If a source circuit is represented as an ideal voltage (VS) in series with an internal resistance (RS), then maximum power is transferred to the load when the load resistor (RL) is equal to the internal resistance of the source.

The source circuit could be a non-ideal battery, shown as an ideal EMF in series with its internal resistance. Alternatively, it might be a Thevenin equivalent circuit, based on a more complex network of voltage sources and resistors, or even the output stage of an electronic circuit, such as an amplifier.

The maximum power transferred occurs when RS = RL and is given by: –

Equation: Pmax = Vs × Vs / (4 Rs)

Worked Example 1

A voltage source of 12 V with an internal resistance of 0.5 Ω is connected in series with an external load resistor.
(a) Identify the value of load resistance which will allow maximum power to be transferred.
(b) Calculate the power dissipated in the load.
(c) Calculate the efficiency of power transfer.
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(a) Maximum power is transferred when RS = RL so RL = 0.5 Ω.

(b) PMax = VS2 / (4 × RS) = 122 / (4 × 0.5) = 144 / 2 = 72 W

(c) % Efficiency = PLoad / (PSource + PLoad) × 100 = 72 × 100 / (72 + 72) = 50%

Effect of Varying the Load Resistance

The power transferred to the load may be calculated for any value of load resistance, by using the following relationships.

A spreadsheet may be used to calculate the power dissipated over a range of values of load resistance. Taking the same values of VS = 12 V and RS = 0.5 Ω used in Worked Example 1, the following results are found.

Notice from the table and graph that a maximum power dissipation of 72 W occurs when the load resistance is equal to 0.5 Ω. This result agrees with the calculated power dissipation from Worked Example 1.

It is also possible to use calculus to prove that maximum power transfer occurs when RS = RL. The technique is to differentiate the general expression for the power dissipated in the load with respect to RL. A local ‘maximum’ or ‘minimum’ (plural maxima or minima) occurs when this expression equals zero, meaning that the gradient of the graph is zero at that point. A second derivative is taken in order to determine whether the identified point is a maximum or minimum, with the latter value being negative or positive if the result is a maximum or minimum respectively.

Worked Example 2

Determine the expression for the power dissipated in the load, when RS = RL.
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If RS = RL then the expression for the current may be written as

I = VS / (2 RS) . . . (1)

Similarly, the power becomes

P = I × I × RS . . . (2)

Substituting (1) into (2) gives

P = VS × VS / (4 RS)

Worked Example 3

Use calculus to prove that maximum power transfer occurs when RS = RL.
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The general expression for the power dissipated in the load resistor is

PLoad = VS2 × RL / (RS + RL)2

Multiplying out the denominator and dividing through by RL gives

PLoad = VS2 / (RS2 / RL + 2 RS + RL)

For PLoad to be a maximum, the demominator must be a minimum. Hence

d / d RL [RS2 / RL + 2 RS + RL] = 0

0 = − RS2 / RL2 + 0 + 1, so RS / RL = 1

A second derivative may be taken with respect to RL, to confirm the above result is a minimum.

d2 / d RL2 [RS2 / RL + 2 RS + RL] = 2 RS2 / RL3

The above result is always positive, so the denominator of the expression for PLoad is a minimum when RS = RL, hence giving a maximum for the power dissipated in the load resistor.

Maximum Power Transfer Versus Maximum Efficiency

It was found in Worked Example 1 that the efficiency of power transfer is 50% at the point of maximum power transfer, when RS = RL. However, it would be incorrect to assume this is the maximum efficiency which may be achieved, as may be confirmed by extending our previous spreadsheet to display both power transferred and efficiency.

The efficiency rises steadily towards 100% as RL increases beyond RS, although the power transferred decreases at the same time. Hence there is something of a trade-off between the two.

Final Thoughts

The maximum power transfer concept is most appropriate in cases where the voltage and internal resistance properties of the source circuit are fixed, and the aim is to maximise the amount of power transferred, even at the expense of reduced efficiency. However, if the source circuit is also under the designer’s control, then efficiency may be improved by reducing or even eliminating the source resistance, while power may be boosted by increasing the source voltage.

An example, illustrating the benefits of innovative design, is the switched mode power supply, where efficiency values well in excess of 90% are achieved. In this case, the output transistor switches on and off many thousands of times each second, alternating between low impedance – high current and high impedance – low current, both of which minimise the losses in the source circuit. This contrasts with the more traditional (and much less efficient) linear power supply, in which the series regulator IC effectively acts as a variable source resistance.

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