Notes › ENGR 2334: Electric Circuits I Lecture 4

Methods of Analysis of Resistive Circuits

calendar_today February 6, 2025 article 587 words access_time 4 min replay Modified April 3, 2025

Table of Contents

Nodal Analysis

A method to obtain element and node voltages in a circuit
One node is the reference node, which supplies GND and is taken to be 0 V
For $n$ nodes, there are $n-1$ nodal equations

Procedure

Choose a good reference node location
Draw the nodes for the locations you want to find
Express the voltage of each circuit element in terms of two node voltages
- Which is just $\Delta v = v_{1} - v_{2}$, assuming the circuit element lies between $v_1$ and $v_2$
Rewrite this expression for current ($i = GV$ or $i = \frac{\Delta v}{R}$) using Ohm’s Law
Apply Current Law to each node (excluding reference node)
Solve the system of linear equations for each node voltage via substitution

Example

Independent Voltage Sources

Adding a voltage source adds a voltage constraint to the system (one less KCL equation needed)
Nodes directly in front of a voltage source have the same nodal voltage as that voltage source

Supernodes

If there exists a voltage source between two nodes, then they combine to form a supernode
The supernode may be analyzed with KCL like normal reference nodes
- However, the sum of the currents entering both reference nodes must equal the sum of the currents exiting both reference nodes

Example

Voltage Source Equation

Recall that voltages in series can be added together
Therefore, the final voltage in a series of unknown voltages will equal the sum of voltages that came become it.
- This is generalizable for any voltage source between two reference nodes; in the diagram below, $V_1 = V_2 + V_A$:
The resulting equation is known as the constraint equation

Dependent Voltage Sources

When a dependent (or controlled) current or voltage source is present, an additional equation is needed to express the controlling variable in terms of node voltages
1. If the dependent source is a current source, include it directly in the current matrix ($\mathbf{i}$)
2. If the dependent source is a voltage source, it either:
  - Creates a supernode, requiring an extra equation
  - Adds a constraint equation relating node voltages

Example

Matrix Method

It is possible to solve for voltages directly by inspection using a conductance matrix
For $n$ nodes, you will have an $n \times n$ symmetrical conductance matrix that is laid as follows:

$$ {\begin{pmatrix}G_{11}&G_{12}&\cdots &G_{1N}\\G_{21}&G_{22}&\cdots &G_{2N}\\\vdots &\vdots &\ddots &\vdots \\G_{N1}&G_{N2}&\cdots &G_{NN}\end{pmatrix}}{\begin{pmatrix}v_{1}\\v_{2}\\\vdots \\v_{N}\end{pmatrix}}={\begin{pmatrix}i_{1}\\i_{2}\\\vdots \\i_{N}\end{pmatrix}} $$

or simply $\mathbf {Gv} =\mathbf {i}$, where: - G is the conductance matrix, constructed as: - Diagonal elements $G_{kk}$ are the sum of conductances connected to node $k$ - Off-diagonal elements $G_{kj}$ are the negative conductance between nodes $k$ and $j$ - $\mathbf{v}$ is the vector of unknown node voltages - $\mathbf{i}$ is the vector of known current injections at each node

Thus, the final solution is $\mathbf{v}=\mathbf{G}^{-1}\mathbf{i}$
Cannot be used when there is a voltage constraint or supernode; therefore, only good when there are independent current sources