Question

You need a capacitance of $50mF$, but you don’t happen to have a $50mF$capacitor. You do have a $75mF$capacitor. What additional capacitor do you need to produce a total capacitance of $50mF$? Should you join the two capacitors in parallel or in series?

Short answer

The capacitance of additional capacitor is$150\mu F$and should be connected in series.

Step-by-step solution

Given information

We have given a total capacitance of $50mF$.

We need to find Should we join the two capacitors in parallel or in series.

Simplify 

As we want$50\mu F$capacitance but we only have a $75\mu F$capacitor. Note, could technically have a $-25\mu F$capacitor if we wanted to install it in parallel.

Since negative capacitance is not possible. So, we have to place them in a series. Considering the series formula for capacitance

$C_{eq}={\left(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+....\right)}^{-1}(30.23)$

or

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...$

We we need to arrive at $50\mu F$capacitance, $C_{eq}=50\mu F$. thus,our setup becomes

$\frac{1}{50\mu F}=\frac{1}{75\mu F}+\frac{1}{C_2}$

From here we solve for $C_2$.

$$\begin{gathered} \frac{1}{50\mu F}=\frac{1}{75\mu F}+\frac{1}{C_2} \\ \to \frac{1}{50\mu F}-\frac{1}{75\mu F}=\frac{1}{C_2} \\ \frac{1}{150}=\frac{1}{C_2} \\ \to C_2={\left(\frac{1}{150}\right)}^{-1} \\ =150\mu F \end{gathered}$$