Question 492: Electrostats & Capacitance

Question

A series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference 6 V . Another parallel combination of $\mathrm{n}_2$ capacitors, each of value $\mathrm{C}_2$ is charged by a source of potential difference 2 V . Total energy of both the combinations is same. The value of $\mathrm{C}_2$ in terms of $\mathrm{C}_1$ is

Accepted Answer

When capacitors are connected in series, the equivalent capacitance is given by:
$ (C_{	ext{eq}})_1 = \frac{C_1}{n_1} $
The potential difference across this series is:
$ V_1 = 6 	ext{ V} $
For the capacitors connected in parallel, the equivalent capacitance is:
$ (C_{	ext{eq}})_2 = n_2 C_2 $
The potential difference across this combination is:
$ V_2 = 2 	ext{ V} $
Given that the total energy for both combinations is the same, we have:
$ U_1 = U_2 $
The energy stored in a capacitor is given by the formula:
$ U = \frac{1}{2} C V^2 $
Therefore, equating the energies, we get:
$ \frac{1}{2} \left(\frac{C_1}{n_1}ight) (6^2) = \frac{1}{2} (n_2 C_2) (2^2) $
Simplifying the above equation:
$ \frac{1}{2} \cdot C_1 \cdot 36 = \frac{1}{2} \cdot n_2 \cdot C_2 \cdot 4 $
Further simplifying:
$ 18 C_1 = 2 n_2 C_2 $
Solving for $ C_2 $:
$ C_2 = \frac{9 C_1}{n_1 n_2} $

Answer

$\frac{3 C_1}{n_1 n_2}$

Answer

$\frac{9 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$

Answer

$\frac{3 \mathrm{n}_2}{\mathrm{n}_1} \mathrm{C}_1$

Answer

$\frac{9 C_1}{n_1 n_2}$

Electrostats & Capacitance

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