01 February, 2007

Nodal Voltage Analysis

In this method, we set up and solve a system of equations in which the unknowns are the voltages at the principal nodes of the circuit. From these nodal voltages the currents in the various branches of the circuit are easily determined.
Steps in the nodal analysis method :
1. Count the number of principal nodes or junctions in the circuit. Call this number n. (A principal node or junction is a point where 3 or more branches join. We will indicate them in a circuit diagram with a red dot. Note that if a branch contains no voltage sources or loads then that entire branch can be considered to be one node.)
2. Number the nodes N1, N2, . . . , Nn and draw them on the circuit diagram. Call the voltages at these nodes V1, V2, . . . , Vn, respectively.
3. Choose one of the nodes to be the reference node or ground and assign it a voltage of zero.
4. For each node except the reference node write down Kirchoff's Current Law in the form "the algebraic sum of the currents flowing out of a node equals zero". (By algebraic sum we mean that a current flowing into a node is to be considered a negative current flowing out of the node.)
Express the current in each branch in terms of the nodal voltages at each end of the branch using Ohm's Law (I = V / R). Here are some examples:
The result, after simplification, is a system of m linear equations in the m unknown nodal voltages (where m is one less than the number of nodes; m = n - 1). The equations are of this form:


where G11, G12, . . . , Gmm and I1, I2, . . . , Im are constants.
Example :

Number of nodes is 4.
We will number the nodes as shown above.
We will choose node 2 as the reference node and assign it a voltage of zero.
Write down Kirchoff's Current Law for each node. Call V1 the voltage at node 1, V3 the voltage at node 3, V4 the voltage at node 4, and remember that V2 = 0. The result is the following system of equations:


The first equation results from KCL applied at node 1, the second equation results from KCL applied at node 3 and the third equation results from KCL applied at node 4.
Solving the above system of equations using Gaussian elimination or some other method gives the following voltages:
V1 = -35.88 volts, V3 = 63.74 volts and V4 = 0.19 volts
For more information on ths topic visit the following sites :

No comments: