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Electric Circuit Analysis/Nodal Analysis - Wikiversity

25 Nov 2018 Create matrix from circuit equations. Solve for Unknown Node Voltages using Cramer's Rule. Part 1: Pre-reading Material. The student is advised‚  What you need to remember from Kirchhoff's Voltage & Current function . If you ever atmosphere setting lost, pull off not be shy to go incite to the previous lesson & go through it again. You can learn by repitition.

This part of the course onwards will collaborate later the Mathematics Department extensively. Mathematical Theory will be kept minimal as mathematical tools are single-handedly used here as a means to an end. associates to relevant Mathematical theories will be supplied to benefits the student.



This Lesson is approximately Kirchhoff's Current Law. The student/User is conventional to endure the following at the decline of the lesson.

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

The following is a general procedure for using Nodal Analysis method to solve electric circuit problems. The get-up-and-go of this algorithm is to expansion a matrix system from equations found by applying KCL at the major nodes in an electric circuit. Cramer's find is later used to solve the unkown major node voltages.

Once the Node voltages are solved, okay circuit analysis methods ( Ohm's law; Voltage and Current Divider principles etc ) can after that be used to judge regard as being 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.

1.) pick select a reference node. ( find of thumb: understand the node subsequent to most branches connecting to it )

5.) Solve matrix for unmemorable node voltages by using Cramer's Rule (Cramer's pronounce announce is simpler, although you can yet nevertheless use the Gaussian method)

The above algorithm is no question basic and useful for 2 x 2 and 3 x 3 size matrices. Generally as the number of major node voltages accumulation and the order of the matrix exceeds 3, numerical methods (beyond the scope of this course) are employed, sometimes taking into consideration the aid of computers, to solve such circuit networks.

Find current through R 3 \displaystyle R_3 using Nodal Analysis method.Solution:

This is the same example we solved in Exercise 6, except that in this act we have further some resistors to growth the complexity of the circuit.

We use node a as the common node (ground if you like). Thus V a = 0 V \displaystyle V_a=0V as we did previously.

Carefully follow the progression of the nodal analysis algorithm explained in share 3 of this lesson as follows in share 5 and 6.

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 comport yourself as follows:

i 1 = i 2 + i 6 \displaystyle i_1=i_2+i_6

  V 1 V b R 1 = V b V c R 2 + V b V d R 6 \displaystyle \beginmatrix\ \frac V_1-V_bR_1=\frac V_b-V_cR_2+\frac V_b-V_dR_6\endmatrix

    V b ( 1 R 1 + 1 R 2 + 1 R 6 ) V c ( 1 R 2 ) V d ( 1 R 6 ) = V 1 R 1 \displaystyle \beginmatrix\ \ V_b(\frac 1R_1+\frac 1R_2+\frac 1R_6)-V_c(\frac 1R_2)-V_d(\frac 1R_6)=\frac V_1R_1\endmatrix (1)

i 3 = i 2 + i 4 \displaystyle i_3=i_2+i_4

  V c R 3 = V b V c R 2 + V d V c R 4 \displaystyle \beginmatrix\ \frac V_cR_3=\frac V_b-V_cR_2+\frac V_d-V_cR_4\endmatrix

    V b ( 1 R 2 ) + V c ( 1 R 2 + 1 R 3 + 1 R 4 ) + V d ( 1 R 4 ) = 0 \displaystyle \beginmatrix\ \ V_b(-\frac 1R_2)+V_c(\frac 1R_2+\frac 1R_3+\frac 1R_4)+V_d(-\frac 1R_4)=0\endmatrix (2)

i 4 = i 5 + i 6 \displaystyle i_4=i_5+i_6

  V d V c R 4 = V 2 V d R 5 + V b V d R 6 \displaystyle \beginmatrix\ \frac V_d-V_cR_4=\frac V_2-V_dR_5+\frac V_b-V_dR_6\endmatrix

  V b ( 1 R 6 ) V c ( 1 R 4 ) + V d ( 1 R 4 + 1 R 5 + 1 R 6 ) = V 2 R 5 \displaystyle \beginmatrix\ V_b(-\frac 1R_6)-V_c(\frac 1R_4)+V_d(\frac 1R_4+\frac 1R_5+\frac 1R_6)=\frac V_2R_5\endmatrix (3)

The neighboring bordering step in this algorithm is to construct a matrix. To get that easily, we the theater all resistances in the above equations 1, 2 and 3 taking into consideration their equivalent admittances - as follows:

G 1 = 1 R 1 \displaystyle G_1=\frac 1R_1 etc consequently in view of that equations 1, 2 and 3 will be re-written as follows:

  V b ( G 1 + G 2 + G 6 ) + V c ( G 2 ) + V d ( G 6 ) = ( V 1 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 G 5 ) . . . . ( 3 ) \displaystyle \beginmatrix\ 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)\endmatrix

[ ( 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 ) ] . [ V b V c V d ] = [ ( V 1 G 1 ) 0 ( V 2 G 5 ) ] \displaystyle \beginbmatrix(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)\endbmatrix.\beginbmatrixV_b\\V_c\\V_d\endbmatrix=\beginbmatrix(V_1\times G_1)\\0\\(V_2\times G_5)\endbmatrix

A . X = Y \displaystyle A.\vec X=\vec Y [ 1.05 0.05 0.5 0.05 0.35 0.2 0.5 0.2 1.2 ] . [ V b V c V d ] = [ 7.5 0 3.5 ] \displaystyle \beginbmatrix1.05&-0.05&-0.5\\-0.05&0.35&-0.2\\-0.5&-0.2&1.2\endbmatrix.\beginbmatrixV_b\\V_c\\V_d\endbmatrix=\beginbmatrix7.5\\0\\3.5\endbmatrix

Now that we have arranged equations 1, 2 and 3 into a matrix we dependence obsession to accomplish determinants of the general matrix, and determinants of alterations of the general matrix as follows:

Remember to right to use *Solutions/CramerAlso, use the provided join for details roughly energetic out the determinant of a 3 x 3 Matrix.

  d e t [ 1.05 0.05 0.5 0.05 0.35 0.2 0.5 0.2 1.2 ] = d e t A     = 0.299 \displaystyle \beginmatrix\ det\beginbmatrix1.05&-0.05&-0.5\\-0.05&0.35&-0.2\\-0.5&-0.2&1.2\endbmatrix&=&detA\\\ \\\ &=&0.299\endmatrix

  d e t [ 7.5 0.05 0.5 0 0.35 0.2 3.5 0.2 1.2 ] = d e t A 1     = 3.498 \displaystyle \beginmatrix\ det\beginbmatrix7.5&-0.05&-0.5\\0&0.35&-0.2\\3.5&-0.2&1.2\endbmatrix&=&detA1\\\ \\\ &=&3.498\endmatrix

  d e t [ 1.05 7.5 0.5 0.05 0 .02 0.5 3.5 1.2 ] = d e t A 2     = 2.023 \displaystyle \beginmatrix\ det\beginbmatrix1.05&7.5&-0.5\\-0.05&0&-.02\\-0.5&3.5&1.2\endbmatrix&=&detA2\\\ \\\ &=&2.023\endmatrix

  d e t [ 1.05 0.05 7.5 0.05 0.35 0 0.5 0.2 3.5 ] = d e t A 3     = 2.665 \displaystyle \beginmatrix\ det\beginbmatrix1.05&-0.05&7.5\\-0.05&0.35&0\\-0.5&-0.2&3.5\endbmatrix&=&detA3\\\ \\\ &=&2.665\endmatrix

Now we can use the solved determinants to arrive at solutions for Node voltages V b ; V c a n d V d \displaystyle V_b;V_candV_d as follows:

1. V b = d e t A 1 d e t A = 3.498 0.299 = 11.717 V \displaystyle V_b=\frac detA1detA=\frac 3.4980.299=11.717V

2. V c = d e t A 2 d e t A = 2.023 0.299 = 6.776 V \displaystyle V_c=\frac detA2detA=\frac 2.0230.299=6.776V

3. V d = d e t A 3 d e t A = 2.665 0.299 = 8.928 V \displaystyle V_d=\frac detA3detA=\frac 2.6650.299=8.928V

Now we can apply Ohm's measure to solve for the current through R 3 \displaystyle R_3 as follows:

I R 3 = V c R 3 = 6.776 V 10 = + 0.678 A \displaystyle I_R_3=\frac V_cR_3=\frac 6.776V10\Omega =+0.678A

As we previously saw, the sure determined sign in the above current tells us that the committed current flowing through R 3 \displaystyle R_3 is in fact in the meting out we chose next drawing stirring the circuit in figure 7.2.

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

As satisfactory the following part is an exercise to test your self roughly speaking the content discussed in this lesson. keep amused refer to ration 11 for other reading material and appealing related external links.

Find the current through R 3 \displaystyle R_3 using the Nodal Analysis method.

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

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