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?

## Comments

Sort by:

TopNewestI think delta - star method can help . – Keshav Tiwari · 2 years, 4 months ago

Log in to reply

– Sophie Crane · 2 years, 4 months ago

Can you give me an explanation/link to an explanation? ThanksLog in to reply

@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, 4 months ago

Log in to reply

– Sophie Crane · 2 years, 4 months ago

Thank you very much. This is an excellent technique and I had never heard of it before. Awesome! :DLog in to reply

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, 5 months agoLog in to reply

– Sophie Crane · 2 years, 5 months ago

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.Log in to reply

very big formula– Aneesh Kundu · 2 years, 5 months agoLog in to reply

– Sophie Crane · 2 years, 5 months ago

It is. That's why I posted this.Log in to reply

– Aneesh Kundu · 2 years, 5 months ago

But u can still assume any value for emf as that wouldn't change the equivalent resistance.Log in to reply

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, 3 months ago

Log in to reply

– Sophie Crane · 2 years, 3 months ago

You are correct on every count. The condition is that the resistors can take on any positive real value.Log in to reply

I got a fraction of 8 terms over 4 terms – Mohamed Ashraf Mostafa · 2 years, 3 months ago

Log in to reply

– Sophie Crane · 2 years, 3 months ago

Did you use delta-wye or something else?Log in to reply