In the circuit as shown in the figure,the eff | vedclass

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(C) The two capacitors of $2 \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 2 \mu F + 2 \mu F = 4 \mu F$.Now,the circuit con

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$4$

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(C) The two capacitors of $2 \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 2 \mu F + 2 \mu F = 4 \mu F$.Now,the circuit consists of three capacitors of $4 \mu F$ each,connected in series between points $A$ and $B$.The equivalent capacitance $C_{AB}$ for capacitors in series is given by $\frac{1}{C_{AB}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.Substituting the values: $\frac{1}{C_{AB}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$.Therefore,$C_{AB} = \frac{4}{3} \mu F$. Wait,re-evaluating the circuit diagram: The two $2 \mu F$ capacitors are in parallel,giving $4 \mu F$. This $4 \mu F$ is in series with the top $4 \mu F$ and bottom $4 \mu F$ capacitors. The total equivalent capacitance is $\frac{1}{C_{AB}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$,so $C_{AB} = 1.33 \mu F$. However,looking at the provided solution image,the circuit simplifies to three $4 \mu F$ capacitors in series. If the question implies a different configuration,based on the provided options,the intended answer is $C = 4/3 \mu F$. Given the options provided,there might be a discrepancy in the question's diagram or options. Assuming the standard interpretation of the provided image,the result is $4/3 \mu F$.

In the circuit as shown in the figure,the effective capacitance between $A$ and $B$ is ........ $\mu F$.

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