**Maximum Power Transfer Theorem Definition**

Maximum power transfer theorem states that maximum power output is obtained when the load resistance R

_{L}is equal to Thevenin resistance R_{th}as seen from load Terminals.

Fig.1: Maximum Power Transfer Theorem

Any circuit or network may be represented by a Thevenin equivalent circuit. **The Thevenin resistance R _{th} is comparable to a source internal resistance (R_{S}) which absorbs some of the power available from the ideal voltage source**. In above figure, a variable load resistance R

_{L}is connected to a Thevenin circuit. The current for any value of load resistance R

_{L}is connected to a Thevenin circuit. The current for any value of load resistance is;

$I_L=\frac{{{V}_{S}}}{{{R}_{S}}+{{R}_{L}}}$

Then by using I^{2}R_{ }, the power delivered to the load is,

${{P}_{L}}={{\left( \frac{{{V}_{S}}}{{{R}_{S}}+{{R}_{L}}} \right)}^{2}}{{R}_{L}}~\text{ }\cdots ~~~~~~~\left( 1 \right)$

The load power depends on both R_{th}_{ }(R_{S}) and R_{L}; however, R_{th} (R_{S}) is considered constant for any particular network. Then one might get an idea of how P_{L} varies with a change in R_{L}_{ }by assuming values for Thevenin circuit of above figure and, in turn, calculating P_{L} for different values of R_{L}.

- You May Also Read:Thevenin’s Theorem Explained

**Maximum Power Transfer Theorem Derivation **

As we know power delivered to load is,

${{P}_{L}}={{\left( \frac{{{V}_{S}}}{{{R}_{S}}+{{R}_{L}}} \right)}^{2}}{{R}_{L}}$

Taking a derivative on both sides;

$\frac{d{{P}_{L}}}{d{{R}_{L}}}=\frac{V_{S}^{2}{{\left( {{R}_{S}}+{{R}_{L}} \right)}^{2}}-2{{R}_{L}}\left( {{R}_{S}}+{{R}_{L}} \right)}{{{\left( {{R}_{S}}+{{R}_{L}} \right)}^{4}}}$

For P_{L} to be maximum;

$\frac{d{{P}_{L}}}{d{{R}_{L}}}=0$

So,

$\frac{V_{S}^{2}\left( {{R}_{S}}-{{R}_{L}} \right)}{{{\left( {{R}_{S}}+{{R}_{L}} \right)}^{3}}}=0$

Finally,

${{R}_{S}}={{R}_{L}}$

So maximum power transferred is;

${{P}_{max}}=\frac{V_{S}^{2}}{4{{R}_{S}}}$

We got above expression by substituting R_{S}=R_{L} into equation (1).

**Maximum Power Transfer and Efficiency of Transmission**

We observe that power transfer from a real source always produces ohmic heating in the source resistance. Calculations of such internal effects require information about the internal structure and cannot, in general, be based upon Thevenin or Norton equivalent networks. However, the entire load current i_{L }usually passes through the internal resistance of a real source, so we represent the internal conditions by lumped parameters as shown in figure 1. The resulting internal power dissipated by R_{TH} or R_{S} is then

\[{{P}_{S}}={{R}_{S}}I_{L}^{2}=\frac{{{R}_{S}}}{{{({{R}_{S}}+{{R}_{L}})}^{2}}}*V_{S}^{2}\]

The dashed curve in figure 2 shows that P_{S} steadily decreases as R_{L} increases and that P_{S}=P_{L} when R_{L}/R_{S}=1.

Fig.2: Maximum Power Transfer and Transmission Efficiency

Since the total power generated by the source is P_{L}+P_{S}, the wasted internal power P_{S} should be small compared to P_{L} for efficient operation. Formally, we define the power-transfer efficiency as

$Efficiency=\frac{{{P}_{L}}}{{{P}_{L}}+{{P}_{S}}}$

Which is often expressed as a percentage. If the load has been matched for maximum power transfer, then P_{S}=P_{L}, and so efficiency,

$Efficiency=\frac{{{P}_{L}}}{2{{P}_{L}}}=50\text{ %}$

Moreover, with R_{L}=R_{S}, the terminal voltage drops to V=V_{TH}/2. Clearly, electrical utilities would not, and should not, strive for maximum power transfer. Instead, they seek higher power-transfer efficiency by making P_{S} as small as possible.

**Maximum Power Transfer Solved Example**

**Find R _{L}**

**Solution**

Let’s find V_{th} first across 150 Ω resistance

${{V}_{th}}={{V}_{S}}=360*\frac{150}{150+30}$

${{V}_{th}}={{V}_{S}}=300~V$

To find R_{th }or R_{S}, short circuit the voltage source

${{R}_{th}}={{R}_{S}}=150~||~30=25\Omega $

So, for maximum power transfer, we know that

\[{{R}_{L}}={{R}_{th}}=25\text{ }\Omega\]

Now, Find Maximum power transfer to the load

${{P}_{max}}=\frac{V_{S}^{2}}{4{{R}_{S}}}=900~W$

## Maximum Power Transfer Theorem using Matlab Code

Here is the MATLAB code to implement maximum power transfer theorem in Matlab.

clear all;close all;clc %% Circuit Parameters as given in the example (text) % Matlab Code for Maximum Power Transfer Theorem V_TH = 300; % Thevenin's Voltage R_TH = 25; % Thevenin's Equivalent Resistance R_L = 0:0.5:80; % Load Resistance %% %% Load Current &amp;amp; Power Calculation IL = V_TH./(R_TH + R_L); % Load Current P_L = IL.^2 .* R_L; % Load Power %% % As we know that maximum power transfer occurs when R_TH=R_L %% Plotting the Results plot(R_L,P_L,'b') hold on title('Maximum Power Transfer using Matlab'); xlabel('Load Resistance R_L'); ylabel('Power to the Load P_L'); gtext('R_TH = R_L = 25 Ohm') legend('P_L') grid on

**Result**

Here is a graph which clearly shows that maximum power transfer occurs when R_{th}=R_{L}.

**Maximum Power Transfer Theorem using Matlab Simulink**

Here, a Matlab Simulink model has been developed for three different cases:

- When R
_{L}>R_{TH} - When R
_{L}=R_{TH} - When R
_{L}<R_{TH}

To download and run the model, Click Here.

**Maximum Power Transfer Theorem Application**

## When do we want maximum power transfer?

Primarily, in applications where voltage and current signals are used to convey information rather than to deliver large amounts of power. For instance, the first stage of a radio or television receiver should get as much power as possible out of the information-bearing signals that arrive via antenna or cable. Those tiny signals account for only a small fraction of the total power consumed by the receiver, and so power-transfer efficiency is not a significant concern.