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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433001978519:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now, as for the physicswhat 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where :dw:1433002153076:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where 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)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1So do you want a strictly physical explanation? (A+ for effort, BTW.)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1The math behind this is simple though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you. It's just the math part that I'm after, I don't know if there's any other way to solve this.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1First 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Assume that wire is circular throughout. dw:1433003575341:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Let 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\[R_{\rm eff} (x) = \cdots \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1For 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's...something, yeah! Never thought about taking it per unit length though.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Uniformity is an important condition. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yes, that is correct. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let R=(R1+R2)/2 and d=R1R=RR2 (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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Also, just an idea. Symmetry is always critical when considering maxima and minima. Consider the AMGMHM inequality below. Equality holds iff all numbers are equal. That's intuitive. \[\dfrac{R_1 + R_2}{4} \ge \dfrac{\sqrt{R_1 R_2}}{ 2} \ge \dfrac{R_1 R_2}{R_1 + R_2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since 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)*(Rd) / (R1+R2)  the equivalent resistance when R1=/=R2 is equivalent to proving that R^2 > (R+d)*(Rd)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And R^2 is always greater than (R+d)*(Rd)=R^2d^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The minimum is found is a similar manner  suppose the equivalent resistance is (R+d)*(Rd)=R^2d^2 where R=(R1+R2)/2 and d=R1R=dRR2 (when R1>R2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since R is constant (because R1+R2 is constant) then the difference R^2d^2 is minimum when d goes to infinity.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm sorry for being confusing and not making much sense here.
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