Q52 / 75 · JEE Main 2025 (4 Apr Shift 2) · Physics
Three parallel plate capacitors $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ each of capacitance $5 \mu \mathrm{~F}$ are connected as shown in figure. The effective capacitance between points A and B, when the space between the parallel plates of $C_1$ capacitor is filled with a dielectric medium having dielectric constant of 4, is :
- A$22.5 \mu \mathrm{~F}$
- B$7.5 \mu \mathrm{~F}$
- Correct answerC$9 \mu \mathrm{~F}$
- D$30 \mu \mathrm{~F}$
Solution
After dielectric
$\begin{aligned}<br/>& \mathrm{C}_1=4 \mathrm{C} \\ & \mathrm{C}_1=4 \times 5=20 \mu \mathrm{~F} \\ & \mathrm{C}_2=\mathrm{C}_3=5 \mu \mathrm{~F}<br/>\end{aligned}$
$\mathrm{C}_1 \& \mathrm{C}_2$ are in series which is parallel to $\mathrm{C}_3$ So
$\begin{aligned}<br/>& C_{c q}=\frac{C_1 C_2}{C_1+C_2}+C_3 \Rightarrow \frac{20 \times 5}{20+5}+5 \\ & =4+5=9 \mu \mathrm{~F}<br/>\end{aligned}$
Correct Option (3)
$\begin{aligned}<br/>& \mathrm{C}_1=4 \mathrm{C} \\ & \mathrm{C}_1=4 \times 5=20 \mu \mathrm{~F} \\ & \mathrm{C}_2=\mathrm{C}_3=5 \mu \mathrm{~F}<br/>\end{aligned}$
$\mathrm{C}_1 \& \mathrm{C}_2$ are in series which is parallel to $\mathrm{C}_3$ So
$\begin{aligned}<br/>& C_{c q}=\frac{C_1 C_2}{C_1+C_2}+C_3 \Rightarrow \frac{20 \times 5}{20+5}+5 \\ & =4+5=9 \mu \mathrm{~F}<br/>\end{aligned}$
Correct Option (3)
Hence, the correct answer is option: 'C'.