State and prove the Maximum power transfer theorem. Derive the formula of Maximum power delivered.

Statement: A DC network (linear, bilateral) delivers maximum power to a load resistance R_L when R_L equals the Thevenin (internal) resistance R_th of the network as seen from the load terminals.

      R_th
  +--/\/\/--+------ A
  |          |   I
 V_th       R_L
  |          |
  +----------+------ B

Proof: replace the network by its Thevenin equivalent (V_th in series with R_th). Load current \( I = \dfrac{V_{th}}{R_{th}+R_L} \).

Power in load: \[ P_L = I^2R_L = \frac{V_{th}^2R_L}{(R_{th}+R_L)^2} \]

For maximum P_L, \( \dfrac{dP_L}{dR_L} = 0 \):

\[ \frac{dP_L}{dR_L} = V_{th}^2\,\frac{(R_{th}+R_L)^2 - R_L\cdot2(R_{th}+R_L)}{(R_{th}+R_L)^4} = 0 \]

\( \Rightarrow (R_{th}+R_L) - 2R_L = 0 \Rightarrow \) R_L = R_th. (The second derivative is negative, so it is a maximum.)

Maximum power: substituting R_L = R_th: \[ P_{max} = \frac{V_{th}^2R_{th}}{(2R_{th})^2} = \mathbf{\frac{V_{th}^2}{4R_{th}}} \]

(In terms of Norton: \( P_{max} = I_N^2R_N/4 \).) At maximum power transfer, half the power is lost in R_th, so efficiency is only 50%. Used in communication/electronic circuits (e.g. matching a speaker to an amplifier).