4;V વિદ્યુતસ્થિતિમાનથી ચાર્જ કરેલા C_1 કેપેસી | vedclass

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Question

A series combination of $n_{1}$ capacitors each of capacitance $C_{1}$ are connected to $4 V$ source as shown in the figure.

Total capacitance of t

Vedclass · Accepted Answer

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

Vedclass · Answer

A series combination of $n_{1}$ capacitors each of capacitance $C_{1}$ are connected to $4 V$ source as shown in the figure.

Total capacitance of the series combination of the capacitors is

$\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{1}}+\frac{1}{C_{1}}+\ldots \ldots 	ext { upto } n_{1} 	ext { terms }=\frac{n_{1}}{C_{1}}$

or $C_{s}=\frac{C_{1}}{n_{1}}.........(i)$

Total energy stored in a series combination of the capacitors is

$U_{s}=\frac{1}{2} C_{s}(4 V)^{2}=\frac{1}{2}\left(\frac{C_{1}}{n_{1}}ight)(4 V)^{2} \quad(	ext { Using }(\mathrm{i})).........(ii)$

A parallel combination of $n_{2}$ capacitors each of capacitance $C_{2}$ are connected to $V$ 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 $\quad C_{p}=n_{2} C_{2}.........(iii)$

Total energy stored in a parallel combination of capacitors is

$U_{p} =\frac{1}{2} C_{p} V^{2}$

$=\frac{1}{2}\left(n_{2} C_{2}ight)(V)^{2}.........$ (Using $(iii))...(iv)$

According to the given problem,

$U_{s}=U_{p}$

Subst tuting the values of $U_s,$ and $U_p,$ 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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