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# Maximum Power Transfer Theorem

The *maximum power theorem (MPT)* is used to find the value of load resistance for which there would be maximum amount of power transfer from the source to load.

## Statement of MPT

A resistive load that is connected to a DC source, receives maximum power when the load resistance is equal to the internal resistance of the source as seen from the load terminals.

## Explanation of MPT

Consider the following circuit diagram to determine the value of R_{L} such that it receives maximum power from the DC source.

The load current is,

$$I=\frac{V_{Th}}{R_{Th}+R_{L}}$$

Thus, the power delivered to the resistive load is,

$$P_{L}=I^{2}R_{L}=(\frac{V_{Th}}{R_{Th}+R_{L}})^{2}R_{L}$$

As we know, the load resistance is variable. Thus, P_{L} can be maximised by varying R_{L}.

$$\Rightarrow\:P_{L}=\frac{V_{Th}^{2}R_{L}}{(R_{th}+R_{L})^{2}}=\frac{V_{Th}^{2}R_{L}}{R_{Th}^{2}+R_{L}^{2}+2R_{Th}R_{L}}$$

$$\Rightarrow\:P_{L}=\frac{V_{Th}^{2}}{(\frac{R_{th}^{2}}{R_{L}}+R_{L}+2R_{Th})}=\frac{V_{Th}^{2}}{D}$$

In order to PL be maximum the denominator (D) term should be minimum, i.e.,

$$\frac{dD}{dR_{L}}=0$$

$$\frac{d}{dR_{L}}(\frac{R_{Th}^{2}}{R_{L}}+R_{L}+2R_{Th})=0$$

By solving the above differential equation, we get,

$$R_{L}=R_{Th}$$

Hence, maximum power will deliver to the load if the load resistance is equal to internal resistance of the source network i.e.

Load Resistance = Internal Resistance of the source

## Amount of Maximum Power (Pmax)

$$P_{max}=\frac{V_{Th}^{2}}{4R_{Th}}$$

Here, Pmax is the amount of maximum power consumed by the load.

The ** total power supplied** by the source is

$$P=2\frac{V_{Th}^{2}}{4R_{Th}}=\frac{V_{Th}^{2}}{2R_{Th}}$$

The *efficiency* of the circuit during maximum power transfer becomes,

$$\eta=\frac{P_{max}}{P}\times\:100=50$$%

## Steps for Solution of Network using MPT

**step 1** – Remove the load resistance and replace all the independent sources by their internal resistance and determine the R_{Th} of the source network looking through the open circuited load terminals.

**Step 2**S – As per MPT, the value of R_{Th} gives the value of load resistance R_{L} i.e. R_{Th} = R_{L}, that allows maximum power transfer.

**Step 3** – Find the value of V_{Th} across the open circuited load terminals.

**Step 4** – The amount maximum power transferred is given by,

$$P_{max}=\frac{V_{Th}^{2}}{4R_{Th}}$$

## Numerical Example

Find the value of load resistance R_{L} in the circuit given below such that maximum power transfer takes place. Also, calculate the amount of maximum power.

**Solution**

**Step 1** – Remove the load resistance and replace all the independent sources by their internal resistance (in this case, 12 V ideal voltage source is short circuited) and determine the value of RTh that give RL corresponding to the maximum power transfer.

$$R_{Th}=1\:\Omega\:\lVert\:6\:\Omega=\frac{1\times\:6}{1+\:6}=\frac{6}{7}\:\Omega$$

As per maximum power transfer theorem,

$$R_{L}=R_{Th}=\frac{6}{7}\:\Omega$$

**Step 2** – Determine the VTh across the open circuited load terminals,

Here, the current in the circuit is,

$$I=\frac{12}{1+6}=\frac{12}{7}\:Ampere$$

As, the load terminal is open circuited, hence the V_{Th }is the voltage drop across the 6 Ω resistor. Thus,

$$V_{Th}=I\times\:6\Omega=\frac{12}{7}\times\:6=\frac{72}{7}\:Volt$$

The amount of maximum power delivered to the load is,

$$P_{max}=\frac{V_{Th}^{2}}{4R_{Th}}=\frac{(72/7)^{2}}{4\times\:(6/7)}=30.857W$$

Hence, the value of load resistance for maximum power transfer is equal to (6/7 Ω ) and the amount of maximum power delivered to the load is 30.857 Watt.

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