Answer
We need an additional capacitor of $150~\mu F$ combined in series with the $75~\mu F$ capacitor.
Work Step by Step
If we combine another capacitor with the $75~\mu F$ capacitor in series, then the equivalent capacitance is less than $75~\mu F$.
If we combine another capacitor with the $75~\mu F$ capacitor in parallel, then the equivalent capacitance is greater than $75~\mu F$.
Therefore, to produce an equivalent capacitance of $50~\mu F$, we need to combine another capacitor in series.
We can find the required capacitance of the additional capacitor:
$\frac{1}{50~\times 10^{-6}~F} = \frac{1}{75\times 10^{-6}~F}+\frac{1}{C}$
$\frac{1}{C} = \frac{1}{50\times 10^{-6}~F}-\frac{1}{75\times 10^{-6}~F}$
$\frac{1}{C} = \frac{3}{150\times 10^{-6}~F}-\frac{2}{150\times 10^{-6}~F}$
$\frac{1}{C} = \frac{1}{150\times 10^{-6}~F}$
$C = \frac{150\times 10^{-6}~F}{1}$
$C = 150~\mu F$
To produce an equivalent capacitance of $50~\mu F$, we need an additional capacitor of $150~\mu F$ combined in series with the $75~\mu F$ capacitor.
