bibhash jha
Last Activity: 15 Years ago
Nodal Analysis of Electric Circuits
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.
The steps in the nodal analysis method are:
- 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.)
- Number the nodes N1, N2, . . . , Nn and draw them on the circuit diagram. Call the voltages at these nodes V1, V2, . . . , Vn, respectively.
- Choose one of the nodes to be the reference node or ground and assign it a voltage of zero.
- 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.)
For example, for the node to the right KCL yields the equation:
Ia + Ib + Ic = 0
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 current downward out of node 1 depends on the voltage difference V1 - V3 and the resistance in the branch.
In this case the voltage difference across the resistance is V1 - V2 minus the voltage across the voltage source. Thus the downward current is as shown.
In this case the voltage difference across the resistance must be 100 volts greater than the difference V1 - V2. Thus the downward current is as shown.
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.
Delta to Star Network.
Compare the resistances between terminals 1 and 2.

Resistance between the terminals 2 and 3.

Resistance between the terminals 1 and 3.

This now gives us three equations and taking equation 3 from equation 2 gives:

Then, re-writing Equation 1 will give us:

Adding together equation 1 and the result above of equation 3 minus equation 2 gives:

From which gives us the final equation for resistor P as:

Then to summarize a little the above maths, we can now say that resistor P in a Star network can be found as Equation 1 plus (Equation 3 minus Equation 2) or Eq1 + (Eq3 - Eq2).
Similarly, to find resistor Q in a Star network, is equation 2 plus the result of equation 1 minus equation 3 or Eq2 + (Eq1 - Eq3) and this gives us the transformation of Q as:

And again, to find resistor R in a Star network, is equation 3 plus the result of equation 2 minus equation 1 or Eq3 + (Eq2 - Eq1) and this gives us the transformation of R as:

When converting a Delta network into a Star network the denominators of all of the transformation formulas are the same: A + B + C, and which is the sum of ALL the Delta resistances. Then to convert any Delta connected network to an equivalent Star network we can summarized the above transformation equations as:
Star to Delta Network.
The value of the resistor on any one side of the Delta, Δ network is the sum of all the two-product combinations of resistors in the Star network divide by the Star resistor located "directly opposite" the Delta resistor being found. For example, resistor A is given as:

with respect to terminal 3 and resistor B is given as:

with respect to terminal 2 with resistor C given as:

with respect to terminal 1.
By dividing out each equation by the value of the denominator we end up with three separate transformation formulas that can be used to convert any Delta resistive network into an equivalent Star network as given below.