MHT CET · Physics · Capacitance
Five capacitors each of capacity 'C' are connected as shown in figure. If their resultant capacity is \(2 \mu \mathrm{F}\), then the capacity of each condenser is

- A \(2.5 \mu \mathrm{F}\)
- B \(2\mu \mathrm{F}\)
- C \(10\mu \mathrm{F}\)
- D \(5\mu \mathrm{F}\)
Answer & Solution
Correct Answer
(C) \(10\mu \mathrm{F}\)
Step-by-step Solution
Detailed explanation
Given, resultant capacity, \(\mathrm{C}_{\mathrm{eq}}=2 \mu \mathrm{F}\)
The equivalent capacity of the circuit is given by \(\frac{1}{C_{a q}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C} \ldots(\because\) all capacitors are in series)
\(\begin{array}{l}
\frac{1}{C_{e q}}=\frac{5}{C} \\
\frac{1}{2 \mu F}=\frac{5}{C} \\
\Rightarrow C=10 \mu F
\end{array}\)
The equivalent capacity of the circuit is given by \(\frac{1}{C_{a q}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C} \ldots(\because\) all capacitors are in series)
\(\begin{array}{l}
\frac{1}{C_{e q}}=\frac{5}{C} \\
\frac{1}{2 \mu F}=\frac{5}{C} \\
\Rightarrow C=10 \mu F
\end{array}\)
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