Question

. Suppose you have a 9V battery, a 2μF capacitor, and a 7.40μF capacitor. 

(a) Find the charge and energy stored if the capacitors are connected to the 

battery in series. (b) Do the same for a parallel connection.

Answer 1

(a) Series capacitors

$$\dfrac{1}{C_{TS}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}=\dfrac{1}{2\times 10^{-6}F}+\dfrac{1}{7.4\times 10^{-6}F}$$

$$=\dfrac{4.7}{7.4\times 10^{-6}F}$$

$$Q_S=C_{TS}U=\dfrac{7.4\times 10^{-6}F}{4.7}\cdot9V$$

$$\approx14.17\times 10^{-6} C=14.17\mu C$$

$$E_S=\dfrac{C_{TS}U^2}{2}=\dfrac{7.4\times 10^{-6}\cdot(9V)^2}{2(4.7)}$$

$$\approx63.766\times 10^{-6} J=63.766\mu J$$

(b) Parallel  capacitors

$$C_{TP}=C_1+C_2=2\times 10^{-6}F+7.4\times 10^{-6}F$$

$$=9.4\times 10^{-6}F$$

$$Q_P=C_{TP}U=9.4\times 10^{-6}F\cdot9V$$

$$=84.6\times 10^{-6} C=84.6\mu C$$

$$E_P=\dfrac{C_{TP}U^2}{2}=\dfrac{9.4\times 10^{-6}\cdot(9V)^2}{2}$$

$$=0.3807\times 10^{-3} J=0.3807m J$$