In the circuit above, can someone please help me understand how to find the voltage differences?
I already converted the delta connection in the middle into WYE to no avail.
I assumed that current will flow through the delta connection in the middle.
Can you please help me understand why no current will flow through it?
Current won't flow unless a source (or time variant magnetic field) drives charges.
In the circuit depicted, each square containes a source; current will flow.
If one connects two of the squares with a resistor (or short), the nodes such connected will be at the same potential. There is no source driving charges between squares (else there would be a current, resulting in a voltage "drop" at the resistor).
Nothing changes connecting the third square.
Make the connection the reference potential out of convenience.
Think about this with superposition. Kill the 80V and the 40V sources and look only at the 60V source. The current going into one end of the source is the same as the current going out of the other end of the source. Think about how that works when applying KCL to the nearest node in the delta. There are four branches connected to the node. Two branches are carrying equal and opposite currents from the 60V source. There's no current left for the other two branches.
You can apply the same logic to the 80V and the 40V sources. By superposition, all three sources together will also not drive any current through the delta. Thus, the voltage drops across the delta resistors are zero.
Now the problem is easy. Calculate the voltage drops across the resistors in each loop. Be sure to draw the polarities correctly. Then follow the path between two nodes to find the voltage difference.
If you redesign the circuit as this one ... it would be "evident".
It is three independent circuits which are wired with a "single" wire, thus no current through the wires.
No current flows in the resistors composing the central triangle because there is no potential difference between the terminals of each of them. Therefore, the meshes with the three independent sources carry currents that do not interfere with each other, so they could easily be electrically separated from each other without any changes in the currents and voltages. This is also demonstrated by performing an analysis based on the network topology and the node-pair voltage method, as follows (by assumption all resistors are equal to R1.But in fact even with all different resistors, the voltages and currents of the central triangle are zero. ):
The topology of your circuit makes it impossible for any current to enter or leave the central triangle of resistors in any of its corners, because what is connected to those corners has no path for this current to go, neither to the outside world, nor to the other corners. And in a triangle of resistors with no current fed into any corner, the resistors will remain current-less, with no voltage between their terminals. This means we can replaced them by shorts, giving just one node in the center.
While we're at it, we can just as well choose that as our reference node, since the circuit didn't have one yet, leading to this circuit: (where the voltages are trivial to compute!)
The voltage differences are now easy to read off the circuit. Note that choosing another reference node (or another reference voltage than 0V) would not change those differences, so the answer obtained in this way is the correct one.
