The Thevenin impedance of a source is ZTh= 120+j60 ohms , while the peak Thevenin voltage is VTh=165+j0V. Determine the maximum available average power from the source

Question

The Thevenin impedance of a source is ZTh= 120+j60 ohms , while the peak Thevenin voltage is VTh=165+j0V. Determine the maximum available average power from the source.

Answer 1

... maximum power is transferred when the load impedance is made to be the complex conjugate of the source impedance. The source impedance is 120+j60 thus the load impedance must be 120-j60 for max power transfer. The current is then (165/240) = .6875 amps and the power into the load is 120*(.6875)^2=56.7 watts