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'Impedance Can Be Complex!' printed from https://nrich.maths.org/
Part 1:
$P = IV$
I = $\frac{E}{Z_1 + Z_2} = \frac{E}{Z_1cos\theta + Z_2cos\phi +
i(Z_1sin\theta + Z_2sin\phi)}$
|I| = $\frac{E}{\sqrt{(Z_1cos\theta + Z_2cos\phi)^2 + (Z_1sin\theta
+ Z_2sin\phi)^2}}$
The real power transferred to $Z_2$: P = I^2 R
P = $ I^2 Z_2cos \phi = \frac{E^2 Z_2cos \phi}{Z_1^2 + Z_2^2 +
2Z_1Z_2(cos \theta cos \phi + sin \theta \phi)}$= $ \frac{E^2
Z_2cos \phi}{Z_1^2 + Z_2^2 + 2Z_1Z_2cos( \theta - \phi )} $
Part 2:
To find the maximum power; we can differentiate the power
expression with respect to $Z_2$ and set the derivative equal to
zero.
$\frac{dP}{dZ_2} = \frac{d}{dZ_2}\frac{E^2 Z_2cos \phi}{Z_1^2 +
Z_2^2 + 2Z_1Z_2cos( \theta - \phi )}$
Let U = $E^2 Z_2cos \phi$
Let V = $ Z_1^2 + Z_2^2 + 2Z_1Z_2cos( \theta - \phi ) $
$\frac{dP}{dZ_2} = \frac{VU' - UV'}{V^2}$
V' = $ 2Z_2 + 2Z_1 cos ( \theta - \phi)$
U' = $E^2 cos \phi$
Substituting thses values into the expression and setting it equal
to zero we find that:
$Z_1 = Z_2$
For maximum power transfer between load and source we must
therefore match the internal impedance of the load with the
impedance of the source