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(A) The circuit consists of a $3 \mu F$ capacitor in series with a parallel combination of two $1.5 \mu F$ capacitors,which is then in series with ano

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$1$

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(A) The circuit consists of a $3 \mu F$ capacitor in series with a parallel combination of two $1.5 \mu F$ capacitors,which is then in series with another $3 \mu F$ capacitor.$1$. First,calculate the equivalent capacitance of the two $1.5 \mu F$ capacitors in parallel: $C_p = 1.5 \mu F + 1.5 \mu F = 3 \mu F$.$2$. Now,the circuit simplifies to three capacitors of $3 \mu F$ each,connected in series.$3$. The equivalent capacitance $C_{eq}$ for capacitors in series is given by: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.$4$. Substituting the values: $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \mu F^{-1}$.$5$. Therefore,$C_{eq} = 1 \mu F$.

What is the equivalent capacitance between the points $P$ and $Q$ in the combination of capacitors shown in the figure (in $\mu F$)?

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