Linear Circuit Analysis

1. Introduction
2. Basic Concepts
3. Simple Circuits
4. Nodal and Mesh Analysis
5. Additional Analysis Techniques
6. AC Analysis
7. Magnetically Coupled Circuits
8. Operational Amplifiers
9. Laplace Transforms
10. Time-Dependent Circuits
11. Two-port networks
Appendix

Alternative Current (AC) Methods

As disccussed in the previous section, most of the methods for the study of DC circuits can be used to study AC circuits (Ohm's law, current division, voltage division, nodal analysis, mesh analysis, superposition, Thévenin's' theorem, Norton's' theorem, and source transformation) if we perform the analysis in frequency (or complex) domain. We summarize these techniques below.

Ohm's Law in AC Circuits

Ohm's law can be extended to AC circuits, in which case the resistor is replaced with a frequency-dependent impedance. $$\begin{equation}V=ZI\end{equation}$$ or $$\begin{equation}I=\frac{V}{Z}\end{equation}$$

− + V I Z
Fig. 1. Applying Ohm's law to an impedance.

Current Division in AC Circuits

Current division can be extended to AC circuits, in which case the resistors are replaced with frequency-dependent impedances. In general, if we have $n$ impedances connected in parallel and the total current going through them is $I$, the current going through impedance $Z_i$ is equal to $$\begin{equation}I_i=I\frac{\frac{1}{Z_i}}{\frac{1}{Z_1}+\frac{1}{Z_2}+...+\frac{1}{Z_n}}\end{equation}$$

If we have only two impendances connected in parallel, the above equation gives $$\begin{equation}I_1=I\frac{Z_2}{Z_1+Z_2}\end{equation}$$ $$\begin{equation}I_2=I\frac{Z_1}{Z_1+Z_2}\end{equation}$$

Voltage Division in AC Circuits

Voltage division can be extended to AC circuits, in which case the resistors are replaced with frequency-dependent impedances. In general, if we have $n$ impedances connected in series and the total voltage across them is $I$, the voltage across impedance $Z_i$ is equal to $$\begin{equation}V_i=V\frac{Z_i}{Z_1+Z_2+...+Z_n}\end{equation}$$

Impedance Simplification in AC Circuits

The complex impedance of an two-port network containing resistors, inductors and capacitors can be computed in the same manner as the resistance of DC networks, provided that the the real values of the resistors are now replaced with complex values corresponding to each impedance. The same rules for the simplification of series and parallel connections that we learned for the resistive networks can be applied to calculation of the complex impedance.

Nodal and mesh analysis in AC circuits

The same algorithm that we used for the calculation of potentials and currents in DC circuits can now be used to compute the complex values of the potentials and currents in AC circuits. Please look at the examples provided in the Sample Solved Problems.

Superposition in AC circuits

The superposition method can also be applied for the calculation of complex potentials and currents in AC circuits. Please look at the examples provided in the Sample Solved Problems.

Source transformation, Norton and Thévenin equivalent AC circuits

The Norton and Thévenin theorems remaing valid for the calculation of complex potentials and currents in AC circuits. Please look at the examples provided in the Sample Solved Problems.

Table 1 shows possible ways to compute the Norton and Thévenin components.

Table 1. How to compute $V_{Th}$, $I_{N}$, $Z_{Th}$ and $Z_{N}$ based on the elements that exist in the AC circuit.
If the circuits contains only... You should...
Resistors, inductors and capacitors
  1. Compute $Z_{Th}$ and $Z_{N}$ using (complex) impedance simplification techniques. Alternatively, you can can use the $V_{test}/I_{test}$ method.1)
  2. $I_{N} = 0$ and $V_{Th}=0$
Resistors, inductors, capacitors and independent sources
  1. Deactivate all the independent sources in the circuit2) and comptue $Z_{Th}$ and $Z_{N}$ using impedance simplification techniques or the $V_{test}/I_{test}$ method.1)
  2. Compute the short-circuit current $I_{sc}$, which gives you $I_{N} = I_{sc}$. Alternatively, if the Norton resistance is not zero, you can compute the open-circuit voltage $V_{oc}$, and set $I_{N} = V_{oc}/Z_{N} = V_{oc}/Z_{Th}$
Resistors, inductors, capacitors and dependent sources
  1. Compute $Z_{Th}$ and $Z_{N}$ using the $V_{test}/I_{test}$ method.1)
  2. $I_{N} = 0$ and $V_{Th}=0$
Resistors, inductors, capacitors and independent and dependent sources
  1. Deactivate all the independent sources in the circuit2) and compute $Z_{Th}$ and $Z_{N}$ using the $V_{test}/I_{test}$ method.1)
  2. Compute the short-circuit current $I_{sc}$, which gives you $I_{N} = I_{sc}$. Alternatively, if the Norton resistance is not zero, you can compute the open-circuit voltage $V_{oc}$, and set $I_{N} = V_{oc}/Z_{N} = V_{oc}/Z_{Th}$

1) You can use either the test voltage or test current method. They should both give you the same result.
2) To deactivate the independent sources in the circuit, you need to replace all the independent voltage sources with short circuits (wires) and all the independent current sources with open circuits (remove them). Make sure you do not modify the dependent sources.

Sample Solved Problems
The examples below are randomly generated.
See also