Problem + Hard 5-1    Source:   exerc.: 26 - section 17.3 - page 727 - - HAYT, William H. Jr. , KEMMERLY, Jack E. , DURBIN, Steven M. - Book:    Analysis of Circuits in Engineering - Ed. McGrawHill - 8th Edition - 2012.

Using Delta-Star transformations for the circuit below, calculate the resistance equivalent between the points a-b

Solution of the Problem + Hard 5.1

Initially, we must name the nodes so that you can follow the transformations carried out step by step. We must calculate the equivalent circuit for points 1-3-b, i.e. the circuit highlighted in green, as shown below.

To transform Triangle to Star, use the equations below:

We must start from the left circuit and reach the right circuit.

Below, we see the values ​​of the resistors of the Star circuit that will replace the Triangle circuit.

Note in the figure below, the insertion of the Star circuit, highlighted in green, and what the new circuit configuration looks like.

Redrawing the circuit between points 2 and b, as shown below, we can see that it is possible to calculate the equivalent resistance between points 2 and o, since there is a 14 ohm resistor (2 and 12 ohm series) in parallel with a 5 ohm resistor (series of 1 and 4 ohms), finding the value of 70 /19 ohms . The figure below shows the circuit with a new topology, highlighted in green, to highlight what was said above.

Now, with this result, adding the resistor value of 2 /3 which is in series with the calculated one (70 /19 ohms ), we find the value of the resistance between points 2 and b, that is 4.35 ohms, as shown below.

Next step, we must calculate the Triangle-Star transformation of the circuit between the points 2 - 4 - 5, highlighted in green, as shown in the figure above. Applying the equations shown at the beginning of the solution, we find the new values of the resistors that make up the Star circuit, highlighted in green, as shown the figure below.

In the figure above, the circuit highlighted in blue will be the next transformation to do. The calculated values ​​and the new configuration of the circuit, already with the transformation, are shown in the figure below.

Note that in the figure below, the circuit of the figure above was redone to show that between the points 5 - u - k, there is a Triangle circuit. The resistors that appear between the points 5 - k are in series, so adding their values, we find 11.94 ohms. Those between the dots u - k, are also in series and add up to 2.93 ohms. Between the points 5 - u, the fraction 60 / 19 has been transformed into the decimal 3.16.

Finally, we have the last Triangle-Star transformation. Therefore, doing the calculation, we find the values ​​that are represented in the picture below.

In the figure above, looking at the branch x - k - b, it can be seen that we have two resistors in series, being able to replace them with a single value equal to 4.27 ohms. and by the branch x - u - 2 - b there are three resistors in series resulting in a single resistor of value equal to 5.39 ohms.

So between the points x and b there are two resistors that are in parallel. Applying the equation that calculates the equivalent resistance of a parallel circuit, we find the value of 2.38 ohms. In the figure below, we can see how the value was equivalent resistance and the final circuit.

Thus, we can see that a very complex circuit can be replaced by a single one. resistor that will behave like the original circuit.