KCET · Physics · Capacitance
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 \(3 \mu \mathrm{F}\)
- C \(4 \mu \mathrm{F}\)
- D \(6 \mu \mathrm{F}\)
Answer & Solution
Correct Answer
(C) \(4 \mu \mathrm{F}\)
Step-by-step Solution
Detailed explanation
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,
\(\frac{1}{C_{s}} =\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{C}+\frac{1}{C} \)
\( \Rightarrow \frac{1}{C_{s}} =\frac{2}{C} \Rightarrow C_{s}=\frac{C}{2}\)
According to question,
\(C_{p}-C_{s}=6 \mu \mathrm{F}\)
\(\Rightarrow 2 C-\frac{C}{2}=6 \mu \mathrm{F} \Rightarrow \frac{3 C}{2}=6 \mu \mathrm{F}\)
\(\Rightarrow C=4 \mu \mathrm{F}\)
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,
\(\frac{1}{C_{s}} =\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{C}+\frac{1}{C} \)
\( \Rightarrow \frac{1}{C_{s}} =\frac{2}{C} \Rightarrow C_{s}=\frac{C}{2}\)
According to question,
\(C_{p}-C_{s}=6 \mu \mathrm{F}\)
\(\Rightarrow 2 C-\frac{C}{2}=6 \mu \mathrm{F} \Rightarrow \frac{3 C}{2}=6 \mu \mathrm{F}\)
\(\Rightarrow C=4 \mu \mathrm{F}\)
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