>>11717111Simplest answer is that it doesn’t need to. The 20V source needs to put out 1A in order to drop 10 volts across the lower load. So you have 1A going into the node on the left. You need 1A across the top load in order to drop the remaining 10 volts, which you already have, so the 10V source doesn’t need to add any current. Using either superposition or loop analysis will show that the current through the 10V source is indeed 0.
In reality though, you don’t know how much current either source will truly put out. Using circuit b as an example, there will be 1A through the load, but there’s no way of telling which source is providing how much, and no, it’s not necessarily half like you may think. Going back to a, voltage sources are regarded as shorts and they will try to push current through each other. Both sources could theoretically be putting out 2A, with the 20V current going through the 10V source without ever touching the upper load, and the 10V current splitting at the node. It would still match the analysis. The arbitrary currents is because the 10V source is redundant. It’s removal would not effect the circuit analytically, but it’s inclusion changes the physical nature in a way that’s not easy, or even possible to consider. Same on the right.
As a side note, the circuits are not equivalent. Circuit a does not reduce to b. You can’t just replace a voltage source and resistive load with a lower voltage source, that changes the loading of the circuit.
