The effective capacitance between the points | vedclass

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(B) The circuit can be simplified step-by-step:$1$. The first part consists of a $2\,\mu F$ capacitor in series with a $5\,\mu F$ capacitor,which is i

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(B) The circuit can be simplified step-by-step:$1$. The first part consists of a $2\,\mu F$ capacitor in series with a $5\,\mu F$ capacitor,which is in parallel with another $2\,\mu F$ capacitor. The equivalent capacitance of the series part is $C_1 = \frac{2 	imes 5}{2 + 5} = \frac{10}{7}\,\mu F$. This is in parallel with the $2\,\mu F$ capacitor,so $C_{eq1} = \frac{10}{7} + 2 = \frac{24}{7}\,\mu F$.$2$. This $C_{eq1}$ is in series with the next $2\,\mu F$ capacitor. Let this be $C_{eq2} = \frac{(\frac{24}{7}) 	imes 2}{(\frac{24}{7}) + 2} = \frac{48/7}{38/7} = \frac{48}{38} = \frac{24}{19}\,\mu F$.$3$. The last part consists of two $1\,\mu F$ capacitors in parallel,which is $1 + 1 = 2\,\mu F$. This is in series with the previous combination. However,looking at the provided simplified diagram,the final circuit reduces to two $2\,\mu F$ capacitors in series.$4$. The equivalent capacitance $C_{PQ} = \frac{2 	imes 2}{2 + 2} = \frac{4}{4} = 1\,\mu F$.

The effective capacitance between the points $P$ and $Q$ of the arrangement shown in the figure is......$\mu F$

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