(This post is kind of therapeutic for me, I'm trying to understand 
things as I go along; it helps to feel like I've an audience when I'm 
writing something)

Having studied basic Physics, one of the first things I learned about 
was the conservation of energy:
    "Energy can neither be created nor destroyed, but rather only 
converted from one form to another"

The main gem of information I take from this quotation is basically that 
you can't just pull energy out of thin air, it has to come from 
somewhere. "There's no free lunches" as they say.

So let's say you have an AA battery, which is 1.5 V. You say to me that 
you're going to step the voltage up to 5 V. Well my first reaction to 
this is that a 5 V battery is more "powerful", so where the hell is that 
extra energy going to come from?

Anyway here's what went on in my head when I was trying to get a grip on it:
    * Power = Energy per second
    * Power is measured in watts, which is joules per second
    * Power = Voltage x Current = volts x amperes
    * 1 ampere = 1 coulomb per second
    * Therefore Power = volts * coulombs / seconds
    * Therefore Energy = volts * coulombs

Now if I'm right, the energy contained in a battery is equal to the 
amount of charge in it (i.e. coloumbs) multiplied by the voltage (i.e. 
volts).

So let's say I have an AA battery rated as follows:
    Voltage = 1.5 V
    Capacity = 500 mA hr = 1800 coulombs

The energy contained in this battery would then be 2700 joules.

So... if I follow the whole "energy conservation principle", then that 
means that if this battery were to somehow become a 5 V battery, then 
it's capacity would have to drop to 540 coulombs, which is 150 mA hr.

Let's say you give me a 200 ohm resistor and a 1.5 volt battery. You say 
to me "Put 25 mA through that resistor". My first reaction would be "I 
can't, the battery isn't powerful enough".

How much truth is there in saying "The battery isn't powerful enough"? 
Is it possible to take that 1.5 volts, step it up to 5 volts, and make 
it put a steady 25 mA through the resistor?

If so, then I imagine it would have to work something like as follows:
    * The current through the resistor would be 25 mA
    * This means the output current from the voltage regulator is 25 mA.
    * If we pretend that the regulator is 100% efficient, then "power 
in" has to be equal to "power out".
    * So if "power out" = 25 mA x 5 V = 125 mW, then the "power in" has 
to be 125 mW also.
    * "power in" = 125 mW = current x 1.5 V
    * Therefore, current coming from the battery is 83 mA
    * I haven't a clue how efficient step-up regulators are, but I'll 
take a wild guess of 95 %, so that would bring the battery current to 
about 87 mA.

So is that how it all works? If you want to use a single AA battery to 
put 25 mA through a 200 ohm resistor then:
    * 87 mA has to come from the AA battery
    * The regulator steps up to 5 V and gives out 25 mA

That right?

I just have one other unrelated question, but it's to do with batteries. 
Let's say I have two AA batteries, each of which is 500 mA hr. If I put 
them in series to form a 3 V battery, then what's the capacity of this 3 
V battery? Is it 500 mA hr at 3 V? (The energy conservation principle is 
telling me that yes it is 500 mA hr at 3 V, but I just want to be sure)

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