Load resistance adding with minimized power and voltage constraint [closed]

Question

Consider the simple circuit below:

Suppose $V$ and $R$ have set voltage and resistance values respectively, and you are to choose the resistance values for $R_1$ and $R_2$. However, you must choose $R_1$ and $R_2$ such that:

• The combined power dissipated by $R_1$ and $R_2$ is minimized
• The voltage drop across the $R_2$ resistor is as close to $V/3$ as possible

My approach is the following:

To minimize the power dissipated, we first get the power dissipated as a function of $R_1$ and $R_2$, and then get the critical points of that function.

$$I^2R=\left(\frac{V}{\sum R}\right)^2(R_1+R_2)=\frac{V^2(R_1+R_2)}{(R+R_1+R_2)^2}$$

Taking the partial derivative with respect to either $R_1$ or $R_2$, we get

$$\frac{V^2(R+R_1+R_2)-2V^2(R_1+R_2)}{(R+R_1+R_2)^3}$$

Setting that equal to zero, we find a critical point when $R_1+R_2=R$.

But we have the constraint that $IR_2\approx V/3$. If we assume equality, we get that

$$\begin{align*} IR_2&=\left(\frac{V}{\sum R}\right)R_2\\ &=\frac{V}{3} \end{align*}$$

Cancelling the $V$'s, we get

$$\frac{R_2}{R+R_1+R_2}=\frac{R_2}{2(R+R_2)}=\frac{1}{3}$$

$$\boxed{R_2=2R}$$

And that is where my problem is - that doesn't make sense. Could you suggest a fix to this, or maybe an alternate approach?

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EDIT: As pointed out, the problem was in the "cancelling the V's" in the final step.

Is this the only way to arrive at the optimal result? Do you know of any other way?

Answer 1

You just made an error in the last equation (after "Cancelling the V's"). You should get:

$$ \frac{R_2}{R+R_1+R_2} = \frac{R_2}{R+R} = \frac{R_2}{2R} = \frac{1}{3}$$

and therefore $ R_2 = \frac{2}{3}R$.