We have solved the problem of maximizing the power delivered by a power-supplying resistive network to a load R

Consider the circuit in Figure.(1), where an ac circuit is connected to a load

The load is usually represented by an impedance, which may model an electric motor, an antenna, and so forth. In rectangular form, the Thevenin impedance

The current through the load is

The average power delivered to the load is

Our objective is to adjust the load parameters

Setting ∂P/∂X

and setting ∂P/∂R

Combining Equations.(5) and (6) leads to the conclusion that for maximum average power transfer,

In a situation in which the load is purely real, the condition for maximum power transfer is obtained from Equation.(6) by setting X

This means that for maximum average power transfer to a purely resistive load, the load impedance (or resistance) is equal to the magnitude of the Thevenin impedance.

First we obtain the Thevenin equivalent at the load terminals. To get

To find

The load impedance draws the maximum power from the circuit when

According to Equation.(8), the maximum average power is

We first find the Thevenin equivalent at the terminals of R

By voltage division,

The value of R

The current through the load is

The maximum average power absorbed by R

_{L}. Representing the circuit in Thevenin equivalent circuit, we proved that the maximum power would be delivered to the load if the load resistance is equal to the Thevenin resistance R_{L}= R_{Th}. We now extend that result to ac circuit.Consider the circuit in Figure.(1), where an ac circuit is connected to a load

**Z**and is represented by its Thevenin equivalent._{L}Figure 1. Finding the maximum average power transfer : (a) circuit with a load, (b) the Thevenin equivalent |

**Z**and the load impedance_{Th}**Z**are_{L}(1a) |

(1b) |

(2) |

(3) |

**R**and_{L}**X**so that_{L}*P*is maximum. To do this we set ∂P/∂R_{L}and ∂P/∂X_{L}equal to zero. From Equation.(3), we obtain(4a) |

(4b) |

_{L}to zero gives(5) |

_{L}to zero results(6) |

**Z**must be selected so that X_{L}_{L}= -X_{Th}and R_{L}= R_{Th}, i.e,.(7) |

ForThis result is known as themaximum average power transfer, the load impedanceZmust be equal to the complex conjugate of the Thevenin impedance_{L}Z._{Th}

*maximum average power transfer theorem*for the sinusoidal steady state. Setting R_{L}= R_{Th}and X_{L}= -X_{Th}in Equation.(3) gives us the maximum average power as(8) |

_{L}= 0; that is,(9) |

## Maximum Average Power Transfer Examples

For better understanding let us review examples below :**1. Determine the load impedance****Z**_{L}**that maximizes the average power drawn from the circuit of Figure.(2). What is the maximum average power?**Figure 2 |

*Solution :*First we obtain the Thevenin equivalent at the load terminals. To get

**Z**, consider the circuit shown in Figure.(3a). We find_{Th}Figure 3. Finding the Thevenin equivalent of the circuit in Figure.(2) |

**V**, consider the circuit in Figure.(3b). By voltage division,_{Th}The load impedance draws the maximum power from the circuit when

According to Equation.(8), the maximum average power is

**2. In the circuit in Figure.(4), find the value of R**_{L}that will absorb the maximum average power. Calculate that power.Figure 4 |

*Solution :*We first find the Thevenin equivalent at the terminals of R

_{L}.By voltage division,

The value of R

_{L}that will absorb the maximum average power isThe current through the load is

The maximum average power absorbed by R

_{L}is
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