Calculate the equivalent capacitance of two capacitors C1 = 4 μF and C2 = 2 μF connected in series.

Here's the solution to the problem.

Step 1: Calculate the equivalent capacitance of the capacitors when connected in series.
Given capacitances are $C_1 = 4 \, \mu\text{F}$ and $C_2 = 2 \, \mu\text{F}$.
For capacitors in series, the equivalent capacitance $C_{eq}$ is:

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$ $$\frac{1}{C_{eq}} = \frac{1}{4 \, \mu\text{F}} + \frac{1}{2 \, \mu\text{F}} = \frac{1+2}{4 \, \mu\text{F}} = \frac{3}{4 \, \mu\text{F}}$$ $$$$