Maximum Power Transfer Theorem

Basic Electrical & Electronics Engineering — Network Theorems, NEC licence examination syllabus (Nepal Engineering Council).

Maximum Power Transfer Theorem

A load draws the most power when it matches the source's internal resistance — and at that point exactly half the energy is wasted.

The idea: why a car battery would destroy your phone

A car battery can deliver hundreds of amps. Your phone charger cannot. Yet the charger is right for a phone and the car battery would ruin it.

Every source has resistance inside it, and how well a source and load work together depends on how those two resistances compare.

The theorem: a source delivers maximum power to a load when R_L = R_th. Not larger, not smaller — equal.

Why equal, and not "as small as possible"

A smaller load draws more current, so surely more power? It draws more current — but the voltage across it collapses, because most of the supply is being dropped inside the source. Go the other way with a very large load and the voltage is healthy but almost no current flows.

Source: 18 V with 3 Ω inside. R_L = 1 Ω: I = 4.5 A → P = 4.5²×1 = 20.25 W R_L = 3 Ω: I = 3.0 A → P = 3.0²×3 = 27.00 W ← most R_L = 9 Ω: I = 1.5 A → P = 1.5²×9 = 20.25 W Too small: voltage collapses. Too big: current dies. The middle is the best available.

💡 Note that 1 Ω and 9 Ω give exactly the same power, and that the curve is flat near the top — so being roughly matched is nearly as good as being perfectly matched. Real designs rely on that.

Worked example

A network reduced to V_th = 18 V and R_th = 3 Ω. Find the load for maximum power, that power, and the efficiency there.

Step 1 — the load

The condition is simply R_L = R_th, so the answer is 3 Ω. No working needed.

Step 2 — the current at that load

I = V_th/(R_th + R_L) = 18/(3+3) = 3 A

Step 3 — the power delivered

P = I² × R_L = 3² × 3 = 27 W

Step 4 — the efficiency, which almost everyone gets wrong

The same 3 A flows through R_th as through R_L, and the two are equal in size — so they burn the same power.

delivered to the load = 3² × 3 = 27 W burned inside R_th = 3² × 3 = 27 W total from the source = 54 W efficiency = 27/54 = 50%

At maximum power transfer the efficiency is exactly 50%. Half the energy is lost inside the source. That is not a mistake in the sum — it is what maximum power transfer means.

Maximum power is not maximum efficiency

Compare the 9 Ω load: it receives less power but wastes only 6.75 W inside the source — an efficiency of 75%.

⚡ Two different goals

Maximum POWERWhen the signal is tiny and precious — a radio antenna, an audio input. Wasting half a microwatt is irrelevant; losing the signal is not.

Maximum EFFICIENCYWhen energy costs money — a power station, a motor, a battery. Here you want R_L far larger than R_th, so little is wasted internally.

💡 This is why the national grid deliberately does not match. If it did, half of every power station's output would be burned in the wires. Not matching is the correct engineering decision there — a question asking for "the best load" is incomplete until it says best for what.

Exam notes

📝 The three question types

Given V_th and R_thState R_L = R_th, compute I, then P = I²R_L. Two lines.

Given a raw circuitDo Thevenin first, then apply the condition. Most of the marks are in the reduction.

"Comment on efficiency"Say 50%, and say why — R_L and R_th are equal so they dissipate equally. This part is frequently dropped.

💡 A useful shortcut for the maximum power itself, if you have V_th and R_th: P_max = V_th²/(4·R_th). Here 18²/(4×3) = 324/12 = 27 W, matching the long way.

Syllabus points

Condition & max power (numerical)

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