Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
The equivalent capacitance become
Solution:
1157 Upvotes
Verified Answer
The correct answer is:
$\frac{5 \mathrm{C}}{7}$
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
$$
\begin{aligned}
C_{\mathrm{eq}} &=\frac{\mathrm{C}_{\mathrm{eq}} \times \mathrm{C}_{1}}{\mathrm{C}_{\mathrm{eq}}+\mathrm{C}_{1}}=\frac{\frac{5 \mathrm{C}}{2} \times \mathrm{C}}{\frac{5 \mathrm{C}}{2}+\mathrm{C}} \\
&=\frac{5 \mathrm{C}^{2}}{7 \mathrm{C}}=\frac{5 \mathrm{C}}{7}
\end{aligned}
$$
$$
\mathrm{C}_{2}^{\prime}=\frac{3}{2} \mathrm{C}
$$
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
$$
\begin{aligned}
C_{\mathrm{eq}} &=\frac{\mathrm{C}_{\mathrm{eq}} \times \mathrm{C}_{1}}{\mathrm{C}_{\mathrm{eq}}+\mathrm{C}_{1}}=\frac{\frac{5 \mathrm{C}}{2} \times \mathrm{C}}{\frac{5 \mathrm{C}}{2}+\mathrm{C}} \\
&=\frac{5 \mathrm{C}^{2}}{7 \mathrm{C}}=\frac{5 \mathrm{C}}{7}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.