Three capacitors $C_1=2\mu \text{F}$, $C_2=3\mu \text{F}$ and $C_3=6\mu \text{F}$ are connected as shown: $C_1$ is in series with the parallel combination of $C_2$ and $C_3$. Find the equivalent capacitance of the whole network.

Step-by-step solution

1. First find the capacitance of the parallel part: $C_p=C_2+C_3=3\mu \text{F}+6\mu \text{F}=9\mu \text{F}$.
2. Now $C_1$ is in series with $C_p$. For series combination, $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_p}$.
3. Substitute the values: $\frac{1}{C_{eq}}=\frac{1}{2}+\frac{1}{9}=\frac{9+2}{18}=\frac{11}{18}$.
4. Invert to obtain $C_{eq}=\frac{18}{11}\mu \text{F}=\frac{12}{5}\mu \text{F}=2.4\mu \text{F}$.
5. Thus the equivalent capacitance is $2.4\mu \text{F}$.

Concept tested: Combination of capacitors

Published 6 September 2026 · How this solution was prepared and checked