Nodal analysis

From Wikiversity

Jump to: navigation, search
Wikiversity Electrical Engineering School
The Lessons in
ELECTRIC CIRCUITS ANALYSIS COURSE

Lesson Review 5 & 6:

What you need to remember from Kirchhoff's Voltage & Current Law . If you ever feel lost, do not be shy to go back to the previous lesson & go through it again. You can learn by repitition.

  • Remember what was learned in Passive sign convention, You can go back and revise Lesson 1.
  • Define Kirchhoff's Voltage Law ( word-by-word ).
  • Kirchhoff's Voltage Law:
    vn = v4 + v1 + v2 + v3 = 0
    n
  • Define Kirchhoff's Current Law ( word-by-word ).
  • Kirchhoff's Current Law:
    in = i1 + i2 + i3 + i4 = 0
    n

This part of the course onwards will collaborate with the Mathematics Department extensively. Mathematical Theory will be kept minimal as mathematical tools are only used here as a means to an end. Links to relevant Mathematical theories will be supplied to assit the student.

Lessons in Electric Circuit Analysis
Lesson #1:
Passive sign convention
Lesson #2:
Simple Resistive Circuits
Lesson #3:
Resistors in Series
Lesson #4:
Resistors in Parallel
Quiz Test:
Circuit Analysis Quiz 1
Lesson #5:
Kirchhoff's Voltage Law
Lesson #6:
Kirchhoff's Current Law
Lesson #6:
Nodal analysis← You are here
Lesson #6:
Mesh Analysis
Quiz Test:
Circuit Analysis Quiz 2
Home Laboratory:
Circuit Analysis - Lab1

Lesson 7: Preview

This Lesson is about Kirchhoff's Current Law. The student/User is expected to understand the following at the end of the lesson.

  • Use KCL at super nodes to formulate circuit equations.
  • Create matrix from circuit equations.
  • Solve for Unknown Node Voltages using Kramers Rule.

Part 1: Pre-reading Material

The student is advised to read the following resources from the Mathematics department:

The following external link has an excellent summary on using Kramer's rule to solve linear equations:

*Solutions/kramer

After you have satified yourself of the above resources, you can go to Part 2.

Part 2: Nodal analysis

Let's start off with some useful definitions:

  • Node:
A point in a circuit where terminals of atleast two electric components meet. This point can be on any wire, it is infinitely small and dimensionless.
  • Major Node:
This point is a node. A set of these nodes is used to create constraint equations.
  • Reference Node:
The node to which Voltages of other nodes is read with regard to. This can be seen as ground ( V = 0).
  • Branch:
This is a circuit element(s) that connect two nodes.

Basic rule: The sum of the currents entering any point (Node) must equal the sum of the currents leaving.( From KCL in Lecture 6).

Part 3

The following is a general procedure for using Nodal Analysis method to solve electric circuit problems. The aim of this algorithm is to develop a matrix system from equations found by applying KCL at the major nodes in an electric circuit. Kramer's rule is then used to solve the unkown major node voltages.


Once the Node voltages are solved, normal circuit analysis methods ( Ohm's law; Voltage and Current Divider principles etc... ) can then be used to find whatever circuit entity. Remember to consult previous lessons if you are not confident in using normal circuit analysis techniques that will be used in this lesson.


Manual Nodal Analysis Algorithm:

1.) Choose a reference node. ( Rule of thumb: take Node with most branches connecting to it )

2.) Identify and Number major nodes. ( Usually 2 or 3 major Nodes )

3.) Apply KCL to identified major nodes and formulate ciruit equations.

4.) Create Matrix system from KCL equations obtained.

5.) Solve Matrix for unknown node voltages by using Kramer's rule ( It is simpler although you can still use gaussian method as well )

6.) Used solved Node voltages to solve for the desired circuit entity.


The above algorithm is very basic and usefull for 2 x 2 and 3 x 3 size matrices. Generally as the number of major node voltages increase and the size of the matrix exceeds 3 x 3, numerical methods ( Beyond scope of this course ) are employed with the aid of computers to solve such circuit networks.


Let's try an example to illustrate the above nodal analysis algorithm.

Part 4 : Example

Figure 7.1: Example 1

Consider Figure 7.1 with the following Parameters:

V1 = 15V
V2 = 7V
R1 = 2Ω
R2 = 20Ω
R3 = 10Ω
R4 = 5Ω
R5 = 2Ω
R6 = 2Ω

Find current through R3 using Nodal Analysis method.

Solution:

Figure 7.2: Voltages at nodes

This is the same example we solved in Exercise 6, except that in this case we have added extra Resistors to increase the complexity of the circuit.

Figure 7.2 shows Voltages at Nodes a, b, c and d.

We use node a as common node ( ground if you like ). thus Va = 0V as we did previously.


Follow Carefully the construction of the Nodal analysis algorith explained in part 3 of this lesson as follows in Part 5 and 6.

Part 5 : Example (Continued)

Now that we have labelled the currents flowing in this circuit using passive sign convention, and have identified Nodes b; c and d as major nodes, we proceed as follows:

KCL @ Node b:

i1 = i2 + i6

Thus by applying Ohms law to above equation we get.

\begin{matrix}\ \frac{V_1 - V_b}{R_1} =  \frac{V_b - V_c}{R_2} + \frac{V_b - V_d}{R_6} \end{matrix}  

Therefore

\begin{matrix}\ \ V_b(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_6}) - V_c(\frac{1}{R_2}) - V_d(\frac{1}{R_6}) = \frac{V_1}{R_1} \end{matrix}   ...............   (1)


KCL @ Node c:

i3 = i2 + i4

Thus by applying Ohms law to above equation we get.

Part 6 : Example (Continued)

\begin{matrix}\ \frac{V_c}{R_3} =  \frac{V_b - V_c}{R_2} + \frac{V_d - V_c}{R_4} \end{matrix}  

Therefore

\begin{matrix}\ \ V_b(- \frac{1}{R_2}) + V_c(\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}) + V_d(- \frac{1}{R_4}) = 0 \end{matrix}   ...............   (2)


KCL @ Node d:

i4 = i5 + i6

Thus by applying Ohms law to above equation we get.

\begin{matrix}\ \frac{V_d - V_c}{R_4} =  \frac{V_b - V_d}{R_6} + \frac{V_2 - V_d}{R_5} \end{matrix}  

Therefore

\begin{matrix}\ \ V_b(\frac{1}{R_6}) + V_c(\frac{1}{R_4}) + V_d(-\frac{1}{R_4}-\frac{1}{R_5}-\frac{1}{R_6}) = \frac{V_2}{R_5} \end{matrix}   ...............   (3)

Part 7:

The next step in this algorithm is to construct a matrix. Inorder to do that easily we substitute all resistances in above equations 1; 2 & 3 with their equivalent Admittances as follows:

G_1 = \frac{1}{R_1} etc thus equations 1; 2 & 3 will be re-written as follows:

\begin{matrix}\ V_b(G_1+G_2+G_6) + V_c(-G_2) + V_d(-G_6) &=& (V_1 \times G_1) ....(1) \\ \ \\ \ V_b(-G_2) + V_c(G_2+G_3+G_4) + V_d(-G_4) &=& 0 ....(2) \\ \ \\ \ V_b(G_6) + V_c(G_4) + V_d(-G_4-G_5-G_6) &=& (V_2 \times G_5) ....(3) \end{matrix}

Now we can create a matrix with the above equations as follows:

 \begin{bmatrix} (G_1+G_2+G_6) & -(G_2) & -(G_6) \\ (-G_2) & (G_2+G_3+G_4) & (-G_4) \\ (G_6) & (G_4) & -(G_4+G_5+G_6)\end{bmatrix} . \begin{bmatrix} V_b\\ V_c \\ V_d \end{bmatrix} = \begin{bmatrix} (V_1 \times G_1)\\ 0  \\ (V_2 \times G_5)\end{bmatrix}

The following matrix is the above with values substituted:

 A.\vec{X} = \vec{Y}\begin{bmatrix} 1.05 & -0.05 & -0.5 \\ -0.05 & 0.35 & -0.2 \\ 0.5 & 0.2 & -1.2\end{bmatrix} . \begin{bmatrix} V_b\\ V_c \\ V_d \end{bmatrix} = \begin{bmatrix} 7.5 \\ 0  \\ 3.5 \end{bmatrix}


Now that we have arranged equations 1; 2 & 3 into a matrix we need to get Determinants of the General matrix, and Determinants of alterations of the general matrix as follows:

Part 8:

Solving determinants of:

  • Matrix A  : General matrix A from KCL equations
  • Matrix A1 : Genral Matrix A with Column 1 substituted by \vec Y.
  • Matrix A2 : Genral Matrix A with Column 2 substituted by \vec Y.
  • Matrix A3 : Genral Matrix A with Column 3 substituted by \vec Y.

As follows:

\begin{matrix}\ det\begin{bmatrix} 1.05 & -0.05 & -0.5 \\ -0.05 & 0.35 & -0.2 \\ 0.5 & 0.2 & -1.2\end{bmatrix} &=& det A \\ \ \\ \  &=& -0.3\end{matrix}


\begin{matrix}\ det\begin{bmatrix} 7.5 & -0.05 & -0.5 \\ 0 & 0.35 & -0.2 \\ 3.5 & 0.2 & -1.2\end{bmatrix} &=& det A1 \\ \ \\ \ &=& -2.203\end{matrix}
\begin{matrix}\ det\begin{bmatrix} 1.05 & 7.5 & -0.5 \\ -0.05 & 0 & -.02 \\ 0.5 & 3.5 & -1.2\end{bmatrix} &=& det A2 \\ \ \\ \ &=& -0.378 \end{matrix}

Part 9:

\begin{matrix}\ det\begin{bmatrix} 1.05 & -0.05 & 7.5 \\ -0.05 & 0.35 & 0 \\ 0.5 & 0.2 & 3.5\end{bmatrix} &=& det A3 \\ \ \\ \ &=& -0.11\end{matrix}


Now we can use the solved determinants to arrive at solutions for Node voltages Vb;VcandVd as follows:


1. V_b = \frac{det A1}{det A} = \frac{-2.203}{-0.3} = 7.378559V

2. V_c = \frac{det A2}{det A} = \frac{-0.378 }{-0.3} = 1.264657V

3. V_b = \frac{det A3}{det A} = \frac{-0.11}{-0.3} = 0.368509V

Now we can apply Ohm's law to solve for the current through R3 as follows:

I_{R_3} = \frac{V_c}{R_3} = \frac{1.264657V}{10\Omega} = +0.126466A 

As we have seen previously, the positive sign in the above current tells us that the effective current flowing through R3 is in fact in the direction we chose when drawing up the circuit in figure 7.2.

Please use the provided link for details on working out the determinant of a 3 x 3 Matrix.

To apreciate the algorithm we have just used, try solving the above problem using either KVL or KCL as we did in lessons 5 & 6 and see just how cumbersome the process would be.

As usual the following part is an Exercise to test your self on content discussed in this lesson. Please look at Part 11 for further reading material and interesting related External links.

Part 10:Exercise 7

Figure 7.3: Exercise

Consider Figure 7.3 with the following Parameters:

V1 = 9V
R1 = 200Ω
R2 = 20Ω
R3 = 10kΩ
R4 = 5kΩ
R5 = 15kΩ
R6 = 1kΩ

Find current through R3 using Nodal Analysis method.


Part 11:

Further Reading & Other Interesting Links:

References:

  • Add reference here!

Completion List

Once you finish your Exercises you can post your score here! To post your score just e-mail your course co-ordinator your name and score *Click Here.

previous lesson
previous page
next page
lesson intro
next lesson
course menu
Resource type: this resource contains a lecture or lecture notes.
Personal tools