Maximum Power Transfer Theorem (ac)

When applied to ac circuits, the maximum power transfer theorem states that
Maximum power will be delivered to a load when the load impedance is the conjugate of the Thevenin impedance across its terminals.
That is, for Fig. 1,
Fig. 1: Defining the conditions for maximum power transfer to a load.
for maximum power transfer to the load, $$ \bbox[10px,border:1px solid grey]{Z_L = Z_{Th} \text{ and} \theta_L = -\theta_{ThZ}}$$ or, in rectangular form, $$\bbox[10px,border:1px solid grey]{R_L = R_{Th} \text{ and} \pm j X_{load} = \mp j X_{Th}}$$ The conditions just mentioned will make the total impedance of the circuit appear purely resistive, as indicated in Fig. 2:
Fig. 2: Conditions for maximum power transfer to $Z_L$.
$$Z_T = (R \pm jX ) + (R \mp j X)$$ and $$\bbox[10px,border:1px solid grey]{Z_T = 2R}$$ Since the circuit is purely resistive, the power factor of the circuit under maximum power conditions is 1; that is, $$F_p = 1$$ The magnitude of the current I of Fig. 2 is $$I = {E_{Th} \over Z_T} = {E_{Th} \over 2R}$$ The maximum power to the load is $$P_{max} = I^2R = ({E_{Th}^2 \over 2R}) R$$ $$\bbox[10px,border:1px solid grey]{P_{max} = {E_{Th}^2 \over 4R}}$$
Example 1: Find the load impedance in Fig. 3 for maximum power to the load, and find the maximum power.
Fig. 3: Example 1.
Solution: Determine $Z_{Th}$ [Fig. 4(a)]:
$$Z_1 = R - j X_C = 6Ω - j 8Ω = 10 Ω \angle -53.13^\circ$$
$$Z_2 = j X_L = j 8 Ω$$ $$Z_{Th} = {Z_1Z_2 \over Z_1 + Z_2}$$ $$ = {(10 Ω \angle -53.13^\circ)(8 Ω \angle 90^\circ) \over 6Ω - j 8 Ω+ j 8Ω}$$ $$ = {80 Ω \angle 36.87^\circ \over 6 \angle 0^\circ}$$ $$ = 13 Ω \angle 36.87^\circ = 10.66Ω + j 8 Ω$$ and $$ Z_L= 13 Ω \angle -36.87^\circ = 10.66Ω - j 8 Ω$$ To find the maximum power, we must first find $E_{Th}$ [Fig. 4(b)], as follows: $$E_{Th} = {Z_2 E \over Z_1 + Z_2}$$ $$ = {(8Ω \angle 90^\circ)(9 V \angle 0^\circ) \over j 8 Ω + 6 Ω - j 8 Ω}$$ $$ = { 72 Ω \angle 90^\circ \over 6 \angle 0^\circ} = 12V \angle 90^\circ$$ Then
$$ P_{max} = {E_{Th}^2 \over 4R} = {(12 V)^2 \over 4(10.66 Ω)} = 3.38W$$

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