circuit analysis - Superposition problem with $Delta$-Y conversion

I'm trying to use superposition to solve the following linear circuit:

Step I step I considered each active component as:

But results are always wrong. Same procedure on many other circuits give me right answers, but not this one, for some obscure reason.

For example calculating I5

Step 1

Calculate parallel $$RP=\frac{(R5+R4)\cdot R6}{(R5+R4)+R6}=\frac{2\cdot 1}{2+1}=\frac{2}{3}$$ Calculate serie $$R_{tot}=R1+R3+RP=3+\frac{2}{3}=\frac{11}{3}$$ Current for the equivalent circuit (R1 and R3 current) $$I_{tot}=\frac{E1}{R_{tot}}=\frac{10}{\frac{11}{3}}=\frac{30}{11}A$$ Current for R5 $$I_{R5}=I_{tot}\cdot\frac{R6}{(R5+R4)+R6}=\frac{30}{11}\cdot\frac{1}{3}=\frac{10}{11}A$$

Step 2

Quite the same as Step 1 but with half of voltage and opposite sign $$I_{tot}=\frac{E3}{R_{tot}}=\frac{5}{\frac{11}{3}}=\frac{15}{11}A$$ Current for R5 $$I_{R5}=I_{tot}\cdot\frac{R6}{(R5+R4)+R6}=\frac{15}{11}\cdot\frac{1}{3}=\frac{5}{11}A$$

Step 3

Calculate parallel $$RP=\frac{(R5+R4)\cdot (R3+R1)}{(R5+R4)+(R3+R1)}=\frac{2\cdot 3}{2+3}=\frac{6}{5}$$ Calculate serie $$R_{tot}=R6+RP=1+\frac{6}{5}=\frac{11}{5}$$ Current for the equivalent circuit (R6 current) $$I_{tot}=\frac{E6}{R_{tot}}=\frac{10}{\frac{11}{5}}=\frac{50}{11}A$$ Current for R5 $$I_{R5}=-I_{tot}\cdot\frac{(R3+R1)}{(R5+R4)+(R3+R1)}=-\frac{50}{11}\cdot\frac{3}{5}=-\frac{30}{11}A$$

Step 4

\$\Delta\$-Y conversion In circuit 3 resistors have the same value so the equivalent Y have 3 resistors of value $$\frac{1}{3}$$ Calculate parallel $$RP=\frac{(R1+RA)\cdot (R3+RB)}{(R1+RA)+(R3+RB)}=\frac{\frac{4}{3}\cdot \frac{7}{3}}{\frac{4}{3}+\frac{7}{3}}=\frac{28}{33}$$ Calculate serie $$R_{tot}=RC+RP=1+\frac{28}{33}=\frac{13}{11}$$ Current for branch RA $$I_{tot}=A5\cdot\frac{(R3+RB)}{(R1+RA)+(R3+RB)}=10\cdot\frac{\frac{7}{3}}{\frac{4}{3}+\frac{7}{3}}=\frac{70}{11}A$$

So I5 should be $$I5=\frac{10}{11}-\frac{5}{11}+\frac{30}{11}+\frac{70}{11}=\frac{105}{11}A$$

Which is wrong because of the result must be $$\frac{85}{11}A$$

The same if I consider RB branch instead of RA.

What's wrong with my procedure?

Answer

I'm going to assume you were able to handle the currents correctly in all of your circuits except for the last one. I think you laid them out correctly and I'm fairly confident in your ability to work out the magnitude and direction, in those cases.

So here is the final circuit case where I think you messed up by mushing up \$R_5\$ into your \$\Delta\$-Y conversion. You need to do that conversion differently, as shown here:

^{simulate this circuit – Schematic created using CircuitLab}

Note that I didn't destroy \$R_5\$ with a \$\Delta\$-Y conversion. I chose, instead, to convert the other \$\Delta\$ so that \$R_5\$ was left intact.

At this point we don't care at all about \$R_A\$. All the current flows through it. We only care about how it splits, after that. Here, you can see that the two branches are \$1.25\:\Omega\$ and \$1.5\:\Omega\$. This means a \$5:6\$ ratio. In other words, \$\frac{5}{11}\$ths of the current is going to flow through the branch containing \$R_5\$. So the current in \$R_5\$ here is \$\frac{5}{11}\cdot 10\:\textrm{A}=4.\overline{54}\:\textrm{A}\$.