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

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Step I step I considered each active component as:

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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?

LPs
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  • You didn't show the values you got at each stage, for one thing. And I'm particularly curious about your last circuit, too. – jonk Oct 18 '17 at 07:13
  • @jonk Ok I try to edit my post with all calculations. About last I made Y to Alpha transform: resistors have same values so the 3 new resistors are 1/3 ohm each. – LPs Oct 18 '17 at 07:15
  • When you do the Y conversion, you lose R5. It no longer exists. – jonk Oct 18 '17 at 07:22
  • @jonk Ok so that's the problem, I thought (erroneously...) that I can calculate the current on the RA branch of parallel. So: is that the problem? Means that I can't solve it that way? I'm editing the question to show all my calculations. – LPs Oct 18 '17 at 07:26
  • You should get \$I_{R_5}=909.09\:\textrm{mA}\$ in the first circuit; \$I_{R_5}=-454.55\:\textrm{mA}\$ in the second circuit; \$I_{R_5}=2.7273\:\textrm{A}\$ in the third circuit; and finally \$I_{R_5}=4.5455\:\textrm{A}\$ in the last circuit. It is this last one that I wonder about in terms of how you handled it. You might want to do a Y on the other end? – jonk Oct 18 '17 at 07:33
  • @LPs: what is a "Y to Alpha tranform"? – Curd Oct 18 '17 at 07:57
  • @Curd probably something wrong: in Italy we say triangle to star stransform. Is it Delta to Y transform..? – LPs Oct 18 '17 at 07:59
  • @LPs Note that your step two current's sign is wrong. – jonk Oct 18 '17 at 08:02
  • @jonk I corrected, but obviously your answer gave me all I need. Step 4 was wrong. Thank You – LPs Oct 18 '17 at 08:06
  • @LPs Cool. If the answer is what you needed, great! – jonk Oct 18 '17 at 08:07
  • @LPs: also in Germany we call it a "triangle to star transform". The term in English is "Delta Y" (Greek letter capital Delta is a triangle). – Curd Oct 18 '17 at 08:50
  • @Curd I see.I corrected the question title too. ;) – LPs Oct 18 '17 at 08:51

1 Answers1

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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:

schematic

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}\$.

jonk
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  • Ok. Thank you. Is surely what I need. So now I have to understand how I must approach those kind of circuit in the way you shown. Because of all circuits in my exam will be like that and I prefer to use superposition instead of calculate big matrix with Kirchhoff laws. – LPs Oct 18 '17 at 08:35