| Δ-Y and Y-Δ 
                    conversionsIn many circuit applications, we encounter 
                    components connected together in one of two ways to form a 
                    three-terminal network: the "Delta," or Δ (also known as the 
                    "Pi," or π) configuration, and the "Y" (also known as the 
                    "T") configuration.  
                      It is possible to calculate the proper 
                    values of resistors necessary to form one kind of network (Δ 
                    or Y) that behaves identically to the other kind, as 
                    analyzed from the terminal connections alone. That is, if we 
                    had two separate resistor networks, one Δ and one Y, each 
                    with its resistors hidden from view, with nothing but the 
                    three terminals (A, B, and C) exposed for testing, the 
                    resistors could be sized for the two networks so that there 
                    would be no way to electrically determine one network apart 
                    from the other. In other words, equivalent Δ and Y networks 
                    behave identically.  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?  A prime application for Δ-Y conversion is in 
                    the solution of unbalanced bridge circuits, such as the one 
                    below:  
                      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 the add up (or subtract, as is the case 
                    with 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 
                    converted circuit.  At this point, we could take these voltages 
                    and determine resistor currents through the repeated use of 
                    Ohm's Law (I=E/R):  
                      A quick simulation with SPICE will serve to 
                    verify our work:  
                      
 
 unbalanced bridge circuit   
v1 1 0  
r1 1 2 12       
r2 1 3 18       
r3 2 3 6
r4 2 0 18       
r5 3 0 12       
.dc v1 10 10 1  
.print dc v(1,2) v(1,3) v(2,3) v(2,0) v(3,0)    
.end     v1            v(1,2)      v(1,3)      v(2,3)      v(2)        v(3)            
1.000E+01     4.706E+00   5.294E+00   5.882E-01   5.294E+00   4.706E+00 The voltage figures, as read from left to 
                    right, represent voltage drops across the five respective 
                    resistors, R1 through R5. I could have 
                    shown currents as well, but since that would have required 
                    insertion of "dummy" voltage sources in the SPICE netlist, 
                    and since we're primarily interested in validating the Δ-Y 
                    conversion equations and not Ohm's Law, this will suffice.
                     
                      
                      REVIEW: 
                      "Delta" (Δ) networks are also known as 
                      "Pi" (π) networks. 
                      "Y" networks are also known as "T" 
                      networks. 
                      Δ and Y networks can be converted to their 
                      equivalent counterparts with the proper resistance 
                      equations. By "equivalent," I mean that the two networks 
                      will be electrically identical as measured from the three 
                      terminals (A, B, and C). 
                      A bridge circuit can be simplified to a 
                      series/parallel circuit by converting half of it from a Δ 
                      to a Y network. After voltage drops between the original 
                      three connection points (A, B, and C) have been solved 
                      for, those voltages can be transferred back to the 
                      original bridge circuit, across those same equivalent 
                      points.  |