How will you connect four capacitors,each of | vedclass

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(A) The correct configuration is shown in the figure,where two capacitors ($C_2$ and $C_3$) are connected in parallel,and this combination is connecte

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Two in parallel and two in series

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(A) The correct configuration is shown in the figure,where two capacitors ($C_2$ and $C_3$) are connected in parallel,and this combination is connected in series with two other capacitors ($C_1$ and $C_4$).Given: $C_1 = C_2 = C_3 = C_4 = 4 \mu F$.First,calculate the equivalent capacitance of the parallel combination $(C_p)$:$C_p = C_2 + C_3 = 4 \mu F + 4 \mu F = 8 \mu F$.Now,the total equivalent capacitance $(C_{eq})$ is the series combination of $C_1$,$C_p$,and $C_4$:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_p} + \frac{1}{C_4} = \frac{1}{4} + \frac{1}{8} + \frac{1}{4}$.$\frac{1}{C_{eq}} = \frac{2 + 1 + 2}{8} = \frac{5}{8} \mu F^{-1}$.$C_{eq} = \frac{8}{5} \mu F = 1.6 \mu F$.

How will you connect four capacitors,each of capacitance $4 \mu F$,to obtain an equivalent capacitance of $1.6 \mu F$?

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