*The article explains the Maximum Power Transfer Theorem, which states that maximum power output is achieved when the load resistance equals the Thevenin resistance of a circuit. It covers the theorem’s definition, mathematical derivation, efficiency considerations, and practical applications, particularly in signal-processing devices like radio and television receivers.*

**What is the Maximum Power Transfer Theorem?**

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.

**Figure 1**: Maximum Power Transfer Theorem Circuit Diagram

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 the 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 PL varies with a change in RL by assuming values for the Thevenin circuit of the above figure and, in turn, calculating PL for different values of RL.

- You May Also Read: Thevenin’s Theorem Explained

**Maximum Power Transfer Theorem Derivation **

As we know, the power delivered to the 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;

### ${{P}_{max}}=\frac{V_{S}^{2}}{4{{R}_{S}}}$ (Maximum Power Transfer Formula)

We got the 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 iL 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.

**Figure 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 the 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 the 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 that 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**

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 an 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.