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(B) First, we simplify the circuit to find the equivalent capacitance between $A$ and $B$.$1$. The two $3 \, \mu F$ capacitors in the middle are in pa

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$10$

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(B) First, we simplify the circuit to find the equivalent capacitance between $A$ and $B$.$1$. The two $3 \, \mu F$ capacitors in the middle are in parallel. Their equivalent capacitance is $C_p = 3 \, \mu F + 3 \, \mu F = 6 \, \mu F$.$2$. Now, the circuit consists of four capacitors in series: $6 \, \mu F$, $3 \, \mu F$, $6 \, \mu F$ (the parallel combination), and $3 \, \mu F$.$3$. The equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{6} + \frac{1}{3} + \frac{1}{6} + \frac{1}{3} = \frac{1+2+1+2}{6} = \frac{6}{6} = 1 \, \mu F^{-1}$. Thus, $C_{eq} = 1 \, \mu F$.$4$. The total charge $q$ flowing through the series combination is $q = C_{eq} 	imes V = 1 \, \mu F 	imes 60 \, V = 60 \, \mu C$.$5$. Since all capacitors are in series, the same charge $q = 60 \, \mu C$ passes through each capacitor.$6$. The potential difference across the $6 \, \mu F$ capacitor is $V = \frac{q}{C} = \frac{60 \, \mu C}{6 \, \mu F} = 10 \, V$.

In the given diagram, the potential difference $(PD)$ between $A$ and $B$ is $60 \, V$. The potential difference across the $6 \, \mu F$ capacitor is: (in $V$)

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