PHYS208 10/6 Class
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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 (C*emf2)/2 = (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.
Series and Parallel Capacitors
Dielectric Materials and Capacitance
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Last updated Oct. 5, 1997.
Copyright George Watson, Univ. of Delaware, 1997.