What determines current through a VS in parallel with a resistor?

Question

This is more of a conceptual question with ideal voltage sources, I am fully aware of its impracticality, but bear with me for the sake of a theoretical discussion. Consider a simple circuit shown below.

^{simulate this circuit – Schematic created using CircuitLab}

now I know that if V1=4V there is 2A of current going through voltage source V1 because of KCL. But in terms of an intuition, is it more correct to say that V1(VS) gets the leftover current from KCL in the node on top of it or should I say that to maintain 4V at the node, it has to consume power and hence the current through it? or are they equivalent statements?

Answer 1

I agree with the 2 amps flowing into V1 and this is more easily seen by converting the 10 volt source and its series 1 ohm resistor to a current source of 10 amps in parallel with that 1 ohm resistor. It's called Norton's theorum. The redrawn circuit becomes: -

^{simulate this circuit – Schematic created using CircuitLab}

Now you have 10 amps in parallel with 1 ohm (R1) and this is applied to the 4 volt source (in parallel with the other 1 ohm resistor, R2). The two 1 ohm resistors are in parallel and these take a current of 8 amps from the 4 volt source (when the 10 amp current generator is ignored).

When the current generator is restored to the circuit (using superposition theorum), that 10 amp source takes over the role of supplying the 8 amps to the two parallel resistors and the remainder (2 amps) goes into the 4 volt source.

So, I would say that it is slightly more intuitive to say that V1 (4 volts) gets the leftover current. However, there isn't any great intuitive way of seeing it without doing some calcs so I'm unsure about the point of your question.