# why is $P^{max}_{Z_L}=|\hat I_L|^2\times 4$,not $|\hat I_L|^2\times 5$,when $\hat Z_L=\hat Z^*_{th}=4-3jΩ$

What is the max power of $$\\hat Z_L ,P^{max}_{Z_L},\$$ when the $$\\hat Z_{th}=4+3j Ω\$$??

First i know the Imaginary part of impedance will cause Reactive Power,so the real power can't be the same as the apparent Power,so we can know if we want to have a $$\P^{max}_{Z_L}\$$,we have to cancel the Imaginary part of impedance,so obviously we can know the $$\\hat Z_L=\hat Z^*_{th}=4-3jΩ\$$,and $$\\hat I_L=\frac{V_{th}}{\hat Z_L+\hat Z_{th}}=25∠0\$$

The answer show me $$\P^{max}_{Z_L}=|\hat I_L|^2\times 4\$$,however, i think the $$\P^{max}_{Z_L}=|\hat I_L|^2\times \sqrt{4^2+(-3)^2}=|\hat I_L|^2\times 5\$$

So i want to ask why is $$\P^{max}_{Z_L}=|\hat I_L|^2\times 4\$$? or the answer is wrong?

25 A flows through a 4 ohm resistance in series with a 3 ohm reactance. Only the resistance dissipates power, hence the power is $$\25^2\times 4\$$ W.
When $$\\hat Z_L =\hat Z_{th}^{*}\$$ the reactive portions of both impedances cancel each other, so the total impedance seen by the sourve $$\ V_{th}\$$ is only the active (resistive) sum of both impedances, hence $$\P_{Z_L}^{max}\$$ is only determined by the active portion of $$\R_L=4\Omega\$$ and the loop current through it.