
Three parallel plate capacitors $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ each of capacitance $5 \mu…

- $22.5 \mu \mathrm{~F}$
- $7.5 \mu \mathrm{~F}$
- $9 \mu \mathrm{~F}$
- $30 \mu \mathrm{~F}$
Solution
$\begin{aligned}
& \mathrm{C}_1=4 \mathrm{C} \\ & \mathrm{C}_1=4 \times 5=20 \mu \mathrm{~F} \\ & \mathrm{C}_2=\mathrm{C}_3=5 \mu \mathrm{~F}
\end{aligned}$
$\mathrm{C}_1 \& \mathrm{C}_2$ are in series which is parallel to $\mathrm{C}_3$ So
$\begin{aligned}
& C_{c q}=\frac{C_1 C_2}{C_1+C_2}+C_3 \Rightarrow \frac{20 \times 5}{20+5}+5 \\ & =4+5=9 \mu \mathrm{~F}
\end{aligned}$
Correct Option (3)
Asked in: JEE Main 2025 (04 Apr Shift 2)