Q134 / 180 · KCET 2020 · Physics
The difference between equivalent capacitances of two identical capacitors connected in parallel to that in series is $6 \mu \mathrm{F}$. The value of capacitance of each capacitor is
- A$2 \mu \mathrm{F}$
- B$3 \mu \mathrm{F}$
- Correct answerC$4 \mu \mathrm{F}$
- D$6 \mu \mathrm{F}$
Solution
Given, $C_{1}=C_{2}=C$ (identical capacitors)
When both capacitors are connected in parallel, then equivalent capacitance
$\begin{aligned}
&C_{p}=C_{1}+C_{2}=C+C \\
&C_{p}=2 C ...(i)
\end{aligned}$
When both capacitors are connected in series, then equivalent capacitance,
$\begin{aligned}
\frac{1}{C_{s}} &=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{C}+\frac{1}{C} \\
\Rightarrow \quad \frac{1}{C_{s}} &=\frac{2}{C} \Rightarrow C_{s}=\frac{C}{2}
\end{aligned}$
According to question,
$C_{p}-C_{s}=6 \mu \mathrm{F}$
$\Rightarrow \quad 2 C-\frac{C}{2}=6 \mu \mathrm{F} \Rightarrow \frac{3 C}{2}=6 \mu \mathrm{F}$
$\Rightarrow \quad C=4 \mu \mathrm{F}$
Hence, the correct answer is option: 'C'.