Thevenin & Norton for AC Circuits | GATE Notes & Videos for Electrical Engineering - Electrical Engineering (EE) PDF Download

The frequency-domain version of a Thevenin equivalent circuit is drawn in Figure. where a linear circuit is replaced by a voltage source in series with an impedance.
The Norton equivalent circuit is depicted in Figure. where a linear circuit is replaced by a current source in parallel with an impedance.
Figure 1. Thevenin equivalent 

Figure 2. Norton equivalent

Keep in mind that the two equivalent circuits are related as just as in source transformation, V_Th is the open-circuit voltage while I_N is the short circuit current.

V_Th = Z_NI_N,        Z_Th = Z_n
If the circuit has sources operating at different frequencies (will be shown in the example below), the Thevenin or Norton equivalent circuit has to 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.

We will not cover the ‘step-by-step’ using these methods, make sure you learn it from the previous posts about Thevenin’s Theorem and Norton’s Theorem for dc circuit.

It is not that different from the ac circuit.

Thevenin and Norton Equivalent AC Circuit Examples

For a better understanding let us review the examples below.
1. Obtain the Thevenin equivalent at terminals a-b of the circuit in Figure.

We find Z_Th by setting the voltage source to zero. As shown in Figure. the 8 Ω resistance is now in parallel with -j6 reactance, so that their combination gives

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

Solution for Figure.: (a) finding Z_Th, (b) finding V_Th

The Thevenin impedance is the series combination of Z₁ and Z₂; that is,
Z_Th = Z₁ + Z₂ = 6.48 - j2.64Ω
To find V_Th, consider the circuit in Figure. Currents I₁ and I₂ are obtained as

Applying KVL around the loop bcdeab in Figure gives

or

2. Find the Thevenin equivalent of the circuit in Figure. as seen from terminals a-b.

To find V_Th, we apply KCL at node 1 in Figure.(6a).

Applying KVL to the loop on the right-hand side in Figure.(6a), we obtain

or 

Hence, the Thevenin voltage is
The solution of Figure.(a) finding V_Th, (b) finding Z_Th

To obtain Z_Th, we remove the independent source.
Due to the presence of the dependent current source, we connect a 3 A current source (3 is an arbitrary value chosen for convenience here, a number divisible by the sum of currents leaving the node) to terminals a-b as shown in Figure. At the node, KCL gives

Applying KVL to the outer loop in Figure. gives

The Thevenin impedance is

3. Obtain current Io in Figure.(7) using Norton’s theorem.

Our first objective is to find the Norton equivalent at terminals a-b. Z_N is found in the same way as Z_Th.
We set the sources to zero as shown in Figure.(8a). As evident from the figure, the (8 – j2) and (10 + j4) impedances are short-circuited, so that
Z_N = 5Ω
To get I_N, we short-circuit terminals a-b as in Figure. and apply mesh analysis.
Notice that meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1,

Solution of the circuit in Figure: (a) finding Z_N, (b) finding V_N, (c) calculating I_o

For the supermesh,
 ... (3.2)

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

I₃ = I₂ + 3 ......(3.3)

Adding Equations.(3.1) and (3.2) gives
From Equation.(3.3),

The Norton current is

Figure.shows the Norton equivalent circuit along with the impedance at terminals a-b. By current division,