Solution of Node Analysis with Cramer Method and Proteus Representation
In circuit analysis, voltage and current values can be found with the node analysis method. In the example, voltage values were found by solving the matrix and determinant using the Cramer method. To ensure the solution in circuit analysis, the circuit was established in the Proteus simulation program.
The circuit containing current sources is examined.
Analysis method:
1- The nodes (such as a, b, c) in the circuit are determined.
2 - The node with the most branches (Vr in the example) is selected as the reference and given a zero value.
3- Voltage values such as Va, Vb, Vc are assigned to each node. If the number of nodes is N, N-1 equations will be formed.
4- Equations are obtained by applying Kirchoff's flow law. According to the passive sign rule, since the resistor is an element that draws power, the point where the current enters is positive. At each node, an imaginary current vector is drawn outward from the node as it passes over the resistor. For example, the currents that will pass through the resistors at node "a" below are directed outwards. Likewise, even if the currents coming out of node "b" pass through the common resistance with the current coming out of node "a", the current direction should always be chosen outward, regardless of node "a". In Kirchoff's current law, currents entering the node are marked negative and currents leaving the node are marked positive.
5- Matrix solution is done using the Cramer method.
Since current enters the "a" node from the 9A current source, it is marked negative. Since the 3A current source is output from the node, the resistors are marked positive along with the currents that will pass through them.
Node equations are written in terms of currents. Each resistor is taken as 1 ohm.
\[I = \frac{{{V_{ab}}}}{R} = \frac{{Va - Vb}}{R}\]
(Current flows from intense potential to less potential)
\[- 9 + 3 + \frac{{Va - Vb}}{1} + \frac{{Va - Vr}}{1} + \frac{{Va - Vc}}{1} = 0\]
\[Va - Vb + Va - 0 + Va - Vc = 6\]
\[3Va - Vb - Vc = 6\]
Since the current from the 3A current source enters the "b" node, it is taken with a negative sign. Since the current going to the resistors is outward, it is taken positive.
\[- 3 + \frac{{Vb - Va}}{1} + \frac{{Vb - Vc}}{1} + \frac{{Vb - Vr}}{1} = 0\]
\[- 3 + Vb - Va + Vb - Vc + Vb - 0 = 0\]
\[- Va + 3Vb - Vc = 3\]
Since the current input to "c" node is from a 7 ampere current source, it is taken with a negative sign, and since the currents going to the resistors will be outward, it is taken with a positive sign.
\[- 7 + \frac{{Vc - Va}}{1} + \frac{{Vc - Vb}}{1} + \frac{{Vc - Vr}}{1} = 0\]
\[Vc - Va + Vc - Vb + Vc - 0 = 7\]
\[- Va - Vb + 3Vc = 7\]
(equation for node "c")
Equations (N-1=3 equations are formed from 4 nodes)
\[3Va - Vb - Vc = 6\]
\[- Va + 3Vb - Vc = 3\]
\[- Va - Vb + 3Vc = 7\]
Va, Vb, Vc coefficients are written into the "A" matrix. Equations can be transformed into matrix as follows.
To find the "Va" voltage value, the values [6,3,7] are added to the first column of the A matrix. Calculation is made using the Cramer method.
The first two lines are written exactly below the determinant, the right diagonal is marked "+", the leftdiagonal is marked "-", and all the numbers along the "+" path are multiplied and added, and the two values are subtracted from each other by multiplying and adding all the numbers along the "-" path.
+ If multiplied and added in the direction of the diagonal arrow:
\[\left[ {6.3.3 + 3.( - 1).( - 1) + 7.( - 1).( - 1)} \right]\]
\[= (54 + 3 + 7) = 64\]
- if multiplied and added in the direction of the diagonal arrow;
\[\left[ {(7.3.( - 1) + 6.( - 1).( - 1) + 3.( - 1).3} \right]\]
\[= ( - 21 + 6 - 9) = - 24\]
Determinant value
= \[64 - ( - 24) = 88\]
Similarly, if Det A value is taken;
\[DetA = (27 - 1 - 1) - (3 + 3 + 3) = 16\]
\[Va = \frac{{88}}{{16}} = 5,5V\]
Vb voltage is obtained in a similar way. [6,3,7] values are brought to the 2nd column.
\[(27 + 7 + 6) - (3 - 21 - 18) = 76\]
\[Vb = \frac{{76}}{{16}} = 4,75V\]
\[(63 + 6 + 3) - ( - 18 - 9 + 7) = 92\]
\[Vc = \frac{{92}}{{16}} = 5,75V\]
By making a quick and simple drawing with Proteus, you can confirm whether the voltage values in the circuit are correct.
Social Plugin