In Fig. 23.12, let C_{1} = 6µF and C_{2} = 3µF, and ΔV = 18 V. Find C_{eq}, Q, ΔV_{1}, and ΔV_{2}.
The equivalent capacitance of the series combination is:
{\frac{1}{C_{\mathrm{eq}}}}={\frac{1}{C_{1}}}+{\frac{1}{C_{2}}}={\frac{1}{6 \ \mu\mathrm{F}}}+{\frac{1}{3\mathrm{\,\mu F}}}={\frac{1}{2\mathrm{\,\mu F}}}\ \ \Rightarrow\ \ C_{\mathrm{eq}}=2\mathrm{\,\mu F}Consequently: Q=C_{\mathrm{eq}}\,\Delta V=(2 \ \mu{F})(18\,{V})=36 \ {\mu}{F}
ΔV_{1} = \frac{Q}{C_{1}} = \frac{36 µV}{6 µF} = 6 V and ΔV_{2} = \frac{Q}{C_{2}} = \frac{36 µV}{3 µF} = 12 V