In the connections shown in the adjoining fig | vedclass

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(D) The given circuit is a balanced Wheatstone bridge network. Let the capacitors be $C_1 = 6\,\mu F$,$C_2 = 12\,\mu F$,$C_3 = 9\,\mu F$,and $C_4 = 18

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(D) The given circuit is a balanced Wheatstone bridge network. Let the capacitors be $C_1 = 6\,\mu F$,$C_2 = 12\,\mu F$,$C_3 = 9\,\mu F$,and $C_4 = 18\,\mu F$. The central capacitor is $24\,\mu F$. Checking the ratio: $\frac{C_1}{C_3} = \frac{6}{9} = \frac{2}{3}$ and $\frac{C_2}{C_4} = \frac{12}{18} = \frac{2}{3}$. Since $\frac{C_1}{C_3} = \frac{C_2}{C_4}$,the bridge is balanced,and no charge flows through the $24\,\mu F$ capacitor. Thus,we can remove the $24\,\mu F$ capacitor. The upper branch has $6\,\mu F$ and $12\,\mu F$ in series: $C_{up} = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4\,\mu F$. The lower branch has $9\,\mu F$ and $18\,\mu F$ in series: $C_{low} = \frac{9 	imes 18}{9 + 18} = \frac{162}{27} = 6\,\mu F$. The equivalent capacitance between $A$ and $B$ is the parallel combination of $C_{up}$ and $C_{low}$: $C_{eq} = 4 + 6 = 10\,\mu F$.

In the connections shown in the adjoining figure,the equivalent capacitance between $A$ and $B$ will be......$\mu F$.

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