Thevenin and Norton Equivalent Circuits AC circuit

Thevenin and Norton Equivalent Circuits

Thevenin's and Norton's theorems are applied to ac circuits in the same way as they are to dc circuits. The only additional effort arises from the need to manipulate complex numbers. The frequency domain version of a thevenin equivalent circuit is depicted where a linear circuit is replace by a voltage source in series with an impedance. The norton equivalent is a linear circuit replaced by a current source in parallel with an impedance. Two equivalent circuits are related as

The thevenin or Norton equivalent circuit must be determined at each frequency. This leads to entirely different equivalent circuits, one for each frequency, not one equivalent circuit with equivalent sources and equivalent impedances.

Example:

Obtain the thevenin equivalent at terminals a-b of the circuit 

We must find Zth by setting the voltage source to zero. As shown in the figure below, the 8ohm resistance is now in parallel with the -j6 reactance, so that their combination gives

Z1=-j6ll8=-j6(8)/8-j6=2.88-j3.84ohms

Similarly, the 4ohm resistance is in parallel with the j12 reactance and their combination gives

z2=4llj12=j12(4)/4+j12=3.6+j1.2ohms

The thevenin impedance is the series combination of Z1 and Z2 that is:

Zth=Z1+Z2=6.48-j2.64ohms

To find Vth consider the circuit in the figure below. Currents I1 and I2 are obtained as

I1=120angle75/8-j6 A

I2=120angle75/4+j12 A

Applying KVL around loob gives

Vth-4I2+(-j6)I1=0

or

Vth=4I2+j6I10=(480angle75)/4+j12)+(720angle75+90)/8-j6

=37.95angle3.43+72angle201.87

Vth=37.95angle220.31 V

Example:

Obtain current Io using Norton's Theorem

Our first objective is to find the Norton equivalent at terminals a-b. Zn is found in the same way as Zth. We set the sources to zero as shown in the figure above. As evident from the figure, the (8-j2) and (10+j4) impedances are short-circuited, so that 

Zn=5ohms

To get In, we short-circuit terminals a-b as in figure below and apply mesh analysis. Notice that meshes 2 and 3 form a supermesh because of the current source linking them. 

For Mesh 1,

-j40+(18+j2)I1-(8-j2)I2-(10+j4)I3=0

(a) finding Zn,   (b) finding Vn,  (c) Calculating Io

For the supermesh,

(13-j2)I2+(10+j4)I3-(18+j2)I1=0

at node a, due to the current source between meshes 2 and 3,

I3=I2+3

Adding Equations gives

-j40+5I2=0  => I2=j8

From eq where 

I3=I2+3=3+j8

The norton current is 

IN=I3=(3+j8) A

Io= 5/5+20+j15 IN = 3+j8/5+j3=1.465angle 38.48 A