# How to find net capacitance?

Jun 6, 2018

$4 \setminus \mu \text{F}$

#### Explanation:

The two $2 \setminus \mu \text{F}$ capacitors are in parallel, so they can be replaced by their equivalent capacitance of $4 \setminus \mu \text{F}$.

This leaves us with two parallel branches, each having two $4 \setminus \mu \text{F}$ capacitors in series.

The capacitance of each branch is (4\ mu"F"times 4\ mu"F")/(4\ mu"F"+4\ mu"F") = 2\ mu"F".

So, finally we have two $2 \setminus \mu \text{F}$ capacitors in parallel, leading to an overall equivalent capacitance of $4 \setminus \mu \text{F}$

Note
For two capacitors ${C}_{1}$ and ${C}_{2}$ in series, the equivalent capacitance is $\frac{{C}_{1} {C}_{2}}{{C}_{1} + {C}_{2}}$, while the equivalent of parallel capacitors is ${C}_{1} + {C}_{2}$