Basic Electrical Engineering Series
Node Analysis
Node analysis is the counterpart of mesh analysis and uses a similar method. However of course we now solve for node voltages rather than mesh currents. We therefore label the circuit with node potentials with respect to a reference node. See diagram below e1, e2, e3, eref.
Each potential is assumed positive with respect to the reference node.
In node analysis the equations are derived using Kirchoffs Current Law (KCL). That is the currents into and out of each node are summed to zero. The equations are then represented using a matrix equation as in mesh analysis, except in this case we have an admittance matrix multiplying a column vector of unknown node voltages and this is equated to a column vector of known currents supplying the nodes.
[Y]*[E] = [I] Ohms Law
As in mesh analysis there are a set of rules that must be adopted in order to get the signs correct, as follows.
The numbers on
the diagonal of the matrix are positive. These represent the
total admittance connected to each node.
The numbers
off the diagonal represent the total admittance from one node
with respect to another node i.e. Y12, Y13
etc. Note there are always two i.e. the admittance from node one
to node two is the same as from node 2 to node 1. Hence for
example Y31 is the same as Y13.
The numbers
from one node with respect to another node are always negative.
The current
sources are positive when entering a node and negative
when leaving a node.
For this last point observe the diagram above. The left-hand generator is supplying the node to which it is connected with current flow from left to right and is therefore positive. The right hand generator on the other hand should be receiving current from the node to which it is connected. However it is labeled supplying the node, therefore it is negative.
So for the circuit above we first redraw with admittances or conductance’s in this case because we are dealing with pure resistances.
Example 1
So writing down the matrix equation for the above circuit we get the following
[Y]*[E] = [I]
We can now solve this equation as before by using Cramers rule or the matrix inversion method.
Solution using Cramers Rule
The next step is to form the determinant of the three equations. A determinant is a number, which is determined from the array of the coefficients of the unknown node potentials, as follows:
The first thing to note is that the determinant can be arrived at by expanding along any row or column.
Let’s choose column 1 to illustrate the procedure.
First we choose the first number of the column (0.15). We then obtain the ‘minor’ by covering up the row and the column that the number lies upon. The array that is left is the minor, as follows:
The minor is evaluated by multiplying the diagonals and subtracting:
i.e. (0.08 x 0.12) – (-0.02 x -0.02).
This result is called the co-factor and this then multiplied by the number 0.15.
The procedure is repeated for each number in the column or row chosen
Another thing to note is that the sign of the number in the determinant has a pattern as follows:
So to complete the exercise lets evaluate the determinant of the above along column 1.
This is:
DG = (+) (0.15) [(0.08).(0.12) – (-0.02).(-0.02)]
(- ) (-0.05)[(-0.05).(0.12) – (-0.02).(-0.1)]
(+) (-0.1) [(-0.05)(-0.02) – (0.08)(-0.1)]
DG = 80 x 10-6
The next step is to replace the coefficients of e1 with the numbers on the right hand side of the equation, that is the column vector of applied currents, as follows:
D1 = (+) (2)[(0.08).(0.12) – (-0.02).(-0.02)]
(-) (0)[]
(+) (3)[(-0.05)(-0.02) – (0.08)(-0.1)]
D1 = 45.4 x 10-3
The next step is to replace the coefficients of e2 with the numbers on the right hand side of the equation, that is the column vector of applied currents, as follows:
D2 = (-) (2)[(-0.05).(0.12) – (-0.1).(-0.02)]
(+) (0)[]
(-) (3)[(0.15)(-0.02) – (-0.1)(-0.05)]
D2 = 40 x 10-3
The next step is to replace the coefficients of e3 with the numbers on the right hand side of the equation, that is the column vector of applied currents, as follows:
D3 = (+) (2)[(-0.05).(-0.02) – (0.08).(-0.1)]
(-) (0)[]
(+) (3)[(0.15)(0.08) – (-0.05)(-0.05)]
D3 = 46.5 x 10-3
Finally we now apply Cramers Rule, as follows:
e1 = D1/DG= 567.5 v e2 = D2/DG = 500 v e3 = D3/DG = 581.3 v
Solution by Matrix Inversion
We have seen that the system of node equations results in the following:
[Y]*[E] = [I] Ohms Law
This can also be solved as follows:
[E] = [Y]-1 [I]
Which involves finding the inverse matrix [Y]-1
There are several methods of doing this. One method is as follows:
So lets do the example shown below.
Example 2
The steps are as follows:
The determinant of the Y matrix
DY = (+) (7) [(6).(11) – (-2).(-2)]
(- ) (-3) [(-3).(11) – (-2).(-4)]
(+) (-4) [(-3)(-2) – (6)(-4)]
DY = 191 = ¦Y¦
Form a new matrix C consisting of the co-factors of each of the elements of Y i.e. evaluate the minor of each element.
Form the transpose – this is the same in this case
... and hence find the inverse Y matrix
Finally E] = [Y]-1 [I]
