In a network of capacitors, a $5,\mu \text{F}$ capacitor is connected in series with the parallel combination of a $3,\mu \text{F}$ and a $7,\mu \text{F}$ capacitor. The resulting combination is then connected in parallel with a $4,\mu \text{F}$ capacitor. What is the equivalent capacitance of the whole network?

Step-by-step solution

1. Find the equivalent capacitance of the two capacitors in parallel: $C_{\text{p}}=3\mu \text{F}+7\mu \text{F}=10\mu \text{F}$.
2. Compute the series combination of $5\mu \text{F}$ and $10\mu \text{F}$: $C_{\text{s}}=\frac{(5)(10)}{5+10}=\frac{50}{15}=\frac{10}{3}\mu \text{F}\approx 3.333\mu \text{F}$.
3. Add this series combination in parallel with the $4\mu \text{F}$ capacitor: $C_{\text{eq}}=4\mu \text{F}+\frac{10}{3}\mu \text{F}=\frac{22}{3}\mu \text{F}\approx 7.33\mu \text{F}$.

Concept tested: Capacitor network equivalent capacitance

Published 16 September 2026 · How this solution was prepared and checked