CAPACITORS IN SERIES AND PARALLEL
In Figs. 24.8 and 24.9, let C_{1}=6.0 \mu \mathrm{F}, \quad C_{2}=3.0 \mu \mathrm{F}, and V_{ab} = 18 V. Find the equivalent capacitance and the charge and potential difference for each capacitor when the capacitors are connected (a) in series (see Fig. 24.8) and (b) in parallel (see Fig. 24.9).
IDENTIFY and SET UP:
In both parts of this example a target variable is the equivalent capacitance C_{eq}, which is given by Eq. (24.5) for the series combination in part (a) and by Eq. (24.7) for the parallel combination in part (b). In each part we find the charge and potential difference from the definition of capacitance, Eq. (24.1), and the rules outlined in Problem-Solving Strategy 24.1.
C=\frac{Q}{V_{a b}} (24.1)
\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}+\cdots (24.5)
C_{\mathrm{eq}}=C_{1}+C_{2}+C_{3}+\cdots (24.7)
EXECUTE:
(a) From Eq. (24.5) for a series combination,
\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{6.0 \mu \mathrm{F}}+\frac{1}{3.0 \mu \mathrm{F}} \quad C_{\mathrm{eq}}=2.0 \mu \mathrm{F}The charge Q on each capacitor in series is the same as that on the equivalent capacitor:
Q=C_{\mathrm{eq}} V=(2.0 \mu \mathrm{F})(18 \mathrm{~V})=36 \mu \mathrm{C}The potential difference across each capacitor is inversely proportional to its capacitance:
\begin{aligned}&V_{a c}=V_{1}=\frac{Q}{C_{1}}=\frac{36 \mu \mathrm{C}}{6.0 \mu \mathrm{F}}=6.0 \mathrm{~V} \\&V_{c b}=V_{2}=\frac{Q}{C_{2}}=\frac{36 \mu \mathrm{C}}{3.0 \mu\mathrm{F}}=12.0 \mathrm{~V}\end{aligned}(b) From Eq. (24.7) for a parallel combination,
\begin{aligned}C_{\mathrm{eq}} &=C_{1}+C_{2}=6.0 \mu \mathrm{F}+3.0 \mu \mathrm{F} \\&=9.0\mu \mathrm{F}\end{aligned}The potential difference across each of the capacitors is the same as that across the equivalent capacitor, 18 V. The charge on each capacitor is directly proportional to its capacitance:
\begin{aligned}&Q_{1}=C_{1} V=(6.0 \mu \mathrm{F})(18 \mathrm{~V})=108 \mu \mathrm{C} \\&Q_{2}=C_{2} V=(3.0 \mu \mathrm{F})(18 \mathrm{~V})=54 \mu \mathrm{C}\end{aligned}
EVALUATE: As expected, the equivalent capacitance C_{eq} for the series combination in part (a) is less than either C_1 or C_2, while that for the parallel combination in part (b) is greater than either C_1 or C_2. For two capacitors in series, as in part (a), the charge is the same on either capacitor and the larger potential difference appears across the capacitor with the smaller capacitance.
Furthermore, the sum of the potential differences across the individual capacitors in series equals the potential difference across the equivalent capacitor: V_{a c}+V_{c b}=V_{a b}=18 \mathrm{~V}. By contrast, for two capacitors in parallel, as in part (b), each capacitor has the same potential difference and the larger charge appears on the capacitor with the larger capacitance. Can you show that the total charge Q_1 + Q_2 on the parallel combination is equal to the charge Q = C_{eq}V on the equivalent capacitor?