A series combination of n_1 capacitors, each | vedclass

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Question

A series combination of $\mathrm{n}_{1}$ capacitors each of capacitance $\mathrm{C}_{1}$ are connected to $4 \mathrm{V}$ source as shown in the figure

Vedclass · Accepted Answer

$\frac {16C_1}{n_1n_2}$

Vedclass · Answer

A series combination of $\mathrm{n}_{1}$ capacitors each of capacitance $\mathrm{C}_{1}$ are connected to $4 \mathrm{V}$ source as shown in the figure.

Total capacitance of the series combination of the capacitors is

$\frac{1}{{{C_{m{s}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_1}}} + \frac{1}{{{C_1}}} +  \ldots  \ldots \,\,upto\,{n_1}\,\,terms\,\, = \frac{{{n_1}}}{{{C_1}}}$

or ${C_s} = \frac{{{C_1}}}{{{n_1}}}$        ........$(i)$

Total energy stored in a series combination of the capacitors is

${U_s} = \frac{1}{2}{C_s}{(4V)^2} = \frac{1}{2}\left( {\frac{{{C_1}}}{{{n_1}}}} ight){(4V)^2}\quad $         (Using $(i)$)    .......$(ii)$

A parallel combination of $n_{2}$ capacitors each of capacitance $\mathrm{C}_{2}$ are connected to $\mathrm{V}$ source as shown in the figure.

Total capacitance of the parallel combination of capacitors is

$C_{p}=C_{2}+C_{2}+\ldots \ldots \ldots+	ext { upto } n_{2} 	ext { terms }=n_{2} C_{2} $

 or $C_{p}=n_{2} C_{2}$      ...........$(iii)$

Total energy stored in a parallel combination of capacitors is

$U_{p} =\frac{1}{2} C_{p} V^{2} $

$ = \frac{1}{2}\left( {{n_2}{C_2}} ight){(V)^2}\quad $   (Using  $(iii)$) .......$(iv)$

According to the given problem,

$U_{s}=U_{p}$

Substituting the values of $U_{\mathrm{s}}$ and $U_{\mathrm{p}}$ from equations $(ii)$ and $(iv),$ we get

$\frac{1}{2} \frac{C_{1}}{n_{1}}(4 V)^{2}=\frac{1}{2}\left(n_{2} C_{2}ight)(V)^{2}$

or $\quad \frac{C_{1} 16}{n_{1}}=n_{2} C_{2}$ or $C_{2}=\frac{16 C_{1}}{n_{1} n_{2}}$

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