Capacitors in series potential difference change
If you have a 1 farad ($c_1$) and a 2 farad ($c_2$) capacitor connected in series ($c = frac{2}{3},mathrm{F)}$, to a 3 volt battery, the charge that will flow $Q = vc = (3 x frac{2}{3})C = 2C$, and the capacitors will charge to $V_1 = Q/c_1 = …