Three capacitors connected in series have an effective capacitance of $4 \mu \mathrm{~F}$. If one of

Question

Three capacitors connected in series have an effective capacitance of $4 \mu \mathrm{~F}$. If one of the capacitance is removed, the net capacitance of the capacitor increases of $6 \mu \mathrm{~F}$. The removed capacitor bas a capacitance of\

Professor PC Thomas Classes · Answer

Let $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ capacitors are connected in series $$\begin{equation*} 	herefore \frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{1}{\mathrm{C}_{\mathrm{s}}} 	ag{i} \end{equation*}$$ $\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{1}{4} \quad\left(ight.$ Given $\left.\mathrm{C}_{\mathrm{s}}=4 \mu \mathrm{~F}ight)$\ on removal of capacitor $C$, than they combination above combination\ $\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}=\frac{1}{\mathrm{C}}=\frac{1}{6} \quad($ Given $\mathrm{C}=6 \mu \mathrm{~F})$\ According to question $\mathrm{C}=\mathrm{C}_{\mathrm{s}}$\ $	herefore \frac{1}{\mathrm{C}_{1}}=\frac{2}{24}$\