Delta-to-Wye Equivalent Circuits
Introduction: This article focuses on Delta and Wye configurations within circuits and how to simplify these for analysis.
The Essentials
So far we have experience combining resistors in series and parallel. There are other configurations of resistors that are more complex than just series and parallel. Some of these are known as Wye and Delta configurations. Figure 1. has an example of delta connections.
Figure 1: Delta Connections
As you can see, there are two representations of a delta connected set of resistors. The left is seen in analysis, but perhaps the right one is more common within schematics. Both of these representations of delta connections are equivalent since they share the same nodes.
Now for wye connections shown in Figure 2.
Figure 2: Wye Connections
It's a similar scenario here, both of these representations are essentially the same.
When it comes to analyzing a circuit with either of these connections, it can be difficult the ability to convert between delta and wye. Below are equations that convert resistor values from a delta connection to a wye connection.
Now let's explore converting from wye to delta connections:
These equations may seem complicated, but doing some example and practice problems definitely help out
Example
Let's look at an example of a circuit needing a delta-wye conversion in Figure 3.
Figure 3: Delta to Wye Example Problem
For this problem, let's solve for \( R_{eq} \). You may notice that the bottom mesh is a delta connected set of resistors. It makes this hard to simplify, but with delta to wye conversions this makes it significantly easier. Let's solve for the wye resistors.
Now redrawing the circuit we have the following in Figure 4.
Figure 4: Delta to Wye Example Simplified
From this, it is a very simple combination of series and parallel resistors to solve for the resistance equivalence. As we combine these into a single resistor, we should get a resistor value of \( R_{eq} = 8.611 \, \text{k}\Omega \).
Let's do a wye to delta example now. Figure 5. has the example problem for this.
Figure 5: Wye to Delta Example
Let's just simply convert this wye connection to delta. Using equations 4,5, and 6, we get the following:
Now these resistor values should result in a circuit schematic that looks kind of like this:
Figure 6: Wye to Delta Example Simplified
This circuit result means that this delta schematic results in the same resistance as the original wye configuration.
Practice
Problem 1.) Using Figure 7., solve for \( R_{eq} \) using wye to delta conversions.
Figure 7: Problem 1 Schematic
Problem 2.) Using Figure 8., solve for the current being drawn into the system from the power source using delta to wye conversions
Figure 8: Problem 2 Schematic
Problem 3.) Using Figure 9., solve for the power supplied by the power source using wye to delta conversions.
Figure 9: Problem 3 Schematic
Solutions:
Problem 1 Solution.) \( R_{eq} = 2.69 \, \text{k} \Omega \). To start, we recognize the wye connection in the circuit and know that we can convert it into its delta counterpart. Using equations 4, 5, and 6, we can solve for the new delta resistor values.
Figure 10. contains a revised version of the circuit with our newly calculated resistor values
Figure 10: Problem 1 Simplified
From here, we can see that the analysis is simple. The parallel resistors (1.7k and 13.12k) are in series with the 6.27 k\( \Omega \) resistor. That combination is then in parallel with the 4.12 k\( \Omega \) resistor. Do the parallel and series analysis and you should get that \( R_{eq} = 2.69 \, \text{k} \Omega \)
Problem 2 Solution.) \( i = 25.9 \, \text{mA} \). Here let's use delta to wye conversions to solve for \( R_{eq} \). Using equations 1, 2, and 3, we get the following resistor values
With these values, we can reference Figure 11 for a modified schematic with our new resistor values
Figure 11: Problem 2 Simplified
As we can see, this is a similar situation to our practice problem above. Use parallel/series resistor analysis and you should get a resistor \( R_{eq} \) value of 2.315 k\( \Omega \).
With this, we can apply Ohm's law to find the current drawn from the power source.
Problem 3 Solution.) \( P = 0.417 \, \text{W} \). As we look at the circuit, we can replace the wye connection with a delta one using equations 4, 5, and 6.
Figure 12 shows a remodeled schematic with these calculations.
Figure 12: Problem 3 Simplified
Now these resistors are all in parallel in one way or another. Solve for the parallel resistors and you should get a \( R_{eq} = 6 \, k\Omega \).
From this we can solve for current. Using Ohm's Law, we find current to be about 8.3 mA. Finally, power is found by multiplying the voltage provided by the source with the current through it.