# Why does the superposition principle for calculating equivalent resistance work for asymmetrical objects?

The method has been discussed in this question: Effective resistance across 2 adjacent vertices of a dodecahedron with each edge $r$

I used this method for an asymmetrical object to calculate its equivalent resistance.

The method shouldn't have worked as the branches connecting the output/input with adjacent vertices aren't symmetrical:

1. Branches connecting adjacent vertices which are on the same hexagon will have the same current.
2. The branch connecting the two hexagons will have different current flow.

I assumed that the current flow will be the same in all three branches and proceeded with solving the question and reached upon the correct answer.

Why is this working, is it a complete fluke or is there some hidden symmetry that I can't spot?

The solution:

Current flowing through the input will be $I$ since there are 12 edges we will assume that $I/12$ is flowing out of each vertices.

Therefore the net current flowing through input is $I-I/12$.

Now I assumed that the current will be distributed evenly across the three branches therefore the current in each branch will be $\frac{I- I/12}{3}$.

Similarly we can calculate the current flowing out of the output.

Superposing the the two cases will give us a net current across the edge, which will be $2×\frac{I- I/12}{3}$.

Finally using $V = 2×\frac{I- I/12}{3}×R$ we can calculate the value of $I$ after plugging bin the values for $V$ and $R$. $I$ is coming out to be $3A$.

• Why do you think the superposition principle should apply only to symmetrical objects? – sammy gerbil Jun 30 '18 at 19:41
• @SammyGerbil I agree that it will work for assymetrical objects. It's just that I assumed the current distribution incorrectly and reached upon the correct answer – Avyansh Katiyar Jun 30 '18 at 20:21

## 1 Answer

Given my answer to a related question, the equivalent resistance of the circuit is $$R_{eq}=R\parallel\frac{113}{67}R=\frac{113}{180}R=\frac{113}{180}(6)=\frac{113}{30}\Omega$$ From $$I-\frac I{12}=\frac V{R_{eq}}=(11)\frac{30}{113}$$ I am getting $$I=\frac{360}{113}\,\text{A}$$ So I don't see why you are convinced that your answer of $I=3\,\text{A}$ is correct.

• It's not a question I just made up. It's from a test series the answer is given as 3 A. – Avyansh Katiyar Jun 25 '18 at 20:36
• I checked my work by two methods: I used both nodal analysis and mesh analysis and also checked that Kirchoff's current law worked in the nodal analysis picture. Lots of answer keys are faulty which you find out if you have to grade them. Have you tried checking whether your assumed currents (I'm not sure from your description what they are) actually solve Kirchoff's voltage law around any loop? – user5713492 Jun 25 '18 at 22:21
• In fact I just now checked (via nodal analysis) that with $\frac{11}{12}I$ going into a node and $\frac1{12}I$ going out all others, the currents to the adjacent nodes in the same hexagon are $\frac{113}{360}I$ while the current to the adjacent node in the other hexagon is $\frac{13}{45}I$ so the assumption that all $3$ branches get the same current just isn't valid. – user5713492 Jun 25 '18 at 22:48
• Thank you. The question was integer type, after rounding off your answer it's coming as 3A. – Avyansh Katiyar Jun 26 '18 at 5:59
• You should not round off the answer, that's what makes both methods to "appear" to give the same answer! Your method gives 3.000A, while the correct method gives 3.1858A. – Guill Jun 28 '18 at 7:29