How it works
For a DC circuit with Thevenin source Vth and source resistance Rth: maximum power is delivered to load RL when RL = Rth. Maximum power Pmax = Vth²/(4Rth). For AC circuits with source impedance Zs = Rs + jXs, maximum power is delivered when ZL = Zs* = Rs − jXs (conjugate matching); maximum power Pmax = |Vth|²/(8Rs). The efficiency at maximum power transfer is 50% — half the power is dissipated in the source resistance. This 50% efficiency is acceptable in communication systems where maximising received signal power matters more than efficiency; in power systems, efficiency is prioritised and loads operate far from this condition.
Key points to remember
Maximum power transfer efficiency is exactly 50% — always state this in exam answers. The condition RL = Rth for DC, or ZL = Zs* for AC, must be clearly distinguished: in the AC case, the load reactance must be the negative of the source reactance, not just equal in magnitude. Varying only RL (no reactance adjustment) in an AC circuit gives Pmax when RL = |Zs| — a different condition from conjugate matching. In practice, RF matching networks (L-network, π-network) transform the antenna impedance to Zs* without dissipating power. Norton equivalent: maximum power = IN²·RN/4, delivered when RL = RN — equivalent to the Thevenin condition.
Exam tip
The examiner always asks you to prove that RL = Rth gives maximum power by differentiating P(RL) = IL²·RL = Vth²·RL/(RL+Rth)² and setting dP/dRL = 0 — show every differentiation step and state the efficiency as 50% at the end.