Question:

A series combination of $10$ capacitors, each of value ' $C_1$ ' is charged by a source of potential difference ' $4\text{ V}$ '. When another parallel combination of $8$ capacitors, each of value ' $C_2$ ' is charged by a source of potential difference ' $V$ ', it has the same total energy stored in it as in the first combination. The value of ' $C_2$ ' is

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Updated On: May 14, 2026
  • $\frac{C_1}{5}$
  • $\frac{8}{5} C_1$
  • $\frac{64}{5} C_1$
  • $\frac{C_1}{40}$
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The Correct Option is A

Solution and Explanation

Case 1: 10 identical capacitors, each of capacitance \(C_1\), connected in series

For capacitors connected in series,

\[ \frac{1}{C_{\text{eq1}}} = \frac{1}{C_1} + \frac{1}{C_1} + \cdots \text{ (10 terms)} = \frac{10}{C_1} \]

Hence,

\[ C_{\text{eq1}}=\frac{C_1}{10} \]

The combination is connected across a potential difference of \(4\,\text{V}\).

Energy stored in a capacitor is:

\[ U=\frac{1}{2}CV^2 \]

Therefore,

\[ U_1 = \frac{1}{2}\left(\frac{C_1}{10}\right)(4)^2 \]

\[ U_1 = \frac{1}{2}\times\frac{C_1}{10}\times16 \]

\[ U_1 = \frac{16C_1}{20} = \frac{4C_1}{5} \]

\[ U_1=0.8C_1 \]

Case 2: 8 identical capacitors, each of capacitance \(C_2\), connected in parallel

For capacitors connected in parallel,

\[ C_{\text{eq2}} = C_2+C_2+\cdots \text{ (8 terms)} = 8C_2 \]

If this combination is connected across a potential difference \(V\), then the energy stored is:

\[ U_2 = \frac{1}{2}C_{\text{eq2}}V^2 \]

\[ U_2 = \frac{1}{2}(8C_2)V^2 \]

\[ U_2 = 4C_2V^2 \]

According to the question, both combinations store the same energy.

So,

\[ U_1=U_2 \]

\[ \frac{4C_1}{5}=4C_2V^2 \]

Dividing both sides by \(4\),

\[ \frac{C_1}{5}=C_2V^2 \]

Therefore,

\[ C_2=\frac{C_1}{5V^2} \]

If \(V=1\),

\[ C_2=\frac{C_1}{5} \]

Final Answer:

\[ \boxed{\frac{C_1}{5}} \]

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