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?
Best Answer
I believe the reason you are having a zero determinant is because the transformation, if it exists, is not unique.
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\$.