CR · DC analysis

Maximum Power Transfer

To send as much power as possible from a source into a load, what should the load resistance be? Learn to find that sweet spot — bad if too small, bad if too large — through its relation to the internal resistance.

Too small loses, too large loses

Slide the load resistance R_L and watch the power into the load. At zero there is no voltage across it; at infinity no current flows; either way the power is zero. Find the one point between them where the power is maximal, R_L = R_th.

Load resistance R_LR_L = 18.0 Ω

Load power and efficiency

P = 4.50 W  ·  η = 75%

Far from match

You cannot have both voltage and current

Power is the product of the voltage across the load and the current through it. With a small load resistance the current is large but almost no voltage appears; with a large load resistance the voltage is high but almost no current flows. The two trade off against each other, so leaning either way shrinks the product. Power is greatest when they are in balance.

The balance is at R_L = R_th

Differentiate the power P = (V_th/(R_th+R_L))²·R_L with respect to R_L and set it to zero, and the maximum falls exactly at R_L = R_th. Setting the load equal to the source’s internal resistance is called matching. The voltage then splits evenly, putting V_th/2 across the load, and the maximum power is P_max = V_th²/(4R_th).

Maximum power is not maximum efficiency

At the match the load and internal resistances are equal, so the total power splits evenly between them. The load receives exactly as much as the internal resistance throws away as heat, so the efficiency is precisely 50%. Matching is the answer when you want to extract the most power, but where efficiency matters — like power transmission — the load is made deliberately large to cut losses. They are different goals.

ObserveP = I² R_L

Power is voltage times current.

ChooseP_max → R_L = ?

Match the load to the internal resistance.

Fill inP_max = V_th² / ?

At the match the voltage splits in half.

On your ownR_L = R_th → η = ?

Load and internal resistance take equal shares.

Back to the first screen

As you slid the load resistance, the load power started at zero, climbed to a peak, and came back down, and that peak fell exactly at R_L = R_th, where the load equals the internal resistance. There the voltage split evenly and the efficiency was 50%. Once you fold a circuit into one source and one resistor with the Thévenin equivalent, what to attach for the most power becomes visible at a glance like this.

The maximum power transfer theorem says the power delivered to a load on a Thévenin source (V_th, R_th) is greatest when R_L = R_th (matched). The maximum power is then P_max = V_th²/(4R_th), with load and internal resistance splitting the power equally, so the efficiency is 50%. If the load is too small or too large, the voltage–current balance breaks and the power falls.