Electrical

# Maximum Power Transfer Theorem with Solved Examples

## 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 RL 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 Rg 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, Ri = Rg + R

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

RL = Ri

Fig. 2.49

Proof

Let us assume that the circuit is linear, then;

Circuit current, I = E / RL + Ri

Power transferred to the load is;

PL = I2 RL = E2 RL / (RL + Ri)2 … (1)

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

dPL / dRL = 0

dPL / dRL = E2 [ 1/ (RL + Ri)2 + RL ( -2/ (RL + Ri)3]

= E2 [ 1 / (RL + Ri)2 – 2 RL / (RL + Ri)3]

∴ 0 = E2 [ 1 / (RL + Ri)2 – 2RL / (RL + Ri)3]

Or 2RL = RL + Ri or RL = Ri … (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, RL = 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 VL = VS / 2

The load current IL = VS / 2RL

The load power PL = VL IL = (VS’)2 / 4RL

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

Solution;

We will remove RL 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, Vth = 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), Rth = 2 + (3 x 3/3 + 3) +1 = 4.5 Ω. Maximum power transfer to the load will take place when RL = Rth = 4.5 Ω

Figure 2.52

Maximum power drawn by RL = V2th / 4 x RL = 7.52 / 4 x 4.5 = 3.125 W

Since same power is developed is Rth, 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 RL. 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 VAB 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, VAB = Vth = 3 x 0.75 = 2.25 V, with point A positive with respect to point B.

Fig. 2.53

For finding RAB, current source is replaced by an infinite resistance.

RAB = Rth = 5 + 3 (2 + 5) / 3 + 2 + 5 = 7.1 Ω

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

Maximum power transferred = Vth2 / 4 RL = 2.252 / 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 Vth 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

Vth/ 10 + Vth/ 5 = 12 + 6 or Vth = 60 V

Now, for finding Ri or Rth, 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, Ri = Rth = 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.

Pmax = I2R = 92 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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