Question

When an electric current passes through two resistors with resistance r1 and r2, connected in parallel, the combined resistance, R, can be calculated from the equation 1R=1r1+1r2, where R, r1, and r2 are positive. Assume that r2 is constant.
(a) Show that R is an increasing function of r1.

(b) Where on the interval a≤r1≤b does R take its maximum value?

Answer 1

Answer:

a)

The combined resistance of a circuit consisting of two resistors in parallel is given by:

$\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}$

where

R is the combined resistance

$r_1, r_2$ are the two resistors

We can re-write the expression as follows:

$\frac{1}{R}=\frac{r_1+r_2}{r_1r_2}$

Or

$R=\frac{r_1 r_2}{r_1+r_2}$

In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.

The derivative of R with respect to r1 is:

$\frac{dR}{dr_1}=\frac{r_2(r_1+r_2)-1(r_1r_2)}{(r_1+r_2)^2}=\frac{r_2^2}{(r_1+r_2)^2}$

We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.

b)

To solve this part, we use again the expression for R written in part a:

$R=\frac{r_1 r_2}{r_1+r_2}$

We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,

$r_1>0$

So the interval of allowed values for r1 is

$0$

From part a), we also said that the function is increasing versus r1 over the whole domain. This means that if we consider a certain interval

a ≤ r1 ≤ b

The maximum of the function (R) will occur at the maximum value of r1 in this interval: so, at

$r_1=b$