Nigel Goodwin,
Well, I did the simple maths and posted them. I calculated power before and after the source impedance change. I did not calculate or describe any voltage changes. Did you even look at my derivation?
If the concept is hard to grasp, think of it this way. Reducing the impedance of the source will increase the series current. That can only increase the power dissipated in the load because its impedance has not changed while its current has increased.
Ratch
Again - that isn't the point - maximum
power transfer occurs when source and load impedances equal each other. If you vary the source impedance, you need to vary the load impedance to maintain maximum
power transfer.
If the load is higher than the source, you don't get maximum power, if the load if lower than the source you don't get maximum power either.
Here's a simple example:
Now 10V source, 1 ohm impedance, 1 ohm load (matched).
Current = 5A, power in load 25W
Now 10V source, 1 ohm impedance, 2 ohm load (not matched).
Current = 3.333A, power in load 22.217W
Now 10V source, 2 ohm impedance, 1 ohm load (not matched).
Current = 3.333A, power in load 11.109W
Notice that the matched one is highest power.
If you plot all the points, you will get a curve on the graph, which shows a single point of maximum power transfer, which is when source and load are equal.
I'm fully aware that this isn't usually what you want, a low source feeding a high load is much more normal, but it's not the maximum power point.