Maximum power transfer theorem in circuit design: Under what condition does a source deliver maximum power to a load (assume resistive source/load)?

Introduction / Context:
The maximum power transfer theorem is a cornerstone in analog design, audio, and communications. It tells us how to choose or ”match” a load to a real (non-ideal) source so that the load receives the greatest possible power. This principle also underpins impedance matching in RF systems and small-signal sensor interfacing.

Given Data / Assumptions:

• A Thevenin source characterized by V_s and internal resistance R_s.
• A purely resistive load R_L (no reactive components).
• Goal: maximize power dissipated in R_L.

Concept / Approach:
Power in the load is P_L = V_L^2 / R_L. With a Thevenin source, V_L = V_s * R_L / (R_s + R_L). Substituting gives P_L = (V_s^2 * R_L) / (R_s + R_L)^2. Taking the derivative of P_L with respect to R_L and setting it to zero yields the optimal condition R_L = R_s. Physically, this splits the source voltage equally and maximizes the product of current and load voltage at the load.

Step-by-Step Solution:

Verification / Alternative check:
Try numerical values: let V_s = 10 V, R_s = 10 Ω. With R_L = 10 Ω, I = 10 / (10 + 10) = 0.5 A, so P_L = I^2 * R_L = 0.25 * 10 = 2.5 W. Choosing R_L much larger or smaller reduces the load power, confirming the theorem.

Why Other Options Are Wrong:

• Greater or less than: Both move away from the matched condition and decrease delivered power.
• Negligible source resistance: This maximizes voltage delivered, not necessarily power for a fixed R_L landscape; the theorem concerns a given R_s.

Common Pitfalls:
Confusing maximum power delivery with maximum efficiency; at R_L = R_s, only 50% efficiency is achieved because the other half is dissipated in R_s.

Final Answer:
When the source resistance equals the load resistance