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?

#Algebra #AlgebraicManipulation #ElectricityAndMagnetism #Resistance #Bash

Easy Math Editor

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Comments

I think delta - star method can help .

Keshav Tiwari - 6 years, 7 months ago

Can you give me an explanation/link to an explanation? Thanks

Sophie Crane - 6 years, 7 months ago

@Sophie Crane It's trivial using Y- $\Delta$ transformation. Just apply this on any of the ends of $\text{R}_{5}$ .

Ishan Singh - 6 years, 6 months ago

@Ishan Singh – Thank you very much. This is an excellent technique and I had never heard of it before. Awesome! :D

Sophie Crane - 6 years, 6 months ago

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 - 6 years, 7 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.

Sophie Crane - 6 years, 7 months ago

That would be a very big formula

Aneesh Kundu - 6 years, 7 months ago

@Aneesh Kundu – It is. That's why I posted this.

Sophie Crane - 6 years, 7 months ago

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

Aneesh Kundu - 6 years, 7 months ago

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 - 6 years, 6 months ago

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

Sophie Crane - 6 years, 6 months ago

I got a fraction of 8 terms over 4 terms

Mohamed Ashraf Mostafa - 6 years, 6 months ago

Did you use delta-wye or something else?

Sophie Crane - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$