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Non-trivial equivalent resistance.

At school we have been learning the rules for equivalent resistance in series and parallel (which I covered at least two years ago, so I'm bored) and I asked my teacher about the circuit shown, which cannot be decomposed into a collection of series and parallel circuits.

I spent most of a lesson just bashing it with algebra, and in the end I found a three-line fraction for \(R_{eq}\) in terms of \(R_{1}\), \(R_{2}\), \(R_{3}\), \(R_{4}\) and \(R_{5}\).

So I was wondering: is there a nice method?

Note by Sophie Crane
2 years, 2 months ago

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I think delta - star method can help . Keshav Tiwari · 2 years, 2 months ago

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@Keshav Tiwari Can you give me an explanation/link to an explanation? Thanks Sophie Crane · 2 years, 2 months ago

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@Sophie Crane @Sophie Crane It's trivial using Y-\(\Delta\) transformation. Just apply this on any of the ends of \(\text{R}_{5}\). Ishan Singh · 2 years, 1 month ago

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@Ishan Singh Thank you very much. This is an excellent technique and I had never heard of it before. Awesome! :D Sophie Crane · 2 years, 1 month ago

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If u use loop rule instead of junction rule, the expression would have been simpler. For making it even simpler u can assume the circuit to be connected across a good emf(I mean u can take different emfs if values of the resistances are known) Aneesh Kundu · 2 years, 2 months ago

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@Aneesh Kundu I used both. The emf is an arbitrary value. All resistances are unknown, and are to be treated as algebraic variables. I am trying to find a general expression for the equivalent resistance in terms of these variables. Sophie Crane · 2 years, 2 months ago

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@Sophie Crane That would be a very big formula Aneesh Kundu · 2 years, 2 months ago

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@Aneesh Kundu It is. That's why I posted this. Sophie Crane · 2 years, 2 months ago

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@Sophie Crane But u can still assume any value for emf as that wouldn't change the equivalent resistance. Aneesh Kundu · 2 years, 2 months ago

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Correct me if I'm wrong, but isn't this the outline of a Wheatstone bridge? If \(\dfrac{R_2}{R_4}=\dfrac{R_1}{R_3}\), then the equivalent can be easily calculated since the resistance \(R_5\) will be ineffective. Is that condition given, or the variables \(R_1,R_2,R_3,R_4,R_5\) can have any positive real value? Prasun Biswas · 2 years, 1 month ago

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@Prasun Biswas You are correct on every count. The condition is that the resistors can take on any positive real value. Sophie Crane · 2 years, 1 month ago

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I got a fraction of 8 terms over 4 terms Mohamed Ashraf Mostafa · 2 years, 1 month ago

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@Mohamed Ashraf Mostafa Did you use delta-wye or something else? Sophie Crane · 2 years, 1 month ago

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