- #1

fluidistic

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Consider a voltage source as a fixed voltage source in series with an internal resistance ##R_\text{source}##. This voltage source is connected in series with a load. If the load is purely resistive, then the total power generated by the source is worth ##P=I^2R_\text{total}## where ##R_\text{total}=R_\text{source} + R_\text{load}##. In impedance matching ##R_\text{load}=R_\text{source}## and it turns out that the total power generated is split equally into the power source as Joule heat and into the load as Joule heat too. The Joule heat in the load is considered a useful work here. It is worth ##I^2R_\text{source}## (remember, I assumed impedance matching.)

So far so good. I have been told that if the load was not purely resistive, for example purely reactive, then the useful power would still be the one calculated above. But this is not what I get. Let's assume a purely reactive load, so ##Z_\text{load} =i\Im{Z_\text{load}}##. In impedance matching ##|Z_\text{load}|=R_\text{load}## and so ##\Im{Z_\text{load}} = R_\text{load}##. The total power generated is ##P=Z_\text{total}I^2=(R_\text{load}+i\Im{Z_\text{load}})I^2=(R_\text{load}+iR_\text{load})I^2##. The real part of that expression is what gets dissipated as Joule heat and is exactly the same as before, and is also lost through the voltage source. The modulus or absolute value of that expression is worth ##\sqrt{2}R_\text{load}I^2##. If this is equal to the generated power then it means only ##(\sqrt{2}-1)R_\text{load}I^2## is left for the load. If that square root of 2 was a 2, we would get the same as in the first case, but we don't. I can't see anything wrong with my reasoning. I do not see how to reach the same value as before.

I would appreciate comments and a pointer on where I go wrong.