Question

Hi,
For PS6, for the second problem we are supposed to use the Lagrangian right? I'm struggling with this one and I want to know if I'm not totally off.

Answer 1

Yes, we are supposed to use Lagrange multipliers. For example, in part a) of the problem we want to determine what choice of I1 and I2 will minimize energy loss. So we want to minimize
\[f(I_{1},I_{2})=I_{1}^{2}R_{1}+I_{2}^{2}R_{2}\] (this is our total energy loss)
But we have the condition \[g(I_{1},I_{2})=I_{1}+I_{2}=I\]
So \[\nabla f=\lambda \nabla g\] or (in form of the system of equations)
\[2I_{1}R_{1}=\lambda\]
\[2I_{2}R_{2}=\lambda\]
\[I_{1}+I_{2}=I\]
Solvig this we have
\[I_{1}=IR_{2}/(R_{1}+R_{2})\] and \[I_{2}=IR_{1}/(R_{1}+R_{2})\]

Answer 2

Thx for your answer!
I'd like to know where you got the first equation from because to me the equation is IR^2=E (lost) . And it is the equation I was looking for because without it it doesn't make much sense, I had the condition though. Thx again

Answer 3

I used the equation you wrote IR^2=E. For the resistor R1 we have:
\[E_{1}=I_{1}R _{1}^{2}\]
for R2:
\[E_{2}=I_{2}R _{2}^{2}\]
Total energy lost for our circuit is the sum of E1 and E2:
\[E=E_{1}+E_{2}=I_{1}R_{1}^{2}+I_{2}R_{2}^{2}\]

Answer 4

Well I f'ed up, I thought the equation was complete I just plugged I2 + I1 = I but the two should be equal to one another right? But the thing is, I checked and they're not and that's puzzling me a lot.

Answer 5

Damn I made a mistake, it's ok they are as expected. But doesn't it work with the two different equation considering the fact that they're equal?

Answer 6

Do you mean what would be if I1=I2? I1 may be equal I2 if (and only if) R1=R2. And then I1=I2=I/2 (this is based on common sense, also you can get this from the equations for I1 and I2 I wrote in my first reply letting R1=R2). But if I1=I2 there is no problem, we have no need to choose these currents, they are already chosen :)

Answer 7

No, it's not what I meant. The thing is I first to substitute I1+I2 into E=I^2R which is actually equal to I2^2R2 + I1^2R1 = E but it didn't work using the lagragian. My question was Why doesn't it work?

Answer 8

It doesn't work because (I1+I2)^2*R is not equal to I1^2*R1+I2^2*R2. (I1+I2)^2*R equals to (I1^2+2*I1*I2+I2^2)*R, where R=(R1*R2)/(R1+R2) (resistance of parallel curcuit). But we were asked to minimize the sum of energy losts on resistors R1 and R2.

Answer 9

Yes it is equal (I1^2+2*I1*I2+I2^2)*R = I1^2*R1+I2^2*R2=(I1+I2)^2*R , I checked with numbers. So why is the first expression not working?

Answer 10

No, (I1^2+2*I1*I2+I2^2)*R is not equal to I1^2*R1+I2^2*R2 (this is why the first equation doesn't work). I think there is some mistake in your calculations. Can you show me how you checked this with numbers?

Answer 11

Yes they are equal, if you replace R= R1R2/(R1+R2) and you set for example R1=2 R2=1 I2= 3 I1=1 it yields the exact same answer no matter what equation you use . But I think that, in order to use the lagrangian, you can't substitute your g=c into your f, but I don't know why theoretically.

Answer 12

OK, for R1=2, R2=1, I2=3 and I1=1 we have I1^2*R1+I2^2*R2=1*2+9*1=11 and (I1^2+2*I1*I2+I2^2)*R1*R2/(R1+R2)=(1+2*1*3+9)*2*1/(2+1)=16*2/3=32/3. So, acutally, they are not equal. Even in physical sense they are different quantities.
What about substituting g=c into f, I think you can do it because in this method you have to deal with derivatives of f and g and g=c itself, so the system of equations you will get is solvable (theoretically). But I'm not quite sure about it, I need to think.

Answer 13

You're right, I checked twice though. Boh whatever, tell me what you came up with about the difference between the two.

thx for your devotion

Answer 14

Well, this is a delicate moment. The quantity I1^2*R1+I2^2*R2 is just sum of energy losts for resistors R1 and R2. It is as if we have two circuits, one with resistor R1 and current I1 and another with resistor R2 and current I2 (and we also have a condition I=I1+I2), and we are asked to minimize their summary energy lost. The quantity (I1+I2)^2*R is the energy lost of parellel circuit with resistors R1 and R2 (we can replace it by circuit with resistance R=R1*R2/(R1+R2) and current I=I1+I2). I hope, you will agree that these two cases are very different things.
So there is an ambiguity in the phrase "total energy lost" because it is not clear "total energy lost" of what. And this is why you made a mistake, so I think it is not your fault. Honestly speaking I also wantd to plug I=I1+I2 in the equation E=I^2*R, but I thought we were not supposed to know formulae from the electric circuit theory (and we need it for the total resistance R=R1*R2/(R1+R2)) so I decided to minimize the first quantity.

Sorry for my poor English, I'm russian and I do my best :)