KCL in 3 loops circuit and two batteries?

Redrawn in standard topology for clarity...

• node voltages are organized vertically from highest to lowest voltage
• ground is usually at the bottom, but in this case there is a negative voltage lower than ground
• signal flow is organized horizontally from left to right

^{simulate this circuit – Schematic created using CircuitLab}

This circuit has three nodes, and the voltage of each node is fully determined. Each of the branches behaves independently. Since these are ideal voltage sources, there is no way for the branch resistance or mesh/loop currents to affect the node voltages.

If this were built as a practical circuit, then this circuit model is not sufficient to capture everything that would happen:

• wires have resistance
• batteries and other energy sources have internal source resistance
• transient effects can affect the observed results
• temperature can affect real component values

The math is pretty easy on this one, it's just applying Ohms's law; the entire difficulty is in visualizing the schematic. Getting in the habit of putting the schematic in a standard format helps make it easier for other engineers to understand (which is a big part of my job professionally). With some experience you will start to recognize certain standard structures like resistive dividers, Wien bridges, CE amplifiers, and so on.

Here's the standard Q&A about schematics: Rules and guidelines for drawing good schematics

V1 and R1 are parallel, so the voltage across R1 is determined completely and always by V1. Therefore, the current is determined by V1/R1. Always.

V2 and R2 are parallel, so the current through R2 is likewise determined by V2 only.

V1 and V2 are not parallel. Resistor R3 connects between them. The current through R3 will depend on the difference between V1 and V2.

Why the second battery doesn't affect the current that passe through R1?

Because R1 is in parallel with V1. Whatever the voltage is produced by V1, exactly that voltage is across R1.

If I try this with a real circuit, will it behave this way?

Only if you use a form of voltage source that works with current flowing "backwards" through it.

aside

Notice the circuit is very easy to solve.

The voltage at node a is exactly the value of V1.

The voltage at node b is exactly the voltage at a minus the value of V2.

From there you can find all the resistor currents very easily.

Also, since every node has a voltage source connected to it, you can't really use KCL to solve the circuit. You would end up combining all the nodes into a supernode.