As in the figure, a load impedance ZL is connected to terminals a-b of an AC circuit whose Thevenin equivalent voltage is Vth = 10 V RMS and internal impedance Zth = 3 + j4 Ω. If both the resistance and the reactance of ZL can be chosen freely, which of the following is closest to the maximum power consumed by the load?

Answer: C. About 8.3 W

Method 1 (conjugate matching): choosing ZL as the complex conjugate of Zth, 3 − j4 Ω, cancels the reactances, so the circuit impedance is just a 6 Ω resistance. The current is 10/6 ≈ 1.67 A and the load power is (10/6)² × 3 ≈ 8.3 W. In general, Pmax = Vth²/(4Rth) = 100/12 ≈ 8.3 W. Method 2 (maximize the formula): with ZL = RL + jXL, P = Vth² RL/((3 + RL)² + (4 + XL)²); first set XL = −4 Ω to remove the reactive part of the denominator, then choose RL = 3 Ω, which maximizes RL/(3 + RL)², giving the same result.

• A. This is the case ZL = Zth = 3 + j4 Ω. The reactances do not cancel, the current stays at 1 A, and the power is 3 W.
• B. This is the value for a resistive load equal to |Zth| = 5 Ω. It is the maximum when the reactance cannot be chosen, but when it can, conjugate matching gives more.
• C. Correct. Conjugate matching gives Pmax = Vth²/(4Rth) ≈ 8.3 W.