The theorem applies to maximum power, and not maximum efficiency. If the resistance of the load is made larger than the resistance of the source, then efficiency is higher, since most of the power is generated in the load, but the overall power is lower since the total circuit resistance goes up.
If the internal impedance is made larger than the load then most of the power ends up being dissipated in the source, and although the total power dissipated is higher, due to a lower circuit resistance, it turns out that the amount dissipated in the load is reduced.
The condition of maximum power transfer does not result in maximum efficiency. If we define the efficiency $eta$ as the ratio of power dissipated by the load to power developed by the source, then it is straightforward to calculate from the above circuit diagram that
Consider three particular cases:
The efficiency is only 50% when maximum power transfer is achieved, but approaches 100% as the load resistance approaches infinity (or the source resistance approaches zero), though the total power level tends towards zero. When the load resistance is zero, all the power is consumed inside the source (the power dissipated in a short circuit is zero) so the efficiency is zero.
In the diagram opposite, power is being transferred from the source, with voltage $V$ and fixed source resistance $R\_mathrm\{S\}$, to a load with resistance $R\_mathrm\{L\}$, resulting in a current $I$. By Ohm's law, $I$ is simply the source voltage divided by the total circuit resistance:
The power $P\_mathrm\{L\}$ dissipated in the load is the square of the current multiplied by the resistance:
We could calculate the value of $R\_mathrm\{L\}$ for which this expression is a maximum, but it is easier to calculate the value of $R\_mathrm\{L\}$ for which the denominator
is a minimum. The result will be the same in either case. Differentiating with respect to $R\_mathrm\{L\}$:
For a maximum or minimum, the first derivative is zero, so
or
In practical resistive circuits, $R\_mathrm\{S\}$ and $R\_mathrm\{L\}$ are both positive. To find out whether this solution is a minimum or a maximum, we must differentiate again:
This is positive for positive values of $R\_mathrm\{S\}$ and $R\_mathrm\{L\}$, showing that the denominator is a minimum, and the power is therefore a maximum, when
If the source is totally inductive (capacitive), then a totally capacitive (inductive) load, in the absence of resistive losses, would receive 100% of the energy from the source but send it back after a quarter cycle. The resultant circuit is nothing other than a resonant LC circuit in which the energy continues to oscillate to and fro. This is called reactive power. Power factor correction (where an inductive reactance is used to "balance out" a capacitive one), is essentially the same idea as complex conjugate impedance matching although it is done for entirely different reasons.
For a fixed reactive source, the maximum power theorem maximizes the real power (P) delivered to the load by complex conjugate matching the load to the source.
For a fixed reactive load, power factor correction minimizes the apparent power (S) (and unnecessary current) conducted by the transmission lines, while maintaining the same amount of real power transfer. This is done by adding a reactance to the load to balance out the load's own reactance, changing the reactive load impedance into a resistive load impedance.
In this diagram, AC power is being transferred from the source, with phasor magnitude voltage $V\_mathrm\{S\}$ (peak voltage) and fixed source impedance $Z\_mathrm\{S\}$, to a load with impedance $Z\_mathrm\{L\}$, resulting in a phasor magnitude current $I$. $I$ is simply the source voltage divided by the total circuit impedance:
The average power $P\_mathrm\{L\}$ dissipated in the load is the square of the current multiplied by the resistive portion (the real part) $R\_mathrm\{L\}$ of the load impedance:
$P\_mathrm\{L\},!$  = I_mathrm{rms}^2 R_mathrm{L} = {1 over 2} >I^2 R_mathrm{L} = {1 over 2} {{left( 
= {1 over 2}{^2 R_mathrm{L}}over{(R_mathrm{S} + R_mathrm{L})^2 + (X_mathrm{S} + X_mathrm{L})^2}},> In order to determine the values of $R\_mathrm\{L\}$ and $X\_mathrm\{L\}$ (since $V\_mathrm\{S\}$, $R\_mathrm\{S\}$, and $X\_mathrm\{S\}$ are fixed) for which this expression is a maximum, we first find, for each fixed positive value of $R\_mathrm\{L\}$, the value of the reactive term $X\_mathrm\{L\}$ for which the denominator
is a minimum. Since reactances can be negative, this denominator is easily minimized by making
The power equation is now reduced to:
and it remains to find the value of $R\_mathrm\{L\}$ which maximizes this expression. However, this maximization problem has exactly the same form as in the purely resistive case, and the maximizing condition $R\_mathrm\{L\}\; =\; R\_mathrm\{S\}$ can be found in the same way. The combination of conditions
can be concisely written with a complex conjugate (the *) as:
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