Two identical capacitors are connected in series. If the equivalent capacitance is $C_{eq}=2\mu \text{F}$, the capacitance of each individual capacitor is:

Solution

1. The formula for two capacitors in series is $\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}$.
2. This simplifies to $\frac{1}{C_{eq}}=\frac{2}{C}$.
3. Rearranging gives $C=2C_{eq}$.
4. Substitute $C_{eq}=2\mu \text{F}$ to get $$\begin{array}{rl}C& =2\times 2\\ & =4\mu \text{F}\end{array}$$.
5. Hence the answer is (C).