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A series combination of $n_1$ capacitors, each of capacity $C_1$ is charged by source of potential difference $4 \mathrm{~V}$. When another parallel combination of $n_2$ capacitors each of capacity $C_2$ is charged by a source of potential difference $V$, it has the same total energy stored in it as the first combination has. The value of $C_2$ in terms of $C_1$ is then
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Verified Answer
The correct answer is:
$\frac{16 C_1}{n_1 n_2}$
$\frac{16 C_1}{n_1 n_2}$
Equivalent capacitance of $n_2$ number of capacitors each of capacitance $\mathrm{C}_2$ in parallel $=n_2 C_2$
Equivalent capacitance of $n_1$ number of capacitors each of capacitances $\mathrm{C}_1$ in series.
Capacitance of each is $C_1=\frac{C_1}{n_1}$
According to question, total energy stored in both the combinations are same
$$
\begin{aligned}
& \text { i.e., } \frac{1}{2}\left(\frac{C_1}{n_1}\right)(4 V)^2=\frac{1}{2}\left(n_2 C_2 \text { b }^2\right. \\
& \therefore C_2=\frac{16 C_1}{n_1 n_2} \\
&
\end{aligned}
$$
Equivalent capacitance of $n_1$ number of capacitors each of capacitances $\mathrm{C}_1$ in series.
Capacitance of each is $C_1=\frac{C_1}{n_1}$
According to question, total energy stored in both the combinations are same
$$
\begin{aligned}
& \text { i.e., } \frac{1}{2}\left(\frac{C_1}{n_1}\right)(4 V)^2=\frac{1}{2}\left(n_2 C_2 \text { b }^2\right. \\
& \therefore C_2=\frac{16 C_1}{n_1 n_2} \\
&
\end{aligned}
$$
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