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Maximum Power Delivered to a Load

A source has an open-circuit voltage of $24\ \text{V}$ and an internal resistance of $R_s = 8\ \Omega$ .

Find the maximum power that can be delivered to a resistive load.

Given

Vth	24 V	Open-circuit voltage
Rs	8 ohm	Source internal resistance

Hint 1

What load resistance extracts the most power from a source with internal resistance?

Hint 2

The optimum is R_L = R_s. Both extremes — short and open — deliver zero.

Hint 3

At the match, P_max = V²/(4R_s).

Worked solution — try the problem first

Matching condition. Maximum power transfer occurs when the load matches the source resistance:

$R_L = R_s = 8\ \Omega$

Current at the match. The total circuit resistance is $8 + 8 = 16\ \Omega$ :

$I = \frac{24}{16} = 1.5\ \text{A}$

Power in the load:

$P_L = I^2R_L = (1.5)^2(8) = 2.25 \times 8 = 18\ \text{W}$

$\boxed{P_{max} = 18\ \text{W}}$

Closed form. Substituting $R_L = R_s$ in general gives a result worth memorising:

$P_{max} = \frac{V_{th}^2}{4R_s} = \frac{576}{32} = 18\ \text{W} \ \checkmark$

Efficiency is only 50%. At the match, the source dissipates exactly as much as the load — the other 18 W is lost internally. This is why matching is used in signal and communications circuits, where extracting maximum power matters, and deliberately avoided in power distribution, where source resistance is kept as low as possible for efficiency.

Concepts:Maximum power transferThévenin equivalentImpedance matching

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