We are now going to solve for V0 in this net work by, applying Thevenin’s Theorem,…first…we break the network at the load and determine the open-circuit voltage…
Note that there exists only one closed path and the current in it must be 2∠0° Amps. Note also that there is no current in the inductor and therefore no voltage across it. Therefore the Thevenin voltage VTH is the voltage due to the voltage supply minus the voltage drop across the series impedance of 2 ohms minus J1Ω…or, +12 volts -2(2-j) due to the impedance voltage drop…which works out to 8 + j2 Volts
Calculating the Kirchhoff’s impedance ZTH… we zero the voltage source and open circuit the current source. Which is simply the three impedances in series (2-j) + j and that reduces to 2 + j1Ω.
The Thevenin equivalent impedance 2 + j1 Ω…so the Thevenin circuit…looks like this…
VTH = 8 + j2 V…and
ZTH = 2 + j1 Ω
The last step in the process is re-connecting the load…
Let's re-look at the original circuit… and remember that our task was to find the voltage V0 which is the voltage drop across the load.