A series combination of n_1 capacitors, each | vedclass

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Question

In series, the equivalent capacitance is$:$ $C s=\frac{C_{1}}{n_{1}}$ and

Energy stored, $E_{S}=\frac{1}{2} C_{S} V_{S}^{2}=\frac{1}{2}\left(C_{1}

Vedclass · Accepted Answer

$\frac{{16{C_1}}}{{{n_1}{n_2}}}$

Vedclass · Answer

In series, the equivalent capacitance is$:$ $C s=\frac{C_{1}}{n_{1}}$ and

Energy stored, $E_{S}=\frac{1}{2} C_{S} V_{S}^{2}=\frac{1}{2}\left(C_{1} / n_{1}ight)(4 V)^{2}=\frac{8 C_{1} V^{2}}{n_{1}}$

In parallel, the equivalent capacitance is $C_{P}=n_{2} C_{2}$ and

Energy stored $, E_{P}=\frac{1}{2} C_{P} V_{P}^{2}=\frac{1}{2}\left(n_{2} C_{2}ight)(V)^{2}=\frac{n_{2} C_{2} V^{2}}{2}$

Here, $E_{S}=E_{P}$

$\frac{8 C_{1} V^{2}}{n_{1}}=\frac{n_{2} C_{2} V^{2}}{2}$

$	herefore C_{2}=\frac{16 C_{1}}{n_{1} n_{2}}$

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