# Delta-star transformation (transition to initial currents)

I go from the first circuit to the second one using delta-star transformation.

Resistances $$R_1,R_2,R_3,R_4,R_5$$ are given. Then we have $$R_{12}=\frac{R_1R_2}{R_1+R_2+R_3}$$, $$R_{13}=\frac{R_1R_3}{R_1+R_2+R_3}$$, $$R_{23}=\frac{R_2R_3}{R_1+R_2+R_3}$$. Also the source voltage $$U$$ is set. So, we can calculate all currents in the second circuit. But how can we go to the currents in the first circuit, namely express $$I_1,I_2,I_3,I_4,I_5$$ in terms of $$U,R_1,R_2,R_3,R_4,R_5$$?

• I think that most physicists would use Kirchhoff's two circuit laws. I don't think that the star-delta or delta-star transformation helps in this task. May 26, 2023 at 20:45
• Thanks, I thought this task is the "canonical" example to use the delta-star transformation. May 27, 2023 at 15:45
• Puk is right and the second sentence of my comment is wrong. Sorry. May 27, 2023 at 15:52

After you solve the second circuit and determine the voltages of all nodes, you have also determined the voltages of all nodes in the first circuit. For example, the node between $$R_{13}$$ and $$R_4$$ in the second circuit is the same as the node shared by $$R_1$$ and $$R_4$$ in the first circuit.
Since you have all the node voltages, determining currents through $$R_1$$, $$R_2$$ and $$R_3$$ is just straightforward application of Ohm's law. You should already have the currents through $$R_4$$ and $$R_5$$ from solving the second circuit.
• Thanks, actually, I've written $I_1R_1=I_{12}R_{12}+I_{13}R_{13}$ and $I_2R_2=I_{12}R_{12}+I_{23}R_{23}$, and then found all initial currents $I_1,...,I_5$. I checked it and it's right. May 27, 2023 at 15:49