# Y-Δ transform with generators. Equivalence of three-terminal devices

So my book asks me to find $$\i_a, i_b, i_c\$$ (as functions of $$\v_1, v_2, v_3\$$) as well as $$\G_a, G_b, G_c\$$ such that the circuit on the right picture (shown below) is equivalent to the circuit on the left. First he suggests to use Y-Δ transform to find $$\G_a, G_b, G_c\$$ with equivalent resitances $$\R_1, R_2, R_3\$$. This is what I tried (note that this exercise is about so-called shift properties, where you duplicate generators in order to transform them into an equivalent form): As you can see I duplicated each current source adding a short circuit (which it can be shown not to invalidate Kirchhoff's laws, thus obtaining parallel of the current source and resistance, which ultimately can be transformed to a voltage source). However, the resulting linear system has 0 determinant: what is wrong with my reasoning? How would you solve this exercise?

Firstly, I'd like to redefine $$\I_b\$$ in the opposite orientation to restore the symmetry (It's not really a big deal though.)
Thevenin's Theorem says that when we look at the circuit from any 2 terminals, say $$\A\$$ and $$\B\$$, it will be indistinguishable from a resistor connected in series to a voltage source.
We can measure the Thevenin's resistance by using a capacitor and observing it's time constant. The transformation from $$\R\$$'s to $$\G\$$'s should therefore be identical to the Y-$$\\Delta\$$ transform.
The problem arises when we try to solve for the currents $$\I_a\$$, $$\I_b\$$ and $$\I_c\$$. Suppose that $$\(I_a, I_b, I_c)\ = (I_1, I_2, I_3)\$$ is a solution that gives the correct open-circuit voltage between all possible pairs of terminals (i.e. $$\AB, AC, BC\$$). Then you can verify that $$\(I_a, I_b, I_c)\ = (I_1+\Delta I, I_2+\Delta I, I_3+\Delta I)\$$ will also be a solution for any $$\\Delta I\$$.