
#19
Jan3112, 04:18 PM

HW Helper
Thanks
P: 9,818

Do not forget what happens inside the battery during charging the capacitor.
There are chemical reactions. The energy of the battery decreases more than the increase of the energy of the capacitor. Part of "chemical work" transforms into heat  inside the battery. If you shortcircuit a battery it will warm up. ehild 



#20
Jan3112, 04:26 PM

HW Helper
P: 3,337

I found this on the hyperphysics site:
From the definition of voltage as the energy per unit charge, one might expect that the energy stored on this ideal capacitor would be just QV. That is, all the work done on the charge in moving it from one plate to the other would appear as energy stored. But in fact, the expression above shows that just half of that work appears as energy stored in the capacitor. For a finite resistance, one can show that half of the energy supplied by the battery for the charging of the capacitor is dissipated as heat in the resistor, regardless of the size of the resistor. So I guess that half the energy will be lost in the wire, since it has finite resistance. (Unless maybe the wire is superconducting, but then there will still be the internal resistance of the battery). 



#21
Jan3112, 04:55 PM

P: 1,506

There must be electromagnetic radiation. I cannot understand why this is not more widely recognised. The connecting wires carry a changing electric current which results in a changing magnetic field...... Electromagnetic radiation. In teaching charging and discharging of capacitors this means of energy loss should be the first point to consider.....it will always happen 


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