Simple Thevenin and Norton Equivalent AC Circuits

Thevenin’s and Norton’s theorem are applied to ac circuits in the same way as they are to dc circuit.

Thevenin’s and Norton’s Theorem for AC Circuit

The Norton equivalent circuit is depicted in Figure.(2), 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.

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.

Read also : equivalent impedance AC circuits

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.(3).

Figure 3

Solution :

We find Z_Th by setting the voltage source to zero. As shown in Figure.(4a), 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

Figure 4. Solution for Figure.(3) : (a) finding ZTh, (b) finding VTh

The Thevenin impedance is the series combination of Z₁ and Z₂; that is,

To find V_Th, consider the circuit in Figure.(4b). Currents I₁ and I₂ are obtained as

Applying KVL around the loop bcdeab in Figure.(4b) gives

or

2. Find the Thevenin equivalent of the circuit in Figure.(5) as seen from terminals a-b.

Figure 5

Solution :

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

or

Hence, the Thevenin voltage is

Figure 6. The solution of Figure.(5) : (a) finding VTh, (b) finding ZTh

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.(6b). At the node, KCL gives

Applying KVL to the outer loop in Figure.(6b) gives

The Thevenin impedance is 

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

Figure 7

Solution :

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

To get I_N, we short-circuit terminals a-b as in Figure.(8b) and apply mesh analysis.

Notice that meshes 2 and 3 form a supermesh because of the current source linking them. For mesh 1,

(3.1)

Figure 8. Solution of the circuit in Figure.(7) : (a) finding ZN, (b) finding VN, (c) calculating Io

For the supermesh,

(3.2)

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

(3.3)

Adding Equations.(3.1) and (3.2) gives

From Equation.(3.3),

The Norton current is

Figure.(8c) shows the Norton equivalent circuit along with the impedance at terminals a-b. By current division,