How it works
For a DC circuit: maximum power is transferred to load R_L when R_L = R_th (the Thevenin resistance of the source network). Maximum power P_max = V_th²/(4R_th). For an AC circuit with source impedance Z_s = R_s + jX_s: maximum power transfer requires the load impedance Z_L = Z_s* = R_s − jX_s (conjugate of source impedance). This means R_L = R_s and X_L = −X_s. If only R_L is adjustable (X_L fixed at X_L0): R_L_opt = √(R_s² + (X_s + X_L0)²). The condition Z_L = Z_s* cancels the reactive parts and reduces the circuit to two equal resistances.
Key points to remember
Under maximum power transfer, efficiency η = P_load/P_total = 50% — exactly half the total generated power is wasted in the source resistance. This 50% efficiency is acceptable in communication systems where signal power is critical, but unacceptable in power systems where transmission efficiency is the priority. In power systems, the load impedance is much larger than source impedance to maximise efficiency, not matched for maximum power. P_max = V_th²/(4R_th); for V_th = 10V and R_th = 5Ω, P_max = 100/20 = 5W. The power delivered to a mismatched load R_L ≠ R_th is P = V_th²·R_L/(R_L + R_th)².
Exam tip
The examiner always asks you to find maximum power and efficiency for a given circuit — reduce the network to Thevenin form first, state the matching condition R_L = R_th explicitly, then calculate P_max and confirm that efficiency is 50%.