Delta-Wye resistor networks

Sometimes when you are simplifying a resistor network, you get stuck. Some resistor networks cannot be simplified using the usual series and parallel combinations. This situation can often be handled by trying the $\Delta - \text Y$delta, minus, start text, Y, end text transformation, or 'Delta-Wye' transformation.

The names Delta and Wye come from the shape of the schematics, which resemble letters. The transformation allows you to replace three resistors in a $\Delta$delta configuration by three resistors in a $\text Y$start text, Y, end text configuration, and the other way around.

The $\Delta - \text Y$delta, minus, start text, Y, end text drawing style emphasizes these are 3-terminal configurations. Something to notice is the different number of nodes in the two configurations. $\Delta$delta has three nodes, while $\text Y$start text, Y, end text has four nodes (one extra in the center).

The configurations can be redrawn to square up the resistors. This is called a $\pi - \text T$pi, minus, start text, T, end text configuration,

The $\pi - \text T$pi, minus, start text, T, end text style is a more conventional drawing you would find in a typical schematic. The transformation equations developed next apply to $\pi - \text T$pi, minus, start text, T, end text as well.

For the transformation to be equivalent, the resistance between each pair of terminals must be the same before and after. It is possible to write three simultaneous equations to capture this constraint.

Consider terminals $x$x and $y$y (and for the moment assume terminal $z$z isn't connected to anything, so the current in $\text R3$start text, R, end text, 3 is $0$0 ). In the $\Delta$delta configuration, the resistance between $x$x and $y$y is $Rc$R, c in parallel with $Ra +Rb$R, a, plus, R, b .

On the $\text Y$start text, Y, end text side, the resistance between $x$x and $y$y is the series combination $R1+R2$R, 1, plus, R, 2 (again, assume terminal $z$z isn't connected to anything, so $\text R1$start text, R, end text, 1 and $\text R2$start text, R, end text, 2 carry the same current and can be considered in series). We set these equal to each other to get the first of three simultaneous equations,

We can write two similar expressions for the other two pairs of terminals. Notice the $\Delta$delta resistors have letter names, $(Ra$left parenthesis, R, a, etc.$)$right parenthesis and the $\text Y$start text, Y, end text resistors have number names, $(R1$left parenthesis, R, 1, etc.$)$right parenthesis.

After solving the simultaneous equations (not shown), we get the equations to transform either network into the other.

Equations for transforming a $\Delta$delta network into a $\text Y$start text, Y, end text network:

Transforming from $\Delta$delta to $\text Y$start text, Y, end text introduces one additional node.

Equations for transforming a $\text Y$start text, Y, end text network into a $\Delta$delta network:

Transforming from $\text Y$start text, Y, end text to $\Delta$delta removes one node.

Let's do a symmetric example. Assume we have a $\Delta$delta circuit with $3\,\Omega$3, \Omega resistors. Derive the $\text Y$start text, Y, end text equivalent by using the $\Delta \rightarrow \text Y$delta, right arrow, start text, Y, end text equations.

Going in the other direction, from $\text Y \rightarrow\Delta$start text, Y, end text, right arrow, delta, looks like this,

Now for an example that's a little less tidy. We want to find the equivalent resistance between the top and bottom terminals.

Try as we might, there are no resistors in series or in parallel. But we are not stuck. First, let's redraw the schematic to emphasize we have two $\Delta$delta connections stacked one on the other.

Now select one of the $\Delta$delta's to convert to a $\text Y$start text, Y, end text. We will perform a $\Delta \rightarrow \text Y$delta, right arrow, start text, Y, end text transformation and see if it breaks the logjam, opening up other opportunities for simplification.

We go to work on the bottom $\Delta$delta (an arbitrary choice). Very carefully label the resistors and nodes. To get the right answers from the transformation equations, it is critical to keep the resistor names and node names straight. $Rc$R, c must connect between nodes $x$x and $y$y, and so on for the other resistors. Refer to Diagram 1 above for the labeling convention.

When we perform the transform on the lower $\Delta$delta, the black $\Delta$delta resistors will be replaced by the new gray $\text Y$start text, Y, end text resistors, like this:

Perform the transform yourself before looking at the answer. Check that you select the right set of equations.
Compute three new resistor values to convert the $\Delta$delta to a $\text Y$start text, Y, end text, and draw the complete circuit.

And voilà! Check out our circuit. It now has series and parallel resistors where it had none before. Continue simplification with series and parallel combinations until we get down to a single resistor between the terminals. Redraw the schematic again to square up the symbols into a familiar style.

We proceed through the remaining simplification steps just as we did before in the article on Resistor Network Simplification.

On the left branch, $3.125 + 1.25 = 4.375 \,\Omega$3, point, 125, plus, 1, point, 25, equals, 4, point, 375, \Omega

On the right branch, $4 + 1 = 5\,\Omega$4, plus, 1, equals, 5, \Omega

The two parallel resistors combine as $4.375\,||\, 5 = \dfrac{4.375 \cdot 5}{4.375 + 5} = 2.33\,\Omega$4, point, 375, vertical bar, vertical bar, 5, equals, start fraction, 4, point, 375, dot, 5, divided by, 4, point, 375, plus, 5, end fraction, equals, 2, point, 33, \Omega

And we finish by adding the last two series resistors together,

$\Delta - \text Y$delta, minus, start text, Y, end text transformations are another tool in your bag of tricks for simplifying circuits prior to detailed analysis.

Don't memorize the transformation equations. If the need arises, you can look them up.