A voltage source (battery) will provide an energy of Q*emf to move the amount of charge Q to fully charge a capacitor to a final voltage of value emf. The energy stored in the capacitor is only Q*emf / 2 however. Where does the other half of the energy go? It's dissipated by the resistance in the charging circuit! See details.
Last updated March 15, 1998.
Copyright George Watson, Univ. of Delaware, 1997.