Answer: Maximum power is transferred when \(R_L=R_{th}\), and then \(P_{max}=\dfrac{V_{th}^2}{4R_{th}}\).
Statement: In a linear, bilateral DC network, maximum power is delivered 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.
Proof: Replace the network by its Thevenin equivalent: a source \(V_{th}\) in series with \(R_{th}\), feeding \(R_L\).
Load current \(I=\dfrac{V_{th}}{R_{th}+R_L}\), so the load power is
\[P=I^2R_L=\frac{V_{th}^2R_L}{(R_{th}+R_L)^2}.\]
For maximum \(P\), \(\dfrac{dP}{dR_L}=0\):
\[\frac{dP}{dR_L}=V_{th}^2\,\frac{(R_{th}+R_L)^2-2R_L(R_{th}+R_L)}{(R_{th}+R_L)^4}=0\]
\(\Rightarrow (R_{th}+R_L)-2R_L=0\Rightarrow R_L=R_{th}\). (The second derivative is negative here, so this is a maximum.)
Maximum power: put \(R_L=R_{th}\):
\[P_{max}=\frac{V_{th}^2R_{th}}{(2R_{th})^2}=\frac{V_{th}^2}{4R_{th}}.\]
At this condition the source loses an equal power in \(R_{th}\), so the efficiency is only 50%. (For a Norton source: \(P_{max}=\dfrac{I_N^2R_N}{4}\).)
