# Circuit Theory/Maximum Power Transfer

## Maximum Power Transfer

Often we would like to transfer the most power from a source to a load
placed across the terminals as possible. How can we determine the
optimum resistance of the load for this to occur?

Let us consider a source modelled by a Thévenin equivalent (a Norton
equivalent will lead to the same result, as the two are directly
equivalent), with a load resistance, *R~L~*. The source resistance is
*R~s~* and the open circuit voltage of the source is *v~s~*:

![](Maximum_Power_Transfer_Circuit.svg "Maximum_Power_Transfer_Circuit.svg")

The current in this circuit is found using Ohm\'s Law:

$$i=\frac{v_s}{R_s+R_L}$$

The voltage across the load resistor, *v~L~*, is found using the voltage
divider rule:

$$v_L=v_s \,\frac{R_L}{R_s + R_L}$$

We can now find the power dissipated in the load, *P~L~* as follows:

$$P_L=v_Li=\frac{R_L \, v^2_s}{\left(R_s+R_L\right)^2}$$

We can now rewrite this to get rid of the *R~L~* on the top:

$$P_L=\frac{v^2_s}{ \left(\frac{{R_s}}{\sqrt{R_L}}+\sqrt{R_L}\right)^2} = 
\frac{v^2_s}{ R_s \left(\frac{\sqrt{R_s}}{\sqrt{R_L}}+\frac{\sqrt{R_L}}{\sqrt{R_s} }\right)^2}$$

Assuming the source resistance is not changeable, then we obtain maximum
power by minimising the bracketed part of the denominator in the above
equation. It is an elementary mathematical result that $x+x^{-1}$ is at
a minimum when *x=1*. In this case, it is equal to 2. Therefore, the
above expression is minimum under the following condition:

$$\frac{\sqrt{R_s}}{\sqrt{R_L}}=1$$

This leads to the condition that:

:   

    :   {\|style=\" border: solid 2px #D6D6FF; padding: 1em;\"
        valign=\"top\"

\|align=\"left\"\|$R_L=R_s \,$ \|}

### Efficiency

The efficiency, *η* of the circuit is the proportion of all the energy
dissipated in the circuit that is dissipated in the load. We can
immediately see that at maximum power transfer to the load, the
efficiency is 0.5, as the source resistor has half the voltage across
it. We can also see that efficiency will increase as the load resistance
increases, even though the power transferred will fall.

The efficiency can be calculated using the following equation:

$$\eta=\frac{P_L}{P_L+P_s}$$ where *P~s~* is the power in the source
resistor. This can be found using a simple modification to the equation
for *P~L~*:

$$P_s=\frac{v^2_s}{ R_L \left(\frac{\sqrt{R_s}}{\sqrt{R_L}}+\frac{\sqrt{R_L}}{\sqrt{R_s} }\right)^2}$$

The graph below shows the power in the load (as a proportion of the
maximum power, *P~max~*) and the efficiency for values of *R~L~* between
0 and 5 times *R~s~*.

![](Maximum_Power_Transfer_Graph.svg "Maximum_Power_Transfer_Graph.svg")

It is important to note that under conditions of maximum power transfer
as much power is dissipated in the source as in the load. This is not a
desirable condition if, for example, the source is the electricity
supply system and the load is your electric heater. This would mean that
the electricity supply company would be wasting half the power it
generates. In this case, the generators, power lines, etc. are designed
to give the lowest source resistance possible, giving high efficiency.
The maximum power transfer condition is used in (usually high-frequency)
communications systems where the source resistance can not be made low,
the power levels are relatively low and it is paramount to get as much
signal power as possible to the receiving end of the system (the load).
