Physics - Capacitor Question with Solution | TestHub

The combination of three charges in series $$\begin{array}{l}\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\\ \frac{1}{C}=\frac{1}{3}+\frac{1}{6}+\frac{1}{6}\Rightarrow C=\frac{6}{4}=1.5\mu \mathrm{F}\end{array}$$ The charge of this circuit $$\begin{array}{l}q=CV=1.5\times 120\\ q=180\mu \mathrm{C}\end{array}$$ The potential difference across the $3\mu \mathrm{F}$ $$q=CV$$ $$V=\frac{q}{C}=\frac{180}{3}=60\mathrm{V}$$