Physics - Capacitor Question with Solution | TestHub

Given $C_1=C_2=C_3=2\mu \text{F}$ and the required equivalent capacitance $C_{eq}=\frac{4}{3}\mu \text{F}$.

Evaluating the equivalent capacitance when two capacitors are in parallel and the third is in series with them:

The equivalent capacitance of two capacitors in parallel is:

$C_p=C_1+C_2=2+2=4\mu \text{F}$

Connecting the third capacitor $C_3$ in series with $C_p$, the effective capacitance $C_{eq}$ is given by:

$\frac{1}{C_{eq}}=\frac{1}{C_p}+\frac{1}{C_3}$

$\frac{1}{C_{eq}}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\mu {\text{F}}^{-1}$

$C_{eq}=\frac{4}{3}\mu \text{F}$

This matches the required effective capacitance. Thus, two capacitors must be in parallel connection and the third in series to them.

Answer: two of them are in parallel connection and third is in series to them.