In this diagram, AC power is being transferred from the source, with phasor magnitude of voltage (positive peak voltage) and fixed source impedance (S for source), to a load with impedance (L for load), resulting in a (positive) magnitude of the current phasor . This magnitude results from dividing the magnitude of the source voltage by the magnitude of the total circuit impedance: The average power dissipated in the load is the square of the current multiplied by the resistive portion (the real part) of the load impedance : where and denote the resistances, that is the real parts, and and denote the reactances, that is the imaginary parts, of respectively the source and load impedances and . To determine, for a given source voltage and impedance the value of the load impedance for which this expression for the power yields a maximum, one first finds, for each fixed positive value of , the value of the reactive term for which the denominator is a minimum. Since reactances can be negative, this is achieved by adapting the load reactance to This reduces the above equation to: and it remains to find the value of which maximizes this expression. This problem has the same form as in the purely resistive case, and the maximizing condition therefore is The two maximizing conditions describe the complex conjugate of the source impedance, denoted by and thus can be concisely combined to: Notes References H.W. Jackson (1959) Introduction to Electronic Circuits, Prentice-Hall. External links Conjugate matching versus reflectionless matching (PDF) taken from Electromagnetic Waves and Antennas The Spark Transmitter. 2. Maximising Power, part 1. Circuit theorems Electrical engineering