- Introduction
- Maximum Power Transfer Theorem Statement
- Proof of Maximum Power Transfer Theorem
- Power Transfer Efficiency
- Maximum Power Transfer Theorem for AC Circuits
- Applying Maximum Power Transfer Example to DC circuit
- Applying Maximum Power Transfer to AC circuit
- Practical Application of Maximum Power Transfer Theorem

### Introduction

In any electric circuit, the electrical energy from the supply is delivered to the load where it is converted into a useful work. Practically, the entire supplied power will not present at load due to the heating effect and other constraints in the network. Therefore, there exist a certain difference between drawing and delivering powers.

The load size always affects the amount of power transferred from the supply source, i.e., any change in the load resistance results to change in power transfer to the load. Thus, the maximum power transfer theorem ensures the condition to transfer the maximum power to the load. Let us see ‘how’.

### Maximum Power Transfer Theorem Statement

The maximum power transfer theorem states that in a linear , bilateral DC network , maximum power is delivered to the load when the load resistance is equal to the internal resistance of a source.

If it is an independent voltage source, then its series resistance (internal resistance Rs) or if it is independent current source, then its parallel resistance (internal resistance Rs) must equal to the load resistance RL to deliver maximum power to the load.

### Proof of Maximum Power Transfer Theorem

The maximum power transfer theorem ensures the value of the load resistance , at which the maximum power is transferred to the load.

Consider the below DC two terminal network (left side circuit) , to which the condition for maximum power is determined , by obtaining the expression of power absorbed by load with use of mesh or nodal current methods and then derivating the resulting expression with respect to load resistance RL.

But this is quite a complex procedure. But in previous articles we have seen that the complex part of the network can be replaced with a Thevenin’s equivalent as shown below.

The original two terminal circuit is replaced with a Thevenin’s equivalent circuit across the variable load resistance. The current through the load for any value of load resistance is

Form the above expression the power delivered depends on the values of R_{TH} and R_{L}. However the Thevenin’s equivalent is constant, the power delivered from this equivalent source to the load entirely depends on the load resistance R_{L}. To find the exact value of RL, we apply differentiation to P_{L} with respect to R_{L} and equating it to zero as

Therefore, this is the condition of matching the load where the maximum power transfer occurs when the load resistance is equal to the Thevenin’s resistance of the circuit. By substituting the R_{th} = R_{L} in equation 1 we get

The maximum power delivered to the load is,

Total power transferred from source is

P_{T }= I_{L}^{2 }(R_{TH} + R_{L})^{ }

= 2 I_{L}^{2 }R_{L } …………….(2)

Hence , the maximum power transfer theorem expresses the state at which maximum power is delivered to the load , that is , when the load resistance is equal to the Thevenin’s equivalent resistance of the circuit. Below figure shows a curve of power delivered to the load with respect to the load resistance.

Note that the power delivered is zero when the load resistance is zero as there is no voltage drop across the load during this condition. Also, the power will be maximum, when the load resistance is equal to the internal resistance of the circuit (or Thevenin’s equivalent resistance). Again, the power is zero as the load resistance reaches to infinity as there is no current flow through the load.

### Power Transfer Efficiency

We must remember that this theorem results maximum power transfer but not a maximum efficiency. If the load resistance is smaller than source resistance, power dissipated at the load is reduced while most of the power is dissipated at the source then the efficiency becomes lower.

Consider the total power delivered from source equation (equation 2), in which the power is dissipated in the equivalent Thevenin’s resistance R_{TH} by the voltage source V_{TH}.

Therefore, the efficiency under the condition of maximum power transfer is

Efficiency = Output / Input × 100

= I_{L}^{2} R_{L }/ 2 I_{L}^{2} R_{L }× 100

= 50 %

Hence, at the condition of maximum power transfer, the efficiency is 50%, that means a half percentage of generated power is delivered to the load and at other conditions small percentage of power is delivered to the load , as indicated in efficiency verses maximum power transfer the curves below.

For some applications, it is desirable to transfer maximum power to the load than achieving high efficiency such as in amplifiers and communication circuits.

On the other hand, it is desirable to achieve higher efficiency than maximised power transfer in case of power transmission systems where a large load resistance (much larger value than internal source resistance) is placed across the load. Even though the efficiency is high the power delivered will be less in those cases.

### Maximum Power Transfer Theorem for AC Circuits

It can be stated as in an active network, the maximum power is transferred to the load when the load impedance is equal to the complex conjugate of an equivalent impedance of a given network as viewed from the load terminals.

Consider the above Thevenin’s equivalent circuit across the load terminals in which the current flowing through the circuit is given as

I = V_{TH} / Z_{TH} + Z_{L}

Where Z_{L =} R_{L} + jX_{L}

Z_{TH} = R_{TH }+ jX_{TH}

Therefore, I = V_{TH} / (R_{L} + jX_{L} + R_{TH }+ jX_{TH })

= V_{TH} / ((R_{L}+ R_{TH}) + j(X_{L} + X_{TH }))

The power delivered to the load,

P_{L }= I^{2 }R_{L}

P_{L }= V^{2}_{TH ×} R_{L} / ((R_{L}+ R_{TH})^{2 }+ (X_{L} + X_{TH })^{2}) ……(1)

For maximum power the derivative of the above equation must be zero, after simplification we get

X_{L} + X_{TH }= 0

X_{L} = – X_{TH}

Putting the above relation in equation 1, we get

P_{L }= V^{2}_{TH ×} R_{L} / ((R_{L}+ R_{TH })^{2}

Again for maximum power transfer, derivation of above equation must be equal to zero, after simplification we get

R_{L}+ R_{TH }= 2 R_{L}

R_{L }= R_{TH}

Hence, the maximum power will transferred to the load from source, if RL = RTH and XL = – XTH in an AC circuit. This means that the load impedance should be equal to the complex conjugate of equivalent impedance of the circuit,

Z_{L} = Z_{TH}

Where Z_{TH} is the complex conjugate of the equivalent impedance of the circuit.

This maximum power transferred, P_{max} = V^{2}_{TH} / 4 R_{TH} or V^{2}_{TH}/ 4 R_{L}

### Applying Maximum Power Transfer Example to DC circuit

Consider the below circuit to which we determine the value of the load resistance that receives the maximum power from the supply source and the maximum power under the maximum power transfer condition.

1.Disconnect the load resistance from the load terminals a and b. To represent the given circuit as Thevenin’s equivalent, we are to determine the Thevenin’s voltage V_{TH} and Thevenin’s equivalent resistance R_{TH}.

The Thevenin’s voltage or voltage across the terminals ab is V_{ab} = V_{a} – V_{b}

Va = V × R2 / (R1 + R2)

= 30 × 20 /×(20 + 15)

= 17.14 V

Vb = V × R4/ (R3 + R4)

= 30 × 5 /(10 + 5)

= 10 V

Vab = 17.14 – 10

= 7.14 V

V_{TH} = Vab = 7.14 Volts

2.Calculate the Thevenin’s equivalent resistance R_{TH} by replacing sources with their internal resistances (here assume that voltage source has zero internal resistance so it becomes a short circuited).

Thevenin’s equivalent resistance or resistance across the terminals ab is

R_{TH} = Rab = [R1R2 / (R1 + R2)] + [R3R4 /(R3 + R4)]

= [(15 × 20) / (15 + 20)] + [(10 × 5) / (10+ 5)]

= 8.57 + 3.33

R_{TH} = 11.90 Ohms

3. The Thevenin’s equivalent circuit with above calculated values by reconnecting the load resistance is shown below.

From the maximum power transfer theorem, RL value must equal to the RTH to deliver the maximum power to the load.

Therefore, R_{L} = R_{TH}= 11.90 Ohms

And the maximum power transferred under this condition is,

Pmax = V^{2}_{TH} / 4 R_{TH}

= (7.14)^{2} / (4 × 11.90)

= 50.97 / 47.6

= 1.07 Watts

### Applying Maximum Power Transfer to AC circuit

The below AC network consists of load impedance ZL of which both reactive and resistive parts can be varied. Hence, we have to determine the load impedance value at which the maximum power delivered from the source and the value of maximum power.

To find the value of load impedance, first, we find the Thevenin’s equivalent circuit across the load terminals. For finding Thevenin’s voltage, disconnect the load impedance as shown in below figure.

By voltage divider rule, V_{TH} = 20∠0 × [j6 / (4 + j6)]

= 20∠0 ×[6∠90 / 7.21∠56.3]

= 20∠0 × 0.825∠33.7

V_{TH} = 16.5∠33.7 V

By shorting the voltage source, we calculate the Thevenin’s equivalent impedance of the circuit as shown in figure.

Therefore, Z_{TH} = (4 × j6) / (4 + j6)

= (4 × 6∠90) / (7.21∠56.3)

= 3.33∠33.7 0r 2.77 + j1.85 Ohms

Hence, the Thevenin’s equivalent circuit across the load terminals is shown in below.

Therefore to transfer the maximum power to the load, the value of the load impedance should be

Z_{L }= Z×_{TH}

= 2.77 – j1.85 ohms

= 2.77 – j1.85 ohms

The maximum power delivered, Pmax

= V^{2}_{TH} / 4 R_{TH}

= (16.5)^{2}/4(2.77)

= 272.25 / 11.08

= 24.5 W

### Practical Application of Maximum Power Transfer Theorem

Consider the practical example of a speaker with an impedance of 8 ohms is driven by audio amplifier with its internal impedance of 500 ohms. The Thevenin’s equivalent circuit is also shown in figure.

According to the maximum power transfer theorem, the power is maximized at the load if the load impedance is 500 ohms (same as internal impedance). Or else internal resistance has to be changed to 8 ohms to achieve the condition however it is not possible. So it is an impedance mismatch condition and it can be overcome by using an impedance matching transformer with its impedance transformation ratio of 500:8.

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