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(A) The capacitors $2\, \mu F$,$3\, \mu F$,and $6\, \mu F$ are connected in series.The equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} =

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$4$

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(A) The capacitors $2\, \mu F$,$3\, \mu F$,and $6\, \mu F$ are connected in series.The equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = \frac{6}{6} = 1\, \mu F$.The total charge $Q$ flowing through the circuit is:$Q = C_{eq} 	imes V = 1\, \mu F 	imes 24\, V = 24\, \mu C$.Since the capacitors are in series,the charge on each capacitor is the same,$Q = 24\, \mu C$.The potential difference across the $6\, \mu F$ capacitor is:$V_6 = \frac{Q}{C_6} = \frac{24\, \mu C}{6\, \mu F} = 4\, V$.

In the given circuit,what is the potential difference across the $6\, \mu F$ capacitor in volts?

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