Electric Circuit Analysis/Nodal Analysis

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 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: ${\displaystyle \sum _{n}v_{n}=v_{4}+v_{1}+v_{2}+v_{3}=0}$
• Define Kirchhoff's Current Law ( word-by-word ).
• Kirchhoff's Current Law: ${\displaystyle \sum _{n}i_{n}=i_{1}+i_{2}+i_{3}+i_{4}=0}$

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 assist the student.

Lessons in Electric Circuit Analysis
Lesson #1:
Lesson #2:
Lesson #3:
Lesson #4:
Quiz Test:
Lesson #5:
Lesson #6:
Lesson #7:
 Nodal Analysis← You are here
Lesson #8:
Quiz Test:
Home Laboratory:

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 Cramer's Rule.

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

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

*Solutions/Cramer

After you have satisfied yourself with 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. Cramer'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 is required.

Remember to consult previous lessons if you are not confident using the circuit analysis techniques that will be employed in this lesson.

Manual Nodal Analysis Algorithm:

1.) Choose a reference node. ( Rule of thumb: take the 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 circuit equations.

4.) Create matrix system from KCL equations obtained.

5.) Solve matrix for unknown node voltages by using Cramer's Rule (Cramer's Rule is simpler, although you can still use the Gaussian method)

6.) Use solved node voltages to solve for the desired circuit entity.

The above algorithm is very basic and useful for 2 x 2 and 3 x 3 size matrices. Generally as the number of major node voltages increase and the order of the matrix exceeds 3, numerical methods (beyond the scope of this course) are employed, sometimes 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:

${\displaystyle V_{1}=15V}$
${\displaystyle V_{2}=7V}$
${\displaystyle R_{1}=2\Omega }$
${\displaystyle R_{2}=20\Omega }$
${\displaystyle R_{3}=10\Omega }$
${\displaystyle R_{4}=5\Omega }$
${\displaystyle R_{5}=2\Omega }$
${\displaystyle R_{6}=2\Omega }$

Find current through ${\displaystyle R_{3}}$ 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 some 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 the common node (ground if you like). Thus ${\displaystyle V_{a}=0V}$ as we did previously.

Carefully follow the progression of the nodal analysis algorithm 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 the Passive Sign Convention, and have identified nodes b; c and d as major nodes, we proceed as follows:

KCL at node b:

${\displaystyle i_{1}=i_{2}+i_{6}}$

Thus, by applying Ohm's Law to the above equation we get:

${\displaystyle {\begin{matrix}\ {\frac {V_{1}-V_{b}}{R_{1}}}={\frac {V_{b}-V_{c}}{R_{2}}}+{\frac {V_{b}-V_{d}}{R_{6}}}\end{matrix}}}$

Therefore

${\displaystyle {\begin{matrix}\ \ V_{b}({\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+{\frac {1}{R_{6}}})-V_{c}({\frac {1}{R_{2}}})-V_{d}({\frac {1}{R_{6}}})={\frac {V_{1}}{R_{1}}}\end{matrix}}}$ ............... (1)

KCL at node c:

${\displaystyle i_{3}=i_{2}+i_{4}}$

Then, by applying Ohm's Law to the above equation we get:

Part 6 : Example (Continued)

${\displaystyle {\begin{matrix}\ {\frac {V_{c}}{R_{3}}}={\frac {V_{b}-V_{c}}{R_{2}}}+{\frac {V_{d}-V_{c}}{R_{4}}}\end{matrix}}}$

Therefore

${\displaystyle {\begin{matrix}\ \ V_{b}(-{\frac {1}{R_{2}}})+V_{c}({\frac {1}{R_{2}}}+{\frac {1}{R_{3}}}+{\frac {1}{R_{4}}})+V_{d}(-{\frac {1}{R_{4}}})=0\end{matrix}}}$ ............... (2)

KCL at node d:

${\displaystyle i_{4}=i_{5}+i_{6}}$

Applying Ohm's Law...

${\displaystyle {\begin{matrix}\ {\frac {V_{d}-V_{c}}{R_{4}}}={\frac {V_{2}-V_{d}}{R_{5}}}+{\frac {V_{b}-V_{d}}{R_{6}}}\end{matrix}}}$

Therefore

${\displaystyle {\begin{matrix}\ V_{b}(-{\frac {1}{R_{6}}})-V_{c}({\frac {1}{R_{4}}})+V_{d}({\frac {1}{R_{4}}}+{\frac {1}{R_{5}}}+{\frac {1}{R_{6}}})={\frac {V_{2}}{R_{5}}}\end{matrix}}}$ ............... (3)

Part 7:

The next step in this algorithm is to construct a matrix. To do that easily, we substitute all resistances in the above equations 1, 2 and 3 with their equivalent admittances - as follows:

${\displaystyle G_{1}={\frac {1}{R_{1}}}}$ etc thus equations 1, 2 and 3 will be re-written as follows:

${\displaystyle {\begin{matrix}\ V_{b}(G_{1}+G_{2}+G_{6})+V_{c}(-G_{2})+V_{d}(-G_{6})&=&(V_{1}\times G_{1})....(1)\\\ \\\ V_{b}(-G_{2})+V_{c}(G_{2}+G_{3}+G_{4})+V_{d}(-G_{4})&=&0....(2)\\\ \\\ V_{b}(-G_{6})+V_{c}(-G_{4})+V_{d}(G_{4}+G_{5}+G_{6})&=&(V_{2}\times G_{5})....(3)\end{matrix}}}$

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

${\displaystyle {\begin{bmatrix}(G_{1}+G_{2}+G_{6})&-(G_{2})&-(G_{6})\\(-G_{2})&(G_{2}+G_{3}+G_{4})&(-G_{4})\\(-G_{6})&(-G_{4})&(G_{4}+G_{5}+G_{6})\end{bmatrix}}.{\begin{bmatrix}V_{b}\\V_{c}\\V_{d}\end{bmatrix}}={\begin{bmatrix}(V_{1}\times G_{1})\\0\\(V_{2}\times G_{5})\end{bmatrix}}}$

The following matrix is the above with values substituted:

${\displaystyle A.{\vec {X}}={\vec {Y}}}$${\displaystyle {\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 and 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:

Remember to read *Solutions/Cramer Also, use the provided link for details on working out the determinant of a 3 x 3 Matrix.

Solving determinants of:

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

As follows:

${\displaystyle {\begin{matrix}\ det{\begin{bmatrix}1.05&-0.05&-0.5\\-0.05&0.35&-0.2\\-0.5&-0.2&1.2\end{bmatrix}}&=&detA\\\ \\\ &=&0.299\end{matrix}}}$

${\displaystyle {\begin{matrix}\ det{\begin{bmatrix}7.5&-0.05&-0.5\\0&0.35&-0.2\\3.5&-0.2&1.2\end{bmatrix}}&=&detA1\\\ \\\ &=&3.498\end{matrix}}}$

${\displaystyle {\begin{matrix}\ det{\begin{bmatrix}1.05&7.5&-0.5\\-0.05&0&-.02\\-0.5&3.5&1.2\end{bmatrix}}&=&detA2\\\ \\\ &=&2.023\end{matrix}}}$

Part 9:

${\displaystyle {\begin{matrix}\ det{\begin{bmatrix}1.05&-0.05&7.5\\-0.05&0.35&0\\-0.5&-0.2&3.5\end{bmatrix}}&=&detA3\\\ \\\ &=&2.665\end{matrix}}}$

Now we can use the solved determinants to arrive at solutions for Node voltages ${\displaystyle V_{b};V_{c}andV_{d}}$ as follows:

1. ${\displaystyle V_{b}={\frac {detA1}{detA}}={\frac {3.498}{0.299}}=11.717V}$

2. ${\displaystyle V_{c}={\frac {detA2}{detA}}={\frac {2.023}{0.299}}=6.776V}$

3. ${\displaystyle V_{d}={\frac {detA3}{detA}}={\frac {2.665}{0.299}}=8.928V}$

Now we can apply Ohm's law to solve for the current through ${\displaystyle R_{3}}$ as follows:

${\displaystyle I_{R_{3}}={\frac {V_{c}}{R_{3}}}={\frac {6.776V}{10\Omega }}=+0.678A}$

As we previously saw, the positive sign in the above current tells us that the effective current flowing through ${\displaystyle R_{3}}$ is in fact in the direction we chose when drawing up the circuit in figure 7.2.

To appreciate 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 the content discussed in this lesson. Please refer to 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:

${\displaystyle V_{s}=9V}$
${\displaystyle R_{1}=200\Omega }$
${\displaystyle R_{2}=20\Omega }$
${\displaystyle R_{3}=10k\Omega }$
${\displaystyle R_{4}=5k\Omega }$
${\displaystyle R_{5}=15k\Omega }$
${\displaystyle R_{6}=1k\Omega }$

Find the current through ${\displaystyle R_{3}}$ using the Nodal Analysis method.