Maximum Power Transfer

Q: Why is the efficiency exactly 50 percent at the match?

With $R_L = R_S$, the load and the source have equal resistance. They carry the same current, so each drops the same voltage and takes the same power. Half the power ends up in each, and the efficiency is the load's share.

In one sentence: A load receives the most power when its resistance matches the source’s resistance, but then half the power is lost inside the source.

$R_L = R_S \qquad P_{\max} = \frac{V_S^2}{4 R_S}$

Read it out loud: “R sub L equals R sub S. Also, P sub max equals V sub S squared, divided by 4 times R sub S. Read R sub L as R L, the load resistance, the resistance of the device that receives the power. R sub S is the source resistance, the resistance inside the source. P sub max is the largest power the load can receive, in watts. V sub S squared means V sub S times V sub S, where V sub S is the source voltage.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Source and load: A source makes electrical energy, such as a battery, a solar cell or a power pack. A load uses that energy, such as a lamp, a motor or a speaker. In a circuit drawing, a real source is an ideal voltage with a resistance inside it, and the load is the resistance that receives the power. Power flows from the source into the load, and some is also lost inside the source.
Power, the watt (W) and $P = I^2 R$: Power is the rate at which energy is delivered or used. Its unit is the watt, written W, and one watt is one joule of energy per second. A resistance $R$ carrying a current $I$ receives power $P = I^2 R$ , where $I^2$ means $I$ times $I$ . Because the current is squared, doubling the current makes the power four times bigger.
Squares, such as $V_S^2$: The small raised 2 in $V_S^2$ is an exponent, and it means the number multiplied by itself. For a 12 V source, $V_S^2 = 12 \times 12 = 144$ V squared. Squaring makes numbers grow faster: doubling a voltage makes its square four times bigger. The unit volts squared divided by ohms gives watts, so the formula’s units work out.
Efficiency and percent: Efficiency is the fraction of the power that reaches the place where it is wanted. It equals the useful power divided by the total power, and it is often written as a percent. If the source makes 18 W and the load receives 9 W, the efficiency is $\frac{9}{18} = 0.5$ , which is 50 percent. A percent is a fraction out of 100.
The slope of a curve, $\frac{dP_L}{dR_L}$: The letter d in front of a quantity means a very small change in it. So $\frac{dP_L}{dR_L}$ is the slope of the power graph: how steeply the power rises or falls when $R_L$ changes by a tiny amount. At the top of a hill the slope is flat, which means the slope equals zero there. Calculus uses this idea to find a peak exactly.

What each symbol means

Symbol	Meaning
$R_L$	The load resistance, the resistance of the device that receives the power, measured in ohms ( $\Omega$ ). It is set by the device you connect, such as a lamp or a motor. For maximum power, R_L must equal the source resistance $R_S$ .
$R_S$	The source resistance, the resistance inside the source, such as a battery’s internal resistance, measured in ohms ( $\Omega$ ). A small value means the source wastes little power inside itself. The example uses 4 $\Omega$ , a simple practice value.
$V_S$	The source voltage, the open-circuit voltage of the source with no load attached, measured in volts (V). It is the emf of the source. A charged 12 V car battery shows a voltage close to 12 V when nothing draws current from it.
$P_{\max}$	The maximum power delivered to the load, measured in watts (W). It is the largest power any choice of load can receive from this source. One watt is one joule of energy per second, so 9 W means 9 joules each second.

The formula piece by piece

$V_S$: Start with the source voltage, the push the source provides, measured in volts. The bigger it is, the more power is available to the load, and the formula uses it squared.
$V_S^2$: The source voltage squared, multiplied by itself. Power depends on the square because the power in a resistance is the voltage squared divided by the resistance. So doubling the source voltage quadruples the power available, which is why the square appears.
$4 R_S$: Four times the source resistance. The 4 comes from the peak calculation. At the match the total resistance is $2 R_S$ , and squaring it gives $(2 R_S)^2 = 4 R_S^2$ . That is where the factor of 4 comes from when the power formula is tidied up at the peak.
$\frac{V_S^2}{4 R_S}$: The squared source voltage divided by four times the source resistance. This value is the most power that the load can ever receive from this source, measured in watts.
$R_L = R_S \qquad P_{\max} = \frac{V_S^2}{4 R_S}$: The whole rule in one sentence: maximum power reaches the load when its resistance equals the source’s resistance, and then the load receives the source voltage squared divided by four times the source resistance, in watts.

What the formula is doing, and why

A source with voltage $V_S$ and internal resistance $R_S$ drives a load $R_L$ in series, so one path carries the same current through both. The current is $I = \frac{V_S}{R_S + R_L}$ , and the power in the load is $P_L = I^2 R_L$ . Both quantities depend on the size of the load, so the power changes as $R_L$ changes.

Why is there a peak? If $R_L$ is very small, the current is large, but nearly all the voltage is used up inside the source. The load gets little voltage and little power. If $R_L$ is very large, the load gets nearly all the voltage, but the current is tiny, so the power is again small. Between those two extremes the power rises to a peak and then falls.

Where the peak is. Using calculus, the power is largest where its slope with respect to $R_L$ , written $\frac{dP_L}{dR_L}$ , equals zero. The slope is flat at the top of the hill, and that happens exactly when $R_L = R_S$ . At that point the power is $P_{\max} = \frac{V_S^2}{4 R_S}$ . Readers without calculus can check the claim by comparing a few load values, as the example does.

Why is half lost? The source resistance also turns power into heat, $I^2 R_S$ . At the match the load and the source have equal resistances, so they drop equal voltages and carry the same current. Equal current through equal resistances gives equal power, so half of the power the source produces reaches the load, and half is lost inside the source. The efficiency is 50 percent.

When the rule does not apply. It answers one narrow question: how to get the most power into one load. A power supply that must hold a steady voltage for many loads uses a low source resistance, because a large resistance wastes power and lets the voltage fall as the load grows. So the rule is not the design goal for an efficient system.

An everyday picture

Think of a water pump pushing water through a pipe. If the pipe is far too narrow, the pump pushes hard but little water flows. If the pipe is far too wide, lots of water flows but the pump has little push. The most energy per second reaches the far end when the pipe’s resistance matches the pump’s own internal resistance. Even then, half of the pump’s effort is used up inside the pump itself.

Worked example: Choosing a load for a 12 V source with 4 ohm resistance

An illustrative source has a source voltage of 12 V and an internal resistance of $4\ \Omega$ . A load can be connected across it. Compare the power delivered to a 4 $\Omega$ load with the power delivered to an 8 $\Omega$ load, and also to a 2 $\Omega$ load. The numbers are made up for practice and are not taken from a specific product.

Find the current for the matched load. A matched load has $R_L = 4\ \Omega$ , the same as the source resistance. The two resistances in series add to $R_S + R_L = 4\ \Omega + 4\ \Omega = 8\ \Omega$ . The current is $I = \frac{V_S}{R_S + R_L} = \frac{12\ \text{V}}{8\ \Omega} = 1.5\ \text{A}$ . An ampere is one coulomb per second, and this same current flows through both resistances.
Find the power in the matched load. The power in the load is $P_L = I^2 R_L = (1.5\ \text{A})^2 \times 4\ \Omega$ . Squaring 1.5 gives $1.5 \times 1.5 = 2.25$ , so $P_L = 2.25 \times 4 = 9\ \text{W}$ . The units check out, because amperes squared times ohms equals watts.
Check with the peak formula. The formula gives $P_{\max} = \frac{V_S^2}{4 R_S} = \frac{(12\ \text{V})^2}{4 \times 4\ \Omega}$ . The top is $12 \times 12 = 144$ , and the bottom is $4 \times 4 = 16$ , so $\frac{144}{16} = 9\ \text{W}$ . The two methods agree, which confirms the matched load is the peak.
Try a larger load of 8 ohms. The total resistance is now $4\ \Omega + 8\ \Omega = 12\ \Omega$ , so the current falls to $I = \frac{12\ \text{V}}{12\ \Omega} = 1\ \text{A}$ . The load power is $P_L = (1\ \text{A})^2 \times 8\ \Omega = 8\ \text{W}$ . The bigger load gets less current and less power, even though its resistance is twice as large as the source’s.
Try a smaller load of 2 ohms. The total resistance is $4\ \Omega + 2\ \Omega = 6\ \Omega$ , so the current rises to $I = \frac{12\ \text{V}}{6\ \Omega} = 2\ \text{A}$ . The load power is $(2\ \text{A})^2 \times 2\ \Omega = 8\ \text{W}$ . The smaller load also gets less than 9 W. Trying values on both sides of the match shows that 4 ohms sits at the top of the curve.
Find the efficiency at the match. The total power the source produces is $I^2 (R_S + R_L) = (1.5\ \text{A})^2 \times 8\ \Omega = 2.25 \times 8 = 18\ \text{W}$ . The load gets 9 W of it, so the efficiency is $\frac{9\ \text{W}}{18\ \text{W}} = 0.5$ , which is 50 percent. The other 9 W becomes heat in the source’s internal resistance, since $I^2 R_S = 2.25 \times 4 = 9\ \text{W}$ .

Answer: The matched 4 $\Omega$ load receives the most power, 9 W. The 8 $\Omega$ load receives only 8 W, and the 2 $\Omega$ load also gets 8 W. At the match, half of the 18 W produced by the source is wasted inside the source, so the efficiency is 50 percent.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Thinking maximum power means maximum efficiency. At the match the efficiency is only 50 percent, so a design that needs high efficiency uses a source resistance much smaller than the load.
Using $V_S$ where $V_S^2$ belongs in the formula. Maximum power depends on the square of the voltage, so doubling the source voltage quadruples the maximum power.
Forgetting the source resistance in the current. The current is set by the total of $R_S$ and $R_L$ , not by $R_L$ alone.
Applying the rule to every circuit. It is about getting the most power into one load, which is different from holding a steady voltage for many loads.

Where you will meet it

Radio and microwave equipment, where cables and antennas are designed to share a common resistance, often 50 ohms, so that the most power passes into the antenna.
Connecting a weak sensor to an amplifier, where the matching decides how much of the sensor’s signal power arrives at the amplifier.
Analyzing how a battery with internal resistance delivers power to a load, using the same model as the internal resistance formula.
Teaching impedance matching, the general version of this idea for circuits that contain coils and capacitors as well as resistors.

Questions people ask

Does maximum power transfer mean the most efficient setup?

No. At the match, half the power is lost in the source, so the efficiency is 50 percent. A design that needs high efficiency keeps the source resistance much smaller than the load.

Why is the efficiency exactly 50 percent at the match?

With $R_L = R_S$ , the load and the source have equal resistance. They carry the same current, so each drops the same voltage and takes the same power. Half the power ends up in each, and the efficiency is the load’s share.

Is the source resistance the same as internal resistance?

For a battery, yes. The internal resistance is the source resistance in this model. The rule applies to any source that can be drawn as a voltage in series with a resistance.

Is maximum power transfer used in power grids?

Usually not. Grids are built to deliver voltage with small losses, so they keep the source resistance very small compared with the load.

Internal Resistance: A real battery has a small resistance inside it, so its output voltage drops when it pushes a current through a load.
Electric Power: Electric power tells you how fast a device uses energy. You get it by multiplying the voltage, the push, by the current, the flow, and the answer is measured in watts.
Ohm's Law: Ohm’s law says the push on a wire, called voltage, equals the flow through it, called current, times how hard the wire resists that flow, called resistance.
Voltage Divider: A voltage divider uses two resistors in series to give a smaller share of a supply voltage. The share is set by the ratio of the two resistances.