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Maximum power transfer theorem proof

Short answer

Maximum power transfer theorem states that maximum power is delivered to a load when its resistance equals the Thevenin resistance of the source network, i.e., R_L = R_th. By setting derivative of power P = E²R_L/(R_th+R_L)² to zero, we get R_L = R_th.

State Maximum Power Transfer Theorem and also prove "maximum power will be dissipated when R_internal = R_L".

Statement

Maximum power transfer theorem: a load receives maximum power from a source (of fixed EMF E and internal resistance R_int) when the load resistance equals the internal (Thevenin) resistance of the source: R_L = R_int.

        R_int
 o--[ ]----+------o
   (E)     |
          [R_L]
 o---------+------o

Proof

Current: I = E/(R_int + R_L)

Power in the load: P = I² R_L = E² R_L/(R_int + R_L)²

For maximum power, dP/dR_L = 0:

dP/dR_L = E² [(R_int + R_L)² − 2 R_L (R_int + R_L)]/(R_int + R_L)⁴ dP/dR_L = E² (R_int − R_L)/(R_int + R_L)³

Setting this to zero gives R_int − R_L = 0, so R_L = R_int.

Check that it is a maximum: for R_L < R_int, dP/dR_L > 0 (power rises); for R_L > R_int, dP/dR_L < 0 (power falls). So the stationary point is a maximum.

Maximum power

P_max = E² R_int/(2 R_int)² = E²/(4 R_int)

Efficiency

Efficiency = P_load/P_total = R_L/(R_int + R_L) = 50 % at maximum power. So the theorem is used where the power is small and matching matters (communication circuits), not in power systems.

Example: E = 12 V, R_int = 3 Ω gives R_L = 3 Ω, I = 2 A, P_max = 12²/(4 × 3) = 12 W.

✓ Answer checked by Claude Opus 5.5 (high effort)

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