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

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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?

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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.

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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.

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