Lagrange Multipliers and Energy Loss Question

[cwill53]

Homework Statement

In this problem we examine how electricity flows through circuits to minimize energy.
A current I flowing over a resistor R results in an energy loss (in the form of heat/light)
equal to $I^{2}R$ per second. It turns out that, in a sense, ”electricity prefers to flow in
the way that minimizes energy loss to resistance”. For example, when an electric
current comes to a fork, it will divide itself up in such a way that a large portion
of the current flows where the resistance is low and a small portion flows where the
resistance is high (you might think all the electricity would flow where the resistance
is low but the energy loss is proportional to $I^2$ so it is better to spread the current
around).
Suppose we have the following situation where a current I comes to a pair of resistors
in parallel (see picture below)

a)Determine what choice of $I_1$ and $I_2$ will minimize energy loss and hence determine
what the currents will be along the two paths. (Alternatively, if you already are
familiar with resistors in parallel and current flows, verify that the currents $I_1$ and $I_2$
do in fact minimize energy loss).

Relevant Equations

$$P_{diss}=I^{2}R$$
$$\nabla f(x,y)=\left \langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y} \right \rangle$$

Constraint: $I=I_{1}+I_{2}$
$P_{diss,R_{1}}=I_{1}^{2}R_{1}$;$P_{diss,R_{2}}=I_{2}^{2}R_{2}$
We want to minimize $P_{diss,TOT}=I_{1}^{2}R_{1}+I_{2}^{2}R_{2}$
$$f(I_{1},I_{2})=I_{1}^{2}R_{1}+I_{2}^{2}R_{2};g(I_{1},I_{2})=I_{1}+I_{2}=I(constraint)$$
$$\nabla f= \left \langle \frac{\partial f}{\partial I_{1}},\frac{\partial f}{\partial I_{2}} \right \rangle=\left \langle 2I_{1} R_{1},2I_{2}R_{2}\right \rangle$$
$$\nabla g= \left \langle \frac{\partial g}{\partial I_{1}},\frac{\partial g}{\partial I_{2}} \right \rangle=\left \langle 1,1 \right \rangle$$
$\nabla f=\lambda \nabla g;2I_{1}R_{1}=\lambda ;2I_{2}R_{2}=\lambda ;I_{1}+I_{2}=I(constraint)$
$I_{1}=I-I_{2};I_{2}=I-I_{1}$
$$I_{1}=\frac{2(I-I_{1})}{2R_{1}}=\frac{IR_{2}-I_{1}R_{2}}{R_{1}}\Rightarrow R_{1}=\frac{IR_{2}}{I_{1}}-\frac{I_{1}R_{2}}{I_{1}}\Rightarrow I_{1}=I\frac{R_{2}}{R_{1}}-R_{2}$$
Where did I go wrong? The solutions have it that
$$I_{1}=\frac{R_{2}}{R_{1}+R_{2}}I$$

 

[cwill53]

Here's the picture, I didn't want to edit the original post because on my computer LaTeX messes up and I have to retype equations.

 

[TSny]

Check your math steps in going from

$\nabla f=\lambda \nabla g;2I_{1}R_{1}=\lambda ;2I_{2}R_{2}=\lambda ;I_{1}+I_{2}=I(constraint)$
$I_{1}=I-I_{2};I_{2}=I-I_{1}$

to

$I_{1}=\frac{2(I-I_{1})}{2R_{1}}$

Note that the last equation is dimensionally inconsistent.

 

[cwill53]

Check your math steps in going from

to
Note that the last equation is dimensionally inconsistent.

I forgot to put the $R_{2}$ in the TeX. It’s there on my paper and I still got the question wrong.

 

[TSny]

With $R_2$ in place, you should get the correct answer.

 

[cwill53]

With $R_2$ in place, you should get the correct answer.

It appears that I’m not getting the right answer however. Either that or what the solutions say is incorrect.

 

[TSny]

$$R_1 =\frac{IR_{2}}{I_{1}}-\frac{I_{1}R_{2}}{I_{1}}\Rightarrow I_{1}=I\frac{R_{2}}{R_{1}}-R_{2}$$

Check this. The left side of the arrow looks ok to me. But, the right side can't be right. It is dimensionally inconsistent.

 

[cwill53]

Check this. The left side of the arrow looks ok to me. But, the right side can't be right. It is dimensionally inconsistent.

Thank you so much for helping me see that. It’s always a slip up with the algebra!

 

[cwill53]

Part (b) supposes we have 3 resistors in parallel instead, such that

$$f(I_{1},I_{2},I_{3})=I_{1}^{2}R_{1}+I_{2}^{2}R_{2}+I_{3}^2R_{3};g(I_{1},I_{2},I_{3})=I_{1}+I_{2}+I_{3}=I(constraint)$$
$$\nabla f=\left \langle 2I_{1}R_{1},2I_{2}R_{2},2I_{3}R_{3} \right \rangle$$
$$\nabla g=\left \langle 1,1,1 \right \rangle$$
$$\nabla f=\lambda \nabla g$$

$$2I_{1}R_{1}=\lambda ;2I_{2}R_{2}=\lambda ;2I_{3}R_{3}=\lambda ;I_{1}+I_{2}+I_{3}=I(constraint)$$

I've spent about an hour going in circles with algebraic manipulations. How should I approach this?

 

[TSny]

Can you see a way to express $I_2$ in terms of $I_1$ and a way to express $I_3$ in terms of $I_1$?

 

[cwill53]

Can you see a way to express $I_2$ in terms of $I_1$ and a way to express $I_3$ in terms of $I_1$?

I tried that, but I kept going in circles with the algebra.

 

[TSny]

You'll need to show the steps you tried. After you expressed $I_2$ and $I_3$ in terms of $I_1$, what was your next step?

 

[cwill53]

$I_{1}=I-I_{2}-I_{3};I_{2}=I-I_{1}-I_{3};I_{3}=I-I_{1}-I_{2}$

I had the above equations, and I tried to use substitution at various steps to try to get an answer. I had about half a page full of algebra, but I erased it all unfortunately. I ended at the same place I started in terms of the algebra.

 

[TSny]

Consider the following equations

$$2I_{1}R_{1}=\lambda ;2I_{2}R_{2}=\lambda ;2I_{3}R_{3}=\lambda $$

Using only these, can you express $I_2$ in terms of $I_1$ and express $I_3$ in terms of $I_1$?

 

[cwill53]

Consider the following equationsUsing only these, can you express $I_2$ in terms of $I_1$ and express $I_3$ in terms of $I_1$?

I got
$$I_{1}=I-\frac{I_{1}R_{1}}{R_{2}}-\frac{I_{1}R_{1}}{R_{3}}$$

Should I expand I out?

 

[TSny]

I got
$$I_{1}=I-\frac{I_{1}R_{1}}{R_{2}}-\frac{I_{1}R_{1}}{R_{3}}$$

Should I expand I out?

I'm not sure what you mean by "expand out". Is there something you can multiply both sides of the equation by in order to eliminate the fractions? That might make it easier to then solve for $I_1$.

 

[cwill53]

I'm not sure what you mean by "expand out". Is there something you can multiply both sides of the equation by in order to eliminate the fractions? That might make it easier to then solve for $I_1$.

What I meant was, should I express I as the sum of $I_1+I_2+I_3$?

What I see is $$I_{1}=I_{1}R_{1}(-\frac{1}{R_{2}}-\frac{1}{R_{3}})+I$$

 

[TSny]

What I meant was, should I express I as the sum of $I_1+I_2+I_3$?

What I see is $$I_{1}=I_{1}R_{1}(-\frac{1}{R_{2}}-\frac{1}{R_{3}})+I$$

That's not incorrect, but you can see it's getting messy.

Going back to the equation $$I_{1}=I-\frac{I_{1}R_{1}}{R_{2}}-\frac{I_{1}R_{1}}{R_{3}}$$ you see that you have two denominators: $R_2$ and $R_3$. The product of these, $R_2 R_3$, represents a "common denominator" of the fractions in the equation. See what happens if you multiply both sides of the equation $I_{1}=I-\frac{I_{1}R_{1}}{R_{2}}-\frac{I_{1}R_{1}}{R_{3}}$ by $R_2 R_3$.

 

[cwill53]

That's not incorrect, but you can see it's getting messy.

Going back to the equation $$I_{1}=I-\frac{I_{1}R_{1}}{R_{2}}-\frac{I_{1}R_{1}}{R_{3}}$$ you see that you have two denominators: $R_2$ and $R_3$. The product of these, $R_2 R_3$, represents a "common denominator" of the fractions in the equation. See what happens if you multiply both sides of the equation $I_{1}=I-\frac{I_{1}R_{1}}{R_{2}}-\frac{I_{1}R_{1}}{R_{3}}$ by $R_2 R_3$.

I can't thank you enough for that. It looks like one big problem I have is my algebra. Are there any books that you have in mind that could help me take it to the next level?

 

[TSny]

I don't know of any particular books to recommend. If you are having a lot of these "algebra snags", then I suggest that you have a conversation with your (calculus?) professor for recommendations. Sorry I can't be of much help here.