In given circuit, $\mathrm{C}_{1}=\mathrm{C}_{2}=\mathrm{C}_{3}=\mathrm{C}$ initially. Now, a dielectric slab of dielectric constant $\mathrm{K}=\frac{3}{2}$ is inserted in $\mathrm{C}_{2}$.
The equivalent capacitance become

• Correct answer
  A

  $\frac{5 \mathrm{C}}{7}$
• B

  $\frac{7 \mathrm{C}}{5}$
• C

  $\frac{2 \mathrm{C}}{3}$
• D

  $\frac{\mathrm{C}}{2}$

Solution

When a dielectric slab of dielectric constant $\mathrm{K}=\frac{3}{2}$ is inserted between the plates of $\mathrm{C}_{2}$ its new capacitance $\left(\mathrm{C}_{2}^{\prime}\right)$ becomes $$ \mathrm{C}_{2}^{\prime}=\frac{3}{2} \mathrm{C} $$ Equivalent capacitance of $\mathrm{C}_{2}^{\prime}$ and $\mathrm{C}_{3}$ is $$ \mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{2}^{\prime}+\mathrm{C}_{3}=\frac{3}{2} \mathrm{C}+\mathrm{C}=\frac{5 \mathrm{C}}{2} $$
Now, $\mathrm{C}_{\text {eq }}$ and $\mathrm{C}_{1}$ are in series. Therefore, their equivalent capacitance is $$ \begin{aligned} C_{\mathrm{eq}} &=\frac{\mathrm{C}_{\mathrm{eq}} \times \mathrm{C}_{1}}{\mathrm{C}_{\mathrm{eq}}+\mathrm{C}_{1}}=\frac{\frac{5 \mathrm{C}}{2} \times \mathrm{C}}{\frac{5 \mathrm{C}}{2}+\mathrm{C}} \\ &=\frac{5 \mathrm{C}^{2}}{7 \mathrm{C}}=\frac{5 \mathrm{C}}{7} \end{aligned} $$

Hence, the correct answer is option: 'A'.