(4) convertea to a y RA Re mn 10 V = B R. 4 18 0 Rs 12 0 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...

Extracted text: (4) convertea to a y RA Re mn 10 V = B R. 4 18 0 Rs 12 0 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. Analyzing the circuit now as a series/parallel combination, determine all the resistors value and fill out the tables with the correct figures: RA Rc R4 R5 E Volts Amps Ohms Rg + R4 Rg + R4 Rc + R5 Rc+ R5 Total E Volts Amps R Ohms Upload the resulting tables on the comment section.


Extracted text: A prime application for A-Y conversion is in the solution of unbalanced bridge circuits, such as the one below: R, 12 0 R, 18 0 R3 10 V Rs 18 0 12 0 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 Rg as our load, but what fun would that be? If we were to treat resistors R1. R2, and Rg as being connected in a A configuration (R. Rar and Rar, respectively) and generate an equivalent Y network to replace them, we could turn this bridge circuit into a (simpler) series/parallel combination circuit: Selecting Delta (A) network to convert: A RAR 12 0 RAC 18 Q RBc 10 V B R4 18.0 12 0 After the A-Y conversion...