Y to Δ Network Conversion

Y to Δ (T to π) Network Conversion

Before reading this page I strongly suggest that you have first read and understood the complementary page on Δ to Y Network Conversion. The working out on this page is based on the results of that page.

Previously we looked at how to calculate the values for a ‘Y’ network to match an existing Delta (Δ) network. Now we will look at how to go the other way. Assuming there is an existing ‘Y’ network, this page looks at how to calculate appropriate values for an identical (electrically speaking) Delta network.

To re-cap here are schematics for each of the configurations, followed by the final equations for converting from Δ to ‘Y’ networks.

Equations for Delta (Δ) to ‘Y’ Conversion

\(R_a = \frac{R_{ab}R_{ac}}{R_{ab} + R_{ac} + R_{bc}}\\\) \(R_b = \frac{R_{ab}R_{bc}}{R_{ab} + R_{ac} + R_{bc}}\\\) \(R_c = \frac{R_{ac}R_{bc}}{R_{ab} + R_{ac} + R_{bc}}\)

Deriving ‘Y’ to Delta (Δ) Equations

What follows involves quite a lot of algebra. If you uncomfortable with how to transpose equations. I suggest that you first take a quick refresher here.

That all said, let’s dive right in…

First rearrange the Δ→Y equations to make their denominator the target:

\(R_a = \frac{R_{ab}R_{ac}}{R_{ab} + R_{ac} + R_{bc}}\) becomes \(R_{ab} + R_{ac} + R_{bc} = \frac{R_{ab}R_{ac}}{R_a}\\\)

\(R_b = \frac{R_{ab}R_{bc}}{R_{ab} + R_{ac} + R_{bc}}\) becomes \(R_{ab} + R_{ac} + R_{bc} = \frac{R_{ab}R_{bc}}{R_b}\\\)

\(R_c = \frac{R_{ac}R_{bc}}{R_{ab} + R_{ac} + R_{bc}}\) becomes \(R_{ab} + R_{ac} + R_{bc} = \frac{R_{ac}R_{bc}}{R_c}\\\)

so:

\(R_{ab} + R_{ac} + R_{bc} = \frac{R_{ab}R_{ac}}{R_a} = \frac{R_{ab}R_{bc}}{R_b} = \frac{R_{ac}R_{bc}}{R_c}\\\)

What we are really wanting to find is R_ab, R_ac, and R_bc in terms of R_a, R_b, and R_c. We’ll start off by finding R_bc.

\(\frac{R_{ab}R_{ac}}{R_a} = \frac{R_{ac}R_{bc}}{R_c}\\\) \(R_{ab} = \frac{R_{ac}R_{bc}}{R_c} \frac{R_a}{R_{ac}} = \frac{R_a R_{bc}}{R_c}\\\) \(\frac{R_{ab}R_{ac}}{R_a} = \frac{R_{ab}R_{bc}}{R_b}\\\) \(R_{ac} = \frac{R_{ab}R_{bc}}{R_b} \frac{R_a}{R_{ab}} = \frac{R_a R_{bc}}{R_b}\\\)

Now substituting R_ab and R_ac from above in:

\(R_a = \frac{R_{ab} R_{ac}}{R_{ab} + R_{ac} + R_{bc}}\\\) \(R_a = \frac{\frac{R_a R_{bc}}{R_c} \frac{R_a R_{bc}}{R_b}}{\frac{R_a R_{bc}}{R_c} + \frac{R_a R_{bc}}{R_b} + R_{bc}} = \frac{\frac{R_a^2 R_{bc}^2}{R_b R_c}}{R_{bc} [\frac{R_a}{R_c} + \frac{R_a}{R_b} + 1]} = \frac{\frac{R_a^2 R_{bc}}{R_b R_c}}{\frac{R_a R_b + R_a R_c + R_b R_c}{R_b R_c}} = \frac{R_a ^2 R_{bc}}{R_a R_b + R_a R_c + R_b R_c}\\\)

and transpose to make R_bc the target:

\(R_{bc} = \frac{R_a (R_a R_b + R_a R_c + R_b R_c)}{R_a^2}\\\) \(R_{bc} = \frac{R_a R_b + R_a R_c + R_b R_c}{R_a}\\\)

Well, that was a lot of algebra and finding R_ab and R_ac is just more of the same. So, I’ll leave the working up to you and just give a reminder of the schematics and the resulting three transformation equations below:

\(R_{ab} = \frac{R_a R_b + R_a R_c + R_b R_c}{R_c}\\\) \(R_{ac} = \frac{R_a R_b + R_a R_c + R_b R_c}{R_b}\\\) \(R_{bc} = \frac{R_a R_b + R_a R_c + R_b R_c}{R_a}\\\)

Aide-Mémoire

Clearly no-one in their right mind wants to go through all of that each time the conversion is required, and memorising yet another three (non-trivial) equations is not an appealing prospect. However, things are not a bad as they first appear.

It can be seen that the numerator is common across all three equations and is just the sum of products for each pair of resistors in the network. Meanwhile the appropriate denominator is found by looking which resistor (in the original circuit) is connected to the node opposite the new resistor that we wish to find the value for.