Δ and Y Conversion Equations
There are several equations used to convert one network to the other:
Δ and Y networks are seen frequently in 3-phase AC power systems (a topic covered in volume II of this book series), but even then they’re usually balanced networks (all resistors equal in value) and conversion from one to the other need not involve such complex calculations. When would the average technician ever need to use these equations?
Application of Δ and Y Conversion
A prime application for Δ-Y conversion is in the solution of unbalanced bridge circuits, such as the one below:
The solution of this circuit with Branch Current or Mesh Current analysis is fairly involved, and neither the Millman nor Superposition Theorems are of any help since there’s only one source of power. We could use Thevenin’s or Norton’s Theorem, treating R3 as our load, but what fun would that be?
If we were to treat resistors R1, R2, and R3 as being connected in a Δ configuration (Rab, Rac, and Rbc, respectively) and generate an equivalent Y network to replace them, we could turn this bridge circuit into a (simpler) series/parallel combination circuit:
After the Δ-Y conversion . . .
If we perform our calculations correctly, the voltages between points A, B, and C will be the same in the converted circuit as in the original circuit, and we can transfer those values back to the original bridge configuration.
Resistors R4 and R5, of course, remain the same at 18 Ω and 12 Ω, respectively. Analyzing the circuit now as a series/parallel combination, we arrive at the following figures:
We must use the voltage drops figures from the table above to determine the voltages between points A, B, and C, seeing how they add up (or subtract, as is the case with the voltage between points B and C):
Now that we know these voltages, we can transfer them to the same points A, B, and C in the original bridge circuit:
Voltage drops across R4 and R5, of course, are exactly the same as they were in the converter circuit.
At this point, we could take these voltages and determine resistor currents through the repeated use of Ohm’s Law (I=E/R):
proof
Solve with Thevenin’s
