solve C=K(Rr/R-r), solve for R
Hola!
I suppose que es $\displaystyle C = K\left( {\frac{{Rr}}
{{R - r}}} \right).$ Then
$\displaystyle \begin{aligned}
\frac{K}
{C} &= \frac{{R - r}}
{{Rr}}\\
\frac{K}
{C} &= \frac{1}
{r} - \frac{1}
{R}\\
\frac{1}
{R} &= \frac{1}
{r} - \frac{K}
{C}.
\end{aligned}$
You can do it desde ahí.
Cheers
The thing you need to remember when you transform formulas is that whatever you do to one side of the equation, you need to do the same to the other side. This maintains the equality of the equation.
So for your formula -
$\displaystyle C=K(Rr/R-r)$, solve for $\displaystyle R$
First, divide both sides by $\displaystyle K$ which gives -
$\displaystyle C/K=(Rr/R-r)$
Now, multiply the top and bottom of the fraction on the right hand side by $\displaystyle 1/R$, giving -
$\displaystyle C/K=r/(1-r/R)$
This is equivalent to -
$\displaystyle K/C=(1-r/R)/r$
Multiply both sides by $\displaystyle r$, giving -
$\displaystyle Kr/C=1-r/R$
Subtract $\displaystyle 1$ from both sides -
$\displaystyle (Kr/C)-1=-r/R$
Multiply both sides by $\displaystyle -1$,
$\displaystyle 1-(Kr/C)=r/R$
Multiply both sides by $\displaystyle R$, giving -
$\displaystyle R(1-(Kr/C))=r$
Now, finally divide both sides by $\displaystyle (1-(Kr/C))$ to get $\displaystyle R$ on its own -
$\displaystyle R=r/(1-(Kr/C))$
There you go. It is a bit long winded, but I wanted to spell out the steps so you understood what I was doing.