Electrical – Distribution of Current in Kirchoff’s Junction Law

You can use the Shockley equation, which is another idealization, but a better one than a fixed diode drop. To apply it, you have to know, for the given diode, all the parameters:

$$I=I_\mathrm{S}\left( e^{V_\mathrm{D}/(n V_\mathrm{T})}-1 \right)$$

I would suggest to replace the transistor with an actual diode model, from which you can pull out the parameters and calculate by hand.

In this answer to a question, I was working with the formula, in relation to the behavior of a diode model in CircuitLab. See the question's "appendix" at the bottom.

I have redrawn the circuit so I can name the nodes and give each component a name. I have also chosen the bottom of Bat1 as our reference 0V so I can assign voltages to each node.

^{simulate this circuit – Schematic created using CircuitLab}

Kirchoff's current law states that the net current into (or out) of any node is zero.

Now because the current in R1 flows out of the positive terminal of BAT1 and the current in R2 flows into the negative terminal BAT2 then \$I_{R1} = I_{R2}\$. You can't have two different currents in a series circuit. This means the current in R3 must be zero.

We can no work out the Voltages at each node:

\$V_{N1}=6 \text{ V } \$, \$V_{N2} = 6 \text{ V } \cdot \dfrac{2 \Omega}{1 \Omega + 2 \Omega} = 4 \text{ V }\$

There is no current in R3 so \$ V_{N3} = V_{N2} = 4 \text{ V }\$

We can see from the loop involving BAT2, R4 and R5 that R4 and R5 each have 5V across them so:

\$V_{N4} = 4 \text{ V } + 5 \text{ V } = 9 \text{ V } \$ and \$V_{N5} = 4 \text{ V } - 5 \text{ V } = -1 \text{ V } \$