anonymous one year ago Challenging simple application of math in physics. I've encountered this problem during my study of simple circuits. The physics behind it is simple and I'll explain it below. What's curious is that the only explanation I found for it was purely a mathematical one. Given the following circuit, asume that only the circular wire has resistance and the rest of the circuit has an electrical resistance of 0. Knowing that you can choose to relocate points A and B explain and prove where these points should be placed so that the equivalent resistance of the circular wire is minimal/maximal

1. anonymous

|dw:1433001978519:dw|

2. anonymous

Now, as for the physics-what we're dealing with is a parallel resistances. The formula for the equivalent resistance when dealing with 2 parallel resistances is: equiv. resistance=(R1*R2)/(R1+R2)

3. anonymous

Where :|dw:1433002153076:dw|

4. anonymous

In this case, we have a circle with resistance and we are free to relocate points A and B as we please. In this equivalence, one arc is a resistance the the other arc is another resistance. |dw:1433002287521:dw|

5. anonymous

The rest of the circuit is of no importance to us because it has no resistance. You can see how relocating points A and B changes what values R1 and R2 have: |dw:1433002398171:dw|

6. anonymous

Where should these two points be so that our equivalent resistance is maximal and/or minimal and why ? Let me remind you that only the circle has resistance and the formula the equivalent resistance in this case is: (R1*R2)/(R1+R2)

7. ParthKohli

So do you want a strictly physical explanation? (A+ for effort, BTW.)

8. ParthKohli

The math behind this is simple though.

9. anonymous

Thank you. It's just the math part that I'm after, I don't know if there's any other way to solve this.

10. ParthKohli

First of all, the answer is that the resistance is maximum when $$R_1 = R_2$$. It is minimum when you do not connect the circular wire to the circuit.

11. ParthKohli

Assume that wire is circular throughout. |dw:1433003575341:dw|

12. anonymous

I guess I forgot to specify that unplugging everything isn't an option lol. But yes, you're right - it is maximum when R1=R2 and minimum when either R1 or R2 tends to 0. Any thought on how to prove it ?

13. ParthKohli

Let the resistance per unit length be $$\dfrac{R}{\ell }$$. Then that of the upper part is $$\dfrac{Rx}{\ell}$$ and that of the lower part is $$\dfrac{R(\ell - x)}{\ell}$$ The effective resistance is$R_{\rm eff} (x) = \cdots$

14. ParthKohli

For simplicity, you can just call the resistance per unit length $$\lambda$$.$R_{\rm eff}(x) = \dfrac{\lambda^2 x(\ell - x)}{\lambda\ell }$It is maximum when $$x = \dfrac{\ell}{2}$$ using calculus or completing the square.

15. anonymous

That's...something, yeah! Never thought about taking it per unit length though.

16. ParthKohli

Uniformity is an important condition. :)

17. anonymous

The idea is that R1+R2 is always the same no matter where you move those two points (the length of the circle stays the same after all).

18. ParthKohli

Yes, that is correct. :)

19. anonymous

Let R=(R1+R2)/2 and d=R1-R=R-R2 (in the case where R1>R2 - what I mean by that is that d is that difference between either resistances and R, which is the same no matter where you put those two points.

20. ParthKohli

Also, just an idea. Symmetry is always critical when considering maxima and minima. Consider the AM-GM-HM inequality below. Equality holds iff all numbers are equal. That's intuitive. $\dfrac{R_1 + R_2}{4} \ge \dfrac{\sqrt{R_1 R_2}}{ 2} \ge \dfrac{R_1 R_2}{R_1 + R_2}$

21. anonymous

For instance, if R1+R2=10, suppose R1=8 and R2=2 - then R=5 and d=3 here. If R1=7 and R2=3 then R=5 and d=2.

22. anonymous

Since R1+R2 is constant no matter what, to prove that R^2 / (R1+R2) - the equivalent resistance when R1=R2 that is - is greater than (R+d)*(R-d) / (R1+R2) - the equivalent resistance when R1=/=R2 is equivalent to proving that R^2 > (R+d)*(R-d)

23. anonymous

And R^2 is always greater than (R+d)*(R-d)=R^2-d^2

24. anonymous

The minimum is found is a similar manner - suppose the equivalent resistance is (R+d)*(R-d)=R^2-d^2 where R=(R1+R2)/2 and d=R1-R=dR-R2 (when R1>R2)

25. anonymous

Since R is constant (because R1+R2 is constant) then the difference R^2-d^2 is minimum when d goes to infinity.

26. anonymous

But since physically you can't have a negative resistance - 0 is your bottom limit so I guess the correct answer in this case would be that it's minimal when d^2 tends to R^2 or when d tends to R (since there are only positive values because we're dealing with resistances) - which basically translates to: "The further you are from (R1+R2)/2 the lower the equivalent resistance will be"

27. anonymous

I'm sorry for being confusing and not making much sense here.