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

**Contents**show

*Make sure to read what is ac circuit first.*

- Node and supernode for ac circuit
- Mesh and supermesh for ac circuit
- Superposition for ac circuit
- Source transformation for ac circuit
- Thevenin and Norton for ac circuit

## Thevenin’s and Norton’s Theorem for AC Circuit

The frequency-domain version of a Thevenin equivalent circuit is drawn in Figure.(1), where a linear circuit is replaced by a voltage source in series with an impedance.

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.

## 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 *-j*6 reactance, so that their combination gives

Figure 4. Solution for Figure.(3) : (a) finding Z_{Th}, (b) finding V_{Th} |

The Thevenin impedance is the series combination of **Z**_{1} and **Z**_{2}; that is,

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

**V**

_{Th}, we apply KCL at node 1 in Figure.(6a).

Figure 6. The solution of Figure.(5) : (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.(6b). At the node, KCL gives

**3. Obtain current I _{o} 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 – *j*2) and (10 + *j*4) 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 Z_{N}, (b) finding V_{N}, (c) calculating I_{o} |

For the supermesh,

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