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  • A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference 4 V. When another parallel combination of n2 capac

A series combination of n1 capacitors, each of value C1, is charged by a source of potential difference 4 V. When another parallel combination of n2 capacitors, each of value C2, 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 C2 , in terms of C1, is then

Option: 1

\frac{2C_{1}}{n_{1}n_{2}}


Option: 2

16\frac{n_{2}}{n_{1}}C_{1}


Option: 3

2\frac{n_{2}}{n_{1}}C_{1}


Option: 4

\frac{16C_{1}}{n_{1}n_{2}}


Answers (1)

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In series, the equivalent capacitance is: \mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}_1}{\mathrm{n}_1} and
Energy stored, \mathrm{E}_{\mathrm{S}}=\frac{1}{2} \mathrm{C}_{\mathrm{S}} \mathrm{V}_{\mathrm{S}}^2=\frac{1}{2}\left(\mathrm{C}_1 / \mathrm{n}_1\right)(4 \mathrm{~V})^2=\frac{8 \mathrm{C}_1 \mathrm{~V}^2}{\mathrm{n}_1}
In parallel, the equivalent capacitance is \mathrm{C}_{\mathrm{P}}=\mathrm{n}_2 \mathrm{C}_2 and
Energy stored, \mathrm{E}_{\mathrm{P}}=\frac{1}{2} \mathrm{C}_{\mathrm{P}} \mathrm{V}_{\mathrm{P}}^2=\frac{1}{2}\left(\mathrm{n}_2 \mathrm{C}_2\right)(\mathrm{V})^2=\frac{\mathrm{n}_2 \mathrm{C}_2 \mathrm{~V}^2}{2}
Here, E_S=E_P

\frac{8 \mathrm{C}_1 \mathrm{~V}^2}{\mathrm{n}_1}=\frac{\mathrm{n}_2 \mathrm{C}_2 \mathrm{~V}^2}{2} \therefore \mathrm{C}_2=\frac{16 \mathrm{C}_1}{\mathrm{n}_1 \mathrm{n}_2}

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