# Maximum Power Transfer Theorem with Solved Examples

Table of Contents

## Maximum Power Transfer Theorem

**Maximum Power Transfer Theorem with Solved Examples-** According to Maximum Power Transfer theorem, a resistive load gets maximum power from a network, when load resistance value equals to the network resistance value, and removing all energy sources, network is viewed from the output terminals, with their internal resistances left behind. The maximum power transfer theorem may be stated as follows;

A resistor load will abstract maximum power from a network when the load resistance is equal to the resistance of the network as viewed from the output terminals, with all energy sources removed leaving behind their internal resistances.

In fig. 2.49, a load resistance R_{L} has been shown connected to the terminal A and B of a network. This network comprises a generator, which has its e.m.f resistance as E, internal resistance R_{g }and a series resistance R. Let us assume that if viewed from terminal A and B, the network’s internal resistance will be as follows;

Internal resistance of the Network, R_{i }= R_{g} + R

According to this theorem, R_{L} will abstract maximum power from the network at a time when its value is equal to the network’s internal resistance, i.e.,

R_{L} = R_{i}

Fig. 2.49

Proof

Let us assume that the circuit is linear, then;

Circuit current, I = E / R_{L} + R_{i}

Power transferred to the load is;

P_{L} = I_{2} R_{L} = E_{2} R_{L} / (R_{L} + R_{i})^{2 }… (1)

We differentiate the above- mentioned equation for maximum power transfer with respect to R_{L}, and then we keep it equivalent to 0, i.e.,

dP_{L} / dR_{L} = 0

dP_{L} / dR_{L = }E_{2} [ 1/ (R_{L} + R_{i})^{2 }+ R_{L} ( -2/ (R_{L }+ R_{i})^{3}]

= E_{2} [ 1 / (R_{L} + R_{i})^{2} – 2 R_{L} / (R_{L} + R_{i})^{3}]

∴ 0 = E_{2} [ 1 / (R_{L} + R_{i})^{2} – 2R_{L} / (R_{L} + R_{i})^{3}]

Or 2R_{L} = R_{L} + R_{i }or R_{L} = R_{i }… (2)

The above-mentioned equation (2) reflects that maximum power will be transferred to the load when the load resistance value becomes equal to the source resistance.

In situation of maximum power transfer, voltages (or terminal potential difference) found parallel to the load tends to become half compared to the open circuit voltages found on terminals A and B, i.e.,

V _{A.B} = E / 2

Thus, circuit’s efficiency also declines and becomes 50 %. Therefore, this system does not prove beneficial from economical point of view. Therefore, this system is used very rarely in places where high efficiency is required instead of maximum power transfer (e.g., transmission and distribution networks, etc.). Maximum power transfer theorem is mostly used in electronics and communication networks, because maximum power transfer has greater relevance and significance in communication than efficiency.

## Maximum Power Transfer Theorem Solved Examples

**Example 1;** What is the value of load resistance RL which will result in maximum load power in the circuit in fig. 2.50?

Fig. 2.50

**Solution;**

The Thevenin equivalent circuit is as in fig. 2.51

Fig. 2.51

For maximum load power, R_{L }= R’ = 87.97 Ω

Because maximum load power exists when the load resistance equals the Thevenin equivalent resistance of the remainder of the circuit it naturally follows that in these circumstances,

The load voltage V_{L} = V_{S }/ 2

The load current I_{L} = V_{S} / 2R_{L}

The load power P_{L} = V_{L} I_{L} = (V_{S}’)^{2} / 4R_{L}

**Example 2;** In the network shown in fig. 2.52 (a), find the value of R_{L}, such that maximum possible power will be transferred to R_{L}. Find also the value of the maximum power and the power supplied by source under these conditions

**Solution;**

We will remove R_{L} and find the equivalent Thevenin’s source for the circuit to the left of terminals A and B. As seen from fig. 2.52 (b), Vth equals the drop across the vertical resistor of 3 Ω because no current flows through 2 Ω and 1 Ω resistors, since 15 V drops across two series resistors of 3 Ω each, V_{th} = 15/2 = 7.5 V. Thevenin’s resistance can be found by replacing 15 V source with a short – circuit. As seen from fig. 2.52 (b), R_{th} = 2 + (3 x 3/3 + 3) +1 = 4.5 Ω. Maximum power transfer to the load will take place when R_{L} = R_{th} = 4.5 Ω

Figure 2.52

Maximum power drawn by R_{L} = V^{2}_{th} / 4 x R_{L }= 7.5^{2} / 4 x 4.5 = 3.125 W

Since same power is developed is R_{th}, power supplied by the source = 2 x 3.125 = 6.250 W

**Example 3;** In the circuit shown in fig. 2.53 (a), obtain the condition from maximum power transfer to the load R_{L}. Hence, determine the maximum power transferred.

**Solution;**

We will find Thevenin’s equivalent circuit to the left of terminals A and B for which purpose we will convert the battery source into a current source as shown in fig. 2.53 (b). By combining the two current sources, we get the circuit of fig. 2.53 (b). it would be seen that open circuit voltage V_{AB} equals the drop over 3 Ω resistance because there is no drop on the 5 Ω resistance connected to terminal A. Now, there are two parallel paths across the current source each of resistance 5 Ω. Hence, current through 3 Ω resistance equals 1.5 / 2 = 0.75 A. Therefore, V_{AB} = V_{th} = 3 x 0.75 = 2.25 V, with point A positive with respect to point B.

Fig. 2.53

For finding R_{AB}, current source is replaced by an infinite resistance.

R_{AB }= R_{th} = 5 + 3 (2 + 5) / 3 + 2 + 5 = 7.1 Ω

The Thevenin’s equivalent circuit along with R_{L} is shown in fig. 2.54, the condition for MTP is that R_{L }= 7.1 Ω

Maximum power transferred = V_{th}^{2} / 4 R_{L} = 2.25^{2 }/ 4 x 7.1 = 0.178

Fig 2.54

**Example 4;** Calculate the value of R which will absorb maximum power from the circuit of fig. 2.55 (a). Also, compute the value of maximum power.

**Solution;**

For finding power, it is essential to know both I and R. Hence, it is essential to find an equation relating I to R.

Let us remove R and find Thevenin’s voltage V_{th} across A and B as shown in fig. 2.55 (b).

Figure 2.55 – a, b, c, d, e.

It would be helpful to convert 120 V, 10 Ω source into a constant current source as shown in fig. 2.55 (c). Applying KCL to the circuit, we get

V_{th}/ 10 + V_{th}/ 5 = 12 + 6 or V_{th} = 60 V

Now, for finding R_{i }or R_{th}, the two sources are reduced to zero. Voltage of the voltage–source is reduced to zero by short-circuiting it whereas current of the current source is reduced to zero by open–circuiting it. The circuit which results from such source suppression is shown in fig. (d). Hence, R_{i} = R_{th} = 10 x 5 / 10 + 5 = 50 / 15 Ω. The Thevenin’s equivalent circuit of the network is shown in fig (e).

According to Maximum Power Transfer Theorem, R will absorb maximum power when it equals 10 / 3 Ω. In that case, I = 60 + 20 / 3 = 9 A.

P_{max} = I^{2}R = 9^{2} x 10 / 3 = 270 W

## Frequently Asked Questions – FAQs

### Question 1: what is the Maximum Power Transfer Theorem?

According to the **Maximum Power Transfer **theorem, a resistive load gets maximum power from a network when the load resistance value equals to the network resistance value, and removing all energy sources, the network is viewed from the output terminals, with their internal resistances.

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