alt.energy.homepower

-> David Williams wrote:
-> > -> Try to operate your alternator into a big variable resistor. With a
-> > -> voltmeter and ammeter attached. When the V drops to .707 of the Open
-> > -> Circuit value, note that load resistance... power in the load will
-> > -> decrease above and below that load
-> >
-> > Shouldn't that be 50% of the open circuit value? The load resistance
-> > has to equal the internal resistance of the generator, which means that
-> > 50% of the EMF will be across each of them.

-> power = V x I

-> but I is proprtional to V

-> so P is proportional to V^2

-> ..707 is the square root of 50%

So what?

Suppose we imagine the generator to be a battery with open-circuit
voltage V, and internal resistance Rb. We connect a load resistance,
Rl, across the battery's terminals. The current that flows will be:

I = V / (Rb + Rl)

The power in the load resistor will be:

P = I^2 . Rl

which is

P = V^2 . Rl / (Rb + Rl)^2

Okay. A bit of calculus coming up...

Assuming V and Rb are constants:

dP / dRl = [V^2 (Rb + Rl)^2 - V^2 . Rl . 2 . (Rb + Rl)] / (Rb + Rl)^4

which will be zero, . the power in Rl is a maximum, when:

V^2 (Rb + Rl)^2 = V^2 . Rl . 2 . (Rb + Rl)

so:

Rb + Rl = 2 . Rl

so Rb = Rl !!!

In other words, the power is a maximum when the load resistance equals
the internal resistance of the battery.

In that situation, the voltage V is equally distributed across Rb and
Rl. So the voltage across Rl is V/2.

It has nothing to do with power being proportional to V^2.

dow