Chapter 16 - Electric Potential, Energy, and Capacitance - Learning Path Questions and Exercises - Multiple Choice Questions - Page 590: 29

(b) 1.5 C

We can find the combination of capacitors 1 and 2 which are in series connection by using the relation $\frac{1}{C_{12}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}$ But $C_{1}=C_{2}=C_{3}=C$ $\implies \frac{1}{C_{12}}=\frac{1}{C}+\frac{1}{C}=\frac{2}{C}$ Or $C_{12}=\frac{C}{2}$ Now $C_{12}$ and $C_{3}$ is in parallel. Therefore, $C_{eq}=C_{12}+C_{3}=\frac{C}{2}+C=1.5\,C$