Question

A 2.0 microF capacitor is charged by a 12V battery. It is disconnected from the battery and connected to an uncharged 5.0 microF capacitor. Determine the total stored energy:

i. before the two capacitors are connected

ii. after they are connected

Answer 1

i.

$$W_1=\frac{CV^2}{2}=\frac{2.0*10^{-6}*12^2}{2}=1.44*10^{-4}\:\rm J$$

$$q_1=q_2=CV=2.0*10^{-6}*12=2.4*10^{-5}\:\rm C$$

ii.

$$C'=C_1+C_2=7*10^{-6}\:\rm F$$

$$W_2=\frac{q_2^2}{2C'}=\frac{(2.4*10^{-5})^2}{2*7*10^{-6}}=4.11*10^{-5}\:\rm J$$