Maximum power transfer and reciprocity theorems

Why it matters

Some sensors and signal sources are so weak that every microwatt counts: antennas, piezoelectric pickups, thermocouples feeding long lines. For them, the load must be matched to the source to extract the most power. Power systems do the opposite and run far from matching for efficiency. Knowing when to match and when not to is basic engineering judgment. Reciprocity tells you when a measurement taken "from the other end" gives the same answer, which matters for passive sensor networks and two-port testing.

Key ideas

Maximum power transfer (DC). Replace the network driving the load by its Thevenin equivalent (Vth, Rth). The load power is P = Vth²·RL/(Rth + RL)². Setting dP/dRL = 0 gives RL = Rth, and then Pmax = Vth²/(4·Rth).

• The theorem fixes the source network and varies the load.
• If instead the load is fixed and you can choose the source resistance, make Rs as small as possible (ideally zero), not equal to RL.

Efficiency at matching. With RL = Rth, half of the power taken from the Thevenin source is lost in Rth, so η = 50% with respect to the Thevenin model. The efficiency with respect to the real sources inside the network can be lower or higher, because Rth is a model, not a physical resistor. Matching suits communication and instrumentation front ends. Power supplies and the grid operate with RL ≫ Rs for high efficiency and good regulation.

Maximum power transfer (AC). With Thevenin impedance Zth = Rth + jXth and RMS voltage Vth:

• If the load ZL can be any impedance, choose the complex conjugate ZL = Rth − jXth. The reactances cancel, and Pmax = |Vth|²/(4·Rth).
• If the load is a pure resistance, choose RL = |Zth|, which gives less power than conjugate matching.
• If only RL can vary while XL is fixed, choose RL = |Rth + j(Xth + XL)|.

Reciprocity theorem. In a linear, bilateral (reciprocal) network with a single independent source, the ratio of excitation to response is unchanged when the positions of excitation and response are interchanged.

• Voltage-source form: a voltage source V in branch 1 produces current I in branch 2 (shorted). Moving V to branch 2 produces the same current I in branch 1.
• Current-source form: a current source I between nodes 1 produces voltage V across nodes 2. Moving it to nodes 2 produces the same V across nodes 1.

Networks of R, L, C, coupled coils and ideal transformers are reciprocal. Dependent sources in general destroy reciprocity, because a controlled source is unilateral. They are still linear; they are simply not reciprocal. For a two-port, reciprocity means z12 = z21, y12 = y21, h12 = −h21 and AD − BC = 1.

Formulas

• Load power: P = Vth²·RL/(Rth + RL)² (W), with V in volts and R in ohms.
• Matching condition (DC): RL = Rth. Maximum power: Pmax = Vth²/(4·Rth).
• Efficiency with respect to the Thevenin source: η = RL/(Rth + RL), which is 50% at matching.
• AC conjugate match: ZL = Zth*. Maximum power: Pmax = |Vth|²/(4·Rth), with Vth as an RMS phasor.
• Resistive load in AC: RL = |Zth| = √(Rth² + Xth²).
• Reciprocity: I2/V1 = I1/V2 (transfer admittance), V2/I1 = V1/I2 (transfer impedance).

Worked examples

Example 1 (standard): matching a resistive network. Given: a 24 V source in series with 4 Ω feeds node a. Node a is connected to ground through 12 Ω. The load RL connects between a and ground. Find RL for maximum power, Pmax, and the efficiency with respect to the 24 V source.

1. Thevenin voltage: Vth = 24·12/(4 + 12) = 18 V.
2. Thevenin resistance: Rth = 4 ∥ 12 = 3 Ω, so set RL = 3 Ω.
3. Maximum power: Pmax = 18²/(4·3) = 27 W.
4. Load voltage: VL = 18/2 = 9 V. Load current: IL = 9/3 = 3 A. The current in the 12 Ω resistor is 9/12 = 0.75 A.
5. Source current: 3 + 0.75 = 3.75 A. Source power: 24·3.75 = 90 W.
6. Efficiency: η = 27/90 = 30%. This is below 50%, because the 12 Ω shunt also wastes power.

Answer: RL = 3 Ω, Pmax = 27 W, η = 30%.

Example 2 (GATE level): AC matching. Given: Vth = 20∠0° V (RMS), Zth = 4 + j3 Ω. Find the maximum load power (a) for any ZL, and (b) for a purely resistive load.

1. (a) Conjugate match: ZL = 4 − j3 Ω, so Pmax = 20²/(4·4) = 25 W.
2. (b) Resistive load: RL = |4 + j3| = 5 Ω.
3. Load current: |I| = 20/|(4 + 5) + j3| = 20/9.487 = 2.108 A.
4. Load power: P = 2.108²·5 = 22.2 W.

Answer: (a) 25 W; (b) 22.2 W with RL = 5 Ω.

Example 3 (reciprocity check). Given: a T network with series arm 2 Ω (port 1 side), shunt arm 4 Ω and series arm 6 Ω (port 2 side). Apply 12 V at port 1 with port 2 shorted, then swap the source and the short.

1. Source at port 1: the input resistance is 2 + (4 ∥ 6) = 4.4 Ω, so the input current is 12/4.4 = 2.727 A.
2. Current divider into the short: I2 = 2.727 × 4/(4 + 6) = 1.091 A.
3. Source at port 2: the input resistance is 6 + (4 ∥ 2) = 7.333 Ω, so the input current is 1.636 A.
4. Current divider into the short: I1 = 1.636 × 4/(4 + 2) = 1.091 A.

Answer: both are 1.091 A, as reciprocity predicts.

Common mistakes

• Choosing RL equal to a physical source resistor instead of the Thevenin resistance of the whole network seen by the load.
• Making Rs = RL when the load is fixed. The best source resistance is then the smallest possible.
• Claiming efficiency is always 50% at maximum power. That holds only relative to the Thevenin model.
• In AC, setting ZL = Zth instead of its conjugate, or using peak instead of RMS values in |V|²/(4R).
• Applying reciprocity to circuits with dependent sources, or with more than one independent source.
• Swapping source and ammeter positions but changing the source type (V to I). The form of the theorem must be kept.

For GATE IN

Typical questions: find RL for maximum power and Pmax in a network that may contain a dependent source (find Rth by test source first); conjugate or resistive matching in AC; a two-measurement reciprocity problem where a response in one configuration must be inferred from another. Short MCQs test the 50% efficiency condition and which networks are reciprocal. Practise the Thevenin-then-match workflow until it is automatic.

Quick check

1. Vth = 10 V and Rth = 5 Ω. What is Pmax?
2. Zth = 6 + j8 Ω. What is the best purely resistive load?
3. Does reciprocity hold for a network containing a CCCS?
4. The load resistor is fixed at 8 Ω and the source resistance is adjustable. What source resistance gives maximum load power?

Answers: 1. 5 W. 2. 10 Ω. 3. Generally no. 4. Zero (the smallest possible).