In a.c. network, the maximum power transfer theorem in AC circuit stated as follows:

*In a linear network having energy source and impedances, maximum amount of power is transferred from source to load impedance if the load impedance is the complex, conjugate of the total impedance of the network, i.e. if the source impedance is , to have maximum power transfer, the load impedance must be .*

To prove it, we assume a circuit (Figure 1) where load impedance is Z_{L} and source impedance is Z_{g}, connected in series with the source V_{g}. Let *I* be the current through the load.

Obviously,

Therefore, Power (real power) =

……..(i)

Let us first find the condition of maximum power flow from source to load when X_{L} is varied keeping X_{g} constant. This is possible when mathematically

However, from equation (i),

……..(ii)

Setting , we get from equation (ii),

.

Next, substituting on equation (i),

……..(iii)

It may be seen that P_{L} would attain maximum value provided R_{g} = R_{L}. Thus, maximum power transfer takes place in a.c. network provided R_{g} = R_{L} and X_{g} = X_{L} or in other words (R_{g} + jX_{g}) = (R_{L} -jX_{L}) i.e., Z_{g} = Z_{L}^{*}. This means that the load impedance is the complex conjugate of the source impedance. Consequently, the amount of maximum power transfer becomes , the efficiency being 50%.

In solving the problems, the internal impedance of the network across the load is to be determined as done for problems dealing with application of Thevenin’s theorem. The load impedance, for which power transfer becomes maximum is then the complex conjugate of the source impedance. Next, the open circuit voltage (V_{o.c}, the Thevenin’s Voltage) is determined across the open circuited terminals and amount of maximum power transfer is watts.