# How To Use Delta-to-Wye Transformations In Circuit Analysis

## Theory

In the last Hub, we analyzed a circuit consisting of both series and parallel elements and successfully found the voltage across and current through every resistor using Ohm's Law, Kirchhoff's Voltage and Current Laws, and the formulas to find equivalent resistances for series and parallel circuits. In this Hub, we'll see a circuit orientation that cannot be solved using only these techniques--a delta circuit--and we will see the transformation that makes it possible to solve this type of circuit.

The derivations of the transformations can be found here. To simply state the results, though, if you have a delta circuit and wye circuit as shown above, then the following equations apply:

- R
_{a}= (R_{1}R_{3}) / (R_{1}+ R_{2}+ R_{3}) - R
_{b}= (R_{1}R_{2}) / (R_{1}+ R_{2}+ R_{3}) - R
_{c}= (R_{2}R_{3}) / (R_{1}+ R_{2}+ R_{3})

A pattern is easily observable. The value of a given limb resistor in the wye configuration is equal to the product of the touching corner resistors divided by the sum of the three.

## Application

Let's look at an example circuit shown below:

What we can do to simplify this resistive network down to a single resistor value is convert the Delta network formed by R_{1}, R_{2}, and R_{3} into an equivalent Wye network as shown in the following circuit:

This circuit is easily solvable using the principles covered in previous hubs. R_{a} and R_{4} are in series, and this subbranch is in parallel with the series subbranch of R_{c} and R_{5}. And these two branches in parallel are in series with R_{a}.

Using the notation '||' to represent the product-over-sum operation, we have R_{eq} = R_{a} + ( (R_{b} + R_{4}) || (R_{c} + R_{5}) ).

Now we use the Delta-Wye transformation. We can't use the exact formulas as written above, since I labeled R_{a}, R_{b}, and R_{c} slightly differently relative to R_{1}, R_{2}, and R_{3} compared to how I labeled them in the paragraphs at the top of this Hub. But remember the way to arrive at the proper formula: for a given limb on the Wye circuit, divide the two "touching" resistors of the Delta formation by the sum of the three Delta resistors.

So we have:

- R
_{a}= (R_{1}R_{2}) / (R_{1}+ R_{2}+ R_{3}) - R
_{b}= (R_{1}R_{3}) / (R_{1}+ R_{2}+ R_{3}) - R
_{c}= (R_{2}R_{3}) / (R_{1}+ R_{2}+ R_{3})

That's it! Substituting these calculated values in for R_{a}, R_{b}, and R_{c} in the equation for R_{eq} above gets us R_{eq}, which we can use to calculate the total current draw I_{T} in the circuit.

## An Example Circuit

Let's use the techniques described above to fine the calculated values of the voltage across and current through every resistor in a real circuit, and then compare those calculated values to actually measured values. Shown below is a real circuit in the same configuration as the circuit above with resistors R_{1}-R_{5}:

In this circuit, the nominal battery and resistor values are V = 1.5 V, R_{1} = 100 kΩ, R_{2} = 1.2 kΩ, R_{3} = 1.5 kΩ, R_{4} = 210 Ω, and R_{5} = 210 Ω (I didn't purposely the bottom two resistors the same value; I just picked what was already on my breadboard. Also, it doesn't make the solution any kind of simplified special case.)

I like using a spreadsheet program like Microsoft Excel or OpenOffice Calc to lay out circuit equations because if you ever change a resistor value, all the others that depend on the one you just changed automatically recalculate.

To start, I insert the resistor values R_{1} - R_{5} into some cells as shown:

Next, we want the spreadsheet to calculate the values of R_{a}, R_{b}, and R_{c} according to the formulas given above. Given the current state of the spreadsheet, we have

- Ra = B1 * B2 / (B1 + B2 + B3)
- R
_{b}= B1 * B3 / (B1 + B2 + B3) - R
_{c}= B2 * B3 / (B1 + B2 + B3)

The spreadsheet now looks like this:

And the spreadsheet value for R_{eq} comes out to be

- R
_{eq}= B6 + ( (B7 + B4) * (B8 + B5) ) / (B7 + B4 + B8 + B5)

Now the spreadsheet looks like this:

Next, we add V_{T} = 1.5 V to a cell in the spreadsheet. Now we can work through the regular series-parallel circuit shown in the second schematic above. I_{T}, the total circuit current, is V_{T} / R_{eq}. With the current values, this comes out to approximately 1.1 mA. This is the current that flows through R_{a}, making the voltage drop across R_{an} equal to R_{a}I_{T}, which is approximately 1.28 V (much of the battery's 1.5 V ).

We can label the remainder voltage that exists across the two branches V_{Rem}. This is of course V_{T} - V_{a}, which is approximately 0.22 V.

What we want to figure out are the voltage levels at the point between R_{b} an,d R_{4}, which we can call R_{b4}, and at the point between R_{c} and R_{5}, which we can call R_{c5}.

We can calculate the current flowing through the branch of R_{b} and R_{4} , which we can label I_{b4}, by dividing V_{Rem} by R_{b} + R_{4}, which comes out to approximately 0.13 mA. With this much current flowing through this branch, and therefore through R_{4}, the voltage of R_{b4} with respect to the battery's negative terminal is I_{b4}R_{4}, which we can call V_{R}_{4}, is approximately 0.028 V.

We can calculate the current flowing through the branch of R_{c} and R_{5}, which we can label I_{c5}, by dividing V_{Rem} by R_{c} + R_{5}, which comes out to approximately 0.97 mA. (Notice the sanity check we can do on our equations at this point: because of Kirchhoff's Current Law, I_{c5} and I_{b4} should add up to I_{T}, which they do). With this much current flowing through this branch, and therefore through R_{5}, the voltage of R_{c5} with respect to the battery's negative terminal is I_{c5}R_{5}, which we can call V_{R5}, is approximately 0.20 V.

Now since the Wye version of the circuit is equivalent to the Delta version, we can say that the voltage potential across R_{3} is equal to the potential difference between V_{R}_{4} and V_{R}_{5}. Current flows from a higher voltage to a lower voltage, so in this given circuit the current through R_{3} flows in the direction from right to left since V_{R5} > V_{R4}. The voltage across R_{3}, which we can call V_{R3}, is V_{R5}- V_{R4}, which is approximately 0.17 V. And the current through R_{3} is V_{R3} / R_{3}, which is approximately 0.12 mA.

The last things to calculate are the voltages across and the currents through R_{1} and R_{2}.

The voltage at the very top node with respect to the negative battery terminal is of course V_{T}, which is 1.5 V, and we know that the voltage at the node at the top of R_{4} with respect to the negative battery terminal is approximately 0.028 V, so the difference between these two voltages is the voltage V_{R1}, which is approximately 1.47 V. The current I_{R1} is simply the voltage across it divided by the resistance, which is approximately 1.47E-05 A (not surprising that this current is so small, given that this resistor is so large).

The voltage across R_{2} is V_{T} - V_{R5} which is approximately 1.30 V, and the current through R_{2} is that voltage divided by its resistance, which comes out to approximately 1.1 mA.

An image of my final spreadsheet is shown to the right.

All the calculated and nominal voltages, resistances, and currents and the measured values are shown in the table below:

Value | Nominal/Calculated | Measured | % Error |
---|---|---|---|

VT | 1.50 V | 1.40 V | 7.1 |

R1 | 100 kΩ | 98.9 kΩ | 1.1 |

R2 | 1.20 kΩ | 1.18 kΩ | 1.7 |

R3 | 1.50 kΩ | 1.48 kΩ | 1.4 |

R4 | 210 Ω | 217 Ω | 3.2 |

R5 | 210 Ω | 216 Ω | 2.8 |

VR1 | 1.47 V | 1.37 V | 7.3 |

VR2 | 1.30 V | 1.20 V | 8.3 |

VR3 | 175 mV | 168 mV | 4.2 |

VR4 | 27.6 mV | 27.7 mV | 0.36 |

VR5 | 203 mV | 197 mV | 3.0 |

IT | 1.10 mA | 967 μA | 14 |

IR1 | 14.7 μA | 13.7 μA | 7.3 |

IR2 | 1.08 mA | 955 μA | 13 |

IR3 | 117 μA | 108 μA | 8.3 |

IR4 | 131 μA | 121 μA | 8.3 |

IR5 | 965 μA | 821 μA | 18 |

Most of the calculated values are pretty close to the measured values.

Hopefully the working out of this sample circuit showed you some useful techniques for circuit analysis. The best way to increase your skills is to get lots of practice!

## Comments

**Svetlana** on February 22, 2015:

Abizar, I had this same problem unsandtendirg resistors. After all, if you connect a 10-ohm resistor directly to your power supply, it will get much hotter than a 10k-ohm resistor. However, this assumes the voltage across the resistor is constant (coming directly from a power supply in this case). If the resistor is part of a larger circuit, and the voltage across it is not fixed, using larger values of resistance will generally cause a larger voltage drop for a given circuit current, and hence more power dissipation. Think of electrical extension cords. A thick, low-resistance cord will generate less heat than a thinner, higher-resistance cord for the same length, powering the same device.For applications, where the circuit current is known, use (I^2)(R) to find power dissipation in a resistor.I hope this helps.