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(D) To find the equivalent capacitance,we analyze the circuit structure. The capacitors of $10 \ \mu F$ and $4 \ \mu F$ are in series,and the capacito

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(D) To find the equivalent capacitance,we analyze the circuit structure. The capacitors of $10 \ \mu F$ and $4 \ \mu F$ are in series,and the capacitors of $5 \ \mu F$ and $6 \ \mu F$ are in series.Let $C_1 = 10 \ \mu F$ and $C_2 = 4 \ \mu F$. Their equivalent capacitance $C_{12}$ is given by:$1/C_{12} = 1/10 + 1/4 = (2+5)/20 = 7/20 \implies C_{12} = 20/7 \ \mu F$.Let $C_3 = 5 \ \mu F$ and $C_4 = 6 \ \mu F$. Their equivalent capacitance $C_{34}$ is given by:$1/C_{34} = 1/5 + 1/6 = (6+5)/30 = 11/30 \implies C_{34} = 30/11 \ \mu F$.These two branches are in parallel across the battery terminals $A$ and $C$. Therefore,the total equivalent capacitance $C_{eq}$ is:$C_{eq} = C_{12} + C_{34} = 20/7 + 30/11 = (220 + 210) / 77 = 430 / 77 \ \mu F \approx 5.58 \ \mu F$.Since this value is not among the options,the correct choice is $D$.

In the circuit shown in the figure,four capacitors are connected to a battery. The equivalent capacitance of the circuit is......$ \mu F$

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