# Resolving this circuit diagram

I feel like I'm missing something really basic here, but why wouldn't V1 have any impact on the voltage at V3 in this diagram?

• Welcome! V2 is shorted to V3. Try to simulate it. Apr 19 at 6:35

V2 is an ideal voltage source. By definition, V2 is is fixed at "V2" independent of all other circuit elements. V3 = V2 because V2 IS the same node as V3. By definition, all nodes have the same potential everywhere on that node.

Of course, these are only theoretical concepts which are what the simulators are built on in the back end. In practice, V1 will influence V2 (to an arbitrary degree of precision) when you consider practical factors like cell impedance and wire resistance.

As an exercise, model your batteries properly and add 100mOhm resistor in series with each one of them. Re run the simulation. Depending on the value of R1 and the precision you have to measure voltage, you may be able to resolve real world effects. Try R1 = 100 ohms to start. V1 should be appreciably different from V2 to see a change in simulation behavior.

• Thanks for the response! So is there anyway I could have arrived at that conclusion via nodal/mesh analysis or is it just a matter of recognizing it as a common node and as such have to have the same voltage? Apr 20 at 4:06
• @BuiltDiff It's a matter of recognizing it as a common node. You only have 3 nodes: Ground, V1, and V2. V3 is the same node as V2. You don't get more nodes simply by adding more names to the same connection (node). Apr 21 at 2:59

Because node 3 (V3) is defined by voltage source V2.

If there would be a resistor in series with V2 connecting to the right terminal of R1, then there would be a contribution of V1 as well.

why wouldn't V1 have any impact on the voltage at V3 in this diagram?

Imagine that V2 was chosen so that it produced 0 volts. I mean, it can have any value you want i.e. all are valid but, imagine it is 0 volts. Do you now see how V1 can have no impact. V2 will remain at 0 volts and, so will V3.