# Maximum Power Transfer Theorem | 3+ Important Steps| Explanations

Maximum Power Transfer Theorem

Image Credit: Iñigo Gonzalez from Guadalajara, Spain, Lamp @Ibiza (624601058)CC BY-SA 2.0

## Introduction to Maximum Power Transfer Theory

In the previous articles related to circuit analysis, we have come across several methods and theories regarding solving problems of a complex network. The maximum power transfer theorem is one of the efficient theories needed to analyze and study advanced circuits. It is one of the primary methods yet important one.

We will discuss the theories, the problem-solving steps, real-world applications, the explanation of the theory. A mathematical problem is solved at last for a better understanding.

## Theory of Maximum Power Transfer Theorem

Maximum power Transfer Theory:

It states that a DC Circuit’s load resistance receives the maximum power if the magnitude of the load resistance is the same as the Thevenin’s equivalent resistance.

The theory is used to calculate the value of load resistance, which causes the maximum power transferred from the source to the load. The theorem is valid for both AC and DC circuits (Point to be noted: For AC circuits, the resistances are replaced by impedance).

## Real world Applications of Maximum power transfer theorem

The maximum power transfer theorem is one of the efficient theorems. That is why there are several real-world applications for this theory. The communication sector is one of its fields. The theory is used for low strength signals. Also, for loud speakers to drain the maximum power from the amplifier.

## Steps for solving problems regarding Maximum Power Transfer Theorem

In general, the below mentioned steps are followed for solving power transfer theory problems. There are other ways, but following these steps will lead to a more efficient path.

• Step 1: Find out the load resistance of the circuit. Now remove it from the circuit.
• Step 2: Calculate the Thevenin’s equivalent resistance of the circuit from the open circuited load resistance branch’s view point.
• Step 3: Now, as the theory says, the new load resistance will be the Thevenin’s equivalent resistance. This is the resistance that is responsible for Maximum power transfer.
• Step 4: The maximum power is derived then. It comes as follows.

PMAX = VTH2 / 4RTH

## Explanations of the Maximum Power Transfer Theory

To explain the theorem, let us take a complex network as below.

In this circuit, we have to calculate the value of load resistance for which the maximum power will be drained from the source to the load.

As we can see in the above images, the variable load resistance is attached to the DC circuit. In the second image, the Thevenin’s equivalent circuit is already represented (both the Thevenin’s equivalent circuit and Thevenin’s equivalent resistance).

From the second image, we can say current (I) through the circuit is:

I = VTH / (RTH + RL)

The power of the circuit is given by P = VI.

Or, P = I2 RL

Substituting the value of I from the Thevenin’s equivalent circuit,

PL = [VTH / (RTH + RL)]2 RL

We can observe that the value of PL can be increased or preferably varied by changing RL‘s value. According to the rule of calculus, the maximum power is achieved when the derivative of the power with respect to the load resistance is equal to zero.

dPL / dRL = 0.

Differentiating PL, we get,

dPL / dRL = {1 / [(RTH + RL)2]2} * [{(RTH + RL)2 d/dRL (VTH2 RL)} – {(VTH2 RL) d/dRL (RTH + RL)2}]

Or, dPL / dRL = {1 / (RTH + RL)4} * [{(RTH + RL)2 VTH2} – {VTH2 RL * 2 (RTH + RL)

Or, dPL / dRL = [VTH2 * (RTH + RL – 2RL)] / [(RTH + RL)3]

Or, dPL / dRL = [VTH2 * (RTH – RL)] / [(RTH + RL)2]

For the maximum value, dPL / dRL = 0.

So, [VTH2 * (RTH – RL)] / [(RTH + RL)2] = 0

From which, we get,

(RTH – RL) = 0 or, RTH = RL

It is now proved that the maximum power will be drawn when the load resistance and internal equivalent resistance are the same.

So, the maximum power which can be drawn by any circuit,

PMAX = [VTH / (RTH + RL)]2 RL

Now, RL = RTH

OR, PMAX = [VTH / (RTH + RTH)]2 RTH

OR, PMAX = [VTH2 / 4RTH2] RTH

OR, PMAX= VTH2 / 4RTH

This is the power drawn by the load. The power received by the load is the same power send by the load.

So, the total supplied power is:

P = 2 * VTH2 / 4RTH

Or, P = VTH2 / 2RTH

The efficiency of the power transfer is calculated as follows.

η = (PMAX / P) * 100 % = 50 %

This theory aims to gain the maximum power from the source by making the load resistance equal to the source resistances. This idea has different and several applications in the field of communication technology, especially the signal analysis part. The source and load resistances are matched previously and decided before the circuit operation started to attain the maximum power transfer condition. The efficiency comes down to 50%, and the flow of power started from source to load.

Now, for electrical power transmission systems, where the load resistances have higher values than the sources, the condition of maximum power transfer is not achieved easily. Also, the efficiency of the transfer is just 50%, which has no good economical values. That is why the power transfer theorem is rarely used in the power transmission system.

## Problems Related to Maximum Power Transfer Theorem

### Observe the circuit carefully and calculate the resistance value to receive the maximum power. Apply maximum power transfer theorem to find out the amount of power transferred.

Solution: The problem is solved by following the given steps.

In the first step, the load resistance is disconnected from the circuit. After disconnecting the load, we mark it as AB. In the next step, we will calculate the Thevenin’s equivalent voltage.

So, VAB = VA – VB

VA comes as: VA = V * R2 / (R1 + R2)

Or, VA = 60 * 40 / (30 + 40)

Or, VA = 34.28 v

VB comes as:

VB = V * R4 / (R3 + R4)

Or, VB = 60 * 10 / (10 + 20)

Or, VB = 20 v

So, VAB = VA – VB

Or, VAB = 34.28 – 20 = 14.28 v

Now, it is time to find out the Thevenin’s equivalent resistance for the circuit.

For that, we short circuit the voltage source and the resistance values are calculated through the open terminal of the load.

RTH = RAB = [{R1R2 / (R1 + R2)} + {R3R4 / (R3 + R4)}]

OR, RTH = [{30 × 40 / (30 + 40)} + {20 × 10 / (20 + 10)}]

OR, RTH = 23.809 ohms

Now the circuit is redrawn using the equivalent values. The maximum power transfer theorem says that to obtain the maximum power, the load resistance = Thevenin’s equivalent resistance. So as per the theory, load resistance RL = RTH = 23.809 ohms.

Formula for the maximum power transfer is PMAX = VTH2 / 4 RTH.

Or, PMAX = 14.282 / (4 × 23.809)

Or, PMAX = 203.9184 / 95.236

Or, PMAX = 2.14 Watts

So, the maximum amount of transferred power is 2.14 watts.

### 2. Observe the circuit carefully and calculate the resistance value to receive the maximum power. Apply maximum power transfer theorem to find out the amount of power transferred.

Solution: The problem is solved by following the given steps.

In the first step, the load resistance is disconnected from the circuit. After disconnecting the load, we mark it as AB. In the next step, we will calculate the Thevenin’s equivalent voltage. VTH = V * R2 / (R1 + R2)

VTH = V * R2 / (R1 + R2)

Or, VTH = 100 * 20 / (20 +30)

Or, VTH = 4 V

Now, it is time to find out the Thevenin’s equivalent resistance for the circuit. The resistances are in parallel with each other.

So, RTH = R1 || R2

Or, RTH = 20 || 30

Or, RTH = 20 * 30 / (20 + 30)

Or, RTH = 12 Ohms

Now the circuit is redrawn using the equivalent values. The maximum power transfer theorem says that to obtain the maximum power, the load resistance = Thevenin’s equivalent resistance. So as per the theory, load resistance RL = RTH = 12 ohms.

Formula for the maximum power transfer is PMAX = VTH2 / 4 RTH.

Or, PMAX = 1002 / (4 × 12)

Or, PMAX = 10000 / 48

Or, PMAX = 208.33 Watts

So, the maximum amount of transferred power is 208.33 watts.

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About Sudipta Roy

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