A network of capacitors consists of a $10,\mu \text{F}$ capacitor in parallel with a series combination of $6,\mu \text{F}$ and $4,\mu \text{F}$ capacitors. This overall combination is then connected in series with an $8,\mu \text{F}$ capacitor. Find the equivalent capacitance of the whole network.

Step-by-step solution

1. Step 1: Find series capacitance of $6,\mu \text{F}$ and $4,\mu \text{F}$: $\frac{1}{C_s}=\frac{1}{6}+\frac{1}{4}\Rightarrow C_s=\frac{6\times 4}{6+4}=2.4,\mu \text{F}$.
2. Step 2: Parallel combination of $10,\mu \text{F}$ and $2.4,\mu \text{F}$: $C_{\parallel }=10+2.4=12.4,\mu \text{F}$.
3. Step 3: Now connect this $12.4,\mu \text{F}$ in series with $8,\mu \text{F}$: $\frac{1}{C_{\text{eq}}}=\frac{1}{12.4}+\frac{1}{8}$.
4. Step 4: Compute reciprocals: $\frac{1}{12.4}\approx 0.08065$, $\frac{1}{8}=0.125$. Sum $\approx 0.20565$.
5. Step 5: Invert to find $C_{\text{eq}}=\frac{1}{0.20565}\approx 4.86,\mu \text{F}$.
6. Step 6: Re‑check calculations: Use exact fractions: $1/12.4=1/12.4$, $1/8=0.125$. Exact solution: $C_{\text{eq}}=\frac{1}{\frac{1}{12.4}+\frac{1}{8}}=\frac{1}{\frac{8+12.4}{99.2}}=\frac{99.2}{20.4}=4.86,\mu \text{F}$.
7. Step 7: The equivalent capacitance is $4.86,\mu \text{F}$.

Concept tested: Capacitor network analysis

Published 20 September 2026 · How this solution was prepared and checked