Talk about variation C

Question

If in the formula $ C = \frac {en}{R+nr}$, $n$ is increased while $ e$, $R$ and $r$ are kept constant, then $C$:

e, R, and r are constants, so we arbitrarily assign values to them. To keep it simple, we make all equal to 1.

$ C = \frac{1*n}{1+n*1}$

$ C = \frac{n}{1+n}$ testing n=1, 2, 3, we get 1/2, 2/3, 3/4, respectively, or 6/12, 8/12, 9/12. With the numerator just 1 less than the denominator, C is progressing towards 1.

One could argue that C is correct if e=0 or R=0

Answer 1

This is not a correct way to study the influence of $n$, because it could be different for other values of $e,R,r$.

In this case,

$$\frac{ne}{nr+R}=\frac er\frac{nr+R-R}{nr+R}=\frac er\left(1-\frac R{nr+R}\right).$$

Assuming that $nr+R>0$, the denominator increases with $n$, then assuming $R>0$ the fraction decreases and, assuming $\dfrac er>0$ the whole expression increases.

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Note that some of the parameters are superfluous (we now assume them all positive).

$$\frac{ne}{nr+R}=\frac er\frac{n}{n+\dfrac Rr}\propto\frac n{n+\rho}.$$

Answer 2

$$ C = \frac er\frac{nr}{R+nr}$$

$$=\frac er\frac{R+nr-R}{R+nr}$$

$$\frac er\Bigl(1-\frac{R}{R+nr}\Bigr)$$

thus if $ n $ increases,

$ C$ will increase to $\frac er$ cause $ \frac{R}{R+nr} $ goes to $0$.

Of course i assumed $e,r,R,n$ positive.