/(1C^2). So that does give an expression in watts, which is always good when we're coming up with a formula for power!
Unfortunately, whilst we do know R (it's R, we just defined it as that!), we do not know V, which is the voltage across the resistor. Notice that it will not be the same as the emf, because some voltage is across the internal resistance, so the voltage across the resistor will in fact be less than the emf. Fortunately, we know how to deal with this: we have a potential divider, so we can apply the formula:
V_out=[V_in*R_1]/[R_1+R_2]
This gives us our value of V for out power formula, so we're making progress!
V=ER/(R+r)
To find the power, we square that and divide by R, which gives us:
P=E^2/[R+2r+(r^2/R)]
That last bit is a little messy. Note that we currently have an expression for power that is in terms of E,r and of course R. This means that they are all important to the expected power output. (Sometimes when doing something like this one of your variables will disappear and it will turn out that it is not necessary to state the value of it. A good example is the proof that all objects fall at the same speed in a vacuum.) The great thing about this formula is that since we know E and r are going to be constants, we only need to vary R. That means the equation is solvable!
To maximise a function, you need to do some calculus. Shock, horror. It is not too hard though. When the above power function, P(R) is maximised, the denominator will be minimised. And at a minimum, the gradient is zero. So we want to find a point in the function
f(R)=R+2r+(r^2)(R^-1)
where the gradient is zero. We can differentiate:
f'(R)=1+0+(-r^2)(R^-2)
Note that we are treating little r as if it is a constant (like k or 2 or pi)
Now, the above must equal zero, so:
1-r^2/R^2=0
r^2/R^2=1
R^2=r^2
R= plus-minus r
R=r (because resistances cannot be negative)
*Whew* That WAS a lot of hard work, in fact! But don't worry, you don't actually need to know this. It is however interesting to think about- if you make a resistor too small, then most of the voltage will be lost in the internal resistance (and give you a very low power and efficiency). If you make it too large, you get almost all of the voltage and it becomes very efficient. But because P=v^2/R... you have made the power very low. Imagine connecting a resistor of infinite resistance to a battery- you don't have any power output at all! (If you did, then the air would heat up when you left a battery near it... which is BAD.)
So if you know the internal resistance, you can find the load resistance that maximises power output. But a warning: the reverse is not true. When you know your resistor value, the internal resistance should NOT be the same as it, you always want an internal resistance of zero. I can't find any references to this, but you ought to be able to prove this by differentiating the same function, but wrt to little r. This shows that power decreases with little r for all positive values (the actual minimum I *think* ought to be at -R, but I don't trust my calculus that much and cannot find a source to back me up) and so is optimised when r=0.
Okay, so that's the maximum power theorem! It takes a little bit of work to prove, but I hope you agree that it's a pretty slick little thing. You wouldn't ever be asked for it at A level though because the maths is beyond what is expected of physics students (not maths students though! If you take both then you really ought to have a go at proving this to yourself.)
See you next time!
